Talk:Monty Hall problem/Archive 17

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I suggest we get back to the content discussion. I think there are at least three proposals still on the table:

1) My rewrite of the "Probabilistic solution" section, see Talk:Monty Hall problem/Archive 16#Suggested change to the Probabilistic solution section above attempting to make the article (at least through the solution sections) scrupulously NPOV, neither endorsing nor criticizing the simple solutions.

Although I dispute the validity of some sources, I see no need to water down what any sources (even Morgan) say so long as they are fairly described, in proportion and in context. We must try and keep out own disagreements out of the article. Martin Hogbin (talk)

And, you're apparently not willing to do this. Many, many sources present a conditional solution saying nothing at all about unconditional approaches. Some present both, without taking a stance about whether one is better than the other. Some present both and explicitly criticize unconditional solutions. Implying approaching the problem conditionally is tantamount to criticizing the unconditional solution is not accurately representing the sources. -- Rick Block (talk) 13:27, 4 June 2010 (UTC)

2) Martin's move of the "Aids to understand" section, which he did and I reverted and he reverted back - and the only reason I haven't reverted again is to avoid an edit war.

There has only been one firm objection to the move. There is support from some editors and I assume tacit approval from others. It is not just text books that are presented this way, I have given two WP pages. I am sure that there are many more. I think if we asked any new editor whether they thought this article pushed any POV, they would not only answer 'no' but be hard pushed to guess what POVs there might even be. Martin Hogbin (talk)

You can't claim tacit approval after you revert a revert. Reverting it again is edit warring. -- Rick Block (talk) 13:27, 4 June 2010 (UTC)

3) My suggestion that the disclaimer that was added to the "Popular solution" section and is now in the "Probabilistic solution" section does not accurately reflect the prevalence of sources that present conditional solutions but instead, as worded, inaccurately implies there are only "some sources" that present conditional solutions and (also inaccurate) that all of these criticize the simple solutions.

I think all POV arguments and problems with accurately reflecting the sources are in your imagination. All the sources are reflected clearly in the article as it is now. I have given my POV above but I am not pushing this. The only opinion that I am promoting is that the article should be sensibly organised to help the reader understand the subject, from all points of view. Martin Hogbin (talk) 10:06, 4 June 2010 (UTC)

Martin - You've done nothing but push your POV for as long as you've been involved with this article. -- Rick Block (talk) 13:27, 4 June 2010 (UTC)

4) My suggestion to include a new level 1 "Solution" header with a very brief NPOV lead-in containing level 2 "Popular solution" and "Probabilistic solution" sections (perhaps with subsections), this is in Talk:Monty Hall problem/Archive 16#Please can somebody explain... above.

It's a little hard to tell, but I think no one has directly objected to #1 (Nijdam's comment is somewhat ambiguous). I clearly strongly object to #2. No one has directly commented on #3. And, although no one seems to object to the structure change in #4 Martin objects to the lead-in text (although he seems to think it is a disclaimer, so I'm not sure he's reading the same words everyone else is).

Is there any way we can reach a resolution on any of these without external assistance (e.g. a mediator)? Can we stay focused on, maybe, one at a time? -- Rick Block (talk) 04:34, 4 June 2010 (UTC)

There Are Three, Better Make That Four, Schools

1. Simple, non-door-specific solutions - Selvin, vos Savant, and many other reliably published sources, plus me

2. [Strike-out this line, I couldn't get strike-out to work properly] Only a conditional solution can address the case where door #3 has been opened - many reliably published sources

2. The solution must consider the specific case where door #3 has been opened

A. Formal conditional solution - many reliably published sources

B. Simple solutions - Rosenthal, Devlin, other reliably published sources

3. #1 is false, #2A is correct, #2B is false - Morgan

If I was giving a test on conditional probability to a class, I would expect them to mention that the host has a decision to make when faced with 2 goats. Otherwise, I consider this trivial, especially within the framework of a game show. I think the reliably published sources who offer simple solutions might agree with me.

Sometimes when making decisions, I have to consider both the likelihood of a risk, and the extent of the damage if that risk occurs.

I consider a Wikipedia reader unlikely to even concern him/herself with how the host chooses between 2 goats. And that the damage done to this reader, if s/he never became aware of this, to be essentially nil.

So, while reliably published sources (and we editors) disagree on the unique validity of a conditional solution, I conclude that the reader will enjoy the paradox more when given the simple solutions first. The published conflicts between the 4 schools discussed above would be reported thoroughly in a 'Controversy' section, with no editorial comment on the validity of the published positions. Glkanter (talk) 20:14, 5 June 2010 (UTC)

Rosenhouse

I don't think it's accurate to portray Rosenhouse as being in Morgan's , or even the conditionalist's camp. In his paper The Monty Hall Problem Jason Rosenhouse April 14, 2008, he writes, among other things:

(What follows is straight cut and paste of just a few portions of his paper. I did have to clean up some words, almost all involving the letters 'fi' which were turned into a special character. Paragraphs and indentations have been lost, sorry.)

Especially noteworthy is the following statement from Selvin's follow-up: \The basis to my solution is that Monty Hall knows which box contains the keys and when he can open either of two boxes without exposing the keys, he chooses between them at random." In writing this he had successfully placed his finger on the two central points of the problem. Alter either of those assumptions and the analysis can become even more complex, as we shall see in the chapters to come. In an amusing coda to this story, Selvin notes in his follow-up that he had received a letter from Monty Hall himself: Monty Hall wrote and expressed that he was not a \student of statistics problems" but \the big hole in your argument is that once the first box is seen to be empty, the contestant cannot exchange his box." He continues to say, \Oh, and incidentally, after one [box] is seen to be empty, his chances are no longer 50=50 but remain what they were in the first place, one out of three. It just seems to the contestant that one box having been eliminated, he stands a better chance. Not so." 36 CHAPTER 1. ANCESTRAL MONTY It would seem that Monty Hall was on top of the mathematical issues raised by his show. It is a pity that more mathematicians were not aware of Selvin's lucid analysis. They might thereby have spared themselves considerable public embarrassment.

Certainly vos Savant's arguments are not mathematically rigorous, and we can surely point to places where her phrasing might have been somewhat more precise. Her initial argument based on the million door case is pedagogically effective, but mathematically incomplete (as we shall see). And there was a subtle shift from the correspondent's initial question, in which the host always opens door three, to the listing of the scenarios given by vos Savant, in which it was assumed only that the host always opens a goat-concealing door. But for all of that it seems clear that vos Savant successfully apprehended all of the major points of the problem, and explained them rather well considering the forum in which she was writing. Her intent was not to provide an argument of the sort a mathematician would regard as definitive, but rather to illuminate the main points at issue with arguments that would be persuasive and comprehensible. In this she was successful.

Four people who were less impressed were mathematicians J. P. Morgan, N. R. Chaganty, R. C. Dahiya and M. J. Doviak (MCDD). Writing in The American Statistician, they presumed to lay down the law regarding vos Savant's treatment of the problem. After quoting the original question as posed by vos Savant's correspondent, they write: [beginning of quote - Glkanter] Marilyn vos Savant, the column author and reportedly holder of the world's highest I.Q., replied in the September article, \Yes, you should switch. The first door has a 1/3 chance of winning, but the second door has a 2/3 chance." She then went on to 46 CHAPTER 1. ANCESTRAL MONTY give a dubious analogy to explain the choice. In the December article letters from three PhD's appeared saying that vos Savant's answer was wrong, two of the letters claiming that the correct probability of winning with either remaining door is 1/2. Ms. vos Savant went on to defend her original claim with a false proof and also suggested a false simulation as a method of empirical verification. By the February article a full scale furor had erupted; vos Savant reported, \I'm receiving thousands of letters nearly all insisting I'm wrong.... Of the letters from the general public, 92% are against my answer; and of letters from universities, 65% are against my answer." Nevertheless, vos Savant does not back down, and for good reason, as, given a certain assumption, her answer is correct. Her methods of proof, however, are not. [end of quote Glkanter] Rather strongly worded, wouldn't you say? And largely unfair, for reasons I have already discussed. Indeed, continuing with their lengthy essay makes clear that their primary issue with vos Savant is her shift from what they call the \conditional problem," as posed by her correspondent (in which it is stipulated that the contestant always chooses door one and the host always opens door three) to the \unconditional problem," in which we stipulate only that after the contestant chooses a door, the host opens one of the goat concealing doors. She did, indeed, make this shift, but this was hardly the point at issue between vos Savant and her angry letter-writers.

MCDD responded to this. They had the audacity to begin with, \We are surprised at the tone of vos Savant's reply." It is unclear what tone they were expecting in response to their bellicose and condescending essay. They then repeated the main points from their earlier essay, emphasizing that the problem vos Savant discussed was not precisely the problem laid out in the initial question. This point is not at issue, but what vos Savant discussed was surely what was intended. Even if it was not, vos Savant made it quite 48 CHAPTER 1. ANCESTRAL MONTY clear what problem she was discussing. Seen in that light, MCDD ought not to have said that her arguments were wrong and contained technical errors when they meant simply that she had altered the problem slightly from what was originally stated. In fairness, MCDD do moderate their tone later on, writing, \None of this diminishes the fact that vos Savant has shown excellent probabilistic judgment in arriving at the answer 2/3, where, to judge from the letters in her column, even member of our own profession failed."

www.math.jmu.edu/~rosenhjd/ChapOne.pdf

Glkanter (talk) 06:31, 6 June 2010 (UTC)

I think Rosenhouse classifies vos Savant's answer very well, she was not wrong, her answer was fine for her intended audience, the general public. Maybe there are some lessons for us to be learned here. We do not want to make this article into a 'bellicose and condescending essay' that most of our audience will not be able to understand. Martin Hogbin (talk) 17:58, 6 June 2010 (UTC)

How's This?

I. Very little mention of controversies (maybe just vos Savant's in Parade magazine, the others exist only in professional journals and text books) until after the solutions are provided.

II. One solution section: vos Savant, Devlin or Rosenthal, conditional. Yes, in that order.

III. Aids to Understanding

IV. The Controversies:

A. vos Savant's 1/2 vs 2/3

B. Door 3 must be revealed

C. Morgan, including simple solutions are false, host bias, and that their's is the (only) 'correct resolution'

A boring litany of who said what, and why, only. Not what they might mean, or who is right.

V. Sources of Confusion

etc...

Maybe switch III and IV...

Glkanter (talk) 17:27, 6 June 2010 (UTC)

Before the solution section, I guess in the Problem statement, I would suggest including Selvin's comment from his 2nd letter (paraphrasing here): '...this solution relies on the host always offering the switch and choosing equally between...' and a mention that the importance of this will be addressed later in the article, with a link to that section. Glkanter (talk) 17:57, 6 June 2010 (UTC)

And I would have a separate, later section for the other simple solutions and any other solutions people want to include. I would put this before the 'Variants' section. Glkanter (talk) 18:00, 6 June 2010 (UTC)

So, anybody else see this as a NPOV way to get a better article and some closure? Glkanter (talk) 02:28, 7 June 2010 (UTC)

Awesome. This is pretty much exactly what I've been suggesting for several months. Martin's response has been, let's say, less than cooperative. -- Rick Block (talk) 02:43, 7 June 2010 (UTC)

How about that. I honestly didn't make the connection. Funny how it looks different when you get there 'by yourself'. Glkanter (talk) 02:57, 7 June 2010 (UTC)

I'll share something with you, Rick. It was not until yesterday that I separated in my mind 'Morgan' from 'other reliably published sources who state that only a solution involving an open door #3 answers the question asked by Selvin and Whitaker.' (I don't need a roster or headcount of each. That they exist is enough.) As per Wikipedia, whether I agree with the 'open door #3' sources or not is irrelevant. That made all the difference in my approach to the article. Glkanter (talk) 03:31, 7 June 2010 (UTC)

Well, I'm glad you're seeing what amounts to the wisdom of Wikipedia's NPOV policy. Do you think vos Savant's explanation about the MHP is correct? Frankly, Wikipedia doesn't give a shit what you think. This is completely the wrong question. Are there sources (not just one, but lots of them) that say vos Savant had her head up her ass? This is a question Wikipedia cares about. If the answer is yes, it really doesn't matter what you think - the article HAS to say what the sources say. It's not about what YOU think, or what I think, or what Martin thinks, or that Nijdam thinks - it's about what the SOURCES think. If the predominant view, among reliable sources, is that vos Savant had her head up her ass then the article must say (whether you agree or not) that vos Savant had her head up her ass. Similarly, if the view that the solution to the MHP must address the case where the player has picked door #1 and the host has opened door #3 is expressed by a significant number of sources the article MUST reflect this view (regardless of whether you, or I, or anyone else agrees with it). If there's disagreement among sources, NPOV says the article CAN'T choose sides based on anything other than the prevalence of published views (and not whether you, or I, or any other editor agrees with any of the published views). -- Rick Block (talk) 06:11, 7 June 2010 (UTC)

So, here's my proposal on how we move forward. Rick, you'll set up a sandbox, which will initially contain only the first section of the current article. We won't add the next section to the sandbox until everybody is happy with the currently-in-process section. Text additions will be in a different font or something, deletions will be striked-out. Just like you did earlier on the talk page, in fact. Anything that gets removed for inclusion later will need to be kept in a separate section of the sandbox so we don't forget it. This is the method that would be best for me (and I think for others), focusing on a manageable portion, rather than trying to consume the whole thing at once. We'll also need a talk page, but you probably had that idea already. Glkanter (talk) 04:04, 7 June 2010 (UTC)

This is pretty much exactly what I've been trying to do, but making edits to the article itself rather than to a sandbox. The only real issue here is Martin's unilateral change (moving the "Aids to understanding" section). His repeated assertion that no one other than me objects to this is, ahem, horseshit. He's clearly trying to put as much distance as he possibly can between what he considers to be the "correct" solutions (i.e. the simple solutions) and the conditional solutions (which he seems to think only address an academic variant). Having the article not choose sides here is not easy - but insisting that "my side is right" is really not a good starting point. -- Rick Block (talk)

Well, editing the article doesn't accomplish at least two of the things I think would help: focusing on one section at a time and saving items for re-insertion until we work on it's later section. Plus, it's so much more emotionally charged, and there's very explicit rules about editing articles that don't affect sandboxes. I don't see any benefit from editing the article directly, and I do see some drawbacks.

There is some merit to your criticism above. But it took me 18 months to understand that Morgan & 'an open door #3' were different. Frankly, I think that was a shortcoming in your presentation of your arguments. And there is understandably some defensive posture taken by Martin and I. I mean, here's Nijdam's previously undiscussed edit at the beginning of the Popular section just last week:

"Several authors present a form of the following popular solutions. They serve as a way of understanding why the probability of gaining the car after switching is 2/3 and not 1/2. Other authors however point to the fact that although the the popular solutions show the correct numerical answer, they are incomplete and actually solve a slightly different problem."

But I've never seen you scold him for violating NPOV. Anyways, while you and I still disagree on some interpretations (hopefully those differences will not need to be dealt with in the article), we agree on more things now than ever before. I think. Hopefully Martin, Nijdam and the others will be on board and contribute. Glkanter (talk) 06:37, 7 June 2010 (UTC)

In what manner is Vos Savant's solution (1990) insufficent to answer Whitaker's question (1990) based on Selvin's MHP problem (1975)?

Popular solution The solution presented by vos Savant in Parade (vos Savant 1990b) shows the three possible arrangements of one car and two goats behind three doors and the result of switching or staying after initially picking Door 1 in each case:

Door 1	Door 2	Door 3	result if switching	result if staying
Car	Goat	Goat	Goat	Car
Goat	Car	Goat	Car	Goat
Goat	Goat	Car	Car	Goat

A player who stays with the initial choice wins in only one out of three of these equally likely possibilities, while a player who switches wins in two out of three. The probability of winning by staying with the initial choice is therefore 1/3, while the probability of winning by switching is 2/3.

END OF COPYING

Glkanter: It is sufficient. More, it is elegant, and celebrates the paradox. It addresses the contestant's door choice, and the door the host revealed, exactly as per Selvin, Whitaker and vos Savant. Glkanter (talk) 15:58, 3 June 2010 (UTC)

If I've picked door 1 and the host has opened door 3, what line or lines of this table help me figure out my probability of winning by switching? Seems like the first two lines apply and the third one doesn't (since door 3 is open and I can see there's no car there). That's the only line that's excluded based on what I know. If the first two lines are equally probable, then why isn't my chance of winning by switching 1/2? If you can explain this simply, using this table, please do so. -- Rick Block (talk) 16:20, 3 June 2010 (UTC)

Yeah, the text should say, "...by staying with the initial choice... winning by switching is 2/3 once door 3 (or door 2) has been opened." Glkanter (talk) 16:32, 3 June 2010 (UTC)

Whitaker says that the host opens door 3. In the final row of that table, the host has opened door 2, a contradiction. -- Coffee2theorems (talk) 16:42, 3 June 2010 (UTC)

Well, now you're getting back to technique, turning it into an incomplete conditional table. Which is not her intent. vos Savant's response encompasses Whtaker's condition, as well as opening door 2. She's solving it simply, or as I call it, 'omniconditionally', using logic. Her solution 'resembles' formal probability because it's in table form. Glkanter (talk) 17:04, 3 June 2010 (UTC)

If the table cannot be taken at face value, then it needs a clear (and sourced) explanation to accompany it, so that the reader can understand how to interpret the table. I haven't seen one yet. -- Coffee2theorems (talk) 17:22, 3 June 2010 (UTC)

The point here is this table does not address the case where the player has picked door 1 AND the host has opened door 3. It addresses all cases where the player has picked door 1. What it says is if you pick door 1, knowing that the host will open a door showing a goat (and not your door) means if you switch after the host opens a door (which will be door 2 or door 3) you'll win with a 2/3 chance by switching. This is a different question from what are the chances of winning AFTER the host has opened a specific door, for example door 3. This difference is the crux of the paradox and is why people get this problem wrong (and is why this solution is completely unsatisfying to many people). The resolution of the paradox requires a conditional probability approach (whether you want to call it that or not). There are fundamentally two ways to do this:

1) argue using this table (or any equivalent simple solution) that if you pick door 1 and then switch (when you're looking at either door 2 or door 3 being open) your chance of winning is 2/3, and since there's nothing special about either door 2 or door 3 your chance in both of these cases must also be 2/3.

2) directly address the conditional probability - i.e. argue that the host must always open door 3 if the car is behind door 2 (1/3 chance of this) but opens door 3 only half the time (and door 2 the other half) if the car is behind door 1 (1/2 * 1/3 = 1/6 chance of this), so you're twice as likely to win if you switch to door 2 when you're looking at the goat behind door 3.

The simple solutions typically omit the part in bold, which is fine but means that somebody thinking about the specific case where they've picked door 1 and the host has opened door 3 can't see the connection - and they'll argue with you. The conditional approach directly resolves the paradox. The simple approach says what the answer must be, but (typically) not why, i.e. doesn't address the paradox. -- Rick Block (talk) 17:28, 3 June 2010 (UTC)

Conversely, Chen's tree diagram that is in the article shows both door #2 and door #3 being opened. It doesn't 'match the problem' any better than the simple solutions. The host opens both doors? It doesn't really solve the puzzle at all. Like Coffee2theorems wrote just a couple of paragraphs above: "If the table cannot be taken at face value..." Chen's tree diagram and accompanying text simply repeat the 'incompleteness' or 'falseness' criticized in the simple solutions. It's no more 'solving Whitaker's problem' than Selvin's table and explanation, and it's a lot more difficult for a reader to understand. The tree diagram *would* match the problem, however, if the 3rd column of the tree, which begins with '1/2', only included the '1/2' and '1' as described by Rosenthal. Which highlights another fact. The problem is ultimately so simple, the various solutions have a lot of overlaps. Rosenthal's solution also bears a lot of similarities to the Combined Doors solution. They have different logic, but ultimately both of the Combined Door images could be used for either solution. Glkanter (talk) 07:31, 7 June 2010 (UTC)

In fact, the tree diagram is captioned 'Tree showing the probability of every possible outcome if the player initially picks Door 1'. Not 'after door #3 has been opened', as it 'should'. So now we're down to relying on the accompanying explanations given by the various sources of how to use their visual aids. I would call that 'semantics', not 'science'. Glkanter (talk) 07:55, 7 June 2010 (UTC)

You could use Grinstead's diagram. It shows just the door 3 path. Which on the same page explains his complaint with the simple solution: "Now suppose instead that in the case that he has a choice, he chooses the door with the larger number with probability 3/4." In other words, the simple solutions only fail without host symmetry. Which is not Selvin's MHP problem. Besides, Grinstead is a text book, he's using the MHP to teach, so he's just contrasting the approaches in order to show why conditional probability may be required, not really criticizing the simple solution. Glkanter (talk) 08:13, 7 June 2010 (UTC)

Sorry for posting so many times in a row, I just got a lot on mind. Of course the 2nd conditional diagram in the article has the same problem, even worse. And Grinstead's diagram will resemble Rosenthal's solution, which Rick criticizes for various reasons in the Rosenthal section below. You could mix Rosenthal's '1/2 and 1' with Chen's tree diagram, but that would be OR. And Grinsteads diagrams are unwieldy, not so friendly for the reader, astute or not. I'm sure you can find a good tree somewhere.

And this shows why it took over 18 months for me to 'understand' the conditional criticism of the simple solutions as separate from Morgan's criticisms. The conditional solutions in the article don't show door #3 open and doors #1 and #2 closed. Well, the Combining Doors solution does. But, the point is the conditional solution chosen for the article does not show 2 closed doors and door #3 open. So that made it very difficult for me to grasp that this was the conditional's complaint about the simple solution. Quite the irony, huh? Devlin's simple Combining Doors solution *does* correctly show the 3 doors as per Whitaker and Selvin, while the conditional Chen's tree diagram and the 2nd table don't come close. Glkanter (talk) 08:58, 7 June 2010 (UTC)

You are only talking about one specific interpretation of Whitaker's question. It is far from clear the Whitaker intended to specify door numbers or even that he actually wanted to know the probability after a door had been opened. If is most likely by far that Whitaker did not appreciate the difference this could possibly make. Martin Hogbin (talk) 19:55, 3 June 2010 (UTC)

And you are only talking about a different specific interpretation - one that virtually no one initially understands the problem to be. Per Krauss and Wang - 97% of subjects in their test were thinking about the specific case of the player initially picking door 1 and the host then opening door 3. But as Kmhkmh says, we're repeating ourselves. I'm willing to compromise and have the article present simple solutions in an NPOV fashion, without immediately saying these solutions don't address the problem as 97% of people understand it - but only as long as the article immediately presents a conditional solution in an NPOV fashion as well. In return, your "compromise" is to continue to insist that the main (and, in your view, only) point of the article is to convince people that the simple solutions are correct, i.e. that the way people initially understand the problem is incorrect (this interpretation is an "academic contrivance") and the "proper" interpretation (based on no sources, whatsoever) is to ignore the which door the host opens. Again, not only is this willfully stupid (since conditional probability is the exact tool for the job) it directly violates core Wikipedia content policies. -- Rick Block (talk) 23:19, 3 June 2010 (UTC)

Rick, in their mail exchange, as described in Selvin's 2nd letter, I'd say it's pretty clear that Selvin and Monty Hall were describing the universal concept of switching being 2/3 v 1/3, not just the specific case of boxes B and A that Selvin used to describe the problem in his 1st letter. Would you agree? Glkanter (talk) 14:33, 4 June 2010 (UTC)

No. They're talking about switching when looking at two closed boxes and an open box which is fundamentally a specific case, i.e. they're not talking about the overall average across all players but a specific player looking at an empty box. The "average" that matters to this player is not the overall average, but the average only of players in the same situation as this player. This is the exact point we've been talking about for years. Since the boxes are identifiable, these two averages (all players vs. all players in the same situation as this player) are logically different and might even have different numeric values. The probability for all players in the same situation as this player is the conditional probability. The probability for all players (averaged across all situations) is the unconditional probability. -- Rick Block (talk) 15:12, 4 June 2010 (UTC)

I wouldn't agree that discussing 'two closed boxes and an open box' is the same as the argument that Whitaker is asking specifically about the exact door 1 and door 3 pairing. I would say that 'two closed boxes and an open box' describes the simple interpretation. Can you save me some (a lot, actually) time and provide the link to the 2nd letter? If not, I'll find it. Thanks. Glkanter (talk) 15:26, 4 June 2010 (UTC)

[1]. -- Rick Block (talk) 16:13, 4 June 2010 (UTC)

How about Selvin himself? His contestant chooses box B, the host shows Box A. His contestant has chosen to switch. For his solution, he makes a table of all 9 equally likely outcomes, then writes:

"Enumeration shows probability of winning is 6/9 = 2/3. If the contestant does not switch boxes, then his probability of winning the car remains unchanged (1/3) after Monty Hall opens an additional box."

No mention of this requiring a conditional solution. http://www.jstor.org/pss/2683689 Glkanter (talk) 00:54, 4 June 2010 (UTC)

I see Rick is anxious to leave the discussion in this section. The point of which is to demonstrate that simple solutions have been used to solve the specific condition exactly like Whitaker's. So having a Simple solution section without any caveats does not do a disservice to either the readers or the other sources. vos Savant uses the same logic as Selvin. Glkanter (talk) 05:10, 4 June 2010 (UTC)

No, I'm anxious to stop repeating ourselves to no apparent purpose. What I'm saying has no source whatsoever is Martin's insistent claim that the "right" way to approach the problem is unconditionally, i.e. AFAIK there is no source that says anything like "of the two ways to approach the problem, unconditionally and conditionally, unconditionally is the right way" (in contrast, there ARE numerous sources that say the opposite). I'm not saying there aren't any sources that approach it unconditionally (there clearly are many), just that there aren't any sources that explicitly mention both approaches and then say that unconditional is the right way.

And, if you read the compromise I'm willing to make I'm not saying the Simple solution must have caveats - so why are you arguing about this? -- Rick Block (talk) 05:27, 4 June 2010 (UTC)

I'm glad you're on board, Rick. If Nijdam decides to add his caveats back into the Popular solution section, Selvin's solution will be my support for reverting it. Glkanter (talk) 10:52, 4 June 2010 (UTC)

Why Isn't The Door 1 and Door 3 Pairing a Subset of 'Always'?

Another simple solution says, 'Your door choice has a 1/3 chance of being the car. Switching always gives the opposite of your original choice. Switching after a door is opened and Monty offers the switch therefore gives you a 2/3 chance of winning the car.'

How can it be argued that this solution does not apply to the specific doors 1 and 3 pairing? Glkanter (talk) 15:04, 4 June 2010 (UTC)

To be a valid statement, the "1/3" and "2/3" in this solution must be probabilities from the same sample set. The 1/3 is clearly from the initial 1/3 + 1/3 + 1/3 = 1 view of the situation (i.e. before we know which door the host has opened). If we've seen the show 900 times and of these 900 shows players initially picked door 1 300 times, this is talking about these 300 airings of the show. "Your choice has a 1/3 chance of being the car" means of these 300 players who have initially picked door 1 we'd expect about 100 to have picked the car. "Switching after a door is opened and Monty offers the switch therefore gives you a 2/3 chance of winning the car" is perfectly true, but it is talking about all 300 of these players (if 100 have picked the car, 200 have not and for these 200 switching wins). This solution isn't saying anything at all about how many players will see the host open door 3 and of those how many win the car by switching. Maybe only 100 players see the host open door 3 and all of them win by switching, while 200 players see the host open door 2 and only 100 of them win by switching (this still totals 200 of the 300). You might think, well of course the 300 and the 200 split evenly by which door the host opens (so 150 see the host open door 3 and of these 100 will win by switching), but this solution doesn't say anything about this. See the issue? -- Rick Block (talk) 15:40, 4 June 2010 (UTC)

A solution isn't required to address only one pairing. If it addresses them all at once, it's inclusive. I didn't make up this solution, it's published. You're arguing about a requirement to use a formal conditional probability solution, a requirement which does not exist, and nothing else. You're not arguing about the problem statement, nor the validity of the solution. Just your preferred technique. You have not adequately answered the question posed. Glkanter (talk) 15:54, 4 June 2010 (UTC)

I presume 'symmetry' rules out most of your explanation, anyways. But maybe one of the experts could weigh in on this. I'm pretty sure Boris already did. Glkanter (talk) 16:04, 4 June 2010 (UTC)

I don't understand your response here. You ask how it can be argued that the simple solution doesn't apply to a specific case. I've answered that. Is there something about this explanation you don't understand? Symmetry is an additional argument, that when added to the simple solution says it applies equally in all specific cases. But the solution you quote above says nothing about symmetry. -- Rick Block (talk) 16:11, 4 June 2010 (UTC)

Well, I'm no expert, but I think where all the premises state 'randomly placed' and 'chooses randomly', etc. provides the basis for invoking symmetry, and that it's simply understood without having to state it. This explanation would be better coming from any of the more qualified editors, though. Glkanter (talk) 16:26, 4 June 2010 (UTC)

Are you saying it's OK to assume symmetry without stating it, or asking whether it's OK? In either case we don't need experts here to weigh in on this - it's in published sources, specifically about the MHP. I've quoted them repeatedly. Presenting a solution that assumes this, without saying anything about it, is what Morgan et al., and Gillman, and Rosenthal, and Lucas, Rosenhouse, and Schepler are criticizing. -- Rick Block (talk) 17:57, 4 June 2010 (UTC)

Here's how I think the (astute) reader will read the solution you quoted:

"Your door choice has a 1/3 chance of being the car."

OK, the car is placed at random, so the chance is 1/3 alright, even in my case where I chose door 1. The solution text means: Door 1 has a 1/3 chance of hiding the car in a situation where Door 3 is still closed. (let's call this situation A, i.e. "door 3 is closed" situation)

"Switching always gives the opposite of your original choice."

Opposite? Oh, right, the car and the goat are "opposites" here, kind of like win and loss. The solution text means: If you picked the car, switching in a situation where Door 3 has been opened gives you a goat, and the other way around. (let's call this situation B, i.e. "door 3 is open" situation) That's always true, OK.

"Switching after a door is opened and Monty offers the switch therefore gives you a 2/3 chance of winning the car."

Or in other words: Switching in situation B gives you 2/3 chance of winning.

Huh? Can you explain this to me a little bit more slowly, please? Where did that number come from? Oh, right, 2/3 = 1-1/3, I guess that's where the author got 2/3 from. This equation is only valid when all the probabilities involved apply to the same situation. The answer 2/3 obviously must apply to situation B, because that's what the question is about. Therefore 1 and 1/3 must also apply to situation B. The total probability is 1 in any situation, so that's OK. But where did 1/3 come from? I was only given a probability equaling 1/3 in situation A, not situation B! What gives? Did the author make a mistake? Is this a sleight of hand by The Man, who tries to stop me from believing in the obvious Truth (I see what you did there!)? The answer must be 1/2 after all, Wikipedia is just wrong as usual.

One needs to explain that this particular probability (that the car is behind door 1) is the same in situations A and B, or the reader will be confused. That's what's missing from the solution. It's enough to say that "It can be shown that this particular probability is the same in situations A and B." (i.e. "the unconditional and the conditional probabilities are equal"), or something to the effect. Even better would be to provide an argument about why it is so, but that's optional. The important thing is that the reader can verify that every step that is given is correct, or they will be left doubting that either the solution is incorrect or that they are interpreting the solution text wrong. -- Coffee2theorems (talk) 13:37, 5 June 2010 (UTC)

Of course, some of the discussiants already understand this point, but strangely enough others, although asking for this explanation, do not accept or even understand it. My main goal is to have this point explained directly in the simple solution section. Nijdam (talk) 11:31, 7 June 2010 (UTC)

Which published sources will you derive this explanation from? Will you also be finding a source that explains the conditional solution? That would be the NPOV way, per Rick. But it takes college professors and text books in lecture halls to teach conditional probability. How can that be done in a Wikipedia article? Glkanter (talk) 11:41, 7 June 2010 (UTC)

I'm paraphrasing a published solution. The question is "why wouldn't the door 1 and 3 pairing be a subset of 'always'", not "does this solution convince you, or in your opinion, an (astute) reader?". Glkanter (talk) 14:15, 5 June 2010 (UTC)

The proof does not apply to the case where the door pair is (1, 3), because the hypothetical example reader I used tried to follow it for that case, and failed. If the proof were applicable to that case, then it would convince the astute reader, because an astute reader believes all valid proofs (or that is a part of my definition of "astute" anyway). Proofs are about convincing a reader, and a "valid proof" must not omit so many steps that the reader cannot follow it, or you will end up accepting that simply saying "The answer is 2/3." is a valid proof, and anyone who cannot follow it is just stupid, which isn't an appropriate stance for an encyclopedia to take. -- Coffee2theorems (talk) 15:26, 5 June 2010 (UTC)

The other way to look at this is that this solution is valid, but it's not talking about situation B (the door 1 door 3 pair), but both of situations B and C (where C is the door 1 door 2 pair) without distinguishing which of these situations you've ended up in. This is what I'm saying in my explanation above. If 300 players have picked door 1, then some of them see the host open door 3 and some of them see the host open door 2. If all of these players switch they "always" get the opposite of their initial choice. So, if you are in a population of players who have picked door 1 and everyone in this population switches (regardless of which door the host opens) 2/3 of this population wins a car. What this means is if you decide to switch BEFORE the host opens a door, you have a 2/3 chance of winning (regardless of which door you see the host open - and sometimes you'll see the host open door 3 and sometimes you'll see the host open door 2). But, this isn't saying anything at all about the sub-population who find themselves in situation B. -- Rick Block (talk) 15:59, 5 June 2010 (UTC)

Well, those are probably sources not familiar with Selvin's 2nd letter that clearly states without any ambiguity that the host chooses randomly when faced with two goats. With that premise in place, the symmetry exists without any need for a 'statement'. Glkanter (talk) 00:03, 5 June 2010 (UTC)

And one of those sources (probably Rosenhthal, but maybe Rosenhaus) called the simple solutions 'shaky'. But he meant that the random goat doors premise was a requirement, that couldn't be overlooked. It was a dramatic flair on his part, to grab his students' attention. Not a criticism. Glkanter (talk)

"With that premise in place" - exactly. The usual statement of the problem does not include this premise (and isn't referenced to Selvin, so the fact that Selvin clarified this really doesn't help much). What the "Problem" section of the article says is that the "standard analysis" is based on this assumption - which makes the simple solutions correct (even for the conditional interpretation). This is actually pretty generous, since whether the usual simple solutions are assuming it or not is not obvious (because they don't say anything about it). For example, vos Savant never said anything about it - including in her detailed experimental procedure (which included explicit randomization steps for every other aspect of the problem that is assumed to be randomized). I've quoted Morgan's comment about this before (in his rejoinder to Seymann): "... one is tempted to conclude that vos Savant does not understand that the conditional problem (of interest to the player) and the unconditional problem (of interest to the host) are not the same, and that 2/3 is the answer to the relevant conditional problem only if p = q = 1/2". The non-generous view is that those presenting simple solutions do not understand the significance of this premise. The generous view is that they do, i.e. they know that it is what forces the problem to be symmetric, and they assume it without stating it.

The difference between this generous and non-generous view is what I think the POV issue we've been discussing is about. I believe Martin's stance is that the article should take the POV that those presenting simple solutions understand the significance of this premise and are obviously interpreting the question to be about the overall average probability rather than the conditional probability - in which case this premise is irrelevant. I believe Nijdam's stance is that since these sources don't say any such thing the article can't imply this. The NPOV compromise I've been suggesting is for the article not to take a stance on this either way. I think it's a fact that the usual academic interpretation of the problem includes this premise. I think it's a fact that the sources presenting simple solutions don't say anything about this premise or how they're interpreting the problem, so we don't know whether they're assuming this premise without mentioning it or interpreting the problem to be about the overall average probability rather than the conditional probability. Either way, presenting simple solutions and saying nothing about this premise accurately reflects these sources. I think it's a fact that sources presenting conditional solutions generally include this premise so presenting a conditional solution including this premise accurately reflects these sources. The criticism that the simple solutions are assuming this premise without stating it, or that they aren't addressing the conditional problem, needs to be somewhere - but I don't think it has to be in the Solution section. -- Rick Block (talk) 01:28, 5 June 2010 (UTC)

Agreed. Not in the Solution section, not preceding it, and only calling it a 'controversy' elsewhere in the article, not a 'fact' that the simple solutions are 'false' or 'incomplete'. That's all I've ever asked since I understood that reliable sources must be referenced in a Wikipedia article. Glkanter (talk) 01:46, 5 June 2010 (UTC)

So, does this mean you're OK with this outline (from above)

<lead> (as is)

Problem

as is

Solution

There are multiple approaches to solving the Monty Hall problem. Most popular sources present solutions based on simple probabilistic reasoning. A different approach to solving the problem, used primarily in academic sources, is to treat it as a conditional probability problem.

Popular solution

mostly or entirely as is

Aids to understanding (under "Popular solution", not same level header)

Probabilistic solution

rewritten to say nothing about the simple solutions, per the draft at User:Rick Block/Monty Hall problem (draft)

Mathematical formulation

Sources of confusion

etc.

Where the Problem section (like it currently does) includes that the standard analysis assumes the host picks randomly between two goats (making the problem symmetrical) and all sections up to "Sources of confusion" are based on this understanding? I don't know whether either Nijdam or Martin are OK with this, but are you? -- Rick Block (talk) 02:59, 5 June 2010 (UTC)

Actually, Rick, the purpose of these two sections was to elicit the views of the other editors to my questions about the simple solutions. I'm still very interested in their responses to the questions posed.

I just read Grimstead. As I read it, he offers his own 'overall' solution 'make your decision before a door is open' with a result of 1/3 and 2/3. He doesn't address how the host opens doors. Then he adds the random host premise to the conditional solution he details and comes up with 1/3 and 2/3. He doesn't really say whether these are the same or not, only that with a host bias (he uses 3/4 for the larger door #) they return different results. I don't read this as a criticism of any published simple solutions (unlike Morgan), more as a way for an instructor to demonstrate when conditional probability is required. Glkanter (talk) 05:57, 5 June 2010 (UTC)

So, is that a "yes", or a "no"? -- Rick Block (talk) 06:05, 5 June 2010 (UTC)

Thank you both, Rick and Coffee2theorems. While, I disagree of course, your responses give me a clearer understanding of where our editorial differences remain. Glkanter (talk) 17:00, 5 June 2010 (UTC)

Is there some reason you're avoiding saying "yes" or "no" here? -- Rick Block (talk) 17:25, 5 June 2010 (UTC)

If you do understand that, then it would be nice to tell us about the differences. I for one am still in the dark. What things do you disagree with, and why? I find it much more difficult to understand your (and Martin's) position than the others'. There's something to be said for making one's position absolutely clear to others. -- Coffee2theorems (talk) 17:36, 5 June 2010 (UTC)

Coffee2theorems, it's not that I don't understand. It's that, like many published sources on the MHP, beginning with Selvin's 1st letter, I consider the simple solutions as presented as being valid for any specific door pairing.

Rick, if what you proposed would be 'the final result', I'd be OK with it. Maybe propose changing a couple of words in the intro. But, other editors don't agree with your proposal, so, in a more general sense, I just can't say 'yes' or 'no'. Glkanter (talk) 19:49, 5 June 2010 (UTC)

Coffee2theorems, I think you are slightly misunderstanding the situation when you talk of Glkanter's and my postion vs the position of the others regarding the structure of the article. The argument about the article structure is now essentially between Rick and the others.

I moved the 'Aid to understanding' to immediately after the simple solutions section because this 'Aids to understanding' section referred to the simple solutions only, thus this order was most logical (I might add that I have no objection to another 'Aids to understanding the conditional solution' being added after the conditional solutions). There have been no strong objections to this structure from anyone but Rick.

Rick then proposed altering the heading levels. I have not objected to this and I do not expect any strong objections from elsewhere. Rick also proposed adding an introductory paragraph, which I do not much like. Could we have something along the lines of:

There are multiple approaches to solving the Monty Hall problem all giving the same result, that a player who swaps has a 2/3 chance of winning the car in the standard version of the problem. For some variations of the problem more complex methods of solution are required

Regarding your comments on people's understanding of various solutions, I would be happy to discuss this subject but suggest we do so on the arguments page.Martin Hogbin (talk) 20:25, 5 June 2010 (UTC)

Martin - you continue to misrepresent things. Nijdam reverted the very same change some time ago [2], so saying I am the only one who strongly objects is simply not true. I am hopeful that the mediation will become more active. If it doesn't relatively soon, I will solicit input about this at Wikipedia:Neutral point of view/Noticeboard. IMO, your continued insistence that the article must endorse the POV that the simple solutions are entirely correct (your POV) and must represent the POV that the problem is inherently conditional as applying only to a more complex variant is reaching the point of wp:disruption. -- Rick Block (talk) 21:46, 5 June 2010 (UTC)

Rick, what are you getting so mad about? You proposed a change the the heading levels - I accepted it. You proposed an introductory paragraph, I suggested a change in wording. If you feel this warrants input from the Wikipedia:Neutral point of view/Noticeboard please go ahead and seek it. Martin Hogbin (talk) 21:59, 5 June 2010 (UTC)

Who's mad? I'm just pointing out that you're not only misrepresenting the situation here but continuing to push your POV. -- Rick Block (talk) 23:18, 5 June 2010 (UTC)

In what way push my POV? I gave agreed your proposed heading level change and suggested a slight change in wording for your intro. Martin Hogbin (talk) 23:23, 5 June 2010 (UTC)

So you're admitting you're misrepresenting things (since you haven't objected - I assume you're tacitly agreeing)? How you're pushing your POV here is your suggested wording, in context, will clearly be read such that "multiple approaches" refers to the multiple simple solutions presented in the "Popular solution" section and the "some variations" refers to the problem addressed by the "Probabilistic solution" section. What this ends up saying is these solutions address DIFFERENT problems and that the simple solutions address the "standard version" while conditional solutions address a variant - consistent with the POV you've been pushing forever. IMO, this is hardly an accident. It's exactly what you're intending it to say. -- Rick Block (talk) 00:01, 6 June 2010 (UTC)

Rick, please just read what I have written.

There are multiple approaches to solving the Monty Hall problem... Your exact words.

...all giving the same result, that a player who swaps has a 2/3 chance of winning the car in the standard version of the problem. This is universally agreed. For what is described in the article in several places as the standard version of the problem (host chooses random goat door), all sources and all WP editors agree the answer is 2/3. Do you want to give the impression that some solutions result in a different numerical answer?

For some variations of the problem more complex methods of solution are required. This is true, and exactly what many sources point out. In the case that the host is known to choose non-uniformly between goat doors the conditional solution is absolutely necessary to solve the problem, the simple solutions fail.

I though we were close to complete agreement here. I have stated only undisputed facts in an attempt to end this dispute. Martin Hogbin (talk) 09:01, 6 June 2010 (UTC)

The "for some variations" bit is NOT talking about the problem being addressed by any of solutions presented in the "Solution" section. Adding it has the (IMO, deliberate) effect of characterizing the conditional solutions as unnecessarily complex and required only for variants, while simultaneously reinforcing the (disputed) notion that the simple solutions are sufficient for the "standard problem". You are not stating undisputed facts - you're making what is disputed (that the simple solutions are sufficient) seem undisputed and what is undisputed (that conditional solutions are a completely standard way to address the problem) seem disputed. As far as I can tell you have absolutely no interest in ending this dispute in any manner other than having the article endorse your POV. All of your so-called compromises do this. -- Rick Block (talk) 16:05, 6 June 2010 (UTC)

So it is just the last sentence you do not like. What do you suggest? Martin Hogbin (talk) 16:59, 6 June 2010 (UTC)

Martin - you already know what I suggest - an NPOV lead-in, i.e.

There are multiple approaches to solving the Monty Hall problem. Most popular sources present solutions based on simple probabilistic reasoning. A different approach to solving the problem, used primarily in academic sources, is to treat it as a conditional probability problem.

If you think there might be some confusion about whether the multiple approaches all say the answer is 2/3 chance of winning by switching, it would be fine to emphasize this, i.e.

There are multiple approaches to solving the Monty Hall problem all giving the same result—that a player who swaps has a 2/3 chance of winning the car. Most popular sources present solutions based on simple probabilistic reasoning. A different approach to solving the problem, used primarily in academic sources, is to treat it as a conditional probability problem.

The previous section has clarified the problem, so there's really no point in saying "in the standard version of the problem" - although if you really insist on adding this that would be OK as long as you stop trying to imply the conditional approach is NOT addressing the standard problem. -- Rick Block (talk) 18:04, 6 June 2010 (UTC)

I am amazed. I said the answer is 2/3 in the standard (symmetrical) version of the problem because for some non-standard versions of the problem (host is defined to always open door 2 if possible within the rules) the answer is not 2/3. This is where it all started. We all agree this do we not? But I am happy to leave this (reference to the standard formulation not the 2/3 answer) out if you are.

I have never tried to imply that the conditional approach is not addressing the standard problem. If course it is. My complaint is that it is unnecessarily complicated.

Could we say 'certain academic sources' or 'statistical sources' or something. I believe the simple solution is common in many 'psychology' academic sources. Martin Hogbin (talk) 21:50, 6 June 2010 (UTC)

You're amazed? Perhaps you haven't been paying attention to what I've been suggesting as a compromise (which is a single solution section including both simple and conditional solutions to the standard problem that effectively complement each other - simple being easier to understand, conditional being clearly technically correct). And, (FYI) it seems to me like you've been arguing the conditional approach doesn't address the standard problem for two years. I'm amazed you're not only admitting the conditional approach addresses the standard problem but saying of course it does. Are you getting less opposed to having the simple and conditional solutions next to each other?

"Certain academic sources" implies a very limited number - it's just as accurate to say "numerous academic sources". If we're trying to be NPOV about this we don't want to exaggerate the number in either direction. How about "Most popular and non-mathematical academic sources ..." and then "used primarily in academic mathematical sources"? -- Rick Block (talk) 22:57, 6 June 2010 (UTC)

Rick, I have never argued that the conditional sources do not address the standard problem, simply that they are not necessary to solve it. I want to keep the simple and conditional solutions separate simply because the conditional solutions ore more complicated to explain and understand. Martin Hogbin (talk) 23:10, 6 June 2010 (UTC)

I looks as though we agree quite closely on this now, Rick. Why do you not make the heading level changes that you proposed and add in your intro:

There are multiple approaches to solving the Monty Hall problem all giving the same result—that a player who swaps has a 2/3 chance of winning the car. Most popular sources present solutions based on simple probabilistic reasoning. A different approach to solving the problem, used primarily in academic sources, is to treat it as a conditional probability problem.

I slip in my comment on this type of formulation here. I'm not happy with it. It seems to say that there are other solutions than the calculation of the conditional probability and that the latter is just academic. Every reader should be made aware that the simple solutions are not complete, but may serve as a way of understanding.Nijdam (talk) 11:44, 7 June 2010 (UTC)

As you know, I do not agree with you that the simple solutions are not complete in the symmetrical case but I am putting that disagreement to one side for the moment. Let us take it that you are correct. So, yes we should have something in the article that says that, but where should it go. I do not object to making the reader aware of this point, all I am trying to do is to prevent the reader from seeing another obstacle in their path towards understanding this difficult paradox. For that reason I would prefer to put anything which says that the simple solution is wrong after the simple solution.

Rick and I have just tried very hard to come up with a compromise statement and I have just agreed Rick's wording. If you absolutely must have something stronger before the simple solution section, what would you propose? Martin Hogbin (talk) 13:13, 7 June 2010 (UTC)

You may not have seen Rick's eloquently worded and highly and relevant posting from yesterday. I'll paste a portion of it here for your benefit, Nijdam:

"Do you think vos Savant's explanation about the MHP is correct? Frankly, Wikipedia doesn't give a shit what you think. This is completely the wrong question. Are there sources (not just one, but lots of them) that say vos Savant had her head up her ass? This is a question Wikipedia cares about. If the answer is yes, it really doesn't matter what you think - the article HAS to say what the sources say. It's not about what YOU think, or what I think, or what Martin thinks, or that Nijdam thinks - it's about what the SOURCES think." - Rick Block 6/6/2010

Glkanter (talk) 12:54, 7 June 2010 (UTC)

I would still like to consider making changes to the intro to address the following points: The answer is not 2/3 for some non-standard versions of the problem; many 'psychology' academic sources deal only with the simple problem and solution; and, after saying there are multiple approaches we only quote two of them. But I suggest that you add your wording and I propose changes until we find a form of words acceptable to all. Martin Hogbin (talk) 09:00, 7 June 2010 (UTC)

Rosenthal

I don't think it's accurate to describe Rosenthal as in Morgan's camp, or even in the conditionalist's camp. Monty Hall, Monty Fall, Monty Crawl Jeffrey S. Rosenthal (June, 2005; appeared in Math Horizons, September 2008, pages 5{7.)

Shaky Solution: When you first selected a door, you had a 1/3 chance of being correct. You knew the host was going to open some other door which did not contain the car, so that doesn't change this probability. Hence, when all is said and done, there is a 1/3 chance that your original selection was correct, and hence a 1/3 chance that you will win by sticking. The remaining probability, 2/3, is the chance you will win by switching. This solution is actually correct, but I consider it \shaky" because it fails for slight variants of the problem. For example, consider the following: Monty Fall Problem: In this variant, once you have selected one of the three doors, the host slips on a banana peel and accidentally pushes open another door, which just happens not to contain the car. Now what are the probabilities that you will win the car if you stick with your original selection, versus if you switch to the remaining door?

4 Monty Hall Revisited The Proportionality Principle makes the various Monty Hall variants easy. However, first a clarification is required. The original Monty Hall problem implicitly makes an additional assumption: if the host has a choice of which door to open (i.e., if your original selection was correct), then he is equally likely to open either non-selected door. This assumption, callously ignored by the Shaky Solution, is in fact crucial to the conclusion (as the Monty Crawl problem illustrates). With this additional assumption, the original Monty Hall problem is solved as follows. Originally the car was equally likely to be behind Door #1 or #2 or #3, and you selected Door #1 (say). The probabilities of the host then choosing to open Door #3, when the car is actually behind Door #1, Door #2, and Door #3, are respectively 1/2, 1, and 0. Hence, the updated probabilities of the car being behind each of the three doors are respectively 1/3, 2/3, and 0. That is, your chance of winning the car is 1/3 if you stick with Door #1, and 2/3 if you switch to Door #2.

http://probability.ca/jeff/writing/montyfall.pdf

Also, this solution presented above seems both simple and fully compliant with the host opening door #3 specifically:

The probabilities of the host then choosing to open Door #3, when the car

is actually behind Door #1, Door #2, and Door #3, are respectively 1/2, 1, and 0. Hence,

the updated probabilities of the car being behind each of the three doors are respectively

1/3, 2/3, and 0. That is, your chance of winning the car is 1/3 if you stick with Door #1,

and 2/3 if you switch to Door #2.

Glkanter (talk) 07:29, 6 June 2010 (UTC)

This solution is a simple conditional solution using Rosenthal's "proportionality principle" (which is really a consequence of Bayes' theorem). The reason this isn't a "featured" conditional solution in the article is because Rosenthal's "proportionality principle" is not a well known principle presented in most probability textbooks - unlike using a decision tree or directly using Bayes' Theorem (which are both completely standard). There are dozens (possibly hundreds) of sources that present conditional solutions to the MHP using a decision tree or Bayes' Theorem - and (AFAIK) only one using Rosenthal's "proportionality principle". However, it is definitely conditional, and he's definitely in "Morgan's camp". -- Rick Block (talk) 18:27, 6 June 2010 (UTC)

Rick, you offered a rant yesterday that features:

"Frankly, Wikipedia doesn't give a shit what you think. This is completely the wrong question. Are there sources (not just one, but lots of them) that say vos Savant had her head up her ass? This is a question Wikipedia cares about."

How is that consistent with your explanation above for not including this simple, conditional solution in the article? It's relates to door #3 being open, which is more than can be said for the current conditional solutions in the article. And it's equivalent to Grinsteads's last diagram of the solution. Glkanter (talk) 11:26, 7 June 2010 (UTC)

It's ONE source presenting this solution. If this was the only source about the problem and the only solution anyone had ever published, it would be included. However, there are many, many sources that present a more traditional conditional solution (either directly using Bayes' theorem or using a decision tree). Given the choice between this solution (one source) or the other solutions (many sources), the other solution wins. It's not based on whether editors agree with one source vs. another, but the preponderance of a view among sources. -- Rick Block (talk) 12:42, 7 June 2010 (UTC)

Morgan is only one source, and every other source you claim backs them up I have proven is not true. Also, other than the implied survey of all the literature revealing a count of 1, everything you wrote is merely your opinion. Rosenthal's method doesn't conflict with any source, and is similar to Grinstead's. Glkanter (talk) 12:59, 7 June 2010 (UTC)

So now, Rick, you're introducing the 'Wikipedia theory' that a source can be reliable in one aspect on a topic, but not another. You're more than happy to incorrectly use Rosenthal as a reliable source in support of Morgan, but then claim Rosenthal is not a reliable source worthy of having his solution in the article. You must be making this stuff up as you go along. Glkanter (talk) 13:38, 7 June 2010 (UTC)

As usual, Rick, you and I read the exact same words and come to different conclusions. Selected text and the link to the entire paper are provided, allowing any other interested editors to decide for themselves.

If we follow my suggestion below, there would be no 'labeling' of solutions at all. They're just reliably published solutions that, hopefully, guide the reader to understanding why it's not 1/2. Camps wouldn't be mentioned at all prior to any solutions. That sounds NPOV to me. Glkanter (talk) 18:48, 6 June 2010 (UTC)

Let us tackle the potential conditional nature of this problem head on.

I sincerely hope that this does not inflame the conditionalist more, it is not intended to, but suppose we treat the 'Simple solution' section as an 'idiots guide' to the problem and then, after an 'Aids to understanding section' specifically aimed at the idiots guide, proceed to tackle the whole thing from the start properly afterwords.

We could start then with a clear explanation of the difference between the unconditional (player chooses at the start) and the conditional problem.

I would want to say something like:

From the wording of Whitaker's statement it would seem that he wanted to know the probability of winning after a specific door had been opened by the host. This is not the same question as asking what would be the best strategy for the player to adopt without knowing which door the host has opened. In other words, the player decides whether to swap or not at the start of the game (ref Morgan and maybe Morgan unconditional statement). This is the, so called, unconditional formulation of the problem, It may be that this was what Whitaker actually wanted to know (ref Seymann) although it is not the way the question is interpreted in most sources. The popular solution given above answers this question perfectly.

If the host has to decide on whether to swap or not after the host has opened a door, it is, in principle, possible that that additional information might be revealed by the host's choice of door to open. This is the, so called, conditional problem that many academic sources address. These sources which argue that the popular solutions, do not properly address the conditional formulation of the problem (although the they get the correct numerical answer for the standard problem) and that a complete solution of the conditional problem requires the use of conditional probability

This is more like the older (and conditionalist) versions of the article but with an 'idiots guide' at the start. This format allows us to be more rigorous in defining the exact questions that we then answer, since the simplest case has already been dealt with. Martin Hogbin (talk) 13:39, 7 June 2010 (UTC)

Am I The Only Sane One Left?

Glkanter (talk) 20:07, 7 June 2010 (UTC)

Simple

For me a formulation of the simple "solution" along the following lines is all that is needed.

A simple way of understanding is to reason that the player initially picks the car with probability 1/3. Opening of one of the other doors with a goat, doesn't influence this probability, hence in this new situation, the car is also with probability 1/3 behind the chosen door. Switching thus must win the car with probability 2/3.

Note: It is not a given fact, but needs some proof, that in the new situation after a door has been opened the probability for the chosen door to hide the car is also 1/3.

(I extended it a little) Nijdam (talk) 07:37, 9 June 2010 (UTC)

You might think so, and in principle that is all I would want in the 'Popular solution' section, however, the experience of many people who have tried to explain the problem shows that, for some reason, people just do not believe the answer when they are told it. How much we can do about that I am not sure but I suggest that we need at least two different approaches to the simple solution, with diagrams or pictures.

What are your thoughts on my suggestion above, that with the simple solution out of the way we can get on with a more serious detailed discussion of the problem? I suppose you have just given them here. Martin Hogbin (talk) 18:37, 7 June 2010 (UTC)

Morgan Camp vs Conditional Camp

I think to be considered in the Morgan Camp, a reliably published source must make these 3 statements:

All simple solutions are false

The problem must be solved with a formal conditional proof

Only Morgan's paper provides what Morgan calls the 'correct resolution' to the problem

Anything short of that, and I would say the source insists on a solution where door #3 has been opened. Glkanter (talk) 19:36, 6 June 2010 (UTC)

Don't be ridiculous. Even Morgan wouldn't be in this camp. Your suggestion that Morgan insists their paper is the only "correct resolution" to the problem is completely your own invention. What we've been calling this "camp" before is:

1. Unconditional solutions don't address the stated problem (this is what Morgan et al. mean by "false")

2. The problem requires a conditional solution (Morgan et al. say nothing about "formal" or "proof")

Others (specific papers or books) in this camp are

Gillman (full citation in the article)
Rosenthal Monty Hall, Monty Fall, Monty Crawl
Grinstead and Snell (full citation in the article)
Eisenhauer The Monty Hall Matrix
Lucas, Rosenhouse, and Schepler The Monty Hall Problem, Reconsidered
Falk (full citation in the article)

Note that there's another, much larger camp of sources who simply present a conditional solution without any comment about unconditional approaches (i.e. #2 from above, but not #1) - this camp includes many (if not most) probability textbooks. Here's a VERY incomplete list:

J. Gill (full citation in the article)
Gelman, http://books.google.com/books?hl=en&lr=&id=TNYhnkXQSjAC&oi=fnd&pg=PR16&dq=monty+hall+bayesian&ots=5G6NeDAxI6&sig=busR1cY8NblIhAMDvX8dYImrRsw
G Koop, DJ Poirier, JL Tobias, http://books.google.com/books?hl=en&lr=&id=1EmjUH7fjJcC&oi=fnd&pg=PA1&dq=monty+hall+bayesian&ots=lIQPivVhhs&sig=8ZcFHSwDhWFXX2WYgqkVFvRUcac
G de Cooman, M Zaffalon - Artificial Intelligence, 2004 - Elsevier, http://linkinghub.elsevier.com/retrieve/pii/S0004370204000827
LA Wasserman - 2004 - stevereads.com, http://stevereads.com/papers_to_read/all_of_statistics.pdf
T Slembeck, JR Tyran - Journal of Economic Behavior & Organization, 2004 - Elsevier, http://linkinghub.elsevier.com/retrieve/pii/S0167268103001719
B Fitelson - Synthese, 2007 - Springer, http://www.springerlink.com/index/Q19N116NL777N361.pdf
PR Mueser, D Granberg - Retrieved December, 1999 - Citeseer, http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.42.6708&rep=rep1&type=pdf
M Carlton - Journal of Statistics Education, 2005 - amstat.org, http://www.amstat.org/publications/JSE/v13n2/carlton.html
SD Hales - Southern Journal of Philosophy, 1995 - bloomu.edu, http://www.bloomu.edu/departments/philosophy/pages/content/hales/articlepdf/closure.pdf
BP Kooi - Journal of Logic, Language and Information, 2003 - Springer, http://www.springerlink.com/index/l418q1628018g4v6.pdf
M Hall - 2004 - uni-konstanz.de, http://www.uni-konstanz.de/ppm/summerschool2004/slides/Korb1.pdf

Rick Block (talk) 22:12, 6 June 2010 (UTC)

I thought we were not going to count refs? Because if so I would like to include every psychology paper that uses a simple solution when defining the problem. Colincbn (talk) 03:52, 7 June 2010 (UTC)

That being said I don't have a problem with the current state of the article in a larger sense. I have not read through each sentence in awhile so there might be some wording I would change but I think it is generally OK. I might expand the sources of confusion section with new info from some of the psych papers I have read recently but that's about it. Oh, and that weird Nash Equilibrium bit in the variants section. Cheers, Colincbn (talk) 04:11, 7 June 2010 (UTC)

Colin, can you suggest names for the two types of academic paper. I can only think of 'statistics' and 'psychology'. If we can agree on appropriate names for the two source types we can give an accurate description of the true situation. Martin Hogbin (talk) 14:23, 7 June 2010 (UTC)

@Rick, I looked at the last item on your list(M Hall - 2004). It gives a very good explanation of why the probability that the car is behind door 1 is unchanged by the opening of a door by the host, thus supporting the simple solution. Can you tell me where it uses a conditional solution please. Martin Hogbin (talk) 23:27, 8 June 2010 (UTC)

You're presumably talking about page 31. On this page, the difference between P(A) and P'(A) is that P'(A) is the conditional probability the car is behind door A after the host opens door B, as opposed to P(A) which is the probability the car is behind door A before the host opens a door. The argument on this page is the symmetry argument for why P'(A) must be the same as P(A) (which no one has ever said is an invalid argument - the issue is the sources typically presenting simple solutions never use it). This argument is immediately followed up (on pages 32-33) with a full Bayesian analysis as well. If you're not reading this as only presenting a conditional solution, you're not reading it correctly. -- Rick Block (talk) 16:37, 9 June 2010 (UTC)

There is always the trivial condition that a door has been opened. If you want to say that that makes the problem conditional that is fine. The conditional solutions that I object to are those that indicate that it might make a difference to the answer which door the host opens. You have now presented two sources which clearly state that it does not matter which door the host opens if the chooses a goat door randomly. In other words there is just a trivial condition (what I have called a null condition) that the host opens any unchosen door to reveal a goat. The point is that we do not need to consider which door he opens. In other words the fact that the host has in fact opened door 3, rather than door 2, is unimportant and does not need to be considered in any calculation or explanation. Martin Hogbin (talk) 22:54, 9 June 2010 (UTC)

But this is STILL a conditional argument. -- Rick Block (talk) 03:23, 10 June 2010 (UTC)

Flawed Conditional Solutions In The Article

My understanding of the conditionalists' criticism of the simple solutions is that they do not address Selvin's and Whitaker's actual scenario: Two closed doors and one open door.

In this section of this talk page, a couple of editors explain to me how both Selvin's and vos Savant's tables fail in this regard, as they do not exactly meet that criteria.

Please take a look at the current Probabilistic solution section where the door situation requirement is also stated. Neither Chen's tree diagram or the table that follow show the 'required' situation. Chen's tree diagram shows two open doors and no closed door, while the second (OR?) diagram shows all three doors open.

Interestingly, Devlin's Combined Door solution in the Popular solution section shows exactly the scenario as described by Selvin and Whitaker, two doors closed, and door #3 open.

The Probabilistic portion of the article then, is not meeting the very standard called for by the conditionalists. And it makes me wonder just what their criticism of the Combining Doors solution is all about.

But, let's just presume everything is reliably published, except for maybe the 2nd diagram in the Probabilistic section. Wouldn't it be better for the reader if one or more of the reliably published conditional solutions (Grinstead's and Rosenthal's come to mind) that *do* properly reflect Selvin's and Whitaker's scenario were in the article instead of the 2 'false' solutions provided? Glkanter (talk) 07:43, 8 June 2010 (UTC)

I see that the editors who are most vocal about the conditional solution have posted elsewhere on this page since I posted the above, but have chosen not to offer their solutions to this problem. I guess I'll replace the current erroneous 'Probabilistic solution' section in the article with Rosenthal's reliably sourced simple conditional solution tomorrow. Glkanter (talk) 21:11, 8 June 2010 (UTC)

The existing solutions absolutely address the case of interest. -- Rick Block (talk) 00:26, 9 June 2010 (UTC)

Rick, I read your brief, unsupported comment above as being directly contradictory with your earlier comment here:

"The point here is this table does not address the case where the player has picked door 1 AND the host has opened door 3. It addresses all cases where the player has picked door 1."

It's obvious that Chen's table suffers from exactly the same flaw you describe. That you can't see, or perhaps, admit this, is another example of the irrationality that I have been struggling with. Glkanter (talk) 01:05, 9 June 2010 (UTC)

Actually, it's obvious Chen's diagram shows ALL possibilities assuming the player has initially picked door 1. For example, following the top fork at every fork we see the case where the player has picked door 1 with the car behind door 1, and the host then opens door 2. What this diagram shows is that if you've picked door 1 there are exactly 4 different cases:

1. car is behind door 1 and host opens door 2

2. car is behind door 1 and host opens door 3

3. car is behind door 2 and host opens door 3

4. car is behind door 3 and host opens door 2

Nothing else can happen. The diagram also shows the probability of each of these, 1/6, 1/6, 1/3, and 1/3 respectively (because it's everything that might possibly happen these probabilities should and do sum to 1). Using a complete diagram like this to calculate conditional probabilities is a completely standard technique (Grinstead and Snell walk through this same example using a version of the same diagram without assuming the player has initially picked door 1). In this case, to calculate the conditional probability of winning by switching given the host opens door 3, you add up the probability for all cases where you win if the host opens door 3 (there's only one, it's 1/3) and divide by the sum of the probabilities in all cases where the host opens door 3 (there are two, one is 1/6 and the other is 1/3).

Now look at vos Savant's table. It also assumes the player has initially picked door 1 and shows the probability of where the car is in every possible case (1/3 behind each door). Fine so far. But, now the host opens door 3. Hmmm. What lines to I look at and what are the probabilities? The second line clearly applies (car behind door 2) - there's a 1/3 chance of this happening. The first line applies as well (car behind door 1). Is the probability 1/3? This seems to be what the table says. If I use this table the way complete probability diagrams are meant to be used, I might just say the probability of winning is the probability the car is behind door 2 (1/3) divided by the probability the car is behind either door 1 or door 2 (1/3 + 1/3). I assume you aren't going to argue this is the right answer. So, either a completely standard technique for computing conditional probabilities has been shown to not work (0% chance of this), or vos Savant's table is not showing a complete picture of what might happen (100% chance of this). -- Rick Block (talk) 16:23, 9 June 2010 (UTC)

Yes, as I said above, you are intentionally misrepresenting vos Savant's table. Fine. Play the fool, if you choose. There are plenty of other simple solutions to choose from. We already understand that vos Savant's solution is considered flawed by some sources. What about Devlin's Combined Doors solution? How is that flawed? The irony is that the 'conditionalists complaint' about some 'simple' solutions, as you put it above,:

"The point here is this table does not address the case where the player has picked door 1 AND the host has opened door 3. It addresses all cases where the player has picked door 1."

exactly describes Chen's tree and the 2nd diagram. Amazingly, here's your new defense of Chen's tree and the 2nd table:

"Actually, it's obvious Chen's diagram shows ALL possibilities assuming the player has initially picked door 1."

Yes, that is precisely the problem!!!!! YOU HAVE NO ARGUMENT, RICK!!! Selvin's solution shows all possibilities, too. But somehow, it's considered 'false'. How on Earth can you deny that? The 2nd diagram doesn't even consider that door 1 has been chosen by the contestant. I'm trying to make the conditional solution in the article consistent with your words above. A table representing Rosenthal's solution would do exactly that. Why not use a reliable source that meets the requirement, rather than one that merely mimics and perpetuates the alleged flaw? I have considered your argument for a day now, and find it hopelessly nonsensical, without any value, meaning, or merit. I will, accordingly, edit the article for the benefit of the reader, to make it consistent with the conditionalists' published concerns. Glkanter (talk) 17:39, 9 June 2010 (UTC)

vos Savant's table addresses all cases where the player has picked door 1 without distinguishing the cases where the player has picked door 1 AND the host opens door 2, or the player picked door 1 AND the host opens door 3. Sorry if that wasn't clear. Chen's diagram and the figure both show everything that can happen (in any combination), assuming the player has initially picked door 1. vos Savant's table does not. If you want to know what is the probability the host opens door 2 (assuming the player picks door 1), Chen's diagram lets you figure this out. vos Savant's table does not. Similarly, if you want to know the probability of winning given you've picked door 1 and the host has opened door 3, Chen's diagram lets you figure this out. vos Savant's table does not. -- Rick Block (talk) 18:00, 9 June 2010 (UTC)

Why I (regretfully) Post So Often

In a word, it's because Rick Block and Nijdam are so full of shit and have their heads up their respective asses.

I presume it's OK to use those words now. Here's a diff from Sunday where Rick uses those words 6 times. That's a lot more than once, why, it's even a 'preponderance'.

For just one recent and obvious example, Rick claims many other sources support Morgan. Here's what alleged 'Morgan Supporter' Rosenhouse says:

Four people who were less impressed were mathematicians J. P. Morgan, N. R. Chaganty, R. C. Dahiya and M. J. Doviak (MCDD). Writing in The American Statistician, they presumed to lay down the law regarding vos Savant's treatment of the problem. After quoting the original question as posed by vos Savant's correspondent, they write: [beginning of quote - Glkanter] Marilyn vos Savant, the column author and reportedly holder of the world's highest I.Q., replied in the September article, \Yes, you should switch. The first door has a 1/3 chance of winning, but the second door has a 2/3 chance." She then went on to 46 CHAPTER 1. ANCESTRAL MONTY give a dubious analogy to explain the choice. In the December article letters from three PhD's appeared saying that vos Savant's answer was wrong, two of the letters claiming that the correct probability of winning with either remaining door is 1/2. Ms. vos Savant went on to defend her original claim with a false proof and also suggested a false simulation as a method of empirical verification. By the February article a full scale furor had erupted; vos Savant reported, \I'm receiving thousands of letters nearly all insisting I'm wrong.... Of the letters from the general public, 92% are against my answer; and of letters from universities, 65% are against my answer." Nevertheless, vos Savant does not back down, and for good reason, as, given a certain assumption, her answer is correct. Her methods of proof, however, are not. [end of quote Glkanter] Rather strongly worded, wouldn't you say? And largely unfair, for reasons I have already discussed. Indeed, continuing with their lengthy essay makes clear that their primary issue with vos Savant is her shift from what they call the \conditional problem," as posed by her correspondent (in which it is stipulated that the contestant always chooses door one and the host always opens door three) to the \unconditional problem," in which we stipulate only that after the contestant chooses a door, the host opens one of the goat concealing doors. She did, indeed, make this shift, but this was hardly the point at issue between vos Savant and her angry letter-writers.

Yep, that's one of the sources Rick still claims is a 'Morgan Supporter', verbatim. I've successfully disproven his claims about many of the others on these talk pages as well.

So, please don't be a Glkanter hater. In a rational world, the article would have been improved long ago, and all this discussion would never have been needed. Glkanter (talk) 18:55, 8 June 2010 (UTC)

Rick also repeatedly makes the claim that no reliable source has ever challenged Morgan. Obviously, as shown above, that's not true, either. Glkanter (talk) 18:58, 8 June 2010 (UTC)

For old time's sake, here another classic Rick comment :

"...Glkanter asks why I haven't responded about his "Is The Contestant Aware?" question. Why should I? Glkanter has repeatedly demonstrated a complete lack of comprehension of nearly everything I've ever said. It's like trying to explain something to a cat. At some point you just have to give up. However, I'll give it another go. Meow, meeeow, meow, meowww. I'm not sure I have that quite right since I don't speak cat, but it's probably about as comprehensible to him as anything else I could say..." Rick Block (talk) 01:53, 4 December 2009 (UTC)

Like I say, whatever it is that's holding up the improvement of the article isn't Glkanter, and it isn't rational. Glkanter (talk) 19:06, 8 June 2010 (UTC)

Ths Rosenhouse source I still claim is a Morgan supporter is The Monty Hall Problem, Reconsidered, co-authored by Lucas, Rosenhouse, and Schepler, which I've quoted plenty of times. -- Rick Block (talk) 20:11, 8 June 2010 (UTC)

Ah, the one that:

Makes no mention of Morgan whatsoever

Includes this initial premise:

"He chooses his door in accordance with the following rules:"

"3. If Monty can open more than one door without violating rules one and two, then he chooses his door randomly."

Identifies the 'source of confusion' as:

"Why All the Confusion? The trouble, you see, is that most people argue like this: “Once Monty opens his door only two doors remain in play. Since these doors are equally likely to be correct, it does not matter whether you 2 switch or stick.” We will refer to this as the fifty-fifty argument."

Then spends the rest of their time describing different problems, also referred to as 'variants' in regards to conditional probability

So, I'll add this to the ever growing list of your other claimed 'Morgan supporter' sources which you have forced me into debunking on these talk pages. Glkanter (talk) 20:40, 8 June 2010 (UTC)

Yes, the one that says:

Take this as a cautionary tale. Whether we are playing Classic Monty or High-Numbered Monty, it is certain that Monty will open a goat-concealing door. In the former case the probability that our initial choice concealed the car did not change while in the latter case it did. This shows that any proposed solution to the MHP failing to pay close attention to Monty’s selection procedure is incomplete.

Incomplete. As in what a mathematician would call "false". -- Rick Block (talk) 21:10, 8 June 2010 (UTC)

That's exactly the irrationality I describe above, Rick. Rosenhouse calls Morgan's paper 'unfair' among other criticisms in a separately published paper, and says the controversy is only about 1/2 vs 1/3 in the same paper. You ignore these, refer to a simple re-statement of mathematical fact, and against all evidence to the contrary, you erroneously conclude that the 3 authors, including Rosenhouse, support Morgan. Glkanter (talk) 00:55, 9 June 2010 (UTC)

This last quote may be referring to Monty's stated selection procedure, that is to say as given in the problem statement. For example, in the variant, High-Numbered Monty, the problem statement says that, 'This time however, we stipulate that Monty always opens the highest numbered door available to him'. Obviously, if this is stated in the problem statement we must pay attention to it. In the standard problem statement there is no such stipulation. Martin Hogbin (talk) 22:03, 8 June 2010 (UTC)

In fact Rosenhouse makes clear that, 'In the former case [referring to what he calls Classic Monty, the standard problem statement] the probability that our initial choice concealed the car did not change...[when Monty opens a goat-concealing door]'. In other words, in the standard MHP, the probability that the car is behind door 1 remains at 1/3 after the host has opened a door, thus the simple solutions hold good. We now have a good source which confirms this fact. Martin Hogbin (talk) 22:22, 8 June 2010 (UTC)

Martin - no one here is arguing that in the standard version the numeric value of the probability the car is behind the player's door before the host opens a door is different from the numeric value of the conditional probability after the host opens a door. What is at issue is the REASONING. As I read it, what Lucas, Rosenhouse, and Schepler are saying here is that the REASON these two probabilities are numerically equal is NOT simply because we know Monty will open a door revealing a goat. Here's what they say again:

Take this as a cautionary tale. Whether we are playing Classic Monty or High-Numbered Monty, it is certain that Monty will open a goat-concealing door. In the former case the probability that our initial choice concealed the car did not change while in the latter case it did. This shows that any proposed solution to the MHP failing to pay close attention to Monty’s selection procedure is incomplete.

That you apparently want to use this in support of "simple" solutions that use this very reasoning is perverse. -- Rick Block (talk) 16:58, 9 June 2010 (UTC)

Later on LRS make clear whet they mean when they state this principle:

'The second principle is that if the doors different from your present choice are equiprobable, then the probability of your choice does not change when Monty opens a door.' This is quite clear in its support for the simple solution, which says that when Monty opens one of two equiprobable doors, the chance that the door 1 hides the car remains at 1/3. Martin Hogbin (talk) 22:45, 9 June 2010 (UTC)

Martin - they're talking about an n-door variant where the host is constrained to pick randomly among all legal choices (any door with a goat that is not the player's door). With this rule, this principle holds. They are paying CLOSE attention to how the host chooses among losing alternatives - not ignoring how this choice is made (like the simple solutions do). Again - your attempt to use this to "support" the simple solution is perverse. Is there something not clear about "any proposed solution to the MHP failing to pay close attention to Monty’s selection procedure is incomplete". -- Rick Block (talk) 03:15, 10 June 2010 (UTC)

It is no use continually re-quoting a small section of the paper. If you read it all you will see that they are trying to arrive at a general principle for solving Monty Hall type problems. To arrive at this general principle, they consider a variety of variants but eventually conclude that ' ... if the doors different from your present choice are equiprobable, then the probability of your choice does not change when Monty opens a door' as a general principle applicable to all cases. Martin Hogbin (talk) 15:55, 10 June 2010 (UTC)

(OUT)Would the LRS form of solution be a good one to start with? I.e.: The host has opened door No. 3. If the car is behind door No. 2, he is forced to. This happens in 1/3 of the cases. If the car is behind door No. 1, which happens in 1/3 of the cases, he has a choice. Half the times he opens door No. 3, hence in 1/6 of the cases. When door No. 3 is opened the car is twice as often behind door No. 2 than behind door No. 1. Nijdam (talk) 16:37, 10 June 2010 (UTC)

I am not sure I would start with the LRS solution but my opinion is that the more (sourced) angles we can approach the problem from the better. Different explanations work for different people. Martin Hogbin (talk) 18:05, 10 June 2010 (UTC)

And now for something completly different

Ok, I have waited before bringing this up here because I felt there are larger things to worry about on this page. But I have finally cracked! What is up with the statement "The host is rewarded whenever the contestant incorrectly switches or incorrectly stays = Switching wins 1/2 the time at the Nash equilibrium"??? I brought this up on Richard Gill's talk page and on Rick's. Rick tracked the edit down to a mathematician in Pennsylvania (good work that) but he has not responded to Rick's inquiry as to what it means. The way it reads to me seems to say that if the producers give Monty a cookie (or a lollipop or whatever) every time the guest messes up they never have to worry about giving anyone a 2/3 chance to win. That seems totally wrong but I know zero about game theory, except that one of its shining stars talked to imaginary people that is... Colincbn (talk) 14:39, 10 June 2010 (UTC)

Professor Foster has responded in email. He said he presented this as an exercise in class or for homework and doesn't remember seeing it published anywhere (and is fine with deleting it as unreferenced). The Mueser and Granberg paper referenced in the article has an appendix discussing the Nash equilibrium assuming the host can freely choose which door to open and whether to make the offer to switch (and wants the player to lose). The equilibrium in this case is not unique. One is the host only opens a door and makes the offer to switch if the player initially picks the car (in which case, if the player switches she loses). Another is the host opens a door and makes the offer to switch all the time if the contestant initially picked the car, and half the time if she didn't. In this case, the player has a 50% chance of winning by switching. -- Rick Block (talk) 16:09, 10 June 2010 (UTC)

Wow, awesome! I think some more explanatory wording is required in the "Nash Equilibrium" variant to explain that, but at least that makes sense. Thanks for figuring this out!! Colincbn (talk) 23:10, 10 June 2010 (UTC)

As the different solutions are equivalent

As the different solutions are equivalent for the random host who always offers the switch, there is no risk of over-simplification, or coddling, or mis-leading the reader by presenting the simplest solutions first, and without caveats.

All reliable sourced solutions and their controversies would, of course, be addressed subsequently in the article.

Maybe even as I proposed in the section called How's This. Glkanter (talk) 01:48, 11 June 2010 (UTC)

Simple Solution Criticism

It's important to understand that there are (at least) 6 separate arguments about the simple solutions:

1. They answer a different problem than that posed by Whitaker.

Response - Per Wikipedia, it doesn't matter if this is true, false or otherwise. This been reliably published.

2. Presenting simple solutions first in the article does a disservice to the reader.

3. Presenting simple solutions first in the article violates Wikipedia NPOV.

Maybe a mediator could set up an area where we present our 'sides' of these binary issues, in order to facilitate some agreement and editing. Glkanter (talk) 02:23, 11 June 2010 (UTC)

You're missing the essence of all of these arguments, which is that there are multiple reliable sources that say the simple solutions answer a slightly different question than the one asked in the MHP. Presenting these solutions without indicating this to the reader does the reader a disservice, and presenting these solutions in a manner implying they are undisputed violates Wikipedia's NPOV policy. -- Rick Block (talk) 03:10, 11 June 2010 (UTC)

Well, if the mediator likes my suggestions, you'll have your opportunity to present that argument in a forum where we can make progress. I have no idea what you're talking about when you say I'm missing something. I simply reject your Straw Man arguments as meaningless. Glkanter (talk) 03:33, 11 June 2010 (UTC)

Additional arguments:

4. Enumerate the distinct 'camps' of simple solution criticism.

5. Is there value to a survey or headcount of each camp? Who would determinations which camp a source is in? How would such census information affect the article?

Glkanter (talk) 04:30, 11 June 2010 (UTC)

Additional arguments:

6. Enumerate the different 'classes' of simple solutions. (This should probably be item #1)

Glkanter (talk) 06:40, 11 June 2010 (UTC)

Another attempt at a compromise.

As we got so close before and everyone agrees that the article needs improvement, let me put my last proposal again with one important change and one clarification. The changes from before are marked in bold bold. I have also added 'Other stuff', much as we have now. I believe this is mainly non-contentious.

Third proposal

The lead

Fine as it is.

The problem

As it is with but without the paragraphs starting: 'Without a clear understanding...', and 'Suppose you’re on a game show...'. These start to complicate the problem and could be included in the 'Academic solution' section

Clarification I forgot that two paragraphs started with the same words. I am only suggesting removing the W&W problem statement. We just give Whitaker's vague statement, which gives us a little flexibility

Simple solution section

Change An introductory statement, written by Rick (or other conditionalists) in as neutral terms as possible as a temporary measure, where it is clearly understood that this statement may be revised, rewritten, or even removed at a later date, subject of course to a consensus to do so. No doubt there will be endless argument about this but at least the article gets improved while we do this.
This must make every effort to convince the new reader that the answer to the problem is 2/3 and not 1/2 by using a variety of simple solutions, (all supported by sources).
There must be no confusion of the reader with statements along the lines that these solutions are incomplete, or do not answer the question as asked, or answer only the unconditional formulation of the question (even if we accept that this is the case).
There must be no mention of the fact that the host has a choice of door when the player has originally chosen the car, this choice should not be shown in any diagrams.

Aids to understanding

Much as it is now, this section must concentrate on the helping the reader understand why the simple solution is correct, in particular, why the fact that the host knows what is behind the doors matters.

Again there must be no confusion of the reader with statements along the lines that these solutions are incomplete, or do not answer the question as asked or answer only the unconditional formulation of the question (even if we accept that this is the case)

Question from a reader

I am a new to this problem and I am now one of the "stay put" readers you are trying to help. I'll try paraphrasing the solution and the point where I get lost. At that moment of revelation, the revealed goat door changes from being a 1:3 chance of having a car to becoming a 0:3 chance of having a car. Since that door has now changed from a 1:3 to a 0:3, that recovered 1:3 must be reallocated to the remaining 2 doors. Then the solution states this lost 1:3 must be totally allocated to the door I am not next to, and the reason given is I had a 2:3 chance of being wrong before the moment of revelation. This is where I get lost in the argument. Why would I look at historical possibilities of being wrong instead of resetting current (post moment of revelation) possibilities of being wrong which gives a 2 doors, 2 options answer? Isn't the moment of revelation a discontinous point and they are 2 seperate equations? So in summary I think if you can explain why the "old" odds still apply it would help readers. Thanks for your consideration and good luck! Martin Gaskell (talk) 11:44, 11 June 2010 (UTC)

Welcome and thanks for your post. It is always useful to have readers who are new to the problem. I think you have put your finger on the difficulty many people have with this problem. If I understand you correctly, you are asking why, when the host opens a door to reveal a goat, the odds of the car being behind the originally chosen door do not change but the odds of it being behind the other unopened door do change. Is that correct? I have a particular interest in making the explanation in the article clear and convincing.

May I ask you this question, which is relevant to some of the discussion going on here but maybe not of much interest to you. If the host had opened door 2 instead of door 3 would you expect the answer to be any different?

To get back to you question, I am sure that someone here will be able to convince you of the correct answer eventually but, if you do not mind, I would like to elicit some more opinions on the article from you first. The problem with the long term editors here is that none of us can remember what it is like not to understand the problem. So, could I ask you to have a look through the article again and say if there is anything in it anywhere that starts to convince you that the probability of winning by switching is 2/3. Martin Hogbin (talk) 13:37, 11 June 2010 (UTC)

And, if you would be willing, can you read through the text proposed in the show/hide section at Talk:Monty Hall problem/Archive 16#A genuine attempt at compromise.? In particular, do the paragraphs relating the solutions to some number of contestants ("if 900 contestants are on the show, ...") help? -- Rick Block (talk) 17:18, 11 June 2010 (UTC)

Academic solution

Here we give Morgan's solution and fully and completely explain the issues involved and exactly why some sources consider that the simple solution is incomplete, does not answer the question as asked, or deals only with the unconditional formulation.

Further details about the problem and its possible interpretations
A clear note about conditional probability stating why some consider the simple version to be incomplete/incorrect
There is no attempt to bury or hide the points raised by Morgan and other sources.
The effect of various host door choice policies must be considered here.
The prevalence of this solution in certain types of reliable source can be made clear.
Equal space and prominence (including an additional 'Aids to understanding' section if desired) to the simple solutions section will be given to this section.

Other stuff

Much as now we have variations and might also consider a game theory section. Martin Hogbin (talk) 12:32, 26 May 2010 (UTC)

Martin - When you say "Morgan's solution" do you mean specifically the conditional solution to the problem as stated by Whitaker under Morgan's (and Gillman's) interpretation that the host preference be taken as a variable (i.e. the probability of winning by switching is 1/(1+q)), or do you mean ANY solution based on conditional probability? The 1/(1+q) solution has already been moved to the Variants section (in a previous attempt to compromise). Up to this section, the article is already only talking about the symmetric problem (as precisely defined by the K&W problem statement). Many, many sources approach the problem conditionally, taking the question to be about

P({\text{win by switching}}|{\text{player picks door 1 and host opens door 3}})\,

It sounds to me like you're simply continuing to suggest the article relegate ALL of these sources to the "academic solution" section, i.e. that the article endorse your POV. This is not a compromise in my book. -- Rick Block (talk) 14:02, 26 May 2010 (UTC)

I agree. I see no real merit in this suggestion either. If we remove any mentioning of the conditional approach and the more detailed discussion of the difficulties of the problem formulation from the "problem"-chapter, then the chapter becomes pointless, i.e. we might as well remove it completely and just have the lead. The "aids to understanding" or "sources of confusion" chapters need to refer to the conditional solution as well as various sources connect the difficulty to get the correct solution with our difficulties to work with conditional probabilities. The only way that I see to avoid this is splitting those chapters in an simple and conditional parts or have these parts as subchapters in the chapters for the simple and conditional solutions.--Kmhkmh (talk) 16:40, 26 May 2010 (UTC)

This is exactly the same as my earlier proposal, to which your only objection was the intro to the the simple solution section. Martin Hogbin (talk) 14:45, 26 May 2010 (UTC)

When I proposed this earlier the only objection raised was to the wording of the intro to the simple section. Now I have given a little, you object to even more. This is not how negotiation is supposed to work. Martin Hogbin (talk) 17:45, 26 May 2010 (UTC)

Actually that's exactly what you get, if you run a lousy negotiation. People simply get fed up and reevaluate their earlier alleged or real concessions. And Andrevan's suggestion for live discussion between you and Rick in particular was exactly to avoid this constant moving forward and backward.--Kmhkmh (talk) 18:22, 26 May 2010 (UTC)

I am the only person actually trying to do something to reach a compromise. I you think you can do better please have a go yourself. The advantage of a wiki discussion is that everyone can see who has moved backwards. Martin Hogbin (talk) 19:12, 26 May 2010 (UTC)

There a big difference between claiming and actually doing. And in particular regarding your statements people can indeed just read the wiki discussion to judge for themelves, what you claim and what you actually do. Be assured I've done so.--Kmhkmh (talk) 19:35, 26 May 2010 (UTC)

I am not sure what you mean. What have I claimed to have done but not done? Martin Hogbin (talk) 20:10, 26 May 2010 (UTC)

Martin - Would you please answer the question I asked - when you say "Morgan's solution" do you mean specifically the conditional solution to the problem as stated by Whitaker under Morgan's (and Gillman's) interpretation that the host preference be taken as a variable (i.e. the probability of winning by switching is 1/(1+q)), or do you mean ANY solution based on conditional probability? Thanks. -- Rick Block (talk) 23:47, 26 May 2010 (UTC)

I mean any solution based on conditional probability or any mention of the word 'conditional'. It was the clear intent of Selvin to avoid the conditional aspect of the problem and I doubt that Whitaker cared about the subject. The majority of sources do not mention conditional probability. It is a distraction from the essential paradox. Martin Hogbin (talk) 10:20, 27 May 2010 (UTC)

That is false claim that you repeat over and over. Correct is that many/most reputable sources which deal with problem in any detail do use the conditional solution. And yes I didn't not count arbitrary websites, puzzle books, general interest newspapers or the yellow press as they are not reputable sources in this context. Moreover conditional probabilities are at the core for some aspects of the paradox. If you insist on a such a position, there isn't really anything to mediate.--Kmhkmh (talk) 11:39, 27 May 2010 (UTC)

Rosenthal descibes it as...

Rosenthal, an incorrectly alleged 'Morgan Supporter' says the odds of a car begin as (1/2 + 1/2), 1, 1 respectively for doors 1, 2 & 3. This reduces to 1/3, 1/3, 1/3.

The simple solutions says the odds of a car begin as 1/3, 1/3, 1/3 respectively for doors 1, 2 & 3.

Afterward, Rosenthal says it's (1/2 + 0), 1, 0. Which reduces to 1/3, 2/3, 0.

Afterward, the simple solutions say it's 1/3, 2/3, 0.

Glkanter (talk) 13:49, 10 June 2010 (UTC)

Exactly. This is what all the conditional solutions say. -- Rick Block (talk) 15:40, 10 June 2010 (UTC)

Perhaps. Then all of Morgan's rantings about a 'host bias', vos Savant using 'dubious' methods, and 'false solutions' amount to nothing. Certainly in light of Selvin's 2nd letter. Glkanter (talk) 15:53, 10 June 2010 (UTC)

And I don't recall ever seeing it stated Rosenthal's way, either in the article, or on these talk pages. Glkanter (talk) 23:54, 10 June 2010 (UTC)

Well, the conditional solutions all say the answer is 1/3, 2/3, 0 if (and only if) you say or assume the host chooses randomly between two goats. The simple solutions say the answer is 1/3, 2/3, 0, saying nothing about how the host chooses, meaning they say the answer is 1/3, 2/3, 0 even if the host doesn't choose randomly. It's like looking at a watch that's stopped at 1:45 to find out what time it is. If you look when it actually is 1:45 the watch says the right time. But, this doesn't mean the watch is working. Similarly, the simple solutions say the right answer if the host chooses randomly between two goats - but if not they don't. What's actually happening is the simple solutions answer a different, but related, question. -- Rick Block (talk) 17:54, 10 June 2010 (UTC)

No, the simple solution answers only the exact question asked, meaning only the case in which the host chooses a goat door randomly, it fails in other cases. Martin Hogbin (talk) 18:11, 10 June 2010 (UTC)

Are you arguing the simple solutions don't literally address the unconditional probability (ignoring the host preference), i.e. "what is the probability of winning if you decide to switch before seeing which door the host opens" - or, are you saying this is not a different question if this question and the conditional question have the same numeric answer? -- Rick Block (talk) 19:07, 10 June 2010 (UTC)

I am arguing that the simple solutions fully address the unconditional problem in all cases, and the conditional problem in the case that the host chooses a goat door randomly. Whether these two problems are exactly the same I would not like to say, but they are clearly equivalent and the same solutions apply to both. Martin Hogbin (talk) 19:25, 10 June 2010 (UTC)

A stopped watch correctly tells you what time it is if you're very careful about what time you look at it. It sounds like you'd argue it's working at those times. -- Rick Block (talk) 20:04, 10 June 2010 (UTC)

Here's the way to both salvage and discredit Rick's original watch analogy. For Selvin's problem, where he does not mention, but solves the puzzle, assuming the host's random behavior (same as vos Savant) until his 2nd letter, the 'two clocks' (simple/unconditional and conditional solutions) always report the exact same time, at all times. Down to whatever level of precision you can determine. How either one of the 'clocks' can then be called 'broken' is inconceivable. Glkanter (talk) 23:26, 10 June 2010 (UTC)

And despite Rick's dubious, false mischaracterization above: "saying nothing about how the host chooses...even if the host doesn't choose randomly", plenty of simple solutions, beginning with Selvin's, state that the host chooses randomly. And it's a reasonable, unstated premise for a puzzle involving a game show, anyways. Glkanter (talk) 23:54, 10 June 2010 (UTC)

I see no connection between a watch and the MHP. I have given you a much better analogy. Just because Pythagoras' theorem only works for right-angled triangles does not mean that it is incorrect. In just the same way the simple solutions only work for the symmetrical case, that does not mean that they are wrong. The simple solution does not work just by chance, it works because of a simple and obvious symmetry. Martin Hogbin (talk) 21:16, 10 June 2010 (UTC)

The connection with the stopped watch is that in both cases there's a solution that works only if you ask the question for which it (the watch, or the simple solution) produces the correct answer. It's like using Pythagoras' theorem to determine the length of the third side of a 6-8-X triangle without saying anything about whether the triangle is a right triangle. Yes, X is 10 if the triangle is a right triangle, but Pythagoras' theorem only applies if the triangle is a right triangle, so unless it's given that the triangle is a right triangle or you can somehow show the triangle must be a right triangle based on what's given, using Pythagoras' theorem is unjustified. Similarly, showing that the unconditional probability of winning by switching is 2/3 for the MHP (which is what the simple solutions do) when asked about the conditional probability is not a complete answer unless you also say something about why these two probabilities must be the same.

You don't need to take my word for this - it's exactly what Morgan et al., and Gillman, and Lucas, Rosenhouse, and Schepler, and Rosenthal, and Grinstead and Snell, and the others I keep mentioning are saying. And, since Wikipedia doesn't care what you think or what I think, and there are plenty of reliable sources that present unconditional solutions without justifying their answer must be the same as the conditional probability it's perfectly fine to present these sorts of solutions. What I'm saying is NOT fine is presenting these solutions as if they are the undisputed way the problem should be approached. This isn't a difficult concept - I'm perplexed why we have to keep arguing about it. -- Rick Block (talk) 03:51, 11 June 2010 (UTC)

The simple solution works for the conditional symmetrical case. It gives the right answer and not just by chance but because the problem is symmetrical. It only fails when it is applied to a non-symmetrical case. Martin Hogbin (talk) 20:43, 11 June 2010 (UTC)

Exactly. Just like trying to use Pythagoras' theorem for a triangle without saying anything about whether it is a right triangle. Has anyone ever said the simple solutions don't give the right answer if the problem is symmetrical? That they fail for the non-symmetrical case, and that the sources presenting them typically say nothing about whether the problem is assumed to be symmetrical or not is exactly the point. -- Rick Block (talk) 21:30, 11 June 2010 (UTC)

You claim that these sources sources support you but in fact some do not, as has been demonstrated on these very pages. Martin Hogbin (talk) 20:43, 11 June 2010 (UTC)

I guess we're going to have to disagree about this. I'm quoting them exactly. As far as I can tell, you seem to be trying awfully hard to ignore what they say. -- Rick Block (talk) 21:30, 11 June 2010 (UTC)

Yes, this supports the section I just created where I separate your various complaints about presenting simple solutions first, from the reliable sources who say it is not solving the exact problem as stated by Whitaker. This shows that the amount of 'misleading' is negligible, if it exists at all. So your arguments about mis-treating the reader or violating NPOV are, of course, easily shown to be dubious, false and incomplete. Of course, the criticisms from the reliable sources go in the article. I understand that simple Wikipedia policy. Do all the other editors? I've written that countless times. It's in my recently proposed format which you called 'awesome' and 'what I've been saying for some time'. Did you forget that? I've simply proven that your arguments about exactly which order things MUST be presented is baseless. Glkanter (talk) 04:00, 11 June 2010 (UTC)

I'm not saying that the simple solutions can't be presented first. Have you forgotten that every single draft I've proposed has simple solutions first? What I am saying is that they can't be presented as if they are the undisputed way the problem should be addressed, since there are multiple reliable sources that dispute this. -- Rick Block (talk) 21:30, 11 June 2010 (UTC)

A Solution Where Door #2 Is Opened

Technically, this wouldn't be Whitaker's problem, either.

But, the doors are equally likely, so it's 'equivalent'.

So, why isn't a solution, where either equally likely door can be opened, also considered 'equivalent'? Glkanter (talk) 17:26, 10 June 2010 (UTC)

You know that, and I know that. It is quite obvious that if the host chooses randomly the answer if the host chooses door 2 is the same as the answer if the host chooses door 3 is the same as the answer if the host chooses an unknown door (unconditional case). Martin Hogbin (talk) 18:14, 10 June 2010 (UTC)

@Martin: The first two have the same value. P(C=2|X=1,H=3)=P(C=3|X=1,H=2)=2/3, but they are (formally) different probabilities. But what do you mean with the last probability you mention? Formulate it in the shown terminology.Nijdam (talk) 09:27, 11 June 2010 (UTC)

Would the distinction between the 'open door 3 problem' and the 'open door 2 problem' that you made, be meaningful to a typical Wikipedia Monty Hall Problem reader? Glkanter (talk) 10:49, 11 June 2010 (UTC)

16 Years Of Silence

Selvin's original problem, simple solution, statement that the host always offers the switch and chooses randomly when faced with 2 goats, and his conditional proof, all appeared in 2 letters to the peer-reviewed professional journal The American Statistician in 1975.

Selvin's statement about his premises looks like it was prompted by some unseen dialog between him and readers of the 1st letter.

Presumably, for the next 16 years, none of Selvin's peers had a problem with his puzzle or solution.

vos Savant was published in 1990. She never mentions any letters about a conditional solution being required out of the 10,000s she received.

Only when 4 of Selvin's peers, who despite including a 'history' of the problem in their 'paper' in the same journal, are not aware of Selvin, or his 'random host' premise, write a 'paper', does a controversy begin.

How bad could a reader be misled by the simple solutions being shown first, without any prior caveats, if for 16 years, Selvin's professional peers, who understood the the existence of the 'random host' premises, were OK with it? Glkanter (talk) 15:03, 11 June 2010 (UTC)

Second Chances

My point in the following is not to discredit the criticisms of the simple solutions, but to determine, for purposes of editing the article, how much attention the host choosing randomly controversy needs to receive, both before the solutions are given, and after the solutions are given.

The first person to name this problem 'The Monty Hall Problem' was Selvin in 1975. He describes the puzzle in his first letter to The American Statistician peer reviewed journal. I do not know if letters are peer reviewed.

He describes the problem with 3 boxes, and 1 set of keys. He describes the contestant selecting a particular box (named A, B, and C), and the host picking up a particular box. It was A & B, but I forget which was which.

He uses an unconditional solution, namely a table of all 9 equally possible outcomes. He combines the 2 empty boxes on a single line when the host must choose between 2 empty boxes.

He has a second letter published in which he names it the Monty Hall Problem, offers a conditional proof, and states unambiguously that his prior solution relies on the host always offering the switch, and choosing randomly when faced with 2 empty boxes.

So, after his 2nd letter, Selvin has stated the random premise and had offered a simple solution that did not exactly resemble the problem he described, in that he listed every possible combination of doors, and determined which ones were winners and which ones were losers, should the contestant switch.

In his problem, the contestant states his preference to switch, (even though Monty tries to deceive him by saying it's 50/50) and the question posed is does the contestant know something that Monty doesn't know? Has he made the right choice?

So, the first thing I'd like to know is, what is the reliably published 'conditionalists' view of Selvin's solution? But I'm only interested in sources that specifically name Selvin.

In 1990, vos Savant (apparently without knowledge of Selvin's letters) answers Whitakers question (3 doors, 1 car and 2 goats) in a more-or-less similar manner to Selvin. While she doesn't mention that the host chooses randomly, she also provides a solution consistent with the host choosing randomly, that did not have the Whitaker specified (he prefaces the door #s with the word 'say,') door 1 selected and door 3 opened.

Selvin clearly describes the Let's Make A Deal game show, names it The Monty Hall Problem, and even describes a real life conversation with Monty Hall. His answer, with the 2 empty boxes on a single line is consistent with the host choosing randomly. Whitaker's problem begins with 'Suppose you're on a game show...' and vos Savant's answer is consistent with the host choosing randomly.

Is it reasonable to assume then, since the puzzle is about a game show, and without having to state it outright, that the host chooses randomly when faced with two goats?

What if vos Savant had the same second chance as Selvin?

What if every simple solution that doesn't (many simple solutions do) clearly state that the host chooses randomly, had a second chance to state the 'host chooses randomly' premise?

What would be the remaining criticisms of the simple solutions? Does this describe a situation where a reader would be misled by being presented simple solutions before the more complex conditional solutions? Would this approach violate WP's NPOV policy? Glkanter (talk) 09:57, 11 June 2010 (UTC)

Here's an example where someone needed a second chance. Happens to everyone. Glkanter (talk) 08:25, 12 June 2010 (UTC)

So, Selvin, when given a second chance, clearly states his 'host always offers the switch' and 'the host always chooses randomly' premises to the problem.

Morgan, when given their second chance, acknowledge the error in their calculations that Martin and Nijdam pointed out in applying prior and posterior probabilities and say "...had we adopted conditions implicit in the problem, the answer is 2/3, period".

It's funny. With all the controversy for nearly 20 years, none of those guys bothered to re-check their math. Go figure.

Nijdam, you've criticized Devlin as being 'lazy' for publishing a simple solution. What is your opinion of these four gentleman who make, publish and defend against vos Savant and the world, an error in simple maths in a peer-reviewed, professional journal? How about their research skills, where they provide a 'history' of the 'Monty Hall problem', and fail to refer to the guy, Selvin, who first named it the Monty Hall Problem. And also stated outright those very "implied conditions" that make 'the answer is 2/3, period'? All in the very same journal their paper was published in? What kind of grade would you give to 'students' like this? I'll start you off with 'lazy', and you can go from there. Thanks. Glkanter (talk) 17:32, 12 June 2010 (UTC)

Again

I repeat: I don't give a d... if some stupid psychiatrist, sociologist or a simple brick layer likes to understand the MHP and is given the simple solution. I do however care when teachers and students think they "understand" the MHP, and they apparently think so, because of the following reasoning, which they happen to find in Wikipedia: The car is with probability 1/3 behind the chosen door. Clearly it is with probabibility 0 behind the opened door, hence it must be with probability 2/3 behind the remaining door. Nijdam (talk) 17:18, 11 June 2010 (UTC)

Is it your belief that the article should be written to accommodate your personal idiosyncrasies, or for the benefit of the reader? Because they represent two very different, mutually exclusive, approaches to the article.Glkanter (talk) 00:53, 12 June 2010 (UTC)

What, if anything, does your comment above have to do with editing a Wikipedia article as per Wikipedia 'reliable sources' policies? Glkanter (talk) 17:42, 11 June 2010 (UTC)

You have yet to demonstrate anything wrong with your simple solution. If the host chooses a goat door uniformly at random, symmetry considerations are all that are needed to prove your answer correct.

I agree that the MHP can be made into a useful teaching tool if it is presented in the right way, namely the usual story, the usual rules, and the information that the car is initially placed uniformly at random, plus the instruction that the question is to be answered strictly on the information given in the question (or some other form of wording that does not allow the principle of indifference, or real world expectation, to be applied to the host's goat door choice).

In this slightly but significantly changed formulation of the problem, Morgan's answer is the only correct one and it becomes a useful teaching tool, neatly demonstrating the ease with which unjustified assumptions can lead to an incorrect answer in conditional probability problems. I have no problem with saying that clearly in this article. Martin Hogbin (talk) 18:45, 11 June 2010 (UTC)

Martin - you are completely missing the point. The sequence of statements

1. The car is with probability 1/3 behind the chosen door.

2. Clearly it is with probability 0 behind the opened door.

3. Therefore, it must be with probability 2/3 behind the remaining door.

is logically incorrect (false reasoning) and confuses prior and posterior probabilities. #1 is talking about the prior probability (before the host opens a door). #2 is talking about the posterior probability (after the host opens a door). #3 combines these to arrive at what happens to be a true statement (the probability IS 2/3), but using completely incorrect reasoning. The simple solutions are NOT talking about the specific case of player picking door 1 and host opening door 3 and are (instead) talking about the overall chance of winning by switching (vos Savant's solution does this), OR they are using this incorrect reasoning.

And, again, you don't need to take my word for it. It is what numerous reliable sources say. -- Rick Block (talk) 19:07, 11 June 2010 (UTC)

Unlike the authors of a well known paper on the subject, I do understand the difference between a probability before door 3 is opened and the probability after it has been opened. There is nothing wrong with reasoning that the posterior probability that the car is behind door 1 is the same as the prior probability using only symmetry. It is obvious and correct.

It is Morgan (and followers) who answer a slightly different question (as described above) from the one actually asked. Martin Hogbin (talk) 19:28, 11 June 2010 (UTC)

Beautiful! If anyone has documented proof, in a peer-reviewed journal, no less, that they understand the before and after probabilities, it is, indeed, you Martin Hogbin! Well done sir! Conversely, if there is anyone that has provided documented proof, in a peer-reviewed journal, no less, that they DO NOT understand the before and after probabilities, it is Morgan, et al. Shame on them. And shame on them for telling vos Savant to stay away from subjects she doesn't understand. Karma is a bitch. Glkanter (talk) 01:07, 12 June 2010 (UTC)

No, they don't. They showed that indeed some form of reasoning is needed to make clear that the prior and posterior probabilities for the chosen door have he same value.Nijdam (talk) 20:58, 11 June 2010 (UTC)

Given just Whitaker's question and the rules, we have to decide on how to treat the initial car placement and the host's goat door choice. Based strictly only on the information given, the best we can do is to say the probability is from 0 to 1 depending on where the car was placed. Alternatively we could apply the principle of indifference (or some real-world knowledge) and take both the car placement and the host goat door choice to be uniform at random (as would be the case on a real game show). In that case the simple solution (combined with the simple and obvious observation that, by symmetry, it cannot matter which door the host opens) gives the correct answer of 2/3. To get any other answer and to require a more complex method of solution we must specify the initial car placement to be random but the host goat door choice to be non-random; not the question asked. Martin Hogbin (talk) 21:19, 11 June 2010 (UTC)

Martin - what source would you quote for your interpretation of this? Once again, Wikipedia doesn't care what you think or, in this context, how you would approach the problem. Wikipedia cares how reliable sources approach the problem. Your argument here is essentially that we should ignore sources you don't like. This is what I call POV pushing. -- Rick Block (talk) 21:36, 11 June 2010 (UTC)

As usual, when you lose the logical argument you go back to trying to talking about sources. Can you find anything wrong with what I say above? We can discuss sources elsewhere if you really want to. Actually there is one source worthy of mention here, Morgan. They now say, '...had we adopted conditions implicit in the problem, the answer is 2/3, period', in other words their original answer applies only to a variant of the question asked. Martin Hogbin (talk) 22:13, 11 June 2010 (UTC)

No, we've been talking about sources the whole time. Sources that you seem to think it's OK to ignore if you disagree with what they say. We're not arguing about whether they say the answer is 2/3, but whether they say things like vos Savant's solution "does not address the problem posed" (Gillman says this), or a simple solution "does not quite solve the problem Craig posed" (Grinstead and Snell say this), or "any proposed solution to the MHP failing to pay close attention to Monty’s selection procedure is incomplete" (Lucas, Rosenhouse, and Schepler say this). And there are more. -- Rick Block (talk) 22:52, 11 June 2010 (UTC)

Look at the top of this section. It starts with Nijdam saying that he does not care what other people (sources) think. He then asks a question, not about what any source or sources say but about logic. I have shown that by the application of symmetry the conditional and unconditional problem must have the same solution.

If you want to talk about sources, that is fine with me, but somewhere else. Do you want to talk about exactly what individual sources say or do you just want to count sources and see what the majority view is? It might be an idea to set up a sources page where we can consider the sources in detail and try to agree exactly what they do say. Then perhaps threads that start discussing logic can continue on the same subject without being derailed by irrelevant discussions about sources.

To make it clear, I am happy to talk about what reliable sources say and I am equally happy to discuss the logic or mathematics of the problem, but please can we stick to one subject at a time. Martin Hogbin (talk) 09:50, 12 June 2010 (UTC)

Um, read it again. It starts with Nijdam saying he does not care if the simple solution is presented to some people, but does care if Wikipedia presents a solution using incorrect logic. I'm saying Nijdam's point is supported by sources. You apparently want to argue "logic".

I would suggest this is not the right page for arguing logic, but /Arguments would be. I would also suggest that your WP:OR arguments about the validity of simple solutions are meaningless as far as editing the article is concerned. We can certainly present simple solutions, since there are plenty of reliable sources that do so. However, what the article says about the correctness of these solutions has to come from reliable sources - not from WP:OR arguments made on this page (or the /Arguments page). -- Rick Block (talk) 06:31, 13 June 2010 (UTC)

And I'm sick and goddam tired of alleged mathematics experts pontificating on oh-so-subtle nuances of the English language spoken by unknown third parties as if they were English Language professors and had perfect knowledge of the speakers' intent. And who are not even native speakers of the tongue. How would you like the converse, Nijdam? Stick to your own discipline as well. Glkanter (talk) 06:49, 14 June 2010 (UTC)

Please Criticize This Re-interpretation of vos Savant's or Carlton's Solution

Host always offers the switch, host must choose randomly

...so, if the contestant chooses door #1, and the host opens door #3, the contestant will win the car twice as often if he switches.

Door 1	Original Probability	Host Has Opened Door 2 or Host Has Opened Door 3	Remaining Door	Probability (Unchanged)
Car	1/3	Goat	Goat	1/3
Goat	2/3	Goat	Car	2/3

OK, for comparison purposes, here's the conditional version of the 'always get the opposite when door 3 is opened' solution. It may satisfy somebody somewhere, but does nothing for the reader. Glkanter (talk) 03:44, 12 June 2010 (UTC)

Door 1	Original Probability	Host Has Opened Door 2 or Host Has Opened Door 3	Remaining Door 2 or Remaining Door 3	Probability After Opening Door 2 or Door 3 (Unchanged)
Car	1/3	Goat/Goat	Goat/Goat	1/3, 1/3
Goat	2/3	Goat/Goat	Car/Car	2/3, 2/3

Glkanter (talk) 17:40, 11 June 2010 (UTC)

The table doesn't say what happens if the player picks door 1 and the host opens door 3 - but rather what happens if the player picks door 1 and the host opens either of door 2 or door 3. If there are 300 players who have picked door 1, it is talking about all 300, not the subset who see the host open door 3, and says that of these 300 about 200 will win by switching. -- Rick Block (talk) 19:17, 11 June 2010 (UTC)

You're saying the problem REQUIRES a conditional solution. That's not an answer to my question. That's pontificating. And ignores the solution.

We've discussed this over and over, and yet there it is again. In the formulation of the MHP the host opens a door and only then, that is after opening and really showing a goat to the player, offers the player to switch. This makes sense of course, and is how anyone understands it and pictures it. The simple solution however is a not a correct solution to this formulation. However it is correct for the slightly different form, where the host announces his intention of opening a door with a goat, and before actually opening a door, makes the offer to switch. This last different form is showed in the above table. Nijdam (talk) 21:16, 11 June 2010 (UTC)

And I, and others, have explained over and over that the obvious symmetry of the situation means that the unconditional solution obviously applies to the symmetrical conditional case. To explain what I mean, you agree the above table applies to the situation where the host has yet to open one of two possible doors to reveal a goat. There is only one more fact to revealed when the host does open a door and that fact is the door number that the he opens. It is quite obvious that if the host chooses which door to open uniformly at random that it does not matter which door it turns out to be. Could anything be simpler? Martin Hogbin (talk) 22:21, 11 June 2010 (UTC)

Let us get things straight: I spoke of "the simple solution", you of "the unconditional solution". Are they the same? And if so, why do you call them different? Then you speak of the "symmetrical conditional case"; which case do you mean? Is it one of the cases I mentioned? Nijdam (talk) 23:04, 11 June 2010 (UTC)

The problem is with the word "obvious". It is not obvious, and needs to be pointed out. Although you often mention the symmetry argument on the talk page, the simple solutions presented in the article itself unfortunately make no mention of it: A reader who sees only one of the simple solutions and is asked "What is the symmetry argument used here?" will not be able to answer that question. Many readers wouldn't even be able to answer the essay problem "Does the probability that the car is behind door 1 change when door 3 is opened? How about the probability that the car is behind door 3? Explain.", so it cannot be obvious. -- Coffee2theorems (talk) 03:15, 12 June 2010 (UTC)

I eliminated the ambiguity in the table's column heading you mentioned. Now what's wrong with the solution? While were on the subject, are there sources that specifically call Selvin's simple solution with the 2nd letter premises, 'false' or 'incorrect' or 'solving the wrong problem'? Glkanter (talk) 00:23, 12 June 2010 (UTC)

First - in the form you now have the table what source do you think it matches? And, why is it obvious the probability the door you picked hides the car is 1/3 - isn't this the whole question? The probability is obviously 1/3 BEFORE the host opens a door, but what this table is now saying is that this is the probability AFTER the host opens a door. -- Rick Block (talk) 01:33, 12 June 2010 (UTC)

This table supports the 'You get the opposite if you switch after the host has opened door 3' solution. And your comment is in error. Glkanter (talk) 01:48, 12 June 2010 (UTC)

The 2nd column of this table shows the 'true' condition of the puzzle: Monty will reveal a goat 100% of the time. Not much of a condition, really. The 'opens door #3' is just nit-picking, but has been addressed here. Would it be better if the 2nd column contained 'goat/goat' to show the separate 'conditions' of opening door #2 vs door #3. The next two columns wouldn't change. Seems pretty redundant, as Monty *has* to show a goat. Glkanter (talk) 03:00, 12 June 2010 (UTC)

By solving for door #2 and for door #3, it's no different than the table in the Probabilistic section which concludes with:

"[If the host has opened Door 2, switching wins twice as often as staying] [If the host has opened Door 3, switching wins twice as often as staying].

Just an awful lot easier for the reader to follow. Glkanter (talk) 03:12, 12 June 2010 (UTC)

Assume the problem includes the statement that the host always opens door 3 if possible. Why can't I use the same table to show the probability of winning by switching is 2/3 with this host behavior? Neither the table or any of your arguments mention anything about how the host chooses between two goats, so the same table (and same arguments) apply in this slightly modified problem (right?). Just to be clear, I'm saying the host always opens a door, always shows a goat, but instead of picking randomly between two goats opens door 3 if he can. There's clearly a 1/3 probability the player has initially picked the car (and a 2/3 probability the player has picked a goat). In accordance with the first line (of the first table) the host opens door 2 or door 3, and if the player has picked the car (1/3 probability) the host shows a goat and the remaining door hides a goat. This has probability 1/3. The other line says that if the player initially picks a goat, the host reveals a goat and the remaining door is the car, and this has probability 2/3. Right? So, this table is saying if the player picks door 1 and the host opens door 3 the player's door has a 1/3 chance of being the car while the other door (door 2 in this case) has a 2/3 chance. However, if the host opens door 3 whenever possible, the actual probabilities are 1/2 and 1/2. Can you please explain this?

My explanation is that there's an error here. Specifically, the error is that the probability that the player's initial pick being 1/3 doesn't necessarily mean the probability of this pick is 1/3 AFTER the host opens a door. This table ASSUMES this is true. It actually IS true if the host picks randomly between two goats, but the table doesn't say anything about this. It takes some amount (more than none) of argument to justify why the initial probability and the probability AFTER the host opens a door have the same value. This is not just a "probability fact" (since it's clearly not true in all cases), but something that has to be justified in a way that applies ONLY if the host is constrained to pick randomly between two goats.

Again, back to what reliable sources say. Per Lucas, Rosenhouse, and Schepler: "any proposed solution to the MHP failing to pay close attention to Monty’s selection procedure is incomplete" - i.e. any solution that completely ignores how the host chooses between two goats is incomplete. -- Rick Block (talk) 05:22, 12 June 2010 (UTC)

mu Glkanter (talk) 05:54, 12 June 2010 (UTC)

Rick, if you rotated your thinking about this 90 degrees, I think you would have a very different analysis. Glkanter (talk) 06:45, 12 June 2010 (UTC)

I suppose if, as you describe above, the host and contestant agreed that when the contestant has selected the car the host will send him a signal to that affect, which is NOT the MHP, then a better strategy would be: after the host has opened door #3 you should switch because you always get the opposite, unless you receive the signal from the host that you already have selected the car, in which case you don't switch.

In that case, you are correct. The 'always switch after door 3 is revealed' strategy and supporting table does not include that contrivance. Nor should it be expected to. And that does NOT invalidate the above tables and narrative. Glkanter (talk) 12:05, 12 June 2010 (UTC)

A standard way to show a proof is incorrect is with a counterexample. I'm suggesting this variant in that vein. The conditions necessary for the simple approach still hold - in particular, the initial probability is 1/3, the host MUST open a door, and doesn't open the player's choice. So, because we already know one of the two unchosen doors is a goat when the host opens one he doesn't change the probability of the player's initial choice. This is EXACTLY what the simple solutions are saying. If it's true in the standard MHP it must be true for this variation as well. If we look at vos Savant's table

Door 1	Door 2	Door 3	result if switching	result if staying
Car	Goat	Goat	Goat	Car
Goat	Car	Goat	Car	Goat
Goat	Goat	Car	Car	Goat

everything it says holds (for this variation as well as the standard MHP). If the car is behind door 1 switching loses. If the car is behind door 2 or door 3 switching wins. The probability of each of these cases (the probability of where the CAR is initially placed) is the same. Therefore (??), the probability of winning by switching if the player picks door 1 and the host opens door 3 must be 2/3. However, in this variation the actual probability is 1/2. What's wrong with this analysis is the same thing that's wrong with applying this analysis to the standard MHP. Specifically, this analysis is not saying what the probability is if the player picks door 1 and the host opens door 3, but rather what the probability is if the player picks door 1 (period). This probability (the probability of winning by switching if the player picks door 1) is 2/3 for both the standard MHP and this variation, i.e. this solution is correct - but it is answering a different question than "what is the probability of winning if the player picks door 1 and the host opens door 3". To make this solution complete for the standard MHP, some argument must be made along the lines of "This solution shows the probability of winning by switching if the player picks door 1. This is the same as the probability in the case the host opens door 3 because ...". The "because" could be the "obvious" symmetry Martin keeps bringing up - but AFAIK no popular source presenting a simple solution ever mentions anything like this (or, even, that the host must choose randomly between two goats). Paraphrasing Morgan, one is tempted to conclude that the popular sources do not understand that the conditional problem (player has picked door 1 and host has opened door 3) and the unconditional problem (average for all players who pick door 1 - ignoring which door the host opens) are not the same, and that 2/3 is the answer to the relevant conditional problem only if the host chooses between two goats randomly. -- Rick Block (talk) 14:26, 12 June 2010 (UTC)

Could somebody from the profession please explain to Rick the fallacy of his argument? Martin has tried by referencing the Pythagorean Theory's applicability to right angels only. Maybe a call out the the Wikipedia math group would help here?

I will not, nor should it be expected of me, defend the table I put together from the 'always the opposite' sources, instead based on the table you created. They are not related in any way.

Rick, this table shows why the 1/3 never changes. Because regardless of what door Monty opens, it's always a goat. That's the condition. That Monty reveals a goat. But it's a 100% condition. So splitting doors 2 & 3, like in the 2nd table, is superfluous, because they are both goats. Notice those last few columns have identical results whether door 2 or 3 is open. All the contestant needs to know is that he's seen a goat. But, since some sources say it must have a conditional solution, showing the host revealed door 3, those are in the table. In an appropriate manner. Would showing the door #2 values in red, like Chun's tree, make this a better presentation?

Rick, if you are ever going to believe me, this is the time. Glkanter (talk) 14:45, 12 June 2010 (UTC)

What I *would* appreciate, Rick, is if you went column by column in the conditional table and told me where I've done something in error. Thanks. Glkanter (talk) 15:02, 12 June 2010 (UTC)

OK. Column by column in the conditional table. What does the 3rd column show? For example the row that has "car" in the first column - does "goat/goat" in the 3rd column mean there's a goat behind door 2 and goat behind door 3? Similarly, the 4th column - does "goat/goat" here mean there's a goat behind door 2 and a goat behind door 3? These are just clarifying questions. The problem is the last column "Probability After Opening Door #2, Door #3". First, to clarify, the two numbers here are the probability the car is behind door 1 if the host opens door 2 or door 3, respectively (right?). You're trying to imply the 1/3 in this column (the last column) is the same 1/3 as the 1/3 in the first column (is this right?). What is your argument for where these probabilities come from? Is this simply "here's what the answer is"? Or is this meant to be an explanation of some sort? How can the original 1/3 split into TWO values, each of which is 1/3? Since there are two cases here corresponding to the one original case, the two probabilities here should ADD UP to 1/3, i.e. they can't both be 1/3 - unless this last column is talking about conditional probabilities. But if this column is talking about conditional probabilities there's no explanation for where these conditional probabilities came from. -- Rick Block (talk)

Rick, what do you think of inviting Boris to comment? I think this table actually accomplishes his 'peaceful co-existence' recommendation very well. Anybody else you would like to have comment on this? Glkanter (talk) 15:11, 12 June 2010 (UTC)

If you're really looking for a table that shows Boris's "peaceful co-existence" I think this one (that I proposed a while ago) is actually far better.

Door 1	Door 2	Door 3	total cases	host opens Door 2		host opens Door 3
				cases	result if switching	cases	result if switching
Car	Goat	Goat	100	50	Goat	50	Goat
Goat	Car	Goat	100	0	N/A	100	Car
Goat	Goat	Car	100	100	Car	0	N/A

Changing the numeric entries to probabilities makes this essentially the same as Chun's diagram, i.e.

Door 1	Door 2	Door 3	probability	host opens Door 2		host opens Door 3
				probability	result if switching	probability	result if switching
Car	Goat	Goat	1/3	1/6	Goat	1/6	Goat
Goat	Car	Goat	1/3	0	N/A	1/3	Car
Goat	Goat	Car	1/3	1/3	Car	0	N/A

From this table you can see that if the host opens door 3 the probability of winning by switching is 1/3 out of (1/6+1/3), i.e. 2/3.

If you'd like to try to get Boris to comment on this, go ahead. However, you do realize Nijdam teaches probability at a university level (and, although I'm not sure, I think Kmhkmh may as well). -- Rick Block (talk) 05:57, 13 June 2010 (UTC)

Well, those editors read these pages, I'm sure they would comment if so motivated. I am very interested in Nijdam's response. I think this table would affect his various discussions with Martin about the '1/3' <> '1/3'. Glkanter (talk) 06:40, 13 June 2010 (UTC)

Thank you, Rick. I appreciate your comments. I think the column headings make it clear what the values in the columns are.

The table does nothing more than show:

by using conditional probability (host reveals a goat 100% of the time), and by showing the results after door #3 is open, that the verbal '...1/3...2/3...you always get the opposite...after door #3...' is true, and satisfies each criticism of the simple solutions. I don't think any other conclusion is intended by this table. Not by me, anyways. And since all 'split' values are the same within each cell, I'd much rather show the simpler table in the Popular solutions section, replacing the table that is there now.

It may be helpful if you think of this solution as the probability of car vs the probability of not-car. That's why it only needs two lines. Unlike your table, it is not door-centric, but still encompasses all the door information of the problem.

I wish I could explain to you why the 1/3 and 2/3 don't change. Pretty much, once I drew up the table this way, and that the column that has Monty opening a door is always a goat, per the premise, I never saw a reason the probabilities should change. He always shows a goat. Every time. No matter what. In fact, no matter his 'host bias', he's still showing a goat. Which is consistent with what a lot of us have been saying. Glkanter (talk) 06:22, 13 June 2010 (UTC)

And I wish I could explain to you the difference between unconditional and conditional. Heptalogos and I had a long discussion about this (involving very similar tables) a while ago. Have you read it? It's at Talk:Monty Hall problem/Archive 15#Responses (starting at about where the first table shows up). Does talking about the number of samples help at all? If we're talking about 300 samples, the unconditional solutions are always talking about all 300 - while the conditional solutions are talking about only half (the half where the host opens door 3). -- Rick Block (talk) 06:46, 13 June 2010 (UTC)

"I never saw a reason the probabilities should change" is not a reasonable step in a solution. It's an argument from ignorance. Yeah, I don't see why P!=NP should be false either - it is obviously true! - and I wish I could explain why it is true, too. Do these words constitute a solution to the P!=NP problem? Of course not. -- Coffee2theorems (talk) 08:49, 13 June 2010 (UTC)

That's only a sentence fragment, which you did not indicate. It intentionally mis-states my actual writings. Your comment is certainly "not a reasonable step in a solution" as you say above. Where did you learn a technique like that? And why would you pull it here? Glkanter (talk) 10:34, 13 June 2010 (UTC)

You are confusing my attempt to assist Rick, with the reliably published solution that I have provided a new table for. Whether you agree is immaterial. Glkanter (talk) 10:21, 13 June 2010 (UTC)

I get it. We're defining the 'condition' differently.

I say it's:

Monty reveals a goat with probability 100%.

You say it's:

Monty opens door #3 with probability 50%

Glkanter (talk) 06:56, 13 June 2010 (UTC)

Please keep in mind my goal for the simple solutions:

Be presented without any prior caveats or other 'framing'

Be presented in the article first. Because we call them 'simple'

vos Savant's 1/3 vs 1/2 would be the only controversy presented before the solutions. All the remaining criticisms and controversies from reliable sources would be presented afterwards, as per my recent proposal . Glkanter (talk) 07:09, 13 June 2010 (UTC)

The "condition" that Monty reveals a goat with probability 100% is a null condition. It doesn't change any probability. If this is what you're trying to show with your table it should look like this:

Door 1	Probability	Remaining door after Host opens Door #2 or Door #3
Car	1/3	Goat
Goat	2/3	Car

Are you suggesting adding some table like this to the article? -- Rick Block (talk) 17:37, 13 June 2010 (UTC)

Well, if you just added back in the column for the 100% condition that the host reveals a goat (I mean, you've got to account for 1 car & 2 goats, right?), which needs to be shown, and then added for clarity an end column that repeats that it's still 1/3 and 2/3, then our tables would be identical. My version with the '/' is more detailed to show the critics that it represents the door #1 & door #3 pairing, but otherwise, both of my tables support the 'switching gives you the opposite' solution. What is your abbreviated version of my table purporting to accomplish? Glkanter (talk) 21:54, 13 June 2010 (UTC)

The abbreviated version is a table version of the text that I thought you were trying to represent as a table. Maybe we should start with the exact text. Which source (in particular) are you talking about? -- Rick Block (talk) 22:42, 13 June 2010 (UTC)

Rick, based on your continuing questions, comments, counter-proposals, suggestions, table substitutes, etc. regarding the simple and elegant table supporting the 'opposite' solution, I believe any combination of the following is likely"

You don't understand how the table supports the reliably published solution

You don't want to admit that you understand and agree with the table

You are trying to trick or trap me in a meaningless semantics circle

You are using a delaying tactic

It's 2 rows and the probabilities don't change. How complicated can it be? Accordingly, I do not find it necessary to continue enabling this behaviour. Glkanter (talk) 00:46, 14 June 2010 (UTC)

Well, you can believe what you want, but I assure you I am acting in good faith here. -- Rick Block (talk) 02:26, 14 June 2010 (UTC)

Reliable Sources

I understand and support Wikipedia's policy.

One of the difficulties comes when the rules don't apply equally. Morgan criticizes a variety of simple solutions.

But not a single one is sourced, except vos Savant. They're just these generic concepts, except that vos Savant left the random host unstated, but implied. Selvin does not have that problem, however.

How does one refute their claims? There's no source material. The solutions are wrong, because Morgan created them wrong. Pure BS. And I see an editor on these pages applying the very same technique in some sort of 'challenge' to another editor.

Has Morgan, any other sources directly criticized Selvin's table solution with the random host premise? Glkanter (talk) 00:49, 12 June 2010 (UTC)

Grinstead and Snell clearly define the problem to include the constraint that the host chooses randomly between two goats. They present a "stay and you win 1/3, switch and you win 2/3" simple solution. And proceed to say "This very simple analysis, though correct, does not quite solve the problem that Craig posed. Craig asked for the conditional probability that you win if you switch, given that you have chosen door 1 and that Monty has chosen door 3." -- Rick Block (talk) 01:55, 12 June 2010 (UTC)

Again, from what you wrote, they are commenting on their own solution. Not a published one. Not Selvin's. Is this what all the simple solution critics do? Except the single case where Morgan takes on vos Savant? Who then later says "...had we adopted conditions implicit in the problem, the answer is 2/3, period"? Glkanter (talk) 02:18, 12 June 2010 (UTC)

But they do say that Whitaker has 'asked for the conditional probability'. Another 'camp' of the conditionalists. Different from 'the solution doesn't show door #3 opened' camp. Well that's their reliably published opinion. Another subset of the controversy to document in the article. Well, simple solutions, except Rosenthal's, are never conditional. I wonder if problems are really 'conditional', or if some of the solutions are 'conditional'. It sounds like G & S are saying there is only 1 correct solution to the problem. I doubt they are correct. Glkanter (talk) 02:31, 12 June 2010 (UTC)

Also it is quite obvious that in the symmetrical case the conditional solution is the same as the unconditional solution. We all agree that the unconditional solution applies in the case of the usual rules, the car has been placed randomly, the player has chosen door 1 and the host will open either door 2 or to 3 with equal probability to reveal a goat. In the conditional problem we know in addition that the host has opened door 3. It is quite clear from the natural and obvious symmetry that the result would have been the same had the host opened door 2. Thus we know that the solution that applies when we know that the host will open either door still applies when he has opened a specific one of the doors. If the host chooses a goat door uniformly at random, the conditional problem is obviously equivalent to the unconditional problem. Does anyone dispute this? Martin Hogbin (talk) 09:29, 12 June 2010 (UTC)

Okay Martin, this almost covers the controversy. I wouldn't go along with your: ... in the symmetrical case the conditional solution is the same as the unconditional solution', but for the rest your analysis is right. Notice however the arguments you need to show that the solution in what you call the unconditional case (the door isn't yet opened), is the "same" (what actually is the same?) as in the conditional case (the player sees a door opened). It is about the necessity of this arguing we discuss. Nijdam (talk) 09:40, 13 June 2010 (UTC)

What is quite obvious is that there are a significant number of reliable sources that present simple solutions, and a significant number of reliable sources that present conditional solutions, and a significant number of reliable sources that explicitly criticize the simple solutions saying they don't exactly address the problem posed. We really don't need to agree about anything other than this. -- Rick Block (talk) 05:02, 13 June 2010 (UTC)

I think further discussion of what the sources actually say would be useful, which is why I have set up the page Talk:Monty_Hall_problem/Sources as described below to discus what the sources actually say. Martin Hogbin (talk) 08:48, 13 June 2010 (UTC)

Sources page?

Despite the continued disagreement and argument on this page we have all (just) remained civil and on topic. Again we could all congratulate ourselves on having what is probably the most detailed and comprehensive discussion on the MHP anywhere. At least one academic has reported learning things about the subject that they did not know before, however, within the established policies of WP the talk page cannot be considered a reliable source. We therefore need to agree on what the sources do say.

It is clear that editors here understand the subject and are passionate about it. We have the opportunity to create the most definitive and useful source of information on the MHP anywhere. A page where we list the sources with one section per source and discuss only what that source actually says would be a useful reference source to guide our discussions here on how to improve the article.

I will set up such a page. Of course, no one is obliged to use it, but in my opinion, it might save us going round and round in circles with claims that multiple sources support our own POV. Martin Hogbin (talk) 10:22, 12 June 2010 (UTC)

I have now set up the page Talk:Monty_Hall_problem/Sources. I suggest that all discussion on what the sources say takes place on this page and all discussion of mathematics and logic takes place on the Talk:Monty_Hall_problem/Arguments page, leaving this page free for suggestions for and discussion about changes to the article. Martin Hogbin (talk) 08:52, 13 June 2010 (UTC)

A Checklist Of Open Items For Nijdam:

The following items still await your input:

Criticism of 2 Row Table supporting 'You Always Get The Opposite'

How Can Something That Is Not Even A Premise Be A Required Condition?

Morgan's 'Grade' in light of their recent admissions

Glkanter (talk) 07:03, 14 June 2010 (UTC)

MHP Premises

Do any of the generally recognized statements of the MHP include these 2 premises?:

The contestant must choose door #1

The host must reveal a goat behind door #3

If not, why must a solution incorporate these as if they *were* premises, or otherwise be considered 'false' or 'incomplete' or 'a slightly different problem'? Glkanter (talk) 14:56, 13 June 2010 (UTC)

The most widely known statement of the MHP is the one from Parade. It includes (as a clarifying example) the situation where the player HAS chosen door #1 and the host HAS opened door #3, which is different from saying that the player MUST choose door #1 and that the host MUST reveal a goat behind door #3. Selvin's original formulation [3] similarly includes the specific case where the player chooses box B and the host opens box A. Opinions here seem to vary (although as far as I can tell, published opinions that discuss this not so much), but I think the clear implication is that any solution is meant to address the specific case given as an example in the problem statement - this case being representative of any of the 6 legal combinations of initial player pick and door the host opens.

If we call the probabilities of winning by switching in each specific case p12, p13, p21, p23, p31, p32, the solution should tell you what p13 (player picks door 1 and host opens door 3) is. The simplest solution (you start with a 1/3 chance of picking the car and end up with the opposite) addresses only the average of all of these. Note that "average" here is not the simple numeric average (meaning the sum divided by 6), but the probabilistic average which is the sum over all cases of the probability of being in each case times the probability of winning in that case. The solution vos Savant presented in Parade (if you pick door 1 you win with a 1/3 chance if the car is behind door 1 and lose with a 2/3 chance if the car is behind door 2 or door 3) addresses the average of p12 and p13. If they are all the same (which is what it means for the problem to be symmetrical - which it is if and only if the car is uniformly randomly placed at the beginning and the host chooses randomly between two goats), they all must be the same as the average of all 6 as well as the same as the average of p12 and p13 - but the simple solutions at least typically say nothing about this.

The point here is that p13 (the conditional answer) IS NOT THE SAME THING as the average of all 6, or even the average of p12 and p13. What I keep hearing you say is that you cannot see that these are different. -- Rick Block (talk) 17:13, 13 June 2010 (UTC)

You have jumped straight past the point that Glkanter has made to address a different one. The point is that there is no evidence that Whitaker intended specific door numbers to be relevant in his question, he just gave them as examples to explain what he meant. Whitakers question actually says, '...the host, who knows what's behind them opens another door, say No 3'. There is no suggestion that Whitaker wants to specify a particular door, he says 'say No 3' just to make clear what he means by 'another door'. Thus Whitaker was probably asking the unconditional question. Martin Hogbin (talk) 18:14, 13 June 2010 (UTC)

The critical point is that Whitaker's question is making it clear the player is deciding whether to switch AFTER the host opens a door, so clearly knows which door she initially picked and which door the host opens, meaning she's in ONE of the 6 possible specific cases (not all 6, not 2, ONE). -- Rick Block (talk) 19:38, 13 June 2010 (UTC)

Whitaker is the one who asked the question. There is no evidence that he intended to ask about the case where a specific door had been opened. It is irrelevant whether a player might know what door was opened. We have to address the question that Whitaker asked. Did Whitaker imagine that it might make a difference which door was opened by the host? Almost certainly not. Had he realised the potential importance of this he would not have needed to ask the question.

Morgan, and others, say we must not use information not given in the problem. It is far from clear that Whitaker does give us door number information in the question, thus we cannot use this in the solution. Martin Hogbin (talk) 21:21, 13 June 2010 (UTC)

It is not a matter of door No. 1 chosen and door No, 3 opened. What matters is that a door is chosen and another opened. It is easier to imagine if the numbers of the doors are given. Otherwise, as Rick already said, we have to solve the problem for each of the 6 combination of door numbers. It doesn't however change the fact that the solution depends on the door numbers. The simplest way out would be to say that the solution for all other combination is equivalent to the situation with door No. 1 chosen and door No. 3 opened, and then give the solution for this situation. This brings us back to where we were. It doesn't change a bit. Nijdam (talk) 21:40, 13 June 2010 (UTC)

Which source says that? Martin Hogbin (talk) 08:41, 14 June 2010 (UTC)

@Martin - What we have to address is what sources say about this. Any opinions about what Whitaker's question means are not relevant unless they're backed up by reliable sources. As far as I know, all sources that examine this aspect of what the question means (if you'd like we can list them) say it means what I'm suggesting it means. -- Rick Block (talk) 22:39, 13 June 2010 (UTC)

Well that is something we might discuss on the sources page. When you actually read the sources they often do not support your POV quite as clearly as you suggest. Martin Hogbin (talk) 08:39, 14 June 2010 (UTC)

As it's not even a premise to the problem, regardless of which reliably published source one looks at, there is no way opening door #3 can be a condition. Glkanter (talk) 21:46, 13 June 2010 (UTC)

The premise is that the player picks a door and the host opens a different door, and the doors are numbered. Therefore, the player knows which door she picked and which door the host opened. The question is obviously asking about one of the 6 specific cases, not the average of two of them and not the average of all 6 of them. -- Rick Block (talk) 22:39, 13 June 2010 (UTC)

The question we are trying to answer is not, "What conditions are implicit in the strictly determined Monty Hall Problem as canonically conceived by mathematicians?" It is, "What might anyone mean when they say, Monty Hall problem?" Andrevan@ 05:39, 14 June 2010 (UTC)

What anyone means when they say Monty Hall problem is a problem where a player picks one of three numbered doors, the host opens a different door (necessarily showing a goat), and then (when the player is looking at two closed doors and one open door) the host offers the player the chance to switch to the remaining (third) door. To reinforce the point that the player is deciding when looking at two closed doors and one open door, most phrasings of the problem include a specific example (e.g. the player picks door #1 and the host opens door #3). The question is not

1) what is the average chance of winning by switching (if you decide to switch BEFORE the host opens a door - either before or after you've initially picked a door)

but rather

2) what is the chance of winning by switching AFTER the host opens a door, i.e. given that you are currently looking at only two closed doors and one open door

I really don't think anyone, other than one or two editors here, disagrees with this. What this means mathematically we can discuss elsewhere, but the common understanding of the problem is definitely #2, not #1. The problem is a paradox BECAUSE the question is #2, not #1. You're not deciding to switch when you're looking at three closed doors, but AFTER one of them has been opened when there are only two closed doors. It is because there are only two closed doors at the point of decision that it seems like the probability must be 50/50 (even though it's not). -- Rick Block (talk) 14:20, 14 June 2010 (UTC)

I am not one of those editors. Symmetry says that by solving the 'overall' problem, I am also correctly solving each and every 'specific pairing' problem. Rick, It's long been apparent that you are not familiar with the topic. Please stop wasting everybody's time. Glkanter (talk) 14:33, 14 June 2010 (UTC)

So you're saying it's a premise that the problem is symmetrical (care to offer a source for this?), and it's this that makes the probability of the player's door "not change" when the host opens a door? Seems to me the "Monty forgets" variant is symmetrical, so there's clearly a little more to it. Please listen to what I'm saying:

Here's a source for symmetry, named Morgan: "...had we adopted conditions implicit in the problem, the answer is 2/3, period". That's from The American Statistician, a peer-reviewed, professional Mathematics journal. The most revered reliable source known to mankind. They've admitted they were wrong, and given up. Why won't you? Glkanter (talk) 15:19, 14 June 2010 (UTC)

I'm fine with presenting (referenced) simple solutions.
I'm fine with presenting simple solutions first.
I'm not fine with presenting simple solutions as if they are the universally preferred way to address the problem.
I'm not fine with ignoring the sources that criticize the simple solutions as not quite addressing the problem posed.

What I've been arguing for, for a long time, is that the article not take a stance (either way) on whether the simple solutions are correct. I keep hearing you and Martin arguing that this is not acceptable to you. -- Rick Block (talk) 14:56, 14 June 2010 (UTC)

I am not going to respond to each and every one of your straw man arguments. Nor am I going to restate my position and supporting arguments for it, again, for no apparent purpose.

Instead, I will provide this link to my most recent proposal for the article. Which perfectly meets your requirements. Glkanter (talk) 15:03, 14 June 2010 (UTC)

Now, if Nijdam would just comment on the various open items, and give his Zeus-like proclamations on how he will allow us to proceed...

Glkanter (talk) 15:08, 14 June 2010 (UTC)

First of all, Rick, you're missing the entire point. Opening door #3 cannot be a condition. It's not even a premise.

Secondly, with symmetry, each pairing has the same result: 1/3, 2/3. Which, of course is the same as the average: 1/3, 2/3.

So, due to symmetry, you and the rest are wrong, but they're published. Solving for the average, DOES solve for each pairing.

But the 'you get the opposite' solution table, doesn't even rely on averages. It works explicitly for door #1 and door #3. And every other pairing.

The published sources still get in the article, per WP policy. Even those admitted error makers, Morgan.

But since they are wrong, you and Nijdam are also wrong: offering simple solutions first, and without caveats, other 'framing', or weasel words does not mis-treat or otherwise 'talk down' to the reader, or violate WP NPOV or any other WP policies. So, cut the crap. Glkanter (talk) 06:20, 14 June 2010 (UTC)

Rick, you've seemingly calculated the probabilities of all 6 possible door pairings. And each comes to 1/3. Just like the overall probability. Congratulations! You just re-proved the time-tested-and-true Symmetry principal of probability. Which the simple solutions take advantage of to solve the MHP. Glkanter (talk) 14:50, 14 June 2010 (UTC)

Does anybody know...

Does anybody know if Wikipedia has official editorial polices such as WP:Cowardice and WP: Intellectual Dishonesty? Glkanter (talk) 16:08, 14 June 2010 (UTC)

I'd be willing to start the round of commenting for WP:OneArticleGroupie, WP:DeadHorseBeater and WP:GetALife glopk (talk) 19:55, 14 June 2010 (UTC)

I'll acknowledge those. And I don't think they can intentionally hurt the article.

Will the other editors step up and acknowledge theirs?

Hey, OCD isn't always a bad thing. Glkanter (talk) 20:03, 14 June 2010 (UTC)

Glopk, I know who I am, here's a chance for you to meet me.

Here's how the first posting in that section concludes. Kinda like that table, no?:

"2/3 of the time I will select a goat. Therefore I should switch. Glkanter (talk) 15:32, 7 February 2009 (UTC)"

Glkanter (talk) 21:08, 14 June 2010 (UTC)

And here's how Morgan, et al phrased it just last month, 19 years after their paper was published, in their response to Nijdam's and Martin's letter to the editors of The American Statistician pointing out an error in one of Morgan, et al's first-year-math-level calculations:

"...had we adopted conditions implicit in the problem, the answer is 2/3, period". - Morgan, et al

I wonder which of the many "conditions implicit" they mean? Do they mean Selvin's outright statement that the host chooses randomly, or that it's a game show, and hosts don't reveal 'biases' to contestant's that indicate where the prize is?

Why is there even a 'controversy' anymore?

Since I wrote that first diff and the rest of the section, I learned that it's the sources that matter. So I've doubled my estimate of how much of the current article adds value. Now, my goal is to get to the simple solutions in the article with only vos Savants controversy being discussed in advance of the solutions. None of the Morgan-inspired criticisms until after the various solutions are given. Subsequently, there would be no editorializing on which 'camp' is right. There's a section called How's This on this current talk page where I describe how I'd like to see the article formatted.

Well Rick and Nijdam see that as violating WP:NPOV and doing a dis-service to the readers. This table, based on the 'you always get the opposite' proves them both wrong, and their argument about the section order is shown to have no merit. And the support for their 'concern' for the reader is also found entirely false. Glkanter (talk) 21:17, 14 June 2010 (UTC)

And what about this one?

You pick a door		You pick another door		You pick the host
You open the door	You close the door	You open another door	You open your eyes	You close your eyes	You hit the host

Nijdam (talk) 10:10, 15 June 2010 (UTC)

I guess you do not like my suggestion. Martin Hogbin (talk) 11:11, 15 June 2010 (UTC)

Weren't you joking, just like me? Nijdam (talk) 13:06, 15 June 2010 (UTC)

Don't test me on this, Rick.

Or anyone else. Glkanter (talk) 18:31, 14 June 2010 (UTC)

Each column of that table comes directly from either the premises or the solution's text immediately above it. There is no need to reduce the clarity of the solution. That does not benefit the reader. Glkanter (talk) 18:37, 14 June 2010 (UTC)

The idea that you would intentionally not account for ~~the car and both goats~~ the player's choice of door #1 and the host having revealed door #3 is foreign to me, and to nearly every other word of the article. Your motivation, therefore, escapes me. Glkanter (talk) 18:40, 14 June 2010 (UTC)

What are you talking about? The text doesn't say anything about door 1 or "original" and "unchanged" probabilities. The whole point of this paragraph is its extreme simplicity. Simplifying the table to match the text is entirely appropriate. For reference, what I changed it to is this:

Player's door	Door the host opens	Remaining Door	Probability
Car	Goat	Goat	1/3
Goat	Goat	Car	2/3

And what I changed it from is this:

Door 1	Original Probability	Host Has Opened Door #2, or Host Has Opened Door #3	Remaining Door	Probability (unchanged)
Car	1/3	Goat	Goat	1/3
Goat	2/3	Goat	Car	2/3

I really don't understand what you're so upset about. It's virtually the SAME table. It presents the same information in a slightly more compact form (that is a better match for what the text says as well). -- Rick Block (talk) 18:47, 14 June 2010 (UTC)

The other column and heading text are from the premises. The solution is not required to restate them. Glkanter (talk) 18:48, 14 June 2010 (UTC)

I hope these aren't your reasons for making that unnecessary revert:

You want this new table to resemble the other (criticized) simple solutions, and look less like the 'more rigorous' conditional solutions

By eliminating the door #s, this table wouldn't exactly match the problem

By making the above changes, your arguments about simple solutions mis-leading readers and violating NPOV could conceivably have any merit whatsoever

You do not want the original 1/3, and the ending 1/3 on the same line, or the original 2/3, and the ending 2/3 on the same line as this is definitive proof that Nijdam's never-ending '1/3 <> /13' argument is wrong.

You are willing to sacrifice the readers' understanding and comprehension of the problem and it's solutions in order to further your POV

Please tell me that I'm wrong, Rick. What are your reasons for changing that table? Glkanter (talk) 19:19, 14 June 2010 (UTC)

The column you removed entirely, the inital 1/3, 2/3 column, was in fact prompted by one of your earliest suggestions:

"The probability is obviously 1/3 BEFORE the host opens a door, but what this table is now saying is that this is the probability AFTER the host opens a door." - Rick Block

And it's peculiar, Rick. That is the 5th different table you have proposed since I first introduced this table early in the morning on June 12th. What is so wrong with this interpretation of the 'you get the opposite' solution that you would propose 5 various alternative tables to a solution you don't even think has merit? It really makes me question your motives, Rick. Glkanter (talk) 19:39, 14 June 2010 (UTC)

This table is in support of the solution consisting of the following sentence:

An even simpler solution is to reason that switching loses if and only if the player initially picks the car, which happens with probability 1/3, so switching must win with probability 2/3.

This sentence says nothing about door numbers, but the table does.

This sentence says nothing about "original" or "unchanged" probabilities, but the tables does.

What's peculiar here is your insistence to include a complicated table in the simple solution section. My motive here is to have the table match the solution. Reading between the lines, even Martin doesn't like this table. -- Rick Block (talk) 00:57, 15 June 2010 (UTC)

Rick, I would not call a 5 column, 2 row table 'complicated', as you did above. That's BS. As I pointed out to you earlier, whatever is in that table, or the column headings, that does not come from the solution comes from the problem's premises or narrative. There is no gain from repeating them in the solution. Nor is there a requirement to do so. There is nothing in this table that is OR. How could the reader possibly benefit from the remove of vital door # information from the headers, Rick? And without the door #s, which come directly from Whitaker's letter, the solution no longer matches Whitaker's letter. Is that intentional on your part, Rick? "Reading between the lines", it sure looks like that is your motive.

I would suggest you let Martin speak for himself. Or is speaking for others, like reliably published sources, a uniquely special skill you possess? I have found Martin very capable of expressing his thoughts. Haven't you? Glkanter (talk) 01:30, 15 June 2010 (UTC)

As I said, my motive is to make the table match the solution. That's it. What I, and Glopk, are trying to change the table to matches the solution nearly word for word. -- Rick Block (talk) 01:54, 15 June 2010 (UTC)

Glopk

Maybe you know the way to force the column headings to wrap? Then the table wouldn't be so wide. Yes, I should make the headings consistent either with '#' or without them. Which way do you have a preference for? Why? An 'unchanged' probability? Please, I defer to Martin and Nijdam for that. But that fraction looks the same on the left and the right. Plus, the reader might be surprised that the values didn't change. It is a famous paradox after all. With 'unchanged'; in the heading, they will understand that that's not a typo, or due to any editorial confusion. Glkanter (talk) 20:21, 14 June 2010 (UTC)

Glopk, I have incorporated 2 of your suggestions. I removed the '#'s from the column headings, and I changed column widths for improved readability. I left 'Original' and 'Unchanged' in the column headings for the reasons described above. They are neutral, probability-oriented terms. Thanks. Glkanter (talk) 02:01, 15 June 2010 (UTC)

@Glkanter. Hmm, let's see. First, I have no idea of what you mean by "neutral, probability-oriented" term. And "original probability"? As good'ol' Inigo Montoya would say: "You keep using that word. I do not think it means what you think it means". My instinct would be to map it to "prior probability", which is a term of art. But the story of this page painfully tells us that you refuse to entertain the notions of prior and posterior probabilities, at least in the context of the MHP. So what is it? And what is "unchanged" about the latter probability, if it's not a posterior (i.e. conditional) probability? Mind you, am not saying that a solution for the "canonical" (original, Vos Savant, call it what you want) case of the MHP cannot be expressed coherently within a frequentist formulation of P.T., and without recourse to conditionals - although one really has to twist oneself in knots to do it correctly (as opposed to "intuitively"). Here I am simply objecting to your use of an unnecessarily verbose and ultimately obscure table format. Rick Block's form is faithful to the source (which, I assume, does not show a table itself), clean and complete. I am reverting to it.glopk (talk) 04:50, 16 June 2010 (UTC)

I see, another 'literalist' unconcerned about the readers' experience. So you disagree with the values in the end column? How? Why? Isn't that exactly the point of the 'you always get the opposite after Monty opens door #3' solution? And part of the lovely confusion of the paradox: 'But isn't' it 1/2?' Give me a break. But most importantly, 'Who cares'? It's from Carlton's reliably published solution. I didn't make it up.

Here's Rick's comment to Martin within the last few day:

"Basically, it's a "choice centric" view rather than "car centric" view (very similar to vos Savant's table). The new table Glkanter has added shows the 1/3-2/3, switch and you get the opposite solution (which didn't have any corresponding graphic)."

You see? Rick finally 'got it' that he needed to 'change his view 90 degrees.' Of course, this simple representation of Carlton's elegant solution conflicts with Rick's 5 year mission to defend Morgan, et al. A mission even they have, at last, given up on. So, where's the beef? How else to show that the probabilities didn't change in a reader-friendly table format?

It's nice to see that Rick is on-board with the table's usage. Just for fun, here are his response to that exact same 5 column, 2 row table when I first interpreted it, just a few days ago (well, it was just 4 columns, Rick actually suggested that the probability needed to be in there twice. Quite the irony, eh?):

doesn't address the door 1/door 3 case (Why, those are the very column headings he wants to obscure! Say it isn't so, Rick!)

what source does this match?

any solution that completely ignores how the host chooses between two goats is incomplete

A standard way to show a proof is incorrect is with a counterexample. (This one shows his lack of understanding of how to dis-prove a solution. Classic stuff. Read his denial of Martin's 'Pythagorean theory' explanation.)

about the table, ::::I wish I could explain to you the difference between unconditional and conditional

what I think the table should look like (Oh, maybe he's seen the light?)

which source are you talking about? (Eh, maybe he hasn't seen the light.)

I assure you I am acting in good faith (sadly, not very competently)

Or, here's the whole section.

Of course, Rick never says, 'By golly, you (and Martin) are right. I guess I was mistaken about the clock being right only twice a day. Thanks for staying with me until I got it.' No, he just decides that the table needs his special brand of "fixin'". Thanks, Rick. Thanks for nothing. You think it's fun to have to respond to all of Rick's nonsense for a 5 column, 2 row table that interprets a reliably published solution? It's not fun at all. It's downright tedious.

What about the column headings with the door #s? You didn't even mention those in your comments above. Why can't they reflect Whitaker's question? Only because it's disadvantageous to Rick's and Nijdam's arguments on what order, and with what caveats and weasel words, will the simple solutions be presented in the article. There can be no valid claim that the reader benefits from the loss of specificity to the table, or that the column headings don't come from Whitaker's question.

"...unnecessarily verbose and ultimately obscure table format..." More BS. It's 5 columns and 2 rows. 3 columns of where the car and 2 goats can be, 2 columns showing the probability before and after the goat is revealed. Period. Nothing else.

It's odd that after such a long break from any commentary, you come in with your guns a-blazin'. How did that come about, anyways, Glopk?

Save your techno-mumbo-jumbo and big words for some other sucker. This is a simple solution, as reliably published by Carlton. Got a problem with it? Take it up with him, not me. I ain't buying' it.

What was that other thing Inigo Montoya used to say? Figuratively, with regard to your specious arguments regarding the appropriateness of this table as it represents Carlton's solution, I suggest the same. Glkanter (talk) 07:16, 16 June 2010 (UTC)

"odd that after such a long break from any commentary, you come in with your guns a-blazin" - man, do you need help? glopk (talk) 07:27, 17 June 2010 (UTC)

What do you all think of this one?

You pick a car		You pick a goat		You pick a goat
You swap	You stick	You swap	You stick	You swap	You stick
You get a goat	You get a car	You get a car	You get a goat	You get a car	You get a goat

For what purpose? As a substitute for the new supporting table in the article? Or in support of which other reliably published source? How does it address Whitaker's door 1 and door 3? Do you see a problem with the new table that supports the 'you always switch' solution? Please share your concerns. Glkanter (talk) 00:47, 15 June 2010 (UTC)

I just put it up for general comment as a possible table for the simple section because it seems simple and intuitive to me. Martin Hogbin (talk) 08:59, 15 June 2010 (UTC)

I do not want to address Whitaker's door 1 and door 3 for two reason:

It makes the problem clearly conditional thus raising the issue of how to treat the host policy. (We have argued about this for years and although I agree with you that, in the absence of knowledge to the contrary, it is correct to treat the host goat door choice as uniform at random, others do not agree so I am trying to circumvent this contentious point for the purpose of clarity)

K&W have shown that people understand the problem better if all references to door numbers are removed. Martin Hogbin (talk) 11:20, 15 June 2010 (UTC)

Martin - isn't that roughly the same as this version?

Player's door	Door the host opens	Remaining Door	Probability
Car	Goat	Goat	1/3
Goat	Goat	Car	2/3

I assume you would prefer this to Glkanter's version. Is that right? I find your version a little hard to interpret (why are there two identical columns? why are there 4 cells under each heading? - this just makes it hard to understand). -- Rick Block (talk) 13:29, 15 June 2010 (UTC)

There are three columns to represent three equally likely possibilities. Maybe you could combine the lower cells vertically. I am just offering this as possibility. To me it seems the simplest to understand. Martin Hogbin (talk) 22:30, 15 June 2010 (UTC)

Yes - I puzzled this out. Basically, it's a "choice centric" view rather than "car centric" view (very similar to vos Savant's table). The new table Glkanter has added shows the 1/3-2/3, switch and you get the opposite solution (which didn't have any corresponding graphic). Direct question - of the three tables we're talking about here, which would you consider to be a more faithful representation of the solution we're illustrating? -- Rick Block (talk) 23:50, 15 June 2010 (UTC)

Yeah, Rick, you 'puzzled this out'. Brilliant! It's exactly the table I initially proposed, right down to originally only showing the probabilities once. Of course, you've obscured the door #s to make it look less similar to Whitaker's letter.

"== Please Criticize This Re-interpretation of vos Savant's or Carlton's Solution =="

Host always offers the switch, host must choose randomly

...so, if the contestant chooses door #1, and the host opens door #3, the contestant will win the car twice as often if he switches.

Door 1	Door 2 or 3	Remaining Door	Probability
Car	Goat	Goat	1/3
Goat	Goat	Car	2/3

Boy, that sure looks familiar!!! Wait, it gets better. Rick even had the balls to write this:

I assume you would prefer this to Glkanter's version.

What a skeeve. Kudos, Rick! Glkanter (talk) 11:20, 16 June 2010 (UTC)

@Martin - of the three tables we're talking about here, which would you consider to be a more faithful representation of the solution we're illustrating? -- Rick Block (talk) 14:00, 16 June 2010 (UTC)

The differences between the tables

There are 5 columns, and just 2 rows in the table I designed to support the 'you always get the opposite' solution in the 'Simple solution' section of the article.

Of those 5 columns, Rick changed the column headings on the first 3, and eliminated the 5th entirely.

These are the changes to the wording Rick felt necessary without prior discussion:

Original Text vs Rick's Substitute

Door 1 was replaced by Player's door

Original Probability was replaced by Probability

Host Has Opened Door 2 or Host Has Opened Door 3 was replaced by Door the host opens

Additionally, column 5, Probability (unchanged), was removed entirely. This column held the same values as the 'Original Probability' column. A pretty important column, as we'll discuss later.

Of less significance is Rick deleted bolding on the words 'car' and 'goat' in the 2 columns which were used to indicate the 'opposites' on both rows of the table.

That is the totality of the changes made by Rick.

Notice that each of the eliminated words and values cause the table to look less and less like Whitaker's original question, making it more difficult for the reader, and other editors, to associate the solution and table to Whitaker's question. I believe this was Rick's primary intent. This is also an important item, and will be discussed later.

Finally, nothing in this table violates any WP policies. Each column heading and the values in the column come from either the problem statement, the premises, or the solution. There is no OR or POV in this table.

Glkanter (talk) 03:46, 15 June 2010 (UTC)

Well, actually, the policy that is most relevant is Wikipedia:Verifiability. The source for this solution is Carlton. Here's what he says (section 5):

Before presenting a formal solution to the Monty Hall Problem to my students, I find that it helps to give an intuitive explanation for the 1/3 - 2/3 solution. Imagine you plan to play Let's Make a Deal and employ the 'switching strategy.' As long as you initially pick a goat prize, you can't lose: Monty Hall must reveal the location of the other goat, and you switch to the remaining door - the car. In fact, the only way you can lose is if you guessed the car's location correctly in the first place and then switched away. Hence, whether the strategy works just depends on whether you initially picked a goat (2 chances out of 3) or the car (1 chance out of 3) [bold added].

There's no mention of "original probability" or "probability (unchanged)". If anything, per the bold sentence the "probability" column should perhaps be changed to "initial probability" - but I'm not insisting on this. Changing the other labels from "Door 1" and "Host Has Opened Door 2 or Host Has Opened Door 3" to "Player's door" and "Door the host opens" is partly editorial, but also makes the table more accurately reflect the reference as well. Although you added this table, you don't "own" it - note the disclaimer in small text below the text entry form ("If you do not want your writing to be edited, used, and redistributed at will, then do not submit it here"). Edit warring to preserve your version is ridiculous. Seriously - I don't there there's anyone here other than you who prefers your version (if there is anyone, please speak up). -- Rick Block (talk) 14:28, 15 June 2010 (UTC)

It can certainly be 'verified' that everything in that table is either a premise of the puzzle, from Whitaker's letter, or from the solution. This literalness of each word stuff on your part is once again, a difference in interpretation of Wikipedia policies from mine. And the way you state your interpretation as 'fact' is another difference from myself. You often leave meaningful aspects of your sources out of your 'justifications'. For all I know, maybe he discusses these items a few pages earlier, anyways? Besides, unlike other simple solutions you criticize, he's not talking about an 'average' probability. He's talking about a strategy one may use in every play of the game, including the doors 1 and 3 pairing. You act like that's a bad thing. How utterly perverse! All this couldn't have been discussed first? What's the danger of leaving it? Who would be injured by leaving it? 16:18, 15 June 2010 (UTC)

If there's anyone here who prefers Glkanter's version of this table, please speak up. To be absolutely clear, we're talking about Glkanter's version:

Door 1	Original Probability	Host Has Opened Door 2 or Host Has Opened Door 3	Remaining Door	Probability (unchanged)
Car	1/3	Goat	Goat	1/3
Goat	2/3	Goat	Car	2/3

and this simpler version:

Player's door	Probability	Door the host opens	Remaining door
Car	1/3	Goat	Goat
Goat	2/3	Goat	Car

If no one speaks up in support of Glkanter's version (other than Glkanter), I will change back to the simpler version again. -- Rick Block (talk) 14:12, 16 June 2010 (UTC)

Rick, why do you continue to embarrass yourself this way? Either through ignorance or impunity, you are violating any number of Wikipedia Policies with this comment.

Silence does not mean consent.

Wikipedia doesn't care which table people 'like'. Wikipedia cares about the sources, and the readers.

If you can build a case that the table in it's present form is not supported by the sources, then make the case and build a consensus. Sources, of course includes Whitaker's letter, vos Savant's column and solution, Carltion's solution, etc.

If you can build a case that eliminating the door numbers from the column headings is better for the reader, then make the case and build a consensus.

You're violating NPOV/Bias by trying to make this table look 'generic', and 'not really addressing Whitaker's question'.

You're failing to show good faith with your actions as described above.

Why not be honest and admit that Glkanter designed both of the above tables? I chose to add the 2nd occurance of probabilities to show that they don't change. Otherwise, all you've done is muddy the waters by changing some column headings.

Glkanter (talk) 18:02, 16 June 2010 (UTC)

I'm sorry but personally I don't like either one of those tables. Considering how clear the verbal explanation that this table is supposed to be representing is I am sure we can come up with something better. I know that does not address the question as asked but that is just how I feel about it. Colincbn (talk) 17:12, 16 June 2010 (UTC)

Thanks for your comment. I don't know how to make the table simpler. 3 columns show where the car and 2 goats are. 2 columns show the before and after probabilities. That's all there is. Please tell me how it fails to support the 'you always get the opposite' solution? Glkanter (talk) 18:02, 16 June 2010 (UTC)

I suggest we remove the table in the article. It is no more than OWN RESEARCH, and adds nothing. Nijdam (talk) 20:31, 16 June 2010 (UTC)

If we can't agree on a table, removing it seems like a sensible idea to me. -- Rick Block (talk) 04:55, 17 June 2010 (UTC)

Good idea, the less of this nonsense the better.Nijdam (talk) 05:43, 17 June 2010 (UTC)

It's interesting, but sad, that a so-called 'educator' is so anxious to suppress a simple table supported by reliably published sources rather than engage in honest dialog about it. Solely because it conflicts with his ideology that '1/3 <> 1/3'. It isn't surprising, though. Glkanter (talk) 20:53, 17 June 2010 (UTC)

Rather than the two tables above, what is wrong with the one that is currently there? Seems fairly clear to me. I don't like the color-scheme that much, but I think it represents the intended sentence fairly well. It could be made easier to understand I suppose. Otherwise how about something like this:

Player's potential doors	result if switching	result if staying
Car	Goat	Car
Goat	Car	Goat
Goat	Car	Goat

It matches the table above it which might help people. Granted I don't think it is absolutely necessary either, but it could help to visualize the solution for some readers. Colincbn (talk) 06:14, 17 June 2010 (UTC)

The first table in this section is basically straight from vos Savant's solution published in Parade [4]. She presented two tables each with three lines varying the location of the car from door 1 to door 3, the first showing the result if you switch in each case and the second showing the result if you stay). Given that this source presents door numbers, I think this table should as well. -- Rick Block (talk) 13:21, 17 June 2010 (UTC)

Is The Table That Supports Carlton's Solution OR?

I agree that he didn't include the table, but the 2 row 'choice-centric' design of the table is certainly consistent with his solution.

The column headings and the values in each cell come directly from:

Whitaker's letter (which doors serve which purposes)

Carlton's paper (the before-a-goat-is-revealed probabilities of 1/3 and 2/3),

Simple math (the after-the-goat-is-revealed probabilities) as shown:

(Probability of being a car * Probability of host revealing a goat (aka, a 100% condition) = Probability of being a car after revealing a goat)

1/3 * 100% = 1/3

2/3 * 100% = 2/3

I spent some time looking, and didn't find any WP policy prohibiting editor-created tables, as long as they are reliably sourced, NPOV, and verifiable. I think the table in it's current form meets these requirements. Glkanter (talk) 06:40, 18 June 2010 (UTC)

I don't know, Rick. Maybe there are other editors who see merit in the table. You, yourself proposed 5 different tables for the exact same solution. You must have seen the need for a visual aide. Maybe you could address my issues, above, before you delete the table for reasons other than sources, verifiability, NPOV or non-benefit to the reader? What did Nijdam mean by his edit comment when he 'corrected' the column headings? That's not silence, and it wasn't a deletion. I think your deletion was pretty aggressive and unnecessary. Glkanter (talk) 14:20, 18 June 2010 (UTC)

No one other than you is arguing in favor of this table. Not even Martin. It's time for you to accept that there is a consensus against this table. You are violating one of the basic guidelines of editing behavior, and are obviously aware of the consequences. -- Rick Block (talk) 02:50, 19 June 2010 (UTC)

How's This? Act II

I. Very little mention of controversies (maybe just vos Savant's in Parade magazine, the others exist only in professional journals and text books) until after the solutions are provided.

II. One solution section: vos Savant, Carlton, Devlin, conditional. Yes, in that order.

III. Aids to Understanding

IV. The Controversies:

A. vos Savant's 1/2 vs 2/3

B. Door 3 must be revealed

C. Morgan, including simple solutions are false, host bias, and that their's is the (only) 'correct resolution'

A boring litany of who said what, and why, only. No original research editorializing on who is right or who is wrong.

Maybe switch the order of III and IV...

V. Sources of Confusion

etc...

Before the solution section, I guess in the Problem statement, I would suggest including Selvin's comment from his 2nd letter (paraphrasing here): '...this solution relies on the host always offering the switch and choosing equally between...' and a mention that the importance of this will be addressed later in the article, with a link to that section.

And I would have a separate, later section for the other simple solutions and any other solutions people want to include. I would put this before the 'Variants' section.

So, anybody else see this as a NPOV way to get a better article and some closure? Glkanter (talk) 21:13, 18 June 2010 (UTC)

Unambiguous?

The problem statement, even the supposedly unambiguous one, fails to state the fundamental rule of the game, which is that the player wins the car if he picks the car door. Or that he wins whatever is behind the chosen door and he prefers the car over goat. Ketorin (talk) 15:15, 19 June 2010 (UTC)

The quoted problem statement from Parade includes an editorial note that the goats are booby prizes. This seems like a sufficient clarification. -- Rick Block (talk) 19:30, 19 June 2010 (UTC)

Yes the "booby prize" clarification is within the conventional problem statement, and as such it would probably be tolerable for that statement alone (given the other discussed uncertainties as well). I am bothered with the so claimed "fully unambiguous" one that does not have this clarification at all.

If this was a regular article, I think it would be ok, but this one claims to have mathematical and logical merits so it should be sound on fundamentals. Ketorin (talk) 20:03, 19 June 2010 (UTC)

The point of the Krauss and Wang version is that it clarifies the ambiguities about the host's behavior and the nature of the question (in the standard way). Their description is "A formulation of the Monty Hall problem providing all of this missing information and avoiding possible ambiguities of the expression "say, number 3" would look like this (mathematically explicit version):", where the missing information they're referring to consists of the host's preference between two goats, the requirement that the host open a door to reveal a goat and make the offer the switch, and the initial distribution of the car behind the three doors. Would you prefer some different wording introducing this version? Perhaps by making this change:

According to Krauss and Wang (2003:10), even if these assumptions are not explicitly stated, people generally assume them to be the case~~. A fully unambiguous, mathematically explicit version of the standard problem is~~; their mathematically explicit version of the problem is:

The understanding that the goats are undesirable booby prizes presumably carries forward from the earlier editorial note. -- Rick Block (talk) 21:07, 19 June 2010 (UTC)

That is one weak reference even in the first place, and I don't see it carrying forward over the article. The reformulation is ok but I'm afraid it does not address my actual concern. Ketorin (talk) 07:22, 20 June 2010 (UTC)

Does anyone see a problem with adding a sentence to the paragraph beginning 'Certain aspects...' making clear that the player receives the prize behind the finally-chosen door and that it is generally assumed that the objective of the game is to win the car'? There are sources which say something like that. This may be obvious to most people but this is a very pedantic subject and we should not leave any loopholes. Martin Hogbin (talk) 09:54, 20 June 2010 (UTC)

The "Certain aspects" is specifically about the host's behavior - if we must add something about this I'd prefer it be separate from this (already long) paragraph. We could editorially annotate the K&W problem statement or perhaps add something at the very beginning of the section explaining this aspect of Let's Make a Deal. I'll try adding something. -- Rick Block (talk) 16:57, 20 June 2010 (UTC)

I think it is a good idea to add something somewhere. This is a subject where obvious assumptions sometimes turn out to be wrong (although sometimes they do turn out to be correct). Best make everything crystal clear. Martin Hogbin (talk) 18:31, 20 June 2010 (UTC)

Did anyone ever understood otherwise??Nijdam (talk) 19:38, 20 June 2010 (UTC)

Maybe not, who knows, but this is a rather pedantic topic where unjustified assumptions are frowned upon. Martin Hogbin (talk) 17:35, 21 June 2010 (UTC)

Here's the diff of the change [5]. -- Rick Block (talk) 13:50, 21 June 2010 (UTC)

I do not much like the idea of inserting text, even in square brackets, into a quotation. Also it is not clear that the statement refers to the finally-selected door. Could we add this sentence, 'Also it is generally assumed that the car is the desired prize and that the goats are booby prizes and that the player gets the prize behind the door that they finally select' before the sentence starting 'According to Krauss...'? Martin Hogbin (talk) 17:35, 21 June 2010 (UTC)

This is really an entirely different topic than what the "Certain aspects ..." paragraph is about. Using square brackets is the standard way of annotating quotes. If we're not going to clarify this inline in the quote then I'd prefer saying nothing about this at all or clarifying this at the beginning of the section, i.e. start the section with "In the game show Let's Make a Deal, originally hosted by Monty Hall, contestants won prizes by picking doors or curtains hiding the prize, sometimes winning very valuable prizes like cars or sometimes winning undesirable booby prizes like goats. Steve Selvin described a problem loosely based on this show ...".

@Ketorin - when reading this did you actually not understand that what is behind the player's choice of door is what the player wins and that the car is considered the preferred, or is this a hypothetical concern? I think anyone actually confused about the context would follow the link to Let's Make a Deal, which I think makes both of these points abundantly clear. This is frankly starting to sound like pedantry for pedantry's sake. -- Rick Block (talk) 18:18, 21 June 2010 (UTC)

Yeah like what if a particular episode is being taped at a location where a goat is preferable to a car. Shouldn't this possibility be addressed? :-( hydnjo (talk) 13:28, 22 June 2010 (UTC)

For the record, my comment above was intended to be jocular and whimsical you know, to bring a smile :-) hydnjo (talk) 19:08, 22 June 2010 (UTC)

Actually, this silly discussion exemplifies the way that some editors do not understand the true nature of the MHP. It is essentially a 'simple mathematical puzzle', in which it is customary to make all necessary assumptions to keep the problem simple. Once you start to consider real-world scenarios or the exact words that someone said, the problem looses it whole point.

In other words, we should assume whatever makes vos Savant's notable solution correct. The player wants the car, an unspecified door is opened by the host... You know the rest. Once you start to get stupidly pedantic about a simple puzzle you start a never ending spiral of pedantry, as is shown here. Martin Hogbin (talk) 17:29, 22 June 2010 (UTC)

Alas, there is nothing to be assumed making Marilyn's so called solution correct. Also not you mentioning: ...an unspecified door is opened by the host.... Perhaps this form of the problem may be given as a variant. Face it: a door is opened by the host, so we (the player) knows which door. That's what counts. It need not be door 3, if you like door 2 better, please. Or the player may have chosen door 3, and the host opens door 1. It's fine with me.Nijdam (talk) 20:45, 22 June 2010 (UTC)

The majority of sources that give the simple solution might be assumed to make different assumptions from those that give a conditional solution. Unfortunately, in common with most sources that give conditional solutions, they do not make their assumptions clear, however they do make their solutions clear. Why do we assume here that the majority of sources got the question wrong? Martin Hogbin (talk) 22:26, 22 June 2010 (UTC)

The clarification that the goats are unwanted has actually been requested several times and had recently been (until this edit) a separate sentence in the "Problem" section. I moved it to an editorial note in the quote of the Parade version to keep the clarification, but in a more compact form. I think that the player wins what is behind the chosen door is strongly implied by the wording (isn't this just the standard meaning of "choose"). How about if we keep the "[unwanted booby prizes]" clarification in the K&W quoted problem but ditch the clarification about winning what is behind the door you choose? -- Rick Block (talk) 18:09, 22 June 2010 (UTC)

Whatever. That was not really my point. Martin Hogbin (talk) 18:23, 22 June 2010 (UTC)

Yes, I know what your point is. I'm trying to keep this thread on the topic at hand. -- Rick Block (talk) 19:16, 22 June 2010 (UTC)

A Morgan Checklist

Selvin describes the problem with boxes. The contestant has chosen box 'B'. The host reveals nothing (a goat) inside of box 'A'.

Selvin first solved his problem with a table of 9 entries. The car in all 3 locations times the contestant choosing all 3 doors.

In this table, he showed each of the three 2-empty-boxes (2-goats) scenarios on a single line.

In his second letter he states that the host choosing randomly between goats is a premise of his puzzle.

He solves the puzzle conditionally using 1/2 to represent the host's probability of either door when faced with 2 goats.

Morgan's original paper focuses solely on Whitaker's letter to vos Savant, with the contestant selecting door 1, and the host revealing door 3.

In their response to the correction letter from Martin and Nijdam, they say:

"Simply put, if the host must show a goat, the player should switch." - Morgan, et al

"To wit, had we adopted conditions implicit in the problem, the answer is 2/3, period." - Morgan, et al

The American Statistician, May 2010, Vol. 64, No. 2

Morgan, J. P., Chaganty, N. R., Dahiya, R. C., and Doviak, M. J. (1991),

“Let’s Make a Deal: The Player’s Dilemma,” The American Statistician, :45 (4), 284–287

Comment by Hogbin and Nijdam and Response

I have three comments:

Given that Morgan, et al only talk about Whitaker's letter in vos Savant's column in their original paper, they must be still talking about Whitaker's letter to vos Savant when they made the above 2 comments.

Given that Selvin solves the problem he just originated unconditionally for all outcomes, it's illogical for Morgan and others to later argue that the MHP should be taken literally to mean doors 1 & 3.

There's no way of knowing which portions of Morgan's original paper they still support. There is at least one math error, and two walk-backs.

The various sources are published, and per Wikipedia will be part of the article. But that doesn't mean the article has to slavishly follow this discredited POV. Glkanter (talk) 08:27, 23 June 2010 (UTC)

According To Morgan, There Is No Difference Between Prior and Posterior

"To wit, had we adopted conditions implicit in the problem, the answer is 2/3, period. Morgan, et al ¹

"Simply put, if the host must show a goat, the player should switch." - Morgan, et al ¹

That's "...the answer is 2/3, period."

and

That's not, "...after the host has shown a goat". It's "if the host must show a goat..."

Simply put, to wit, the Queen Bee of 'simple solutions are false' has recanted.

¹ The American Statistician, May 2010, Vol. 64, No. 2

Morgan, J. P., Chaganty, N. R., Dahiya, R. C., and Doviak, M. J. (1991),

“Let’s Make a Deal: The Player’s Dilemma,” The American Statistician,

45 (4), 284–287: Comment by Hogbin and Nijdam and Response

Glkanter (talk) 13:59, 23 June 2010 (UTC); revised order of Morgan's comments - Glkanter (talk) 20:03, 23 June 2010 (UTC)

It really doesn't matter how many times you repeat these non sequiturs about Morgan et al., they are not going to become true by the sole virtue of being repeated ad nauseam on this talk page. -- Coffee2theorems (talk) 19:29, 23 June 2010 (UTC)

Are you suggesting that the conclusions reached by Morgan, et al in their original paper are unchanged by their subsequent comments? That seems contrary to the point of them making the comments. How would *you* interpret their new comments? Glkanter (talk) 19:40, 23 June 2010 (UTC)

The Morgan paper criticizes the simple solutions for the arguments they use, not for the value (2/3) or the decision (the player should switch) that is obtained. Their criticism of the "simple" (i.e. unconditional) solutions is exemplified by statements like "F1's beauty as a false solution is that it is a true statement! It just does not solve the problem at hand." and "F5 is incorrect because it does not use the information in the number of the door shown." Clearly they say nothing that would constitute a retraction of these statements, so your conclusion that "the Queen Bee of 'simple solutions are false' has recanted" does not follow.

Unfortunately, I do not have full access to the response by Morgan et al., only the first page of two at [6]. Their comment that "Simply put, if the host must show a goat, the player should switch." is not new, but is already contained in the original article. They are reiterating it just in case the reader has forgotten it ("In any case, it should not distract from the essential fact that [...]"). The response "To wit, had we adopted conditions implicit in the problem, the answer is 2/3, period." appears to be about an issue unrelated with the correction in the math, and the discussion gets cut short by the page break there (including the references to the letters where the issue was raised). My guess is that this is an admission that the standard assumptions (with the possible exception of the unbiasedness of the host) are implicit in the problem. That, however, again has nothing to do with the criticism of the "simple" (unconditional) solutions. -- Coffee2theorems (talk) 23:54, 24 June 2010 (UTC)

POV and disclaimers

There has been some edit warring over the addition of disclaimers in the various sections. Without doubt there are sources that give simple solutions and sources that say that the simple solutions are incomplete. There are also editors who hold both points of view.

It would seem to me least POV to not have a disclaimer in the simple section saying it is incomplete and also not have assertions in the Probabilistic section saying that some sources disagree. A full and sensible discussion of the reasons why some sources say the simple solutions are incomplete at the start of the Probabilistic section would seem the least POV to me. Martin Hogbin (talk) 15:44, 23 June 2010 (UTC)

If no one can give a good reason why we should have POV disclaimers in the article, I will remove them. Martin Hogbin (talk) 20:06, 23 June 2010 (UTC)

Unreferenced OR

Most of Gill's recent changes are completely unreferenced and therrfore appear to be OR. Also they're not written in the appropriate style (see MOS:MATH. I have very limited connectivity at the moment but would suggest we move the new sections here to work on rather than have them in the article in their current state. Rick Block (talk) 16:59, 25 June 2010 (UTC)

Agree, moving sections below. glopk (talk) 22:31, 25 June 2010 (UTC)

I trust we can all see this move as a temporary step whilst supporting references are added. The additions certainly improve the article and will move it forward towards becoming a modern and comprehensive exposition of the subject rater than a shrine to a narrow, not very notable, and somewhat contrived treatment. Martin Hogbin (talk) 09:17, 26 June 2010 (UTC)

The odds-approach can be found in Rosenhouse's book or Rosenthal's paper if I'm not mistaken. For the other parts I'm not sure. I think it might be a good idea to keep those sections here until we have them properly sourced and maybe optimized style/language a bit as well. Once that is done in sufficient manner, they should be reintegrated into the article.--Kmhkmh (talk) 10:08, 26 June 2010 (UTC)-

Fine, and I suggest we cut out the current proof using Bayes theorem and replace it with a link to the wikipedia section on Bayes which has it too. The article will look a whole lot balanced then and the additions I propose or some of them will fit in much more neatly if only people will take the trouble a) to understand b) to rewrite using short sentences and short words. Sorry I am a professional mathematician and voracious reader and native English speaker so I love writing convoluted sentences which no-one who did not go to Cambridge University in the early 70's can understand Gill110951 (talk) 17:32, 27 June 2010 (UTC)

Alternative Proof: odds form of Bayes law

It is illuminating to use Bayes' theorem in the memorable form, see [7]:

posterior probability is proportional to Bayes prior times likelihood

or equivalently

posterior odds equals prior odds times likelihood ratio, also known as the Bayes factor.

Odds are the ratio of two probabilities. In this case we are look at the ratio of the two probabilities that the car is behind Door 2 or Door 1, first prior to observing a Goat behind door 3, and then posterior to this event. The likelihood ratio, also known as the Bayes factor, is the ratio of the probabilities of what is observed, Goat behind Door 3, given each of the two competing hypotheses (Car behind Door 2;Car behind Door 1). Suppose the Car is initially equally likely to be behind any of the three doors, and suppose the Player has chosen Door 1. We then observed the Host opening Door 3, revealing a Goat. Throughout, we take the Player's initial choice, Door 1, as fixed. Initially the two hypotheses of interest, Car behind Door 2 and Car behind Door 1, were equally likely, hence their prior odds are 1 to 1. We want to know how the odds change by acquiring the information that there is a Goat behind Door 3. The likelihood ratio is by definition the ratio of the probabilities of this information - Host opened Door 3 revealing Goat - under the two competing hypotheses: Car behind Door 2, and Car behind Door 1. If the Car was behind Door 2, the Host was forced to open Door 3, so the probability thereof was 1. If the Car was behind Door 1, the Host had a choice, and if he chooses uniformly at random, the chance that he actually opened Door 3 was 1/2. Hence the likelihood ratio is 1 to 1/2 or 2:1 for Door 2 versus Door 1. Odds of 2 to 1 corresponds to posterior probabilities 2/3 and 1/3 since these are the only two, mututally exclusive, possibilities.

More generally the Host could use any probability between 0 and 1 with which to decide to open Door 3 when he has a choice between Doors 2 and 3. Thus the odds for Door 2 to Door 1 can lie anywhere between 1 to 0 and 1 to 1; i.e., the conditional probability given the initial choice door 1 and the opened door 3 lies anywhere between 1 (odds of 1 to 0) and 50% (odds of 1 to 1). Whichever it is, it is never unfavourable to switch doors (see Morgan et al., 2010). Even if it sometimes makes no difference to switch, there always are situations where it is favourable to switch, since the average probability that switching gives success is 2/3. —Preceding unsigned comment added by Glopk (talk • contribs) 22:33, 25 June 2010 (UTC)

Proof by symmetry

In the case that the Host makes his choice uniformly at random when he has a choice, we can argue using mathematical invariance very rapidly that the conditional probability must be 2/3, since it is, by the law of total probability equal to the average of the conditional probabilities. But in the situation without host bias, the problem remains the same under relabelling of the doors. So all the conditional probabilities (given choices of player and host respectively) are equal, and all must therefore be equal to their average, 2/3. —Preceding unsigned comment added by Glopk (talk • contribs) 22:33, 25 June 2010 (UTC)

Forcing symmetry

According to Marilyn vos Savant the door numbers mentioned in her restatement of Whitaker's question were just added to aid the readers' imagination, and should not play any role in the solution. There is a neat way to mathematically enforce the requirement that the solution should not depend on arbitrary door numberings, as follows. Let the door numbers mentioned in the problem have been introduced by the player for his own personal purposes (semantics: "say Door 1" could mean "I will call this, my first choice, Door 1"), and let him do this (uniformly) at random. First he chooses his own door at random and names that himself "Door 1". He next selects one of the remaining doors at random and names it "Door 2". The third door he names of course, "Door 3". He observes the Host open Door 3 revealing a Goat. By the player's symmetrization of the problem, i.e., by randomization of door names, the host is equally likely to open either of the doors which he himself named Door 2, Door 3, when the car is actually behind the player's initial choice. His conditional probability of finding the car by switching to the door which he named Door 2, is therefore 2/3. —Preceding unsigned comment added by Glopk (talk • contribs) 22:35, 25 June 2010 (UTC)

Symmetry by ignorance

For many people, perhaps starting with Laplace, probability is a measure of subjective degree of belief. From this point of view, it is reasonable to suppose that the player has no idea about whether or not the host has any bias, and that his beliefs about the possibly bias are completely symmetric. For such a player, the host when allowed to choose, chooses with some probability p which lies somewhere between 0 and 1, and about which the player knows nothing. If the player is totally ignorant about this possible bias he will be equally likely e.g. to believe it is below 1/4 as above 3/4. His belief is symmetric. Thus the player expects 0 bias, or in other words, the player's expectation of the host's p is 1/2. Morgan et al., as corrected by wikipedia editors Hogbin and Nijdam (2010), and acknowledged by Morgan et al. (2010), come out with a posterior subjective probability of 2/3 that switching gives the car, in this scenario. The computations using Bayes theorem can be avoided by remarking that the symmetry of the Bayes hyperprior on p makes the problem symmetric in the door numbers, and thereby forces the conditional probabilities to be 2/3 again.

Game theoretic version

According to economists and game theorists, for instance the famous Nalebuff (1987) and Rinnoy Kan (1990), the Monty Hall problem is a problem of game theory or decision theory. According to this formulation, we see this final phase of the quiz show as a game between the Host, who has hidden the Car behind a Door, and later opens a Door to reveal a Goat, and the Player, who first chooses a Door, and later has the opportunity to revise his initial choice. The Host wants to keep the Car, the Player wants to get it. This is a finite two-person zero-sum game so there exists a solution, equilibrium, or saddle-point by the famous minimax theorem of John von Neumann (1926), the founder of game theory and hence of much of mathematical economics. At least, there exists a solution, as long as we allow both persons (Host and Player) to use randomization (i.e., toss coins or dice to determine their choices). In this context, a solution is a strategy for the Host, called his minimax strategy, a strategy for the Player (his minimax strategy), and a number p, such that if the Player uses his minimax strategy he is guaranteed to win the Car with at least probability p, whatever strategy is used by the Host, and if the Host uses his minimax strategy, he is guaranteed to lose the Car with at most probability p, whatever strategy is used by the Player. Now consider the following pairs of strategies: Host: hide Car uniformly at random (with equal probabilities over all possibilities); later open a door uniformly at random if he has a choice; Player: choose Door uniformly at random; later always Switch. With the Player's minimax strategy the Player wins the Car with at least probability 2/3 (in fact, with exactly probability 2/3) whatever strategy is used by the Host. Conversely, with the Hosts' minimax strategy, the Player can't do better, since his initial choice is irrelevant if the car is hidden uniformly at random, and switching gives him 2/3 probability to get the car, not switching only 1/3 probabilitiy, so all he can get (by randomization of these two deterministic strategies) are probabilities between 1/3 and 2/3.

This comes from the Conditional Solution section.

..."The simple solutions correctly show that the probability of winning for a player who always switches is 2/3, but without further assumptions this does not necessarily mean the probability of winning by switching is 2/3 given which door the player has chosen and which door the host opens...

I don't think that's a true statement. What further assumptions are required, and what other results will a simple solution return? Glkanter (talk) 14:57, 26 June 2010 (UTC)

The simple solutions make no mention of the premise that the host choose randomly between two goats. This is the "further assumption". Perhaps this should say "further argument" rather than "further assumptions". -- Rick Block (talk) 04:25, 27 June 2010 (UTC)

Classic weasel wording. I bow before The King. Glkanter (talk) 03:16, 28 June 2010 (UTC)

Now, the Conditional Solution section says this: "...The popular solutions correctly show that the probability of winning for a player who always switches is 2/3, but without additional reasoning this does not necessarily mean the probability of winning by switching is 2/3 given which door the player has chosen and which door the host opens..."

I don't think that's a true statement. What additional reasoning are required, and what other results will a simple solution return? PLease tell me it isn't referring Rick's nonsense about the random host? Besides, it's 'given' when following the K & W formulary. Who decided this was an issue worthy of being called out ion such a weaselly way? Glkanter (talk) 07:29, 28 June 2010 (UTC)

With Selvin's stated premises: "Host always offers switch, host chooses randomly when faced with 2 goats, and the K & W version, I think most of this paragraph from the Conditional solution section is a bunch of hot air. Further, Morgan's later comments contradict a lot of what's written.

Also, 2 of the simple solutions *do* show door 3 being opened: The combined Doors and (with the tree) Carlton's.

Selvin DOES NOT say the problem requires a conditional solution. He simply offered a conditional proof, and made no other comment on his original solution of a table of all outcomes. His citation should be removed from "...That probability is a conditional probability (Selvin 1975b;..."

What is referenced to Selvin is "That probability is a conditional probability" - referring to the probability of winning given which door the player has chosen and which door the host opens. From [8]:

An alternative solution to enumerating the mutually exclusive 
and equally likely outcomes is as follows:

A = event that keys are contained in box B
B = event that contestant chooses box B
C = event that Monty Hall opens box A


Then

P(keys in box B | contestant selects B and Monty opens A)

= P(A | BC) = P(ABC)/P(BC)
            = P(C | AB)P(AB)/P(C | B)P(B)
            = P(C | AB)P(B | A)P(A)/P(C | B)P(B)
            = (1/2)(1/3)(1/3)(1/2)(1/3)
              1/3

If the contestant trades his box B for the unopened box 
on the table, his probability of winning the card is 2/3.

You may or may not understand what this says, but he's saying the probability given the player's choice and the box the host opens is a conditional probability. -- Rick Block (talk) 04:25, 27 June 2010 (UTC)

I have twice offered a proposed format (above) of the article where the differences from the sources are addressed only after the solutions have been given. The sole exception would be vos Savant's Parade controversy would come before the solutions.

"Conditional probability solution (as currently in the article)

The simple solutions show in various ways that a contestant who is going to switch will win the car with probability 2/3, and hence that switching is a winning strategy. Some sources, however, state that although the simple solutions give a correct numerical answer, they are incomplete or solve the wrong problem. These sources consider the question: given that the contestant has chosen Door 1 and given that the host has opened Door 3, revealing a goat, what is now the probability that the car is behind Door 2?

In particular, Morgan et al. (1991) state that many popular solutions are incomplete because they do not explicitly address their interpretation of Whitaker's original question (Seymann). The popular solutions correctly show that the probability of winning for a player who always switches is 2/3, but without further assumptions this does not necessarily mean the probability of winning by switching is 2/3 given which door the player has chosen and which door the host opens. That probability is a conditional probability (Selvin 1975b; Morgan et al. 1991; Gillman 1992; Grinstead and Snell 2006:137; Gill 2009b). The difference is whether the analysis is of the average probability over all possible combinations of initial player choice and door the host opens, or of only one specific case—to be specific, the case where the player picks Door 1 and the host opens Door 3. Another way to express the difference is whether the player must decide to switch before the host opens a door, or is allowed to decide after seeing which door the host opens (Gillman 1992); either way, the player is interested in the probability of winning at the time they make their decision. Although the conditional and unconditional probabilities are both 2/3 for the problem statement with all details completely specified - in particular a completely random choice by the host of which door to open when he has a choice - the conditional probability may differ from the overall probability and the latter is not determined without a complete specification of the problem (Gill 2009b). However as long as the initial choice has probability 1/3 of being correct, it is never to the contestants' disadvantage to switch, as the conditional probability of winning by switching is always at least 1/2."

Glkanter (talk) 15:22, 26 June 2010 (UTC)