Talk:Two envelopes problem/Archive 2

Archive 1

Archive 2

Archive 3

Archive 4

Archive 5

Merge

Definitely a good idea to merge the two pages. I'm a bit new at it all, so if anyone wants to do it, that would be good. Can we make it so that searching for "two envelopes" will find the "envelope problem"?

John

I don't know if they should be merged necessarily, but it might be a good idea to define the difference more sharply and compare the two throughout. Personally I'm still at a loss for why the two envelopes problem is still open considering that the envelope paradox is solved. The difference seems to be that in the latter you're allowed to look into the envelope before making a decision, whereas the first is a more recent variant that asks about the analysis before one takes a peek. If that's correct then readers should be directed to the paradox before they take on the open problem. Personally though I'm confused if that's the only difference. I don't understand why looking would change anything. The amount of money in your envelope might be an arbitrary value, but if you reason about that variable then the reasoning must include every possible value that variable might have, and each assumption of a value is analogous to a hypothetical peek. Davilla 07:35, 8 February 2006 (UTC)

I'm having a long and heated discussion with the two authors of envelope paradox for the purpose of merging the two pages. It's not the case that there are two different problems, one called the "two envelopes problem" and another called the "envelope paradox." It's just two different names of the same thing. Actually, the problem/paradox have gotten many different names over the years. Accordingly, there should be only one page at wikipedia devoted to this problem/paradox. All the different names should be redirected to one and the same page. On this we agree, but we disagree on some key issues about what the merged article should contain and state. INic 03:31, 20 February 2006 (UTC)

Any status updates on the merge? It has been a while, and it really would be nice if the articles were merged.

... I went ahead and merged the pages by making the other page just redirect here -- that article is essentially the same as this one. If you have nitpicks to work out, work them out on one article -- keeping parallel articles for months because we can't settle minor disputes seems a bit absurd.

I agree! INic 00:12, 25 August 2006 (UTC)

Ranking of Problems

Why insist on ranking to problems as "hard", "harder", "hardest"? Any such ranking depends on how you think about them, and is implicitly POV, not to mention un-important. The parallel article has them marked simply as first, second, third. I ask this on the Talk page instead of just fixing it because it seems at least one contributor insists on the distinction. (comment from the history: reverted to "the hardest problem" as this is, in fact, the hardest to solve). Incidentally, the "hardest" problem looks trivial to me.

I agree that it's really not possible to rank the variants of the problem according to their difficulty in any absolute sense. However, when you read the papers published so far you will see that almost all authors think they are more difficult in this increasing order. This wikipedia article only reflects this fact. You are, of course, free to disagree with the majority of the authors in the references. I do. But to include your or mine personal opinions about this in the article would be a violation of NPOV. INic 11:15, 21 August 2006 (UTC)

Thanks for the clarification. Is it possible to cite which articles specifically list the third form as the hardest? I'd like to read a more detailed analysis written by someone who thinks the problem is not trivial, but didn't find it in a brief search through th refs -- Though there are many references listed, the article does nothing to indicate what information comes from which source.

I'll clarify "In fact, the hardest variant" to "A variant often considered hardest" in light of the discussion.

The hardest problem (which is the fourth variant) is hardest just because it doesn't contain any probability reasoning. It is thus immune to all explanations relying heavily on probability theory. I agree that the article could refer to the references to a greater extent. I will improve that shortly. INic 00:53, 25 August 2006 (UTC)

Thanks!

I would clarify by saying that it is immune from subtle errors in its own probabilistic reasoning, since it has none. I doubt that it is immune from explanations utilizing probabilistic reasoning (the A vs A/2 formulation sounds like an abstract example of Simpson's paradox...), although that doesn't mean someone has actually come up with one yet. Baccyak4H 17:49, 29 August 2006 (UTC)

POV is, however, exactly what they have in the parallel article. First of all they say that the problem is solved which it's not by any means. Secondly, they refuse to include the most basic and original version of the problem which here is presented first. Thirdly they have their very own invented "solution" to the hardest problem—nowhere to be found in the literature. INic 11:15, 21 August 2006 (UTC)

On reading the solution in the other article, I appreciate that this version does not include it -- it looks, at best, poorly stated -- but it would be nice if we could have some referenced discussion of the reasoning behind the third paradox.

The last (fourth) variant was invented by the logician Raymond Smullyan. As far as I know no one so far has managed to extend one of the proposed solutions for the other cases to this purely logical case. If anyone find that somewhere it would be great to have it included here as a proposed solution. INic 00:53, 25 August 2006 (UTC)

Thanks again. I'll keep an eye out, but it's probably not happening. As I read it, any extension of the proposed solutions wouldn't really work because Smullyan has fundamentally changed where the paradox is introduced -- it's not really the same problem anymore, so why would it have the same solution? (specifically: his version is just sweeping details under the rug and comparing things that aren't the same as if they were supposed to be ... but that's, of course, out of place 'original research,' so I won't bother to expand)

The point Smullyan wanted to make was to question if the paradox really has anything to do with probabilities at all. But if this is an entirely different problem someone has to show that. It's not obvious. I agree that it's very unlikely someone would manage to extend his/her other solutions to this case easily, but chances are that a good solution to this problem automatically solves the other variants. BTW, you are allowed to speak about your own ideas for a solution here at the talk page. INic 14:30, 27 August 2006 (UTC)

Cool -- so, if you don't mind, I'll spew some hopefully coherent ideas about solving the hardest problem:

The way I was thinking about the Smullyan variant, his first statement is saying that "if you hypothesize about what values you could get in the other envelope, compared to the value in the current envelope, you get two possible values -- one larger, one smaller. The distance to the larger hypothetical value, from the middle known value, is larger than the distance to the lower hypothetical value." His second statement is saying that, "if you consider the two actual values in the envelopes, (they have this relation to each other and) the distance between them is the same going either way." These two statements aren't contradictory at all. The only way you eek out a contradiction is by using loose enough language such that the two distances considered in the first statement are thought to refer to the same thing conceptually as the single distance in the second statement.

So, that's where I think the paradox comes in. ... did that make any sense? I hope so.

I don't quite understand your solution here. However, I just added a summary of the only solution to this variant I know of in the article. Please have a look and see if it's the same idea that you have. INic 03:16, 15 October 2006 (UTC)

It's a rather different way of stating it, but I think it's saying what I was grasping for (and saying it much better than I could!). (Or maybe it's saying something a bit different, but I like it better anyway.) So yeah. Thanks for that!

The introduction

I don't like the introduction:

The Two Envelopes Problem is a puzzle within the subjectivistic interpretation of probability theory. This is still an open problem among the subjectivists as no consensus has been reached yet.

Because the subjectivistic interpretation of probability is closer to the layman's conception of probability this paradox is understood by almost everybody. However, for a working statistician or probability theorist endorsing the more technical frequency interpretation of probability this puzzle isn't a problem, as the puzzle can't even be stated when imposing those more technical restrictions. Consequently, all published papers below with different ideas on a solution are written from the subjectivistic or Bayesian point of view.

I don't like it because it tells me the point of view I should have about the problem before expainig what the problem is. Comments about subjectivism should come later as a possible pont of view, not at the very beginning BEFORE everithing else!! --Pokipsy76 16:22, 24 October 2005 (UTC)

Good point! I moved the second paragraph to after the presentation of the puzzle and made it a comment. Is this good enough you think? If not, please feel free to improve it yourself. INic 01:14, 25 October 2005 (UTC)

A Harder Problem

The explanation given doesn't make sense. Let's say I open the envelope and there is a $100. If I switch I expect to gain $125. Now that I switched I do the following calculation: I have 1/2 * 50 + 1/2 * 200 = $125. In the other envelope I see $100. So why would I switch back? The fact that I openned the envelope and found a sum greater than 0 added additional information about the expected value in the game. There is no paradox in this case.

-- Martin

You are right, it should be made more clear in the article how the contradiction appears in this case. In essence it's the same contradiction as in the first case. The only difference is that here the explanation given for the first case doesn't work anymore. I'll try to improve the text. INic 13:28, 2 August 2006 (UTC)

Tad's original research

I have just done a total rewrite of this article. As far as I know this is the first complete solution to the puzzle. I have removed most of the pre-existing material in the interests of brevity. Perhaps the new solution should be abridged eg the examples of oscillating series and the treatment of the finite case in the postscript are not strictly necessary. Part of the problem of presentation is to make it intelligible to non-mathematicians yet to be rigorous at the same time. My solution attempts to cover both bases.

The previous write-up had the tollowing faults. The paradox statement was somewhat clumsy and open to variant interpretations. The article unnecessarily divided up the paradox into various versions. The 3rd paradox statement without probability makes no sense - on what basis could you possibly make a choice? Above all, the old page failed to provide the solution. I have merged "two envelopes problem" with "envelope paradox". I would prefer to call it "two envelopes paradox" but it is probably best to retain the old name so as not to confuse things.

Maybe it is worth keeping the winning strategy under the heading "Discussion"?

Tad Boniecki 15 April 2006 218.214.57.242 21:42, 14 April 2006 (UTC)

Tad, I agree personally with some of the conclusions you have. However, Wikipedia is not the place to publish original research. It's against the policy of this encyclopedia (as it is against the policy of any encyclopedia). If you want your ideas on wikipedia you have to publish your ideas in a peer reviewied journal first. Until that happens you are welcome to make additions to and change this article but based solely on ideas already published. Please see the references. INic 19:59, 17 April 2006 (UTC)

Tad, another option for you is to use Academic Publishing Wiki INic 09:00, 25 August 2006 (UTC)

Many thanks. I have just added my solution plus a discussion to the wikia Philosophy Journal. - Tad

The 'an even harder problem' does not make sense

I have a problem with the 'An even harder problem'. First of all, there are two major mistakes in the calculation of the expected gain:

1) the first 1/2 should be omitted

2) a factor (1 - q)^(n-1) is missing in the final expression. so it should be

(2^n q(1 - q)^n - 2^(n-1)q(1 - q)^(n-1))/R = 2^(n-1)q(1 - q)^(n-1)(1 - 2q)/R

which is still positive if q < 1/2

But most of all, it does not make sense to consider distributions with infinite mean. They don't have practical value and you can't make any conclusion based on calculations with these kind of distributions.

It is easy to prove that the expected gain of switching envelopes before opening them, should be zero for each 'real' (finite mean) distribution. The expected amount in the two envelopes is the same by symmetry which means that the expected gain of switching (difference of the two) is zero, if the expected amount in the envelopes is finite.

This means that for every 'real' distribution that the mean of the gain is zero. If there are amounts, with probability > 0, for which it is better to switch then there are other amounts, with probability > 0, for which it is better not to switch.

The only distribution for which the reasoning of the paradox holds, is the one for which the envelopes contain 0 with probability 1. But then, the gain of switching remains zero even if the amount is doubled in the other envelope.

To conclude, It is better to remove the section about 'An even harder problem'. Distributions with infinite mean really don't make any sense and it just confuses the reader.

Regards, Stefaan

I agree that the example presented here can be better. I will soon replace this with an example found in the literature. But I don't agree that the only admissible distributions are the ones with finite means. Cauchy distributions for example lacks both mean and variance but that doesn't make them 'unreal' or unuseful. INic 21:17, 30 July 2006 (UTC)

Recent Edits - lets' express the probelm clearly before going into Bayesiana -Do we need to START with Bayes?

I recently edited the intro so it is easily understood:

The two envelopes problem is the following:

• You are shown two indistinguishable envelopes and asked to select one, which you do.
• You are then told that each envelope contains money, and one contains twice as much as the other.
• You are given the option to swap the envelope you have chosen for the other one.
• What should you do?

Some writers analyse this as a puzzle or paradox within the subjectivistic interpretation of probability theory; more specifically within Bayesian decision theory. This is still an open problem among the subjectivists as no consensus has been reached yet.

However, this has now been removed so the definition is

a puzzle or paradox within the subjectivistic interpretation of probability theory; more specifically within Bayesian decision theory. This is still an open problem among the subjectivists as no consensus has been reached yet.

IMHO it's far better to explain the problem BEFORE going into how it might be analysed. (I've tried it on lots of people - they have had good analyses without a whiff of Bayesianism!

First of all the two envelopes problem is not the one you have written above ending with the question "What should you do?". Instead the very statement of the problem must include the reasoning leading to the contradiction. That is followed by the question where the flaw in the reasoning is. That is the problem. See also my answer in the section above with title "What's the problem?". INic 00:13, 31 August 2006 (UTC)

Secondly, it is actually considered good wikipedia practice to have a short introduction to set the subject of the article into the right context before going into the meat of the matter. Please see WP:LEAD. INic 00:13, 31 August 2006 (UTC)

Thirdly, wikipedia articles should only be based on the contents of already published papers in reputable magazines. That you have friends that have solved this problem in an entirely different way doesn't matter I'm afraid. All published suggested solutions so far are Bayesian in character, and that is a fact that should be reflected here. When your friends have published their solutions we will rephrase this article accordingly. INic 00:13, 31 August 2006 (UTC)

Thanks - that's helpful. We are I think using 'problem' in two different ways. To me it is the problem' as I posed it - ending "What should you do?". To you, the problem is in the solution.

The problem or puzzle is to solve the paradox or get rid of the contradiction. I haven't seen any disagreement here in the literature. The only disagreement here is about what to call it, not what it is. My personal opinion is irrelevant. INic 02:28, 1 September 2006 (UTC)

What you call the 'problem', I would call the 'paradox'. I hesitate to say which is 'correct', but I think it is useful to distinguish between the two issues.

I can't find any clear distinction in the literature between what you call 'problem' versus 'paradox.' These words are used interchangeably. INic 02:28, 1 September 2006 (UTC)

I feel that the 'problem' is easy to state. The 'paradox' (or dilemma or whatever) is fairly easy to recognise. And the 'solution' is difficult indeed - but Bayes only arises in the solution.

No, as is explained in the comment of the lead section Bayesianism enter the stage already in the set-up of the problem. The only exception to this is the last, hardest, variant of the puzzle which isn't connected to probability at all. INic 02:28, 1 September 2006 (UTC)

You say that "All published suggested solutions so far are Bayesian in character" I don;t know the literature so cannot argue with this. However the following is not Bayesian:

Start off as before:

• You are offered an envelope which you open. It contains £10 say

• You are then shown two other envelopes and told that ONE of them £5 and the other one contains £20.
• Should you swap?

This has a clear non-Bayesian answer.

                                        Johnbibby 13:23, 31 August 2006 (UTC)

This is clearly not the same problem. This is seen by the fact that no contradiction or paradox can arise based on this situation. And that this problem merits an non-Bayesian analysis and solution is thus totally beside the point. INic 02:28, 1 September 2006 (UTC)

I agree I shd not quote my friends in Wikipedia! - but I think it's OK to refer to them in 'Talk'

                                  Johnbibby 13:23, 31 August 2006 (UTC)

Yes that is OK. INic 02:28, 1 September 2006 (UTC)

"solution" -> "discussion"

I suggest to replace the word "solution" in the paragraph titles by "discussion" because what we do is discussing the problem and it is not clear who is proposing the "pèroposed solutions" and who is suggesting that these arguments *are* actually solution.--Pokipsy76 13:32, 15 September 2006 (UTC)

I agree that "discussion" is more fitting, especially for the later paragraphs. However, for the first "proposed solution" there isn't much disagreement that this is the correct solution. Hence to replace that with "discussion" would be incorrect as no discussion exists here really. "Proposed solution" I think is appropriate for covering all cases, from almost total agreement to total disagreement. Another option would be to rename them "solution," "proposed solution" and "discussion" respectively. However, some might argue that this introduces an unnecessary element of POV to the article. INic 01:09, 16 September 2006 (UTC)

One stupid persons viewpoint

I was posed this paradox a while back (the "harder" version), and passed it on to a friend of mine. He then pointed me to this page. I'm no wiki expert (I only registered today) but feel that it is not readily comprehensible to non-specialist readers. Or someone as stupid as me.

My way to explain the paradox is to see that the values are chosen before you even see the envelopes. You having chosen the one with $512 in means that the value of the other envelope is already set. The chances of that envelope having a $1024 or $256 in it are not 50/50, they depend upon the generosity of the person filling the envelopes.

Choosing the envelope with $512 in it was the 50/50 part of it, not whether the other one is higher or lower. That is why the maths doesn't work, it is a failure to interpret the question correctly.

I think that this agrees with the proposed solution, but is there any chance that someone could attempt to make the page a little more accessable to the layperson? Situationist 00:48, 8 November 2006 (UTC)

Math is not psychological or material, it is conceptual. Generosity does not play any part in this, as evidenced by the following two concepts. 1) The envelope filler MUST put two amounts of money, one in each envelope. 2) One MUST ALWAYS be twice as great as the other. Your example, in which you use real dollar amounts, can be contradicted by the assumption the filler is wealthy beyond belief. Or look a it this way: Have him put in $10 and $20. If you pick $10, you would see his as "generous". But if you pick $20 the first time, you would see him as stingy. Things like this are all a matter of opinion anyway. Try using variables when examining the problem, and imagine a computer filling the envelopes as well. It may help you understand it better. MateoCorazon 01:42, 12 November 2006 (UTC)

Arguments page

I will soon clean up this talk page and remove everything that are discussions of the puzzle itself and not discussions of the article. Discussion of the puzzle itself can go on forever and has to be done elsewhere. Maybe someone can put up a link to a blog or discussion board dedicated to this puzzle at the top of this talk page? iNic 01:40, 17 November 2006 (UTC)

You might consider a solution like that chosen at talk:0.999..., with the following box near the top:

--Niels Ø 07:52, 17 November 2006 (UTC)

Yes, that would be a good solution here too! iNic 12:34, 17 November 2006 (UTC)

Proposed solution of the original problem

I suggest to add the following explanation:

"If the smaller amount is consistently denoted as X, the larger amount is 2X. Both envelopes contain either X or 2X with probability 1/2 each. Hence the expected value in both envelopes is 0.5 X + 0.5 · 2 X = 1.5 X and you don't gain on average by swapping."

In my opinion this is not self-evident or beside the point, but fundamental to understand the problem. --M.T. 20:07, 11 December 2006 (UTC)

As the situation is symmetric it's evident that it doesn't matter which envelope we pick. That is the very cause of the paradox as the reasoning in the text leads to a non-symmetric result, which obviously must be wrong. The problem is not to find another way to calculate that doesn't lead to contradictions (that is very easy), but to pinpoint the erroneous step in the presented reasoning leading to the contradiction. That includes to be able to say exactly why that step is not correct, and under what conditions it's not correct, so we can be absolutely sure we don't make this mistake in a more complicated situation where the fact that it's wrong isn't this obvious. Who in the literature claim that your calculation is fundamental to understand the problem? iNic 01:37, 12 December 2006 (UTC)

I agree that it's important and interesting to find the error in the presented reasoning with the result that the error lies in the calculation of the expected value. However, I see two reasons to mention the correct calculation: First, it's just interesting, at least for me, and presumably also for other readers. Second, and even more important, stating that the calculation of the expected value is wrong does not necessarily implicate that the consequence is wrong. In the literature at least Schwitzgebel and Dever who are cited in the article consider the correct calculation of the expected value noteworthy. They write: "The availability of such a non-paradoxical calculation is old news, of course; the novelty here is the identification of the crucial difference between the paradoxical and non-paradoxical calculation." --M.T. 17:06, 12 December 2006 (UTC)

Well, first of all Schwitzgebel and Dever clearly state what is the interesting problem (page 4). Our interest in the two envelope paradox [...] is not in providing a correct analysis of the decision problem — we take it that [...] this is trivial, and in the case of the two-envelope problem, it is well worked-over territory — but rather [...] in explaining what is wrong in the clearly absurd, but intuitively seductive, calculation that, in the paradox, envelope A is better than envelope B (or vice versa). Secondly, if the article should mention the correct calculation we have to be able to refer to some proof that this is the only correct calculation. All you have provided is a calculation that doesn't lead to paradoxes. Unfortunately, this is only a necessary condition and not a sufficient condition for a correct solution. Any calculation preserving symmetry will avoid the paradox. Thirdly, I doubt that there might be readers that doesn't realize the absurdity of the conclusion in the paradox. But if you think that should be stressed you can do that without mentioning the "correct" calculation. iNic 05:38, 15 December 2006 (UTC)

The calculation I have cited above is from a newer paper from Schwitzgebel and Dever which is called "The Two Envelope Paradox and Using Variables Within the Expectation Formula". This calculation of the expected value is also given in "Adom Giffin: The Error in the Two Envelope Paradox" and "Priest and Restall: Envelopes and Indifference". We don't have to proof it. However, if the article says that the presented calculation of the expectation value is wrong, this does not necessarily say that the consequence -- namely to swap -- is wrong. It could be possible that the correct calculation of the expectation value would be that the expected value of the money in the other envelope were 6/5 A, so one would gain on average by swapping. So we have to give a proof that the expected value of both envelopes is the same, so swapping will not lead to a gain on average. --M.T. 22:18, 25 December 2006 (UTC)

It's a little surprising that you view your calculation as the only correct one even after reading those papers. Priest and Restall consider three different calculations and concludes that all of them are correct, depending how the situation is interpreted; and without any interpretation at all the problem is underdetermined, i.e., not strictly solvable. Schwitzgebel and Dever defend their idea that each occurrence of a random variable in an expected value calculation should have the same mean, they do not have to be constants as in your calculation. Adom Giffin adds the assumption that the situation can be repeated an unlimited number of times. This assumption is a natural one to make for a physicist, but here it's actually not justified. You should in fact be glad if you got this generous offer only once. Hence, we can't really be sure how envelopes would be filled if the offer ever would be repeated. Giffin postulates that they'd be filled with the same amount every time. But fact is that we don't know that. iNic 05:33, 29 December 2006 (UTC)

If the correct calculation would lead to 6/5 A as the expected value for the other envelope, and the reasoning was correct otherwise, it leads to a contradiction. So your scenario isn't logically possible as you claim it is. Please feel free to add a comment about that in the article for those that doesn't see it directly. But please don't write that it's wrong to swap. It's no more wrong to swap than to stick. If you think this point should be stressed too for some readers please add a comment about that too. iNic 05:33, 29 December 2006 (UTC)

We are talking about the original problem, when the envelope is not opened. In this case all we know is that one envelope contains twice as much money as the other envelope, but we don't know which one, and the chance to get the envelope with the larger amount by picking one at random is 50 %, as well as the chance to get the envelope with the smaller amount (point 2 of the reasoning in the article). So the only correct calculation of the expectation value in this case is the one I have cited above.

OK, what is in that case wrong with the following: Denote by A what we have in our selected envelope and denote by Y the difference between the contents of the two envelopes. The other (not selected) envelope then either contains A-Y or A+Y. Both options are equally likely, i.e., has probability 1/2, so the expected value for the other envelope must be

${1 \over 2}(A-Y)+{1 \over 2}(A+Y)=A$,

which shows once more that we don't gain anything by swapping envelopes. This is not the same calculation as yours, yet it seem to be equally valid. If your calculation is the only one that is correct, this must be false. iNic 03:59, 3 January 2007 (UTC)

This calculation is indeed wrong. It contains the same error as described in the article, namely that A once denotes the larger amount and once the smaller amount. --M.T. 19:21, 3 January 2007 (UTC)

Yes maybe, but do you think that Priest and Restall would have had the same analysis? Definitely not. Please note that the solution proposed by Schwitzgebel and Dever is only a suggestion; it's not at all at the level of encyclopedic 'truth.' Priest and Restall, for example, have a fundamentally different solution, which is seen more clearly, perhaps, when applied to this equation than the original one. As the article shouldn't violate the NPOV policy, it's a bad idea to promote some authors opinions as the "correct" ones. To tell the readers what the "correct" solution is without qualifying that with who is saying that, and even hide the fact that it's controversial, is to do exactly that. iNic 00:23, 8 January 2007 (UTC)

Priest and Restall describe three different mechanisms for putting money in the envelopes and how the expected value is calculated in those cases. They describe mechanism 1 as follows: "A number, n, is chosen in any way one likes. One of the two envelopes is chosen by the toss of a fair coin, and n is put in that; 2n is put in the other." This describes the situation presented in the article for the case that the envelopes are not opened. We have an amount n and an amount 2n, and we pick one with a probability of 50 %. For this case Priest and Restall calculate an expected value of 3/2 n. --M.T. 13:57, 11 January 2007 (UTC)

Priest and Restall postulate some different mechanisms for putting money in the envelopes. They do not suggest that one of these is the "correct" one as you do. On the contrary they say it's impossible to know what the correct mechanism is and hence the original problem is underdetermined. Please re-read their article. And when it comes to the formula above, 1/2(A-Y) + 1/2(A+Y) = A, do you think that Priest and Restall would claim that this formula is wrong because A has different mean in its two occurrences to the left? iNic 00:58, 15 January 2007 (UTC)

I do not suggest that one of the mechanisms is the "correct" one, but that the mechanism described in this article is equivalent to mechanism 1 described by Priest and Restall. And for each mechanism there is exactly one correct calculation of the expected value. Priest and Restall have intentionally formulated the problem in such a way that different mechanisms can be assumed. Especially, they describe a "mechanism 2" which leads to a formula for the expected value in the other envelope that is given in step 7 of this article: E = 1/2 · 2A + 1/2 · A/2. With this mechanism you really have a 50:50 chance to get either 2A or A/2 in the other envelope, while in your envelope you have the fixed amount A. And yes, I'm sure that Priest and Restall would claim that the formula you gave above is wrong for the problem formulated in this article because A has different mean in its two occurrences to the left, because they write: "In a computation of expectation, the designation of the variables must correctly refer to the elements of the state-space, and not slide around in the process." So the question rather should be: Who actually specifies another formula for the expected value for the case we are talking about? --M.T. 21:11, 15 January 2007 (UTC)

I do understand your frustration that we can't easily state the "correct" expectation in the article. I wish we could, too. The reason we can't do this is that what this problem/paradox really boils down to is a question about probability itself and how it should be interpreted. Different authors have different solutions depending upon what philosophical interpretation they adhere to. For example, Priest and Restall have a totally different solution than Schwitzgebel and Dever. Philosophically speaking they are from different planets. Your impression is that they basically say the same thing, but that is very far from correct I'm afraid. Trust me. And the NPOV policy of course forbids us to claim that one author or philosophy is more correct than any other. So the best thing to do here is to leave it up to the reader to decide what she thinks is the best solution to the problem. iNic 03:54, 21 January 2007 (UTC)

You ask "Who actually specifies another formula for the expected value for the case we are talking about?" Well, if you look further down in the article you find the idea that the correct calculation should depend on how we assign values to the envelopes, i.e., exactly how is n chosen when the envelopes contain n and 2n? One of those possibilities leads to the expected value 11/10x in the "other" envelope, where we have x in our chosen envelope. This holds whether we look at x or not, so it's still relevant for the case we talk about. iNic 03:54, 21 January 2007 (UTC)

If there are different opinions on which calculation is correct, the NPOV policy requires that we present both, so the reader is able to decide what (s)he thinks is the best solution to the problem. Schwitzgebel and Dever write in their first paper on page 6: "Thus, the second method of analyzing the two envelope problem (yielding the indifference result) meets both constraints and is correct." The case you have mentioned is a special case with a special distribution that is known to the player. But we are still talking about the original problem here. --M.T. 21:02, 25 January 2007 (UTC)

To repeat, what you call "the problem" isn't the problem here at all. The problem under discussion here is not what a 'correct' calculation would look like. Instead, what occupies all authors is where the presented argument (that leads to a contradiction) is wrong. This is a big difference. So not only would you have to start a new Wikipedia article for your problem, prior to that you would have to start an academic debate centered at that issue. THEN you can apply the NPOV policy to the potentially different views there. When it comes to the two envelopes problem, we already obey the NPOV policy. iNic 00:37, 30 January 2007 (UTC)

What I call "the problem" or "the original problem" is the problem that is described as "the puzzle" at the beginning of the article: "Let's say you are given two indistinguishable envelopes, each of which contains a positive sum of money. One envelope contains twice as much as the other. You may pick one envelope and keep whatever amount it contains. You pick one envelope at random but before you open it you're offered the possibility to take the other envelope instead." If you state that the only interesting question is to find the error in the following reasoning, this is your personal point of view. But obviously there is an interest on if and how the expected value can be calculated correctly, which is documented in several papers. Furthermore, the main question is often seen as "Should I swap the envelopes or should I not, and under which conditions (envelope opened, know distribution,...)?" So if you try to suppress this documented points of view, this is a violation of the WP:NPOV policy. --M.T. 10:36, 31 January 2007 (UTC)

I've clarified the introduction even more now so it shouldn't be any doubt what this paradox is about. But if you feel that it would be interesting to start a new wikipedia article about the decision problem per se, please do so. iNic 19:52, 31 January 2007 (UTC)

Priest and Restall describe three mechanisms by which money could be assigned to the envelopes. For the original problem (amendment: given in this article) only the first mechanism describes the problem correctly. Actually, already point 6 of the reasoning given in the article is wrong in this case ("6. Thus, the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2"). Priest and Restall write: "The paradox is standardly formulated by employing reasoning of FORM 2 in a context where a mechanism of FORM 1 is deployed, which of course gives counter-intuitive results." I think that this point is not made clear enough. --M.T. 13:55, 31 December 2006 (UTC)

The purpose of the Priest and Restall paper is to show the dangers when using the principle of indifference, and why it often leads to paradoxes. The use of the principle of indifference easily produces results that can't all be right at the same time. This is old news in itself (Bertrand's paradoxes). What is new is that they argue that the two envelopes problem is caused by the (mis)use of the principle of indifference. In other words; the two envelopes problem is just yet another example of a Bertrand paradox, according to Priest and Restall. iNic 03:59, 3 January 2007 (UTC)

Historically, the reactions to Bertrand's paradoxes was to abandon the principle of indifference from probability reasoning altogether. However, it survived to this day in elementary books on probability for children, as it is still considered the simplest approach to probability. Today, the only ones still seriously defending (a refined form of) the principle of indifference are the Bayesians. This is why the two envelopes problem is of interest to Bayesians first and foremost. iNic 03:59, 3 January 2007 (UTC)

If we claim in the article that there is one and only one correct calculation of expected values for the closed envelopes, we will get into trouble further down in the article. In the "even harder problem" for example, some Bayesians in fact claim that it's better to always switch envelopes, even if you don't look in the envelopes! They will claim that your expected value calculation above is correct only for prior distributions of a certain kind. So even if we can find some authors supporting your claim, it will still only be a POV. And considering the fact that this is a sidetrack, i.e., not what this problem is really about, I think we are inviting more trouble than we solve by adding this POV calculation. iNic 03:59, 3 January 2007 (UTC)

Use of 'we' and 'you'

Hi,

Congratulations on creating an informative article. One grips is that the Manual of Style says articles should where possible not address the reader directly using 'we' or 'you'. I think the tone of this article should reflect that using more passive statements and so on. Thanks Andeggs 22:19, 22 December 2006 (UTC)

Personally, I like this direct style and don't think it becomes too unencyclopedic. The disadvantage of using passive statements everywhere is that the text can seem too academic and abstract for many readers. In addition, the common way to express this paradox in the (academic) literature is in this direct way. I don't think wikipedia should try to give a more academic impression of a topic than the academic articles that it's the summary of. iNic 00:50, 23 December 2006 (UTC)

Actually having had a look at the article again and at the manuals of style I think its the use of 'you' that is more of a problem than the use of 'we'. As the main Manual of Style says: "Avoid second-person pronouns". I'm sure this is possible with a bit of rewording but I don't want to do it and then be reverted immediately. I think the point about the academic literature is a little bit weak because we are writing an encyclopedia but they are writing maths/philosophy books and articles - and these are necessarily different. Thanks Andeggs 12:03, 23 December 2006 (UTC)

Please go ahead and do the changes you find appropriate. English isn't my native language so at least I won't revert any language changes. Regarding the academic style I was thinking about another writer at the talk pages here requesting a summary of this article for non-specialists that he could understand. Even the guideline you refer to include skepticism about its own applicability: While opinions vary on how far this guideline should be taken in mathematics articles — an encyclopedic tone can make advanced mathematical topics more difficult to learn — authors should try to strike a balance between simply presenting facts and formulas, and relying too much on directing the reader or using clichés [...] iNic 12:37, 23 December 2006 (UTC)

the danger of concrete examples

The page as it currently exists uses in example "the St. Petersburg lottery has an infinite expected value" (sic) which falls prey to Richard Feynman's old trick of defeating mathematical conundrums by forcing expression in concrete terms. There is no lottery which has an infinite expected value in the real physical world. Lotteries have set values; one method of calculating expected value is to use the chance of winning as related to the set of known or possible payoff values. There are no valid methods of calculating expected value of a lottery ticket that can yield infinity, so the example as stated does nothing to improve understanding. Expecting infinite yield from any real-world monetary instrument is literally insane. —Preceding unsigned comment added by 204.27.178.252 (talk) 23:45, 25 November 2008 (UTC)

Formatting of References

I suggest to convert the formatting of the references ("Published papers") to an automated style instead of manually numbering the references. E.g., if I wanted to add a paper of Schwitzgebel and Dever, I would format it as follows: <ref name="Schwitzgebel_and_Denver_02">Eric Schwitzgebel and Josh Dever. ''The Two Envelope Paradox and Using Variables Within the Expectation Formula.'' 2006-07-07. ([http://www.faculty.ucr.edu/~eschwitz/SchwitzPapers/TwoEnvelope060707.pdf])</ref> at the position where it is cited and add a section with the following text

== References ==
<!-- ----------------------------------------------------------
 See http://wiki.riteme.site/wiki/Wikipedia:Footnotes for a 
 discussion of different citation methods and how to generate 
 footnotes using the <ref>, </ref> and  <references /> tags.
----------------------------------------------------------- -->
<references />

at the end of the article. --M.T. 22:49, 25 December 2006 (UTC)

I considered all different solutions for adding references available, with their pros and cons, when I finally settled for this one. The main advantages are that we can have the reference list in chronological order and don't have to have a reference to every single paper in the text. Your proposed system for references lacks these freedoms. iNic 05:57, 29 December 2006 (UTC)

The particular PDF you want to add is a more popular version of their original text already in the list. As such, I don't think it merits a new entry in the list, and it's in fact very easy to find via the extra link "A Simple Version of Our Explanation" already present (next to the link to their main paper from 2004). iNic 05:57, 29 December 2006 (UTC)

I would say that the paper that I suggested to add is very different from the one already listed, and it's much easier to understand.

What do you mean with "chronological order"? Does it mean that new papers should always be added at the end? --M.T. 14:21, 31 December 2006 (UTC)

The most common suggestions here recently at the talk page has been to either make this article more encyclopedic in style, even if it makes it more difficult to understand, or to make it more popular so that it's accessible to more people. One solution to this dilemma suggested itself when I read the Monty Hall paradox article recently. It has a separate version in Simple English! We could do the same for this article. No doubt, a Simple English version will be far more difficult to write than this one. But it might be worth a try, don't you think? iNic 04:25, 3 January 2007 (UTC)

Yes, the papers are added "in order of appearance." It makes it easy to add new ones at the end as they pop up. The list is already quite long and confusing, and if we add different versions of the same basic ideas from all authors as separate entries we only add to length and confusion, not to content. iNic 04:25, 3 January 2007 (UTC)

Why no game-terminating event?

Doesn't the paradox arise from purposefully leaving out a game-terminating event, i.e. a time limit, or some other contest event that causes the player to end up with zero dollars, and thus that either of the envelopes result in a positive outcome? Using a game-terminating event , the solution is obvious -- keep any envelope and minimize risk for a zero gain. However, without some sort of terminating event, of course the game must continue forever because there is no incentive to pick one or the other envelope since either represents a 50/50 chance at the highest gain. Fine Arts (talk) 18:02, 27 October 2008 (UTC)

If you only have 2 envelopes and you pick one and then take the other for any reason, deosn't the game end? Wouldn't the game end the moment you picked the second envelope? People don't hover indecisevely forever between these sorts of choices, even if they seem equal. Any number of factors outside of the little scenario will cause the game to end one way or another. The solution I propose is that, if the amount of money in the first envelope is an indespensible amount, say abitrarily 1 million dollars, you would take it where as half of that would be a huge loss and double that would be just gravy. The incentive to 'go on' is lessened. Where as, though if the amount in the first envelope is more trivial, say abitrarily, 4 dollars, double it is just as meaningless as half, so you might as well just press your luck.Zavion (talk) 16:34, 14 November 2008 (UTC)

Comments about the Wikipedia Article

The following discussion is related to the Wikipedia article as stated in January 2009. The following topics were split from a more comprehensive discussion that included a detailed discussion of the solution to the paradox as well. The basis for my including a discussion of the solution in my original discussion of the article was my belief that it was probably necessary to understand the solution in order to evaluate and edit the existing article. The solution portion of that original discussion is presently on a separate discussion section page identified as the “Arguments Page.” This page can be accessed from the link at the top of this page.

Questionable Title

The title of the Wikipedia article appears questionable with respect to the use of the word “problem,“ a more general term, rather than the word “paradox.” Presently the title of the Wikipedia article is “Two envelopes problem.” However, in another Wikipedia article titled “List of paradoxes,” the title that is linked to this article is stated as “Two-envelope paradox.” Interestingly, the opening sentence of the article states that the problem is a puzzle or paradox, so according to the article itself, it does not appear that classifying the problem as a paradox is in any way a contested issue.

I also performed a quick Internet search using Google. Search words “two envelope problem” and “two-envelope problem” provided 551 results, while search words “two envelope paradox” and “two-envelope paradox” provided 3870 results. Although popular usage may not be the true or ultimate test for being correct, it does indicate a popularity in usage of nearly eight to one.

It also appears that irrespective of the title containing the word problem or paradox, the title should probably begin with the article “the.” Accordingly, the title should probably be either “The Two-Envelope Problem” or the “The Two-Envelope Paradox.” I am also somewhat confused by this new trend to not capitalize the first letter of each major word in a title. Does anyone have an explanation for this?

General Comments on the Wikipedia Article

In general, I found parts of the article detailed and well done, but overall I found the article poorly organized, poorly worded, and not thoroughly or well explained. When I performed an Internet search of the paradox, I found a completely different wording to the more popularly stated version of the paradox and found at least three popular approaches to disproving the paradox. Interestingly, the popular version of the paradox included the provision that the amount contained in the first enveloped was known to the player, while none of the four versions mentioned the Wikipedia article included that provision. Also, though my research was rather limited, I did not come across any of the various specific versions of the paradox that are discussed in the Wikipedia article. The Wikipedia article also contains at least one comment claiming that there is some contention or dispute over the flaw or the validity of the flaw in the paradox. After reading several of the articles and papers posted on the Internet, I find that fact somewhat difficult to believe without further qualification or explanation.

Wording Used in the Article

My observation while reading the article was that there were instances where the meaning or thought was not clear. Because probability is not my field of expertise, I wasn’t sure whether this was true because the text was simply not worded well or whether the text was stated in terminology that was foreign to me. A good example is the first sentence of the section titled “An even harder problem.” The subject sentence is as follows: “The solution above does not rule out the possibility that there is some non-uniform prior distribution of sums in the envelopes to give the paradox force.” The main problem I found in understanding the sentence was in the meaning of the phrase “to give the paradox force.” At first, I suspected the meaning to be related to that which established a stronger or better-established paradox. However, two sentences later in the discussion the subject is regarding “a sensible strategy that guarantees a win.” It wasn’t until I began reading the following section on the “Proposed solution” that the discussion returned to the paradox. Then, even after reading the section on the solution, I was still unable comprehend the meaning of the subject sentence.

Accordingly, if this readability problem is related to technical terminology, then perhaps the terminology should be replaced by terminology more suited to the general public, the intended readers of an encyclopedia. Perhaps otherwise, the article should be reviewed and the general wording improved.

Four Versions of the Paradox

The present version of the Wikipedia article addresses four versions of the paradox. These are identified as “The problem,” “An even harder problem,” “A non-probabilistic variant,” and “History of the paradox.” I address each of these divisions individually in the following sections. I also included some discussion related of the respective solutions. I did this being fully aware of the notice to include solutions on the “Arguments Page.” My rationale for retaining these discussions was twofold. First, for the most part the existing solutions as provided in the article are problematic and require some discussion for the purpose of editing. Accordingly, these points could not be easily made without some reference to a common understanding of the solution. Second, I also did not wish to expend the time to edit and/or rewrite these sections.

Notes on the Section Titled “The problem”

The most elementary version of the paradox described in the Wikipedia article is stated in 12 enumerated statements. Although the 12 statements may or may not have some basis in the historically correct version of the paradox, this version is not only inconsistent with the popular versions, but appears to be extended beyond the basic version required for gaining a fundamental understanding of the paradox.

Regarding the discussion of the flaw in this version of the paradox, the flaw is explained mathematically in the section identified as “Proposed solution.” In brief, the flaw is attributed to a misrepresented and thereby incorrect calculation of the expected outcome stated in step 7 of the problem. However, I also found it possible to analyze the flaw by the path of reason. Such a solution may be more attractive to readers not having a strong math background.

First, if the player reasons that there are only two envelopes containing two amounts, and for the purpose of reasoning simply calls these amounts C and 2C, then the player can extend this reasoning as follows. If initially the player randomly chooses the envelope containing the smaller amount C, then A equals C by definition. If the player then chooses to swap that amount, amount A, for what could be perceived as being either ½A or 2A, the actual swap can only be for amount 2A, thereby providing an apparent gain of A. In terms of the C, however, the swap is equivalent to having swapped the initial amount C for amount 2C. Because the player initially selected the envelope containing amount C, the smaller amount, there is no such amount equal to ½A in this case.

By comparison, if initially the player randomly chooses the envelope containing the larger amount 2C, then A equals 2C by definition. If the player then chooses to swap that amount, again amount A, for what again could be perceived as being either ½A or 2A, the actual swap this time can only be for ½A, thereby providing an apparent loss of ½A. In terms of C, again, the swap is equivalent to having swapped the initial amount 2C for amount C. Similarly, because the player initially selected the envelope containing 2C, the larger amount, there is no such amount equal to 2A in this case. Accordingly, despite the truth to the statement that the player can gain A and only lose ½A, we can see that this statement is only true because A has twice the value in a loss than in a win. In terms of C, the player can only win or lose the same amount, amount C.

The source of confusion creating the paradox is in being drawn into false reasoning based on believing that because the amount contained in the initial envelope can be represented by A, that A must be a fixed amount. It is easy to be drawn into this trap because the player can visually see and hold the envelope or, at least, imagine doing so. In this context it is then difficult to revert back to the logic and reason that the outcome was predetermined at the time the initial envelope was assigned.

Also worthwhile noting is that statement 6 in what is called “The switching argument” (stated as, “thus, the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2”) is very open to misinterpretation. The truth of this statement is dependent on the precise definition of variable A. As stated previously, although it is true to state that (1) the probability of A being the smaller amount is one-half and (2) the probability of A being the larger amount is also one-half, once the value of A is known, these statements are no longer true for all specific values of A.

Notes on the Section Titled “An even harder problem”

A second version of the paradox is described in this section. This version of the paradox alters the frequency probability of the amount pairs that contain a common amount. For instance, in the primary version of the problem the frequency probability of a pair of envelopes containing $10 and $20 is the same as for a pair of envelopes containing $20 and $40. In this variation, the probability is determined by a mathematical formula that results in the pair containing the lower amount to have a frequency probability of 3/5, while the pair containing the higher amount to have a frequency probability of 2/5. Doing this causes some confusion. Although it might appear that this has an affect on the overall 50:50 win-loss probability of a swap, it only does so in certain conditions. Tracking this distinction appears to add another layer to the confusion already present in the original version of the paradox. My suggestion would be to include this topic under another title. In effect, this discussion is more related to frequency probability than to the Two Envelopes Problem.

Regarding the mathematical formula ( 3/5 * x/2 + 2/5 * 2x = 11/10 x ) that is used to calculate the expected outcome to be 11/10 x, the formula is incorrect for the same reason that the previous mathematical formula ( ½ * 2A + ½ *A/2 = 5/4 A ) was incorrect in the primary version of the problem. Interestingly, in the primary version of the problem the flaw in the formula was uncovered and was explained in terms of the expected amount contained in the two envelopes. The formula for that calculation was ½ * C + ½ * 2C = 3/2 C. For example, if C is equal to $10 and 2C is equal to $20, then the average or expected value is $15.

If we now apply this same reasoning to this version of the problem, the formula becomes 3/5 * C + 2/5 * 2C = 7/5 C. And, if again we use the same values for C and 2C, the average or expected amount contained in the envelopes is $14. Because the frequency distribution of $10 is stipulated to be 3/5 while the frequency of $20 is stipulated to be the remaining 2/5, the expected amount is correct. Accordingly if this reasoning is applicable and sufficient for the basic version of the problem as stated in the Wikipedia article, it should also be applicable and sufficient for the “harder” version The article simply does not address this issue and instead discusses reasoning along a completely different line. However, despite all of the discussion, it certainly appears that the solution should be identical to that of the primary version of the problem.

Regarding a solution to uncovering the flaw in this version of the paradox (if that is the objective), I also found it possible to analyze the flaw by using the same path of reason I used in the primary version. Accordingly, I provided that path of reasoning below.

First of all, to minimize confusion over the frequency of the pairs of amounts and the amounts themselves, let’s assign designations for everything. Accordingly, let’s call the pair containing the lower amount X and the pair containing the higher amount Y. In terms of the common member B, it follows that X represents B/2 and B, while Y represents B and 2B.

Next, suppose that the envelopes contain the pair designated as X. And, as before, if the player reasons that there are only two envelopes containing two amounts, and for the purpose of reasoning simply calls these amounts C and 2C, then the player can extend this reasoning as follows. If initially the player randomly chooses the envelope containing the smaller amount C, then A equals C by definition. If the player then chooses to swap that amount, amount A, for what could be perceived as being either ½A or 2A, the actual swap can only be for amount 2A, thereby providing an apparent gain of A. In terms of the C, however, the swap is equivalent to having swapped the initial amount C for amount 2C. Also worth noting is that in terms of B, C equals B/2 and 2C equals B. Therefore the swap was also equivalent to having swapped the initial amount B/2 for amount B.

By comparison, if initially the player randomly chooses the envelope containing the larger amount 2C, then A equals 2C by definition. If the player then chooses to swap that amount, again amount A, for what again could be perceived as being either ½A or 2A, the actual swap this time can only be for ½A, thereby providing an apparent loss of ½A. In terms of C, again, the swap is equivalent to having swapped the initial amount 2C for amount C. Again, worth noting is that in terms of B, C still equals B/2 and 2C still equals B. Therefore the swap was also equivalent to swapping the initial amount B for amount B/2. Accordingly, in terms of either B or C, the gain (B/2 or C) and loss (B/2 or C) is the same.

Next, we simply repeat the same reasoning for the pair of amounts designated as Y. Again we would expect to get the same matching gain and loss. However, now worth noting is that in terms of B, C equals B and 2C equals 2B. Accordingly, in terms of either B or C the gain (now B or C) and loss (B or C) is the same.

In this version of the problem the source of confusion creating the paradox extends one level deeper than in the classic version. Accordingly, not only is there the confusion of being drawn into false reasoning based on believing that because the amount contained in the initial envelope can be represented by A, that A must be a fixed amount, there is also the draw to confuse the probability of the frequency of the amount pairs with the probability of the frequency of the amounts within a pair.

Notes on the Section Titled “The non-probabilistic variant”

The non-probabilistic version is stated in two enumerated statements as follows: 1. Let the amount in the envelope chosen by the player be A. By swapping, the player may gain A or lose A/2. So the potential gain is strictly greater than the potential loss. 2. Let the amounts in the envelopes be Y and 2Y. Now by swapping, the player may gain Y or lose Y. So the potential gain is equal to the potential loss.

Although the Wikipedia article states that “so far, the only proposed solution of this variant is due to James Chase,” I find this confusing and a somewhat difficult to believe. This variation does not appear sufficiently different from the primary version for the same reasoning to not apply.

Regarding the truth of the two enumerated statements, statement 1 is false and statement 2 is correct. Accordingly, the part of statement 1 that is incorrect is the stated conclusion (“so the potential gain is strictly greater than the potential loss”). The flaw in the conclusion is the result of faulty reasoning between the previous statement (“by swapping, the player may gain A or lose A/2”) and the stated conclusion. The suggested reasoning is that because the player may gain twice the amount that he (or she) may lose, that there must be a potential gain. This conclusion is, in fact, incorrect, because it does not consider the offset created by the fact that swapping larger values of A will result in a greater number of losses than gains, and that swapping smaller values of A will have a greater number of gains than losses. Although neither of the enumerated statements specifically claims the probability for the second envelope to contain either ½A or 2A to be 50:50, it is implied. Accordingly, the implied 50:50 probability for the second envelope to contain either ½A or 2A, however, does also imply that each and every specific values of A must also have that same 50:50 probability. The details of this were discussed in greater detail previously.

Regarding the solution to the paradox, if the player initially chooses the envelope containing the amount Y, then A equals Y as defined in statement number 1. Accordingly, the player can only gain an amount Y in swapping for the remaining envelope containing the amount 2A. In terms of amount A, the player effectively swaps amount A for amount 2A. Because the player initially selected the envelope containing the amount Y, the smaller amount, there is no such amount equal to ½A. By comparison, if the player initially chooses the envelope containing the amount 2Y, then A equals 2Y as defined in statement 1. Accordingly, the player can only lose an amount Y in swapping for the remaining envelope containing the amount ½A. And, in terms of amount A, the player effectively swaps amount 2A for amount A. Similarly, because the player selected the envelope containing 2Y, the larger amount, there is no such amount equal to 2A. According, the player can only gain or lose the amount Y, and the reasoning is an exact parallel to the reasoning in the primary version of the paradox.

Regarding the paradox of the non-probabilistic version being difficult to resolve as claimed in the article, the above explanation does not indicate that to be true. Perhaps there is some misunderstanding or perhaps this issue requires further research.

Notes on the Section Titled “History of the paradox”

The historical version of the problem that is characterized by the narrative of two strangers wagering over the amount of money in their wallets appears to be another version of the Necktie Paradox rather than a version of the two-envelope problem. However, the historical version differs from the Necktie Paradox in that each man may or may not possess information regarding the contents of his own wallet that may alter his own advantage or disadvantage in the wager. Accordingly, if both men are oblivious to the contents of their wallet, the structure of the wager becomes identical to that of the Necktie Paradox. However, if either man is aware of the contents of his wallet, that knowledge can have a significant effect on the decision to partake or not partake in the wager.

For instance, if one of the men were carrying more than an expected amount of money in his wallet, he would be wise to not partake in the wager. Similarly, if one of the men were carrying less than an expected amount of money in his wallet (for example, no money), he would likely have a winning advantage. Also, because either man can have the ability to base the decision of whether to make the wager on his knowledge of the contents of his wallet, the problem provides no rationale for providing credibility to having a 50:50 or random probability for either man winning the wager. However, in the interest of focusing only on the paradox portion of the problem, it would probably be more prudent to play along with the intent of the narrative and not be picky about the unintentional flaws in the narrative.

Despite the history of how the two-envelope problem may have evolved, I disagree that the two-envelope problem is simply another version of the two-wallet problem. First, there is the problem of correlating the parenthetically stated “½” probability to the narrative as I mentioned above. Second, even though neither man knows the amount in the wallet of his rival, either man is not precluded from being fully aware of the amount in his own wallet. And, assuming that this amount is reasonable and thereby does not completely override the presumed 50:50 probability to win or lose, one man knowing the amount does affect the path of reasoning to be distinctly different from the two-envelope problem. Third, the two-envelope problem specifically addresses amounts that differ by a factor of two. Although this does not alter reasoning in the problem, it does simplify the discussion when attempting to represent the amounts as variables. And fourth, because the two-envelope problem involves amounts that are unknown and undisclosed, it allows for the inclusion of the argument that proposed a continuous and endless exchange. That same argument is more difficult to propose in the two-wallet problem because the narrative does not support the exchange of unopened wallets.

Accordingly, it appears out of perspective to relate the two-wallet problem to the two-envelope problem without mentioning the more closely related Necktie Paradox. This is similar to (though perhaps not as important as) misplacing a species in the evolutionary tree.

Bill Wolf (talk) 21:10, 21 February 2009 (UTC)

Suggestion to remove non-probabilistic version

It seems to me that the non-probabilistic version of the paradox does not make any sense, since there is no point in comparing potential rewards if no probabilities are attached to them. It's like asking, is it preferable to enter a lottery where the prize is $1,000 or one where it is $100,000? If you are given no more information then the question is meaningless. Or am I missing something? Soler97 (talk) 15:46, 17 July 2009 (UTC)

Yes you missed the point. The logical variant is not a decision problem. And even if you still don't think this version makes any sense, that is of course not a valid reason for removing the section from the encyclopedia. iNic (talk) 01:18, 2 October 2009 (UTC)

I propose adding a link in external references

I propose adding a link to http://www.opentradingsystem.com/quantNotes/Two_envelopes_.html

It contains a quick experimental resolution argument. Kaslanidi (talk) 20:34, 30 December 2009 (UTC)

I object, per WP:ELNO points 1, 4, and 11. - MrOllie (talk) 20:35, 30 December 2009 (UTC)

I second the objection. OhNoitsJamie ^Talk 20:44, 30 December 2009 (UTC)

Constructive suggestions

Much of the debate here and inclarity in the article stems from the fact that there are two different ways to present the paradox, a mathematical one and a realistic one. The solution to the realistic one is easy, and it involves finite upper limits to the prize sums, individual utility functions and such. The mathematical paradox (where you cannot make such assumptions) is the actual paradox and much harder to crack. This distinction should be made more clear.

Another suggestion: Many of the references seem to be missing the point of the paradox entirely, or are otherwise irrelevant. They should be pruned away. Smite-Meister (talk) 22:42, 24 March 2009 (UTC)

I think it's true that this article is dreadful. I read this and came away confused, believing that there was a serious unsolved problem in statistics. Then I read Devlin's Angle, and found out there is a trivial solution to this. I think that, in its current form, this article is bad for wikipedia. I propose we delete the entire "non-probabilistic variant", this was apparently proposed in a 18 year old logic puzzle book, and is confusing. I propose we delete the entire "An even harder problem" section as well - it is not relevant to the two envelope paradox, and it seems to be confusing. I propose we delete the solutions and use Devlin's solution - which is that it is not correct to calculate the expectation in this manner. Since the article is in such a state, I think a simple statement like this would do. We can link to Devlin's article for the full explanation. If you object to these changes please talk here. Dilaudid (talk) 16:27, 26 May 2010 (UTC)

Having worked through the "even harder problem" I think this is valid and quite interesting. I have tried to add points. Since the original two-envelope problem is now clearer, I don't think including the more complicated example will detract from it - is everyone else happy with this version? Dilaudid (talk) 21:24, 26 May 2010 (UTC)

I object strongly to delete sections describing important versions of the paradox entirely. On the contrary, we need to add more of the opinions and arguments that exists in the literature. iNic (talk) 17:51, 7 June 2010 (UTC)

Could someone please explain the first part of "A Second Problem" which states:

Suppose that the envelopes contain the integer amounts {2ⁿ, 2ⁿ⁺¹} with probability 2ⁿ/3ⁿ⁺¹ where n = 0, 1, 2,…

Why was this probability formula chosen, and what significance does it have for the problem? 96.245.11.102 (talk) 13:08, 26 November 2010 (UTC)

The purpose of "A Second Problem" is to refute the claim, by Menzies and Oppy, that the solution to the paradox is to require a prior probability distribution. It's an explicit construction of a valid prior probability distribution that is still paradoxical in the sense that the expectation would lead you to switch no matter what you choose. --Tkircher (talk) 16:05, 3 December 2010 (UTC)

Equation source

${1 \over 2}2A+{1 \over 2}{A \over 2}={5 \over 4}A$

Does anyone have the exact source were this equation first comes from? I can't see how anyone could take this equation seriously. "In the first term A is the smaller amount while in the second term A is the larger amount." This is the problem summed up perfectly, why it is listed as a 'proposed solution', it IS the solution.--Dacium 04:58, 17 April 2007 (UTC)

If you read on in the article you will find a short history of the paradox. I reverted your addition of a "correct equation" because it belongs to a different discussion. This article is about what is wrong about the switching argument (i.e., how to solve the paradox), not about different ways to reason instead (or different ways to act if put in this situation). However, I welcome a separate article about the decision problem where we can gather all these ideas (all "correct calculations" and strategies). (Please have a look at the archived discussions for this talk page as this question has been brought up before.) iNic 01:16, 18 April 2007 (UTC)

I don't get it. This is what is wrong about the switching argument. The A in 4. is smaller than the A in 5. Therefore the proposed equation makes no sense.--90.179.235.249 (talk) 17:58, 19 January 2010 (UTC)

Dacium has correctly explained the solution to the problem at the beginning of this section. A correct explanation is also given at http://brainyplanet.com/index.php/Envelope (Click on "Solution".) Additionally, the article by Priest and Restall. Envelopes and Indifference (PDF, 79 kB). 2003-02-15 at http://consequently.org/papers/envelopes.pdf gives further explanation which may or may not be helpful. This is not a paradox. It is a SOLVED problem. Dagme (talk) 00:36, 26 December 2010 (UTC)

solution

I'm confused, is there a solution? In the intro, it says, "This is still an open problem among the subjectivists as no consensus has been reached yet." But the article seems to suggest there's a solution (if interpreted another way?). Could someone clarify this for me and amend the article if necessary? 129.120.94.151 22:37, 19 April 2007 (UTC)

I agree, it was confusing. I deleted recent edits that made the article contradictory in this respect. Thank you for pointing this out. iNic 13:13, 20 April 2007 (UTC)

As I pointed out in the previous section (at the end of that section as of now), Dacium has correctly explained the solution to the problem at the beginning of the previous section. A correct explanation is also given at http://brainyplanet.com/index.php/Envelope (Click on "Solution".) Additionally, the article by Priest and Restall. Envelopes and Indifference (PDF, 79 kB). 2003-02-15 at http://consequently.org/papers/envelopes.pdf gives further explanation which may or may not be helpful. This is not a paradox. It is a SOLVED problem. Dagme (talk) 00:39, 26 December 2010 (UTC)

It is not as simple as that. The main problem is the lack of clarity in the problem definition. Priest and Restall give an interesting discussion of some of the issues involved. But there is no universally accepted solution (or problem). Martin Hogbin (talk) 10:41, 26 December 2010 (UTC)

No basis is given for the claim that "It is not as simple as that". The problem is stated clearly in the Wikipedia article under discussion, in Devlin's article, in the Priest - Restall article and in many other places. I personally believe that it IS "as simple as that", but I would be very interested to see a substantive discussion showing otherwise. Absent such discussion, the claim that "It is not as simple as that" is simply a hollow, groundless claim. Dagme (talk) 06:48, 4 January 2011 (UTC)

crazy

I think this discussion is crazy. Just look at the french page on this topic (http://fr.wikipedia.org/wiki/Paradoxe_des_deux_ch%C3%A8ques), if you want the situation properly discussed.— Preceding unsigned comment added by 129.194.8.73 (talk) 13:26, 15 August 2007

---

I've looked at the French page but as far as I can see it doesn't really cover all the bases. The explanation they use does come up with the correct answer, however, that is not required. We already know the correct answer i.e. there is no reason to swap. That is why it is a paradox. The problem is, what is the logical error in the reasoning?

Now, the French page states the 50% term should be 50%*2. But this is right at the heart of the issue and can't just be changed because the answer comes out right. Let me expand...

In the English article the non-probabilistic variant section says

1. Let the amount in the envelope chosen by the player be A. By swapping, the player may gain A or lose A/2. So the potential gain is strictly greater than the potential loss.

2. Let the amounts in the envelopes be Y and 2Y. Now by swapping, the player may gain Y or lose Y. So the potential gain is equal to the potential loss.

Now, it is clear that 1. must be wrong and 2. must be correct. But at point does 1. go wrong?

As far as I can see (and I might be wrong here), the French article is doing the equivalent of saying "Well, this formula (relating to 1.) is wrong but if we use this other formula (relating to 2.) it comes out right." Very true. But what is the logical error in 1.?

Also, I think it specifically does not pass the test where we open our envelope and see that it contains $10. In this case, the paradoxical formula can still be (apparently) justified - gain $10 vs loose $5. Whereas the 50%*2 term now appears to make no sense at all. Fontwell (talk) 14:42, 5 September 2008 (UTC)

Actually, "1" is correct and "2" is wrong. "2" is wrong because "gain Y or lose Y" depends on two different selections of the first envelope--to gain Y by switching I must have chosen the Y envelope, and to lose Y by switching I must have chosen the 2Y envelope. These two things cannot simultaneously be true. "1" is correct whether A is the larger amount or the smaller amount. 76.234.120.116 (talk) 01:15, 21 September 2008 (UTC)

Erm, you seem to have fallen for the original paradox. Which just goes to show what a corker it is!

The overview is this:

The initial choice of envelope was arbitrary and had no way of choosing the better outcome. No new information was presented to us after the choice. If we subsequently swap envelopes it cannot put us in an advantageous position. We should all believe this to be correct.

Thus, any mathematical argument that shows an advantage in swapping is flawed. This is the paradox. Thus 1. must be wrong.

The explanation for why 2. is wrong is flawed. I think there is a confusion between 1. and 2. going on here.

The big difference in 2. is that all it says is that the two amounts of money are Y and 2Y. It doesn't specify which envelope you choose. So, if you don't know which is which, then swapping will either swap you from holding Y to holding 2Y (gain of Y) or swap you from holding 2Y to holding Y (loss of Y). Thus the possible gain matches the possible loss.

The comment:

"...gain Y or lose Y" depends on two different selections of the first envelope--to gain Y by switching I must have chosen the Y envelope, and to lose Y by switching I must have chosen the 2Y envelope. These two things cannot simultaneously be true."

This is all correct. However to say "2 is wrong because" of it is just a non-sequitur. All that is missing from that comment is to finish it off: So what we do in game theory (or in fact probabilities in general) is to say: "These two things cannot simultaneously be true" but we don't know which is actually true. So let us assign a probability (of truth) to each case so that we can determine our 'expected gain'.

In this case (2.) we have no reason to think that holding Y or 2Y is more probable. So we have two equally likely possibilities. One of them gains us Y while the other looses us Y. The overall 'expected gain' is the sum of these two. Strictly, Expected Gain = {Y x 0.5} + {-Y x 0.5} = 0

Fontwell (talk) 17:31, 23 September 2008 (UTC)

This paradox teaches us what expected value really is. Expected value is calculated from either the envelope with the least amount in it, or the other one. The fallacity many of us (including me at first) make is to calculate it from the envelope i pick or the other envelope. I do understand why many bayesians make this mistake, yet a bayesian solution is available. Let the amounts in the envelope a priori be x and 2x. you pick 50%x or 50%2x. By swapping you gain x or lose x. Expected value is zero. Now let us call the amount in the envelope we pick A and substitute the x'es in the above tree. We pick A, we could gain A or we could lose 1/2A. Seems like there is a expected gain. But, also substitue the prior. We now see that the prior could have been {A, 2A} or {1/2A, A}. Meaning we no longer have a prior, rather a subjunctive hyper-prior. This proves the fallacy. We compare leaves from both seperate trees. —Preceding unsigned comment added by 193.29.5.6 (talk) 06:27, 1 April 2010 (UTC)

Let me try to explain the fallacity in a number of steps:

1. I'm offered to play a game. I bet for instance €10, the host throws a coin (50% head, 50% tails). On head, I double up, on tails I half down. However, I'm allowed to withdraw before the host throws. Would I go for it or not ? It is clear I have an expected gain by playing the game (1/2 €+10 + 1/2 €-5 = €2.5).

2. I'm playing the game in a different situation. I'm offered €10. The host tells me he had drawn a coin prior to the current event. On head, he doubled his initial amount, on tails he halfed it. Anyway, the outcome is €10. I'm now offered to either go home with €10 or switch to the initial amount. Would I got for the switch ? When switching the expected value appears to be in my favour (1/2 €+10 + 1/2 €-5 = €2.5). Yet it is not. Expected value is calculated from prior to posterior, not the other way around. So the way I must think is: from my initial value I could have either doubled up or halved down to €10. Switching would mean I either lose my initial amount or gain half my initial amount. So I do not want to switch at all ! The problem we have with reasoning this way (the correct way), is that threat our €10 is if it were the prior situation, and the past events are the posteriors.

3. The 2 envelope paradox is a combination of both steps above. You pick an envelope. The amount inside of it is either the initial amount (step 1). In this case you want to switch. Or the amount inside of it is the result of the experiment of the other envelope (step 2). In this case you want to stay. Both scenarios are equally likely, so it should not matter wheter you switch or stay. —Preceding unsigned comment added by 94.225.129.117 (talk) 07:42, 3 April 2010 (UTC)

This is a solved problem, not an open problem

The intro says "This is still an open problem among the subjectivists as no consensus has been reached yet." but that is not correct. There is a clear consensus about the solution in peer reviewed publications. Just have a look at Samet et al., 2004. Yes, this problem is strongly debated by the public and many solutions are proposed and put online but public debate does not mean there is no scientific consensus. It would be the same to call the Monty Hall problem an open problem. There is a clear solution to the Monty Hall problem with scientific consensus no matter what the public thinks or feels.

The solution is independent of Bayesian or frequency interpretation and can be captured in a single phrase:

"The Two Envelopes problem uses an impossible probability distribution."

The use of an impossible probability distribution is the core of most if not all statistical paradoxes. Just like "proofs" that show 1=2 sneak in a division by zero or set the square root of -1 equal to 1 or -1 ( which non-mathematicians easily accept ) so do statistical paradoxes sneak in an impossible probability distribution. The Two Envelopes problem uses the impossible distribution of an infinite set with every element having the same probability. "Any amount of money ( with no upper limit ) with every amount of money as likely". For a non-mathematician the probability distribution of the Two Envelope problem may sound completely reasonable but is mathematically utter nonsense just like dividing by zero.

For every probability distribution the sum of all the probabilities of every possible outcome has to be 1. If you don't have this statistics completely breaks down. You get infinite means and impossibilities as clearly shown in the Two Envelope problem.

If you change the impossible probability in the Two Envelope problem into a real probability by putting an upper limit to the money or by not giving the every amount of money the same probability, the paradox disappears.

There is a clear scientific consensus about this and any professional mathematician or statistician will confirm this. This is basic knowledge.

The article should clearly state this and use it as main point. Obviously this should mean a complete rewrite of the article.

I will not rewrite the article because I don't like to be sucked into an edit war. I just wanted to make it clear that the article is missing the main point. I hope in the end reason will prevail. Michael Korntheuer 17:37, 15 February 2008

If you read half way down the article you will learn that it's not the case that the paradox disappear for every proper prior distribution. If you read even further down you will discover that we actually don't even need a prior distribution to get the paradox, as we don't have to invoke the probability concept at all. iNic (talk) 22:35, 15 February 2008 (UTC)

True about a distribution summing to 1, although there are still infinities involved, specifically infinite expected values. False about not needing to invoke the probabity concept at all. In particular, how do you make the leap from "Let the amount in the envelope you chose be A. Then by swapping, if you gain you gain A but if you lose you lose A/2. So the amount you might gain is strictly greater than the amount you might lose" to "you should swap" without looking at probabilities? To decide to switch inherently involves an assessment of probabilities. To say otherwise is to say you'd be happy to play roulette as long as you can't see the wheel and are thus ignorant of the probabilities involved. Warren Dew (talk) 01:37, 3 May 2008 (UTC)

The section Two envelopes problem#A non-probabilistic variant does describe a paradox that has nothing to do with probability. (Though I don't see it as a variant.) The solution of that paradox (IMO) is that the term potential gain is ambiguous and used in two different meanings, giving different results. Oded (talk) 18:12, 27 July 2008 (UTC)

Yes, the statement "if you gain you gain A but if you lose you lose A/2" is making exactly the mistake described in the first section! The "A" that you "gain" is the same actual numerical value as the "A/2" that you "lose".

The source of the apparent paradox in "an even harder problem" is similar to what Michael Korntheuer was driving at--the fact that there are situations where the "double" value of one possible set is the same as the "single" value of another set. That is, X^a = 2X^b. Since a and b have equal probability, there is no way to identify whether an envelope has X^a or 2X^b--even if we open the envelope. 192.91.147.35 (talk) 01:14, 11 September 2008 (UTC)

The problem vanishes once the problem is set up correctly : we have only two envelops, and their value is deterministic. Which one will be elected is not determined. The experiment as stated in the paradox is this : there are two envelops : A with value X and B with value 2X. The experimentor will chose A with probability 1/2, and B with probability 1/2. Once the choice has been made (say the choice was A), a fairy destroys the envelop B and replaces it with a new envelop C with value X/2 or 2X (1/2 probability each). This is clearly not the meant experiment. In the experiment, B stays there with its determined content. To modelize the experiment, you must not forget to take into account that both values are determined through the whole experiment. The fact that you don't know which envelop is A and which envelop is B makes the only probabilistic problem in the experiment. This is the only probabilistic variable in the experiment : no amount in the envelops is probabilistic. Subtenante (talk) 11:31, 17 December 2009 (UTC)

I slightly disagree with you there. The amount in envelope A is deterministic, the amount in envelope B is probalistic (50% A*2 50% A/2). The experiment starts at a point the probalistic event has taken place. At this point the only thing you should care about is to know which envelope is A, and which is B. You know that the amount in envelope B has a higher expected value than the amount in envelope A. Because you can't distinguish the envelopes, you cannot make a decision, other than to take a gamble. The fallacity people tend to make is to threat both envelopes as if they were envelope A, and the amount in the other envelope was determined from the one they have picked. That's why switching seems always the best decision.

Let's dig a bit deeper into the fallacious thinking process. When picking an envelope there are two outcomes possible. You pick envelope A, you pick envelope B. If you pick A, you may reason as follows: B = 50%A*2 + 50%A/2 > A. The amount in B is expected to be higher than A, so you want to switch. If you pick B, you reason the exact same way: A = 50%B*2 + 50%B/2 > B. The amount in A is expected to be higher than B, so you want to switch. But, you must not threat A - deterministic - as B - probalistic -. The above reasoning is fallacious. Instead reason as follows: B = 50%A*2 + 50%A/2. I got B, so switching to A would mean: 50%(-A) + 50%(+A/2) < 0. There is a potential loss in switching to A, so you should stay with B. The fact A can take several values doesn't even matter. Assume B = €10, A0 = €5, A1 = €20. Swithing to A gives 50%(-A0) + 50%(+A1/2) = €+2.5. Even though there seems an expected gain by switching to A, there isn't (because: 50%(-A) + 50%(+A/2) < 0). The fact we distributed A by the hyper-prior A0, A1 misleads us. —Preceding unsigned comment added by 193.29.5.6 (talk) 06:21, 6 April 2010 (UTC)

There is no benefit to switching

This is in the article at the moment:

"But in every actual single instant when an envelope is opened, the conclusion is justified: the player should switch! Not many authors have addressed this case explicitly trying to give a solution. Chalmers, for example, suggests that decision theory generally breaks down when confronted with games having a diverging expectation, and compares it with the situation generated by the classical St. Petersburg paradox."

This is wrong, there is no benefit in switching. It is proven by the fact that there does not exist any pair of envelopes for which switching makes sense. Since this result is based on having perfect knowledge about the system it has to be right. The calculation from the point of view of the person who opened one envelope does not assume perfect knowledge of the system (he does not know what is in the other envelope) so this calculation should be put in doubt.
This means switching is NOT beneficial and the above text in the article is misleading and should preferably be removed.
Another thing is that there is no need to rehash St. Petersburg paradox in my opinion, the link to it should suffice.
Enemyunknown (talk) 18:31, 14 September 2008 (UTC)

If you think this is wrong please write a paper about this and get it published in a peer reviewed magazine. When this is done someone else, not you, can add your opinions and ideas in the wikipedia article. What matters in wikipedia is verifiability, not truth. iNic (talk) 18:05, 15 September 2008 (UTC)

I'm not going to bother, but I'd like to see the source of that statement, on page it was only said "See John Norton, 1998, for a suggestion", i fixed it with the link from exchange paradox page. I tried to find the suggestion in the long pdf skimming through it but to no avail, can you maybe point me to the page on which it is stated? http://www.pitt.edu/~jdnorton/papers/Exchange_paradox.pdf

It should also be clearly stated if the advice to switch relates to the version of the paradox described immediately above it or to all versions.

What about suggestion about the lottery? I think inclusion of its description here is confusing as its a completely different paradox, the reader won't understand why its expected value is infinite without reading the full description anyway, so i think it would be better to just leave the link to it.Enemyunknown (talk) 10:26, 18 September 2008 (UTC)

If you notice the levels of the headings, they indicate that this text applies to the specific lottery situation stated. I have not checked sources, but it looks right to me. And the comments about the infinite expected value are highly relevant.

The point of the example is this, in my understanding, this:

The whole paradox has something to do with assumptions about what these envelopes may contain. If you are in a quiz program where the max prize is known to be $1 000 000 and your envelope contains $500 000 or less, you may consider switching, but if it contains $1 000 000, don't switch - you can't win more anyway; you will be sure to loose. If you are given the choice of envelopes in some other situation where you have clues that allow you to assign a priori probabilities to various pairs of amounts, you may have a similar situation where swithcing is favourable for small amounts but not for large amounts - and this may be perceived as solving the paradox: In real situations with real distributions (whether they are known or not), the IS no paradox; it's as fictitious as the Martingale system. But alas, if your a priori probabilities (wherever you've got them from) are as in the lottery given here, you should switch in all cases, and we do have a paradox, contrary to all common sense and reason! That is, until you realise that this "lottery" is NOT a real situation, as the expected value of the contents of the envelopes is infinite. Not the richest bank in the world could finance a lottery which truly had this distribution, and if you are TOLD that this is the distribution, you KNOW you've been lied to. E.g., you could obtain your own a priori distribution by modifying the given distribution: The probability of prizes larger than $1 000 000 000 is not small, as given by the lottery probabilities, but truly zero. (Don't believe me? Then make the cut at $1 000 000 000 000 000 000 instead!) Then, again, there is no paradox; only the baffling question about how to construct your a priori distribution when your information is incomplete.--Noe (talk) 11:40, 18 September 2008 (UTC)

There are lots of sources for this statement, as well as there are many authors that make the connection to the St Petersburg paradox, and they think this connection is central for a proper understanding this paradox. You are of course free to disagree personally that this is a valid connection, but as I said before: wikipedia should only reflect the discussion found in the literature. Of course, there are also authors that denies that these two paradoxes are connected, and this view is also mentioned in the article. iNic (talk) 14:29, 20 September 2008 (UTC)

The situation is symmetrical and there does not exist a single pair of envelopes for which switching gives benefit, the article many times this to say about the decision to switch: "This conclusion is just as clearly wrong as it was in the first and second cases."

However someone included this surprising paragraph:

"But in every actual single instant when an envelope is opened, the conclusion is justified: the player should switch! Not many authors have addressed this case explicitly trying to give a solution.

I looked into the reference and I couldn't find this suggestion, so I asked you for the source, since you failed to provide it I looked once again and this time I found that the author of the provided reference claims something directly opposite to what this paragraph says, here is the quote from the source:

"I consider a variant form of the paradox that avoids problems with improper probabilities and I argue that in it this expectations give no grounds for a decision to swap since that decision has to be based on summation on all the expectations. But this sum yields a non-convergent series that has no meaningful value."

http://www.pitt.edu/~jdnorton/papers/Exchange_paradox.pdf

I therefore remove this paragraph, do not put it in unless you can properly source it.

Enemyunknown (talk) 04:31, 21 September 2008 (UTC)

For example Don Fallis ("Taking the Two Envelope Paradox to the Limit", Southwest Philosophy Review, Volume 25, Number 2, 2009) explicitly tries to solve this case. You can find the statement you don't believe exists on page 18, case 4. iNic (talk) 18:10, 7 June 2010 (UTC)