Talk:Sleeping Beauty problem

Error in Introduction

The current introduction to the article reads "The Sleeping Beauty problem is a puzzle in decision theory in which whenever an ideally rational epistemic agent is awoken from sleep, they have no memory of whether they have been awoken before. Upon being told that they have been woken once or twice according to the toss of a coin, once if heads and twice if tails, they are asked their degree of belief for the coin having come up heads."

This is erroneous, since she is never told the number of past awakenings, per the usual description in the literature. This error dates back to edits done on March 6, 2020. — Preceding unsigned comment added by 98.184.228.73 (talk) 20:46, 24 February 2023 (UTC)

Anthropomorphic Principle

A mention of the anthropomorphic/anthropic principle must be made in this article, since it deals with the exact same issue as this riddle. To rephrase it: Goldie Locks finds herself alive. What is the probability that the fundamental physical constants are the right values for life to exist in the Universe? —Preceding unsigned comment added by 207.38.162.22 (talk) 04:49, 14 April 2009 (UTC)

100:1 probability

I added a paragraph where the difference between the to different lines of reasoning is sharpened by increasing the number of times, Sleeping Beauty is awakened in one of the cases. It's not very well written, but I hope it can serve some pedagogical function.

By the way, Robert Stalnaker discusses the Sleeping Beauty problem at some length in his John Locke Lectures 2006–2007, especially lecture 3 and 4.

90.184.154.143 (talk) 16:41, 29 April 2008 (UTC)

Zuboff

The sleeping beauty paradox was originally formulated in unpublished work by Arnold Zuboff (a lecturer at my university)as Elga acknowledges in his paper - which is available online, the reference is at the bottom of the entry. Shouldn't this be mentioned at the beginning of the page?

82.35.67.216 21:44, 25 April 2007 (UTC)grezo

re-formulation

I've mixed feelings about the re-formulation done by User:Gdr. I'm aware of the style guide preferring prose over lists, but I find my older version easier to follow. Also I'm totally unaware of using credence in a quantitative manner. But I'm not a native speaker. Pjacobi 18:06, 15 Jul 2004 (UTC)

"Credence" isn't the word I would have used, but I think it's okay, and at any rate it seems to be the word the original problem used. I also preferred the list version of the article -- but then, the structure of the list may lead people to mentally categorize the outcomes in a way they're not "supposed" to -- so I don't know. Triskaideka 01:03, 16 Jul 2004 (UTC)

1/2 vs. 1/3

This isn't a fair betting game, because it is equivalent to her making a 50-50 bet and winning $1 if she's right or losing $2 if she's wrong!

The betting argument may sound funny at first, because it sounds unintuitive (at least to some people).

The probability that the coin landed heads up is indeed always 1/2. However, if Beauty bets $1 during each interview that the coin landed heads up, then, after a large number of these experiments, we see that every time the coin lands heads up, Beauty bets $1 once and wins, whereas every time the coin lands tails up, Beauty bets $1 TWICE and loses EACH TIME.

-Dusik (2:40 PM EDT, Thursday, July 15, 2004)

IMHO this paradox isn't actually that difficult and the heated discussion obscured this fact. 1/2 and 1/3 are both correct answers, but to different question: Q What's the probability of a heads experiment? A 1/2 -- Q What's the probability of Beauty being awakened after heads? A 1/3. Pjacobi 18:45, 15 Jul 2004 (UTC)

The probability of Beauty being awakened after heads is 1, not 1/3. — Preceding unsigned comment added by 159.53.78.141 (talk) 16:56, 3 November 2015 (UTC)

Yes, it should be: "What is the probability of this instance of Beauty being awakened happening after the coin came up heads?"75.86.253.146 (talk) 18:33, 6 February 2016 (UTC)

The idea of betting is sort of thrown in at the last minute, as the entry stands now, and I think it's irrelevant to the original problem. The problem never states that Beauty gains anything by correctly guessing which way the coin landed. If she did, then certainly she should guess tails. But in fact she's never asked which way the coin landed at all. She's asked how probable it is that the coin landed heads. And since the problem specifies that she has no way of getting additional information about the coin toss, the question she's being asked, in any interview, is really just "What is the probability that a fair coin that was tossed at some point in the past landed on its head?" Of course the answer to this is 1/2. In fact, I don't think this is a paradox, just an easily misunderstood problem. Triskaideka 19:27, 15 Jul 2004 (UTC)

"The problem never states that Beauty gains anything by correctly guessing..." She gains being correct.75.86.253.146 (talk) 18:36, 6 February 2016 (UTC)

Most mathematical paradoxes are, since mathematics tends not to contradict itself. I think you've summed up the counter-intuitive factor in this one nicely Daibhid C

I edited the article, added a paragraph about how the solution depends on the reader's interpretation, and also added a link to a web page that says about the same thing. People may disagree on just what the interview question means. Triskaideka 01:03, 16 Jul 2004 (UTC)

This sentence at your new link [1] sums it up: Is it based on the percentage of runs of the experiment where the coin comes up tails? Or is it based on the percentage of interrogations where the coin comes up tails? Pjacobi 07:11, 16 Jul 2004 (UTC)

Thanks for all the feedback, guys, and for elaborating on the gambling aspect of the problem in the article. I think I've got a clearer idea of the issue at hand now, and my conclusion is actually that there is only one "correct" answer to the question: 1/2. The answer of 1/3 is not the correct one, but rather the more profitable one, precisely for the reason of being right twice 50% of the time, just because the question is *asked* twice. So, in other words, guessing the probability correctly is not the goal of the contest. So, in conclusion, the paradox here is that a good percentage of the people hearing this paradox will perceive this as a gambling scenario and look for the best rewarded answer. It's one of those instances where "common sense" may get in the way of mathematics. Another reason why it's a paradox is because it's unintuitive, in many cases I would guess because the reader will not completely understand the question logically, but use (untrained) intuition to guess the answer, which should commonly yield answers of both 1/2 and 1/3. Am I making any sense? Dusik 15:55, 11 July 2005 (UTC)

Edit 00:30, 19 Jul 2004 81.152.235.34

I find the "solution" and "analogy" of 81.152.235.34 slightly misleading. I'm still of the opinion, that more than anything else, this paradox highlightens the importance of clearly formulated questions, as 1/2 and 1/3 are both correct answers, but to different questions. In addition I would like to illustrate the possible questios and answers by fair odds when betting, as quite a number of peoply enjoy clearer thinking, if money is involved.

Comments?

Pjacobi 09:42, 19 Jul 2004 (UTC)

Yes, it would be interesting to know what was the original question formulated by Zuboff. Is Beauty asked to say the day? or just the coin?

With today's additions, I'm quite satisfied. Pjacobi 09:55, 19 Jul 2004 (UTC)

---

I restored the gambling argument for ⅓ since I think people do think more clearly about probability when it's expressed as a gamble. I also cut a big chunk of waffle by User:Triskaideka and 81.152.235.34 and replaced it with a concise solution and a spoiler warning. Gdr 23:55, 2004 Jul 22 (UTC)

I reiterate my protest against this usage of the idea of gambling. I would rather see it used to help the reader understand an argument that has already been made. As it stands, the idea of gambling is abruptly inserted, after the problem statement but before the solution, and it seems to create a new solution that wouldn't exist without the idea of gambling, which isn't present in the problem statement.

I also don't understand the objection to my and 81.152.235.34's text, which I thought attempted to explain in plain language how the two answers could be arrived at. I'm not particular about how we approach the explanation, but I do think that User:Gdr's revisions have removed some of that explanation in favor of text that laypeople won't find as accessible. Brevity is a good thing, but not if it comes at the expense of the reader's understanding. Triskaideka 16:03, 23 Jul 2004 (UTC)

I don't like the current version best. The explanation What is the conditional probability of heads, given that Beauty has been awakened and interviewed? The rules for conditional probability give ⅓ as the answer is incompatible with the definition of Conditional probability or needs a completely other formulation, as the probability of Beaty being awakened and interviewed is 1.

But I wasn't fully satisfied with any prior version, so I don't know how to proceed.

Pjacobi 16:36, 23 Jul 2004 (UTC)

by the Pseudonymous 81.152.235.34 - now registered as BernardSumption

Having read the disagreement on the (huge) usenet thread it seems to me that there will never be a total resolution. This is why when suggesting the answer to be 1/2, I made clear the assumption that this result hinges on. The edits by myself and Triskaideka were an attempt to put both sides of the argument in laymen's terms, and Gdr's edits were probably more correct and concise from a mathematical point of view. I see that Gdr reverted those changes, which is good for laymen but means that we now have no links to appropriate probability theory pages. Perhaps a solution is for someone with more knowledge of probability than I to add the appropriate links to pages on probability theory terms to the current version, so that people wishing to dig deeper can do so?

Also, Pjacobi had a good point that the whole paradox arises from the lack of a clear formulation of the question, so I added his sentence to that effect near the top of the article. --BernardSumption 09:40, 26 Jul 2004 (UTC)

Somebody seems to have addressed a philisophical problem without even knowing it. Although guessing tails will make her win twice as much on average, she will remember the reward just the same either way. Does getting the reward an extra time really matter if she just forgets it? Daniel 02:49, 10 Feb 2005 (UTC)

In the current version, I have some difficulty in seeing the motivation for interpreting "credence" as a fraction out of the number of awakenings. (There seems to be a typo in the current version. "1/2" == "Tuesday"?) The final section that indicates that guessing "tails" will gain her more correct answers actually seems to motivate that interpretation better, although in somewhat different terms. I have not read the usenet thread. (I am assuming that 95% of it is simply different posters refusing to accept other posters' interpretations of an ambiguously worded question.) Is there perhaps a better way to explain/present that interpretation? Perhaps "what are the odds ..."? -- Wmarkham

Also, the article credits Adam Elga with the problem, but one of the linked pages states "The Sleeping Beauty problem, and the older Paradox of the Absentminded Driver, were first posed by Piccione and Rubinstein, in 1994." Is the Piccione and Rubinstein problem the same, but older? Is it possible that Elga should be credited with simply a more current wording? -- Wmarkham

I believe much of this article, and most of the posts on the talk page, are misguided. Our goal here should not be to solve the paradox or argue about which point is right -- Wikipeda isn't about original research -- but to state the current and historical (sure, it's only a couple of years old, but still) views. As it is now, there is no consensus in the philosophical literature. There are dozens of papers on the paradox, and they split fairly evenly on the answers. Moreover, the naive arguments offered in article as it is at present are certainly not all arguments, and indeed Elga doesn't use either of them (although he does mention them) to argue for his view in the original paper where the problem was presented. Another thing is that the inclusion of a link to Terry Horgan's paper is rather arbitrary, as it is obscure and rather uninteresting. If any paper should be linked to, it should be Elga's, or perhaps Lewis' response to Elga. Regarding who should be credited with inventing the problem, to answer Wmarkham's query: the Sleeping Beauty problem is indeed present in Piccione and Rubinstein's article, but they do not devote much attention to it, or give it a name. The Paradox of the Absentminded Driver, which the real focus of their paper, is not the same paradox. To make things even more complicated, Elga acknowledges in his original article that Stalnaker is the person to give SB its name, and Stalnaker had in turn read about "similar problems" in unpublished work by Zuboff. So there seems to be several independent discoveries of the paradox. Nonetheless, most credit Elga, as he was in any regard the first person to bring it to the forefront and recognize it as an important problem in its own right. Now, before someone asks why I don't do all these changes myself: hopefully, someday when I have an hour or two to spare I will. Right now I don't. -- Miai

Very sober analysis. Go ahead! Be bold in editing this page. --Pjacobi

I agree with Miai that the jury is still out on this one and the article should not purport to offer a solution. The solution presented says that the deciding factor is "how often each of the branches is sampled" and goes on to say, "By sampling the tails branch more times than the heads branch we guarantee that the probability of tails is higher than that of heads." This assessment appears to assume the principle of indifference, the status of which is not only controversial for independent reasons but seemingly at issue the Sleeping Beauty problem. If so, then a defense of this solution raises and must confront this issue. --Jcblackmon 15:31, 8 July 2007 (UTC)jcblackmon

Re-intuit, halfers!

I'm going to toss a coin. If it lands tails, I will write the name of a popular female singer in the next paragraph. If it lands heads, I will write the name of a less-popular-than-he-used-to-be male politician. Okay, here goes!

Now, remember the question: what is your credence now for the proposition that BRITNEY SPEARS my coin landed heads? Are you really going to tell me it's 50%? I'm not talking about gambling here; I really only want your opinion as to whether it landed heads or tails. Not whether it might have landed heads or tails when I flipped, and not whether a coin normally lands heads or tails. Your credence now.

The intuitive response is the correct one: zero percent. There might have been a 50% chance at one point, but not given the information you have. Similarly, if you flipped a coin and saw heads, you wouldn't analyse the probability as being 50% any more. From your perspective, the probability would have changed to 100%.

This is conditional probability: the probability of some event A, assuming event B (as defined in Wikipedia). Event B in this case is my naming Britney, or in my second example, your witnessing a monarch or president's picture facing up. In the Sleeping Beauty example, SB is assuming an event as well: her being woken up.

Assuming her being woken as outlined in the article, the probability of heads is 1/3. I don't think this is merely a question of how you interpret the question. There are two perspectives: the coin tosser's and SB's. It's intuitive to always want to revert to the coin tosser's perspective, because 50-50 is intuitive. But it's also wrong - her subjective experience affects her perception of the probability.

Experience always affects probability. If two teams of equal ability are almost finished their game, and the score is 8-0, then most people would put the odds of the team with 8 winning at close to 100%. You don't say "there's a 50-50 chance of each team winning."

The article seems to favour the 'halfer' viewpoint. I disagree with the analogy (in which a second coin toss has no effect on the first). It's true that SB's waking had no effect on the toss, but it does have an effect on her analysis, which is what she's been asked for. Worrying about whether she has an effect on the toss is still moving back to the intuitive coin-tosser's perspective, not hers.

To highlight my point: how can anyone possibly think that the gambler's odds are different than the probability?24.64.223.203 08:21, 31 October 2005 (UTC)

The problem with this argument is that "being woken up" is exactly the same result for either flip, unlike with your "Britney Spears" example, so Sleeping Beauty learns absolutely nothing from it. She will experience the same thing either way, and hence, may as well give an answer before the experiment is even conducted. (There's no possible universe in which she can think "Well, I wasn't interviewed, so I guess the probability on heads is X%".) Suppose we ask her beforehand what answer she expects to give? Would it be most rational for her to say 1/3 then? Maybe. But that's the crux of this.

Nick Bostrom argues that SB does have new evidence: namely, that she is in the experimental phase of the experiment. That this is new information can be demonstrated quite easily by altering the problem: Sleeping Beauty is always awakened on both Monday and Tuesday; but in a red room unless it is Tuesday and the coin landed on Heads. In that one case, she is awakened in a blue room. With all four combinations corresponding to an awakening, on Sunday she knows any single future follows P(H&Mon) = P(T&Mon) = P(H&Tue) = P(T&Tue) = 1/4.

She certainly gets “new information” if she awakes to see she is in a blue room, and can update her credence to P(H|Blue) = P(H&Tue)/P(H&Tue) = 1. But then she also must be able to update her credence if she sees she is in a red room, to P(H|Red) = P(H&Mon)/[P(H&Mon)+P(T&Mon)+P(T&Tue)] = (1/4)/(3/4)=1/3. Yet her total information in the latter case is identical to her total information in the actual problem. She knew she would be awakened in a red room, so that fact isn't "new." But when she compares it to the futures she knew could exist on Sunday, she can eliminate one. The same is true in the original problem: she knows her present isn’t the H&Tue. future she knew could exist on Sunday. The answer is 1/3. JeffJor (talk) 20:39, 4 June 2012 (UTC)

The problem with your argument is that you state "(There's no possible universe in which she can think "Well, I wasn't interviewed, so I guess the probability on heads is X%".)"...This is completely incorrect, the probability question stipulates that she has drug-induced amnesia! She should have no recollection of any event except for having gone to sleep at the start of the experiment EACH TIME SHE IS ASKED. The question is specifically asked with the understanding that each time she is asked she has no experience of any previous event. Each time is the 1st time, thus 1/2! — Preceding unsigned comment added by 69.247.124.190 (talk) 14:45, 20 June 2013 (UTC)

And the problem (well, one of them) with yours is that you can't identify the difference in her information when she is simply interviewed (and left to sleep on Heads+Tuesday), and when she is interviewed in a red room (and wakes in a blue room on Heads+Tuesday). You equate SB's inability to observe the event Tuesday&Heads with the absurd assumption that it doesn't exist as a "possible universe." On Sunday, it is just as much a "possible universe" as any other combination. But when she is awake, it is not. This is "new information," because she has ruled out a "possible universe." JeffJor (talk) 11:02, 21 June 2013 (UTC)

To some extent this reminds me of the Boy or Girl paradox, otherwise known as the Two Child Problem. In this case some seemingly uninformative addition, such as knowledge of the child's birthday being a specific season or day of the week or year, alters a certain probability about the sex of the other child. This seems similar to what is going on here, apart from one fact: in the children problem both a simple counting argument and conditional probability formulas easily enough result in the same conclusions. Here, however, it seems that people come up with multiple different methods for reassigning the probabilities, for updating them, and can't agree on what is applicable, what assumptions would be correct, and what not. Two of these main methods correspond to the two main positions. These methods are as follows: either we take the fair coin and branch into equally probable branches, which are a sort of larger states, containing substates, as you will, each of which is a day. In this case, it may seem intuitive that within one branch, each day is equally probable, however each branch is given to have 1/2 probability, we stick to this (which would mean not updating that value, but using it to update the others, to determine unknowns when the situation changes, different information is received) and use it to distribute the probabilities. This means in the case of H there is 1/2 probability given to the only day, Monday, while in case of T 1/2 is distributed over all the days present, being Monday and Tuesday each getting 1/4. The other method is to say that we only need the probabilities to add to one, and rather that days within a branch, for which we know the total probabilities, are all equal, _ALL_ probabilities of all days are equal, the branching does not really matter. This reflects a way of updating our information, because we are no longer in the situation before the branch, rather SB is now 'awake' and the point is that she does not know which branch has been taken, only that she is awake now and that there were multiple possibilities for that to be the case.

It seems, however, that the arguments so far fail to convince that either of these approaches, or perhaps another, has the definitive edge (otherwise this would not be considered an open problem at the time of writing, after all). The coloured room argument is not quite valid, because there is a difference in information there. Despite what was previously stated by someone arguing the matter, Tuesday&Heads is absolutely not a possible universe (in the original statement of the problem) even before awakening, claiming otherwise is at best misleading. I could understand, though, if this was done from a biased point of view attempting to justify using the coloured room argument. In the coloured room argument, however, we begin with four options, only SB has to update her credence based on the colour information she receives. However in the original formulation, there is the amnesia statement, which is there to make clear that there can be no differentiation between the days at the time of awakening. The philosophical question is wether simply 'I am awake' makes an informational difference. The probabilistic question is wether this means Monday&H, Monday&T and Tuesday&T all have equal probability (thirder), or wether the probability must be (re)distributed in some other way (e.g. days having equal probabilities within branches following the 1/2 chance coin flip, leading to the halfer position). If we start out with 4 days, an equal amount on each branch, there is no immediate problem, as each 'strategy'/method for distributing the probabilities gives the same result: each option has probability 1/4. However breaking this symmetry leads to the current problem, in which it is not obvious how to progress. The coloured room argument does not apply, because there never was a fourth option to begin with, so we cannot use the same conditional updating formula without further assumptions about the probabilities, which would essentially be/lead to assuming the answer, usually either 1/2 or 1/3, as far as it has been discussed it at least seems to be the case.

Another argument that is made has to do with betting. Gambling has historically lain at the foundation of probability theory, after all. The problem is that this is an addition to the problem and is rather about 'strategy'. Which strategy, however, depends on the addition far more than on the original sleeping beauty problem. One such addition is Life or Death betting. Each time SB bets wrong, she is killed, making this a rather high stakes game. If this is the case, a fair coin is kind of bad for SB, as it gives her fifty-fifty chances for H or T. Yet in this game the probability of the coin landing H should inform her bet on it being either H or T. There is, however, a less morbid and completely obvious, different gamblification/gamification. In this version there is prize money to be won. Each time sleeping beauty guesses right, she will win a price. However, there is a problem with her 'expected gain' that often goes unaddressed. Consider the prize to be some value and this value must somehow be distributed over the moments she gets to guess. If each guessing opportunity has equal value, no matter which side the coin came up, then SB can win the most if the coin came up T, because it offers more opportunities, which gets to the heart of the intuition of Bostrom's extreme sleeping beauty variation argument. If at each of these awakenings SB guesses T, she gains, so from a 'utility' point of view, this is a rational decision, a good strategy. The expected value is indeed higher this way. However what if the value is not equal independent of the coin flip? What if there is simply a total value to be won, and it is 'smeared out', distributed, over all possible days she can bet? Note that this is not information SB receives, she can just receive whatever she has won, if it is anything at all, and information on wether she won or not, at the end, after the experiment is concluded, and simply has to state 'H' or 'T' upon awakening to 'place her bet'. If the maximal value is decided and the same, though, that means that, say it is 100, if H, SB can gain 100 in one bet on Monday if she says Heads. But she can gain 50 on each of Monday and Tuesday if she says Tails each time she awakens. For the full run, at the end of any branch, she can have gained maximally 100 and she always has the possibility to gain the same full 100. There is no superior strategy, as any strategy consists of saying one, and always one and the same option. Either H or T. The strategy cannot be different, because it will be either H or T and she might only awaken once and each time she does she cannot tell wether it is the first time or not and in what 'branch' she currently is, so all she can do is say her one bet that she thinks would be winning. At least, this is the case for deciding upon one answer and sticking to it as the best one. There may, however, be a better strategy: letting the decision itself rely on chance. Decide with some probability q on T, so with 1-q H on otherwise. The question is what the optimal value of q is for your expected value and we can do this for either problem, with equal total branch value and equal single guess gain. To keep things a little general, it is interesting to work this out for a possibly unfair coin, with probability p of getting H, as described earlier. I think this is already more than enough addition to the talk section though and the working out of this new problem should probably be left for some article, perhaps such articles even already exist. I will just leave a source here that takes a different approach to attempt to begin disentangling the SB problem: https://towardsdatascience.com/the-sleeping-beauty-problem-a-data-scientists-perspective-56223b5128e4 .(I don't know wether this is appropriate, I have not read the entirety of terms and conditions for placing things on wikipedia, specifically the talk page, I simply wanted to make an addition to the discussion. Mainly because people seem to easily get absorbed in their own positions on this problem and yell out "so that's obviously wrong" or imply it by bringing up an argument at one point or another that doesn't actually entirely answer the core issue. I thought pointing out the shortcomings and perhaps bringing in a different perspective would lead to a more healthy continuation of the search for insight concerning this problem.)

?

This question as stated should be an answer of 1/2. The question stipulates that SB is provided a drug that induces amnesia (thus she has no subjective reason to recall anything prior to the current coin toss.)

I would add, that the question should also stipulate that a coin toss is unfamiliar to SB prior to the test AND therefore she has no psychological predisposition to choosing one over the other, AND the question should give her the choice of answering heads or tails because asking someone to confirm an outcome instead of predicting an outcome is biased by nature (some are agreeable and some contrarian by personality.)

Is it a frivolous article?

I think that this is not actually frivolous (as per note on version 2006-03-05 01:57:51 of the article), or at any rate less frivolous than the Monty_Hall_problem. The arguments for both sides given at the external link seem quite serious, and it evidently confuses people.

The Monty Hall problem is of course not a logical paradox, but is one in the sense that is is counter-intuitive. I am actually less sure about this, as it seems to raise the question of what we mean by 'credence'.

PJTraill 14:06, 5 March 2006 (UTC)

There are two clear statements that make the answer to the question 1/3: "What is your credence now for the proposition that our coin landed heads?" and "She knows all aspects of the experiment before hand". Thus she knows that 50% of the time she is awoken on heads, and 50% of the time she is awoken twice, on tails. Resulting in her credence being tails, since 2/3's of the time she is awoken on tails, and 1/3 on heads. If she did not know she was going to be awoken twice on tails, then obviously she should answer 1/2.--155.144.251.120 21:26, 17 April 2007 (UTC)

What is credence?

I have adopted the term credence as it occurs in the original articles, but I was unfamiliar with it.

I get the impression from http://users.ox.ac.uk/~mert0130/teaching/lecture16.doc that it is a subjective estimate of probability, but that seems to eliminate any possibility of discussing the rightness of Beauty's answer.

I see in http://fitelson.org/probability.pdf a lot more discussion, making it plausible that there should be various answers.

It seems to me that there should be a Wikipedia article on Credence (probability theory), or at least that it should come in the the article on probability, but a search threw up nothing much.

PJTraill 15:14, 5 March 2006 (UTC)

Credence is not a subjective estimate of some objective probability, but is rather a measure of partial belief. It often results from an estimate of objective probabilities, but often doesn't. Even if there are no objective chances (for example, if the world is deterministic) then credences will still be important, because people don't know everything for certain. Similarly, even if the actual situation is chancy, one can have evidence that goes against this chance (especially after the fact, as mentioned elsewhere here), so credence is importantly different from objective chance too.

Isn't credence just another word for subjective probability in this context? If that is the case (as is supported by the contributor above as well as in the first link PJTraill refers to) the simplest way to solve the dead crecence-link would be to let it point to the subjective probability page. INic 15:39, 20 September 2006 (UTC)

Duplicate

I see now that there is already an article at Sleeping beauty paradox. Sorry about this. That has been around since 2004-07-15 11:42:50 and gives more explanations, though I'm not sure if it covers all the same ground. If nobody else does, I'd better clear this up (make a redirect or scrap?), and redirect the links, though I'm a bit busy just now. My point about Credence (probability theory) remains. PJTraill 15:28, 5 March 2006 (UTC)

Make a redirect. - Rainwarrior 04:19, 21 June 2006 (UTC)

• I just merged the articles and the talk pages. -- Reinyday, 17:21, 24 June 2006 (UTC)

Thanks PJTraill 00:55, 6 November 2006 (UTC)

Paragraph deleted

A small section titled "Credence" completely denies the 1/3 possibility thus negating the factual accuracy of the entire article explaining a paradox. you are twice as likely to be awaken if it was tails, so 1/3. The section titled "Variations" seems to be a rebuttle to the false section, without being bold enough to delete it.

Example

I was trying to come up with a simpler example (maybe with less sordid drug abuse) and while it's not completely the same, I think it makes the 1/3 answer clearer...

The Mad Coin Tosser has been tossing a single coin each day for several years although lately he has become tired of his daily ritual. Last month he decided that each time the coin lands heads he will toss the coin again the next day, but should the coin land tails, he will take a break the next day and toss the next coin the day after that.

What are the chances that the last coin he tossed was heads?

Since you don't know whether today is a coin tossing day or a break day, you are in the same position as the Sleeping Beauty. Indeed, the only real difference with the cruel SB experiment is whether it is run once or forever, and I'm pretty sure that doesn't change the odds. --88.113.96.178 12:34, 12 March 2007 (UTC)

The problem with that example is that the answer is 3/4 tails and 1/4 heads, and its really very different.--Dacium 12:04, 17 April 2007 (UTC)

New paragraph

user:220.233.90.162 recently added the following last paragraph to "Solutions":

This still does not explain the disparity. The probability of an event may be defined as the theoretical knowledge that an observer has of the various paths leading to the present and their relative frequencies. The experimenter sees 1,000 paths leading to the present, consisting of 500 heads and 500 tails. Beauty sees not 1,000 but 1,500, of which 500 originate in heads and 1,000 in tails.

I think the preceeding paragraphs did explain the disparity, and if they didn't, they should be changed to do so. Comments?--Niels Ø (noe) 10:49, 2 July 2007 (UTC)

I agree that the paragraph quoted above introduces more obscurity than clarity. There used to be a section with the following explanation:

Another way to emphasize this effect is to increase the number of times Sleeping Beauty is being awakened in the tails case. Imagine for instance, that she is still interviewed only once in the heads case, but interviewed e.g. a hundred times in the tails case, with a new case of amnesia after every interview. From her point of view, any interview will then be a hundred times more likely to be a part of a tails case than a heads case simply because more interviews take place during a tails case than during a heads case.

This still seems like a good way to explain the puzzle. Does anyone feel differently, and if so, why? 90.184.0.217 (talk) 22:52, 17 January 2011 (UTC)

Dispute 1/3 answer

"What is your credence now for the proposition that our coin landed heads?" This could mean "What probability should she give to the coin actually landing heads?" and is definitely a question about her knowledge of the ratio of throws H:T that actually occur (or land). It appears to (erroneously) imply that this probability could vary which might explain why it is often interpreted as the quite different question "What probability should she give to her correctly guessing that the coin had landed heads?" to which the article's analysis is directed. The point is this: it would be impossible for SB to calculate that she would correctly guess H one time in three (which the experiment contrives by giving her two guesses for T) unless she knew that the ratio of H:T that actually land is 1:1 (H will land with a probability of 1/2). If with a biased coin, the ratio H:T that land becomes 1:2, then on being awoken she would correctly guess H one time in five. She can only estimate her proportion of correct guesses of H if she knows the probability with which a H actually lands. Due to there being two possible ways to interpret the question I think there IS no answer. (Puzzle Master 15:29, 4 July 2007 (UTC))

My answer to this is that the problem states that it is a fair coin and that Beauty knows all the details of the experiment, except what we are explicitly told that she does not know. (Tad )

Monty Hall

The Monty Hall problem is uncontroversial. I have removed the reference to it from the article. ---Dagme (talk) 03:26, 4 October 2011 (UTC)

False consensus

The wording of this article implies that the 'Thirder' answer has been disproved and that the 'Halfer' answer is known as the correct answer. As I understand it, looking at the range of academic peer reviewed literature on the subject, the solution to the paradox is still disputed by various academic authors, so the article should make clear there is not yet a known answer. This seems to be a neutrality issue... — Preceding unsigned comment added by 82.35.98.140 (talk) 21:37, 11 October 2011 (UTC)

Bernoulli trial

There is no one correct answer. It depends on how you define the Bernoulli trial.

Imagine the coin flip occurs on Sunday and then Beauty is awakened on Monday and maybe on Tuesday as the rules specify. After the coin is flipped on Sunday an upturned bowl is placed over the coin and it is left undisturbed for the rest of the week.

The HALFERS define the Bernoulli trial as the the coin flip. The waking of Sleeping Beauty is a secondary event. When awakened, Sleeping Beauty is being asked to reflect on an event that occurred in the past. The halfers imagine the interviewer asking "Hey we flipped a coin on Sunday, what do you think the chances are that it came up heads?" Beauty will say 1 out of every 2 trials regardless of the number of times she was awakened.

The THIRDERS define the Bernoulli trial as the waking of Sleeping Beauty. Sleeping Beauty is being asked about the present. The thirders imagine the interviewer asking "See that bowl there? What do you think the chances are that the coin under it reads heads?" Beauty will answer 1 out of every 3 trials. It could be argued that these trials are not truly independent.

I see nothing in the problem that specifies which definition of the Bernoulli trial is correct. The debate is a result of built in vagueness.

P.S. I know the talk page is supposed to be for discussing the article, not debating the problem that the issue is presenting, but hey, eveyone else is doing it... — Preceding unsigned comment added by 65.29.106.194 (talk) 03:41, 25 March 2012 (UTC)

I agree. This is similar to Bertrand paradox (probability). --NeoUrfahraner (talk) 07:32, 23 April 2012 (UTC)

Sleeping Beauty isn't asked anything on being awakened, either on Monday or on Tuesday. She is asked before the experiment is run what her degree of belief should be on her first awakening, which will be Monday. Seriously, go back and check the statement of the problem. Simon.hibbs (talk) 13:37, 14 December 2023 (UTC)

Except that, even if we go for this interpretation of the unclear wording in Elga's paper, SB doesn't know, when she is awake, whether it is Tuesday or Monday, therefore does not know if she has been awaken before. The "first awake" wording is in fact not relevant to the problem.

SB should definitely give the same answer, whether the question is asked before the experiment is made or during the experiment, once she has been awaken, because she has the same information.

There are two possible ways to frame this question :

1- what is the probability that the result of a fair coin toss is heads ? In this acception, the answer is trivially 1/2, all this spleep, amnesia and drugs shenanigans are for nothing and the 2000 article is not just confuse, it's idiotic.

2- when she is awaken, knowing that this awaking is linked to the result of the coin toss in the manner described, but not knowing the result of the coin toss was nor what days it is, what should she judge the likelihood of a heads result is ? In this acception, the answer is neither 1/2 nor 1/3, is rather easy to calculate and the 2000 article is pointless, because there is in fact no paradox, just unclear wording and bad math. 2A02:8440:7144:7683:9833:4327:51AE:4235 (talk) 11:33, 29 October 2024 (UTC)

Conditional Probability

The Sleeping Beauty problem is not a mathematical paradox. The two probabilities are the result of calculating probability in different ways, the unconditional probability being 1/2 and the conditional probability (given the condition that Sleeping Beauty be awake) being 1/3. Neither of these is "the right way"; both are valid approaches.

The confusion arises because the question does not make it clear which of these probabilities is being asked for. My take is that the question as phrased is asking for the unconditional probability, but the alternate view is equally valid.

If has been my experience that this is one of those pissy little issues that generate huge amounts of hot air to no effect. At one time this was a six line article that wikipedia featured. Bickering among the "halfers" and "thirders" has turned it into a typical wikipedia wankfest, full of jargon, extended explanations, bipolar paragraphs, and something totally useless as a reference. Would it be possible to cut out all the crap and get it back down to about six lines? 125.255.16.233 (talk) 15:10, 4 July 2012 (UTC)

What part of the question "What is your credence now?" leads you to believe that Sleeping Beauty is not being asked for the conditional probability based on the information she has "now" ? I agree with just about everything you said, except where you said it isn't clear that a conditional probability is being asked for. It most certainly is. But some references, specifically Lewis and his followers, don't agree with you. They say that her information does not allow her to update to a conditional probability of 1/3. And unfortunately, in a Wikipedia article it does not matter what you or I beleive is true, only what can be cited in a reference. JeffJor (talk) 19:41, 6 July 2012 (UTC)

Last time I saw the problem set, the tag was "When asked 'what was the probability that the coin landed heads', what should Sleeping Beauty reply?" As I said, the problem is in the phraseology, not the mathematics. As such, the problem is not a paradox, it's just a question of how you interpret a poorly phrased question. Do you agree with this? 125.255.16.233 (talk) 13:42, 7 July 2012 (UTC)

As far as the article is concerned, you shouldn't be considering the "last time I saw the problem set," you should be considering the problem as stated in the article. But in any properly-formed version, SB is asked for the probability based on her knowledge set. The entire point of the problem is whether that should be different than the probability based on an outsider's knowledge set. If you think you saw a different wording somewhere else, that was the question which was poorly phrased and you should stop considering it.

And no, it isn't a true paradox - it is a veridical paradox. That means a statement that intuitively seems false (at least, to some) but is actually true. The Monty Hall Problem is an example, since it can be demonstrated that there is an advantage to switching, but intuition suggests (to some) that there shouldn't be.

Thirders and Halfers alike think the other group is intuiting the SB problem incorrectly, and not seeing what they think is the mathematically correct solution. But all of them are basing their solution on what they think SB should say based upon her knowledge set when she is awakened. So your point about unconditional vs. conditional probability is invalid. As is your point about it not being a paradox, since no part of the article calls it a paradox. The point is whether the conditional probability should be the same as the unconditional probability.

In other words, please stop approaching this article based on what other treatements of the SB Problem say (or you think they say). Look at this article, and address only the shortcomings in it. You haven't so far. JeffJor (talk) 20:21, 7 July 2012 (UTC)

Could you please climb down off your high horse and approach this with an open mind. I can give you a revert war if you want one, but I think that's childish, and I'd rather come to a reasonable compromise. But I can't do that if you have your back up. Before we can reach a compromise we need to find points of agreement. That is what I am trying to do.

The article on the Monty Hall problem once ran to about six lines and was nominated as an article of the day (or some such). Then the wikiwank set in and now it is a bloated useless monstrosity of the sort that wikipedia prefers to sweep under the carpet. This article is the same. It uses complex precise technical language that the average person cannot understand. Rather than just state the facts, it features lists of justifications. It is not an article intended to inform the casual visitor, it is a battleground between two groups of people each determined that their personal view must not be underrepresented. I think it can be better than that. Do you agree? I would suggest it's worth about two paragraphs, not seven separate sections. 125.255.16.233 (talk) 01:20, 8 July 2012 (UTC)

Maybe you need to review what is allowed in a Wikipedia article. Unless you can find an acceptable source that says "the interviewer is asking for the unconditional probability that the coin landed heads," you can't put it in the article. What you or I believe to be correct (and I agree with more of what you say than you are admitting) is 100% irrelevant. The only "reasonable compromise" we can come to is one that satisfies official policy.

Try reading the Lewis reference, which is the origin of the halfer position and so is a reference that cannot be left out. Any statement you want to include that contradicts it absolutely must come from a good source, and maybe more than one. He addresses the probability at three places, calling them P, P_-, and P₊. P_- is the probability SB assigns before she is put to sleep on Sunday; what you call "unconditional." P is the probability SB assigns based on her knowledge when she is awakened; what you call "conditional." P₊ is a second conditional probability she would assign after being told, on Monday, that it is Monday. Lewis put that in his version of the problem. Lewis says P_-(HEADS)=P(HEADS)=1/2, which contradicts what you want to put in the article (and that P₊(HEADS)=P(HEADS)+1/6). So, where is your source for disagreeing?

You also want to say that it is unclear whether the interviewer is asking for P or P_-. As far as I know, no source has considered that unclear; so again, where is your source? These two points are the underlying points to the change you want to make, and represent your "high horse," not mine. I don’t have one, but Wikipedia does. It doesn't matter whether you or I believe them to be true; references are needed to include them. And if you want to discuss what you and I may believe, this is also not the correct place. Ask a question on my talk page, if your want. JeffJor (talk) 22:45, 8 July 2012 (UTC)

I'll have a look at the source, but consider my answer to the problem:

If someone tossed a coin and asked me what my credence was for the proposition that the coin landed heads, I would respond 50% for obvious reasons. To me, "what is my credence" is just one of many ways of asking what the probability is. So I can see no reason why Sleeping Beauty should answer differently. The business with Mondays and Tuesdays and putting to sleep and amnesia-inducing drugs is just window dressing added by the problem's poser to mislead the solver into giving a false answer. It's not unusual to do that, and in fact considering how people normally try to solve this it's clear why the problem was posed that way.

So why am I wrong and you right? And what has my answer to do with conditional probability? 125.255.16.233 (talk) 07:10, 9 July 2012 (UTC)

1) It doesn't matter if you are right or wrong - you can't put your own original content into the article. But ...

2) You are wrong. Say a six-sided die is rolled to determine teams - odds vs. evens. You join the odds team, and another person on the team asks you to assess the probability that he rolled a five. Do you say 1/3 or 1/6? Since the roll was used, in part, to establish the circumstances that allow the question to be asked, the question can only mean the conditional probability. And ...

3) Every source acknowledges that they are answering based on the information, even those who say it is still 1/2. Finally ...

4) This is not the place to debate the correctness of the answers. It is for discussing how well the article explains the information established by verifiable sources. JeffJor (talk) 11:00, 9 July 2012 (UTC)

If you think that anyone who edits Wikipedia backs up everything they say with a source you are severely mistaken. Sources only get asked for when people are trying to keep material out of an article. And in my experience citing a source doesn't stop people from removing material. In practice, people turf out stuff they consider wrong no matter how well sourced it is. And if something is correct and not controversial no one bothers sourcing it. Shall we go through the article and delete everything from it that is not backed up with a source? In theory everything needs to be sourced; in practice everything needs to be correct. Different things.

Sources "get asked for" when someone tries to put untrue information in an article, as you are. JeffJor (talk) 20:17, 9 July 2012 (UTC)

Now, with regard to my answer, how many people who come up with the same answer as me are going to write a paper about it? It's a trivial problem not worth half a paragraph, let alone a paper. Only people who think it is a problem in conditional probability would bother providing a source.

And with regard to (2), you are presenting one of the typical straw man arguments used with this problem, in that you are assuming that it is a problem in conditional probability and then presenting another problem which is phrased in such a way as to clearly be such a problem. Surely the mathematics in the problem is trivial! It's only interesting at all because it is phrased in such a way as to create confusion. If it were clearly a problem in conditional probability this article would not exist.

Try thinking of it like this. If the problem asked "What is your credence now for the proposition that the coin landed heads, given that you are awake?" the answer would clearly be 1/3. So why hasn't the poser asked it that way? My answer is, because that's not what he is asking for. He has deliberately added the trappings of a conditional probability problem in order to trick the reader into giving the wrong answer. You've fallen for his trick. 125.255.16.233 (talk) 14:11, 9 July 2012 (UTC)

And the opinion of many reputable (but I agree, mistaken) sources is that "SB's credence, given the fact that she is awake, is clearly 1/2." They all agree that the poser HAS asked it like that. In fact, I have found nobody except you who thinks otherwise. Do not put it in the article. JeffJor (talk) 20:17, 9 July 2012 (UTC)

I had a look at the Lewis article. Two things flow from it.

Firstly, there is an obvious flaw in it, as I am sure others noted. At E7, Lewis states that SB has no new relevent uncentred evidence. But that is not true. At P_ she knows she is not in any of the three states. At P she knows she is in H1, T1, or T2. At P+ she knows she is in H1 or T1. The new evidence is that she has gone from being in no state to being in one of three.

Secondly, the problem as stated in the article differs from the problem as stated here. Specifically, in the article when SB is woken on the Monday she is told it is Monday whereas here she is not. Also, Lewis does not conclude that the probability at all three awakenings is 1/2, but only that it is so at the Monday awakening prior to being told it is Monday. Consequently this source is being used to justify a passage that is misciting it. Either the statement should be changed to properly cite the source, or the citation should be removed. One can draw the conclusion that these differences don't matter, but that is new research and can't therefore be put into Wikipedia.

If the citation is removed, there is no citation for anything in the section "Halfer position" and therefore the whole section is original research and should be removed. Will you make these changes, or shall I? :-) 125.255.16.233 (talk) 00:55, 10 July 2012 (UTC)

Neither. Our beleif in a citation is not what makes it valid. Lewis belongs in any article on this topic. Leave it. JeffJor (talk) 10:49, 10 July 2012 (UTC)

Error in the "Thirder position" demonstration

Hello! I just stumbled upon this interesting article. I want to point out that there seems to be a mistake in the demontration in the "Thirder position" part: it calculates probabilities on events without even defining them, which is a pretty bad thing to do. Reading the demonstration I can see two possible interpretations for the events "Monday" and "Tuesday":

1) We take the experiment as a whole (not focusing on a specific interview), and we define these events as follow:

Monday: "SB is woken up on Monday."

Tuesday: "SB is woken up on Tuesday."

(Note that in this case these events are not mutually exclusive, if the coin comes up tails they will both occur.)

2) We consider one specific interview (if the coin comes up heads it is Monday, if it comes up tails, we select Monday or Tuesday with equal probabilities). Then we can define these events as follow:

Monday: "The selected intervew is on Monday."

Tuesday: "The selected interview is on Tuesday."

(In this case the events are mutually exclusive.)

Then, if we choose definition 1), the first two paragraphs of the demonstration are correct (up to and including "P(Tails and Tuesday) = P(Tails and Monday) = P(Heads and Monday)"), but the statement that these events are mutually exclusive is false, and the rest of the demonstration does not hold.

If we choose definition 2), these three events are indeed exhaustive and exclusive, but then the statement " P(Tails and Monday) = P(Heads and Monday)" is false as we can see:

P(Tails and Monday) = P(Tails) x P(Monday | Tails) = 1/2 x 1/2 = 1/4

P(Heads and Monday) = P(Heads) x P(Monday | Heads) = 1/2 x 1 = 1/2

And again the demonstration does not hold.

Hence the importance of definitions... — Preceding unsigned comment added by 125.39.117.52 (talk) 13:44, 31 January 2013 (UTC)

125.39.117.52, first I must tell you that we really aren't supposed to debate the issue here, just the article. Everybody does it anyway, as I will to some extent (although it is because I'm responding to you).

The "mistake" you point out is paraphrased directly from the cited source, so it is "correct" for the article whether or not the statement "P(Tails&Monday)=P(Heads&Monday)" is correct. But it is correct. It follows directly from P(Tails|Monday)=P(Heads|Monday)=1/2, which is supported both in the article and in the reference. And to my knowledge that support has never been disputed. The result has, after making the halfer's mistake as you did, but that is different.

However, you did identify the source of the discrepancy; both because of the lack of definitions, and the specific definition that was overlooked. But it is the mistake in the halfer argument, which is based on the assertion that SB has no "new information" upon which to update probabilities. The argument never defines what "new information" means, nor why it is necessary to update.

In fact, what is necessary to update, is for you to be able to define a non-trivial partition (a set of events, all with non-zero probability, such that every possible outcome belongs to exactly one event) in the posterior frame of reference that is not a non-trivial partition in the prior frame of reference. The prior probabilities of this partition will not sum to 1, so the extended form of Bayes' Theorem will change some probabilities. Usually, this happens because information you have eliminates some events from the non-trivial partition. The halfer's mistake is assuming that is the only way it happens. When they say there is no "new information," they mean that no events are eliminated this way.

But you just pointed out, quite nicely, how an event can get added to the non-trivial partition: "Tails&Monday" and "Tails&Tuesday" represent the same outcome in the prior frame of reference, so only one belongs in the partition. But we need both to form a partition in the posterior. So by defining what "new information" is, we can see that SB does, indeed, have it, and:

P(Heads|Awake) = P(Awake&Monday|Heads)*P(Heads) /

[P(Awake&Monday|Heads)*P(Heads)+P(Awake&Monday|Tails)*P(Tails)+ P(Awake&Tuesday|Tails)*P(Tails)]

= (1)*(1/2)/[(1)*(1/2)+(1)*(1/2)+(1)*(1/2)]

= 1/3.

The right-hand side uses the probabilities in the prior, but must include all of the events in a partition of the posterior. JeffJor (talk) 15:14, 2 February 2013 (UTC)

Hello! Got your answer.

First I would like to point out that my comment was absolutely relevent to the article: I wasn't debating over the solution of the problem, but pointing out a major mistake in a mathematical demonstration of an article claiming to be about probabilities. Calculating probabilities without defining the events is mathematicaly wrong, that is hardly controversial. (The demonstration in the article does worse than that, it uses two different definitions and uses them at different stages to reach its conclusion.)

As far as the source of the demonstration is concerned: I first thought that the article might have misinterpreted/oversimplified it, but after reading it I agree with you that it is not the case. For me that simply means that the source is unreliable. This is less surprising if you consider that its author is not a mathematician but a philosopher (sure, he specializes in epistemology of mathematics, but that's very different from actual mathematics). By the way, althought it seems to me that the content of the "halfer position" part is correct, it is also sourced by a philosopher's work, and is therefore also questionable.

So my point is: the SB problem seems to have interesting philosophical implications that have been debated amongst philosophers, and this aspect can be interesting to develop in the article, but as far as the math is concerned, it should be backed by reliable mathematical sources. As it is clearly not the case right now, I suggest removing the current mathematical content (and focus on the philosophical aspect or wait to have relevant sources).

Else, to answer your comments about the demonstration itself: If you choose one definition for the events "Monday" and "Tuesday" (the number 1 in my previous comment), then it is indeed correct that P(Tails|Monday)=P(Heads|Monday)=1/2, but in this case the events "Tuesday&Tails", "Monday&Tails" and "Monday&Heads" do not constitute a partition of the universe (a necessary condition to apply Bayes' Theorem). If you choose another definition of these events (the number 2 in my previous comment), then they constitute a partition of the universe, but by definition, P(Monday | Tails)=1/2 and P(Monday | Heads)=1 . The main problem of the current demonstration is that it uses definition 1 in the two first paragraphs, and then suddenly switches to definition 2 in the last one. This is obviously wrong, and would have been avoided by defining properly what the events "Monday" and "Tuesday" meant. 218.11.179.37 (talk) 19:03, 2 February 2013 (UTC)

Oh, and by the way, as I agree that definitions are important =): here "new information" about the tossing of the coin would be defined as "Knowing that a event not independant from the result of the toss occured." Or put in another way: "Knowing that an event A, such that P(A|Tails) ≠ P(A|Heads), occured".

For instance the event T:"SB is interviewed on Tuesday." is such an event, as P(T|Tails)=1 and P(T|Heads)=0. On the contrary, the event M:"SB is interviewed on Monday." doesn't give "new information", as P(M|Tails)=1=P(M|Heads).

Whether or not the published paper that originated the subject of this Wikipedia article is correct, in your opinion, is irrelevant to Wikipedia. If it were wrong, as you claim, it would not make for a significant subject. But if it deserves a place on Wikipedia, the Elga reference belongs.

Now, did you read that reference, or just deduce from this article that it is wrong? Because it doesn't deal with "the experiment as a whole," as you describe it. It only deals with "one specific interview." It adequately defines the three possible outcomes, which make for three independent atomic events in a partition of what could apply to the "one specific interview" SB finds herself in. It first argues, correctly based on those definitions and for a specific interview, that P(Monday|Tails)=P(Tuesday|Tails) so P(Monday&Tails)=P(Tuesday&Tails). Then it argues, again correctly, that P(Heads|Monday)=P(Tails|Monday) so P(Heads&Monday)=P(Tails&Monday). Then, since the order of the random variables doesn’t matter, we can indeed conclude that P(Heads&Monday)=P(Tails&Monday)=P(Monday&Tails)=P(Tuesday&Tails). Finally, since the three independent atomic events in the partition Elga defines must all have the same probability, each must be 1/3.

The failure to define events comes with the halfer argument, since it is the only argument that deals with "the experiment as a whole." And exactly as you described, it fails to define the difference in what is independent in the two systems, and gets the wrong answer because of it. If you want to represent "the experiment as a whole," you have to recognize that Tuesday still exists even if SB sleeps through it. So there are three random variables, not two: Heads/Tails, Monday/Tuesday, and Asleep/Awake. Where you evaluated P(Heads&Monday)=P(Heads)*P(Monday|Heads), you really meant P(Heads&Monday|Awake)=P(Heads |Awake)*P(Monday|Heads&Awake). And you can't assume P(Heads|Awake)=1/2 when that is what you are trying to deduce.

The "necessary condition for Bayes Theorem" you refer to, is that you use a partition of the posterior universe. The three events I listed do indeed form a partition of that universe. They don’t form a partition of the prior universe, because two of them represent the same event in that universe, which is what the halfer's failed to define adequately. That is "new information." JeffJor (talk) 17:57, 3 February 2013 (UTC)

It has been years since this discussion started, but some of the things discussed I wanted to ask some clarification on from anyone who could clarify them. It seems relevant as it is relevant to the correctness of the content of the article. Perhaps I don't correctly understand 'uses the probabilities in the prior', if so please explain, then my understanding of the prior and posterior universes is lacking. However above, in the argument with the explicit calculation using the extended version of Bayes' Theorem, shouldn't P(Awake&Monday|Tails) and P(Awake&Tuesday|Tails) have a probability of 1/2 rather than one? Which would result, for the end result, in 1/2. I am almost certain there is something here I am missing, an understanding I lack, but it seems to me that on the odd chance that I am understanding this correctly and the probabilities here were fallacious, then this has consequences for the correctness of the 'Thirder position' assumptions that should be reflected in the article. Could someone please explain? Though I know this is not a forum for discussing these things, I wasn't sure where else to post this as it was relevant here and only here.

Feb. 2013 corrections

I finally made some corrections I've been meaning to. The problem statement should more accurately reflect Elga and Lewis, neither of whom let Sleeping Beauty go after Monday, or ever said the coin was flipped on Sunday. In fact, both allowed that it could (or should) be on Monday. And the work credited to Bostrom never links the halfer position with the Self-Sampling Assumption (in fact, it never mentions that assumption). But it does say the halfer argument must be wrong (page 8, paragraph 2). And, despite how many people phrase it, "awaken" is an intransitive verb: "I awakened because you woke me." JeffJor (talk) 22:29, 10 February 2013 (UTC)

What a stupid question...

The question is only a paradox because of the ambiguity of "degree of belief for the coin having come up heads". The probability of a coin "having come up heads" is either 1 or 0. There's no 1/2 or 1/3. It's either 1 or 0 because the event has already happened. The probability of that coin is, of course, 1/2. If you iterate this experiment many times and the question is posed at every possible instance, clearly 1/3 of the time the question is posed when the coin was heads. — Preceding unsigned comment added by 140.180.250.151 (talk) 00:27, 15 April 2013 (UTC)

Ambiguous statements like "the degree of belief" are used because, when faced with probability conundrums like this one, people always seem to object to some aspect in the definition of probability. Just like you did. Maybe you didn't notice it, but there is no requirement to flip the coin before awakening SB. The event has not, necessarily, "already happened." So this is a grammatical paradox as well: some of the possible awakenings (on Monday) are unaffected by the coin, and so can be considered to happen before it. While others (on Tuesday) must occur after it. So we can't just say "the probability the coin was, or will be, heads, because neither tense is guaranteed to apply in SB's awakened reality.

The resolution of this conundrum is the one you failed to recognize in your complaint. Probability does not apply to a result that will happen, any more then it applies to a result did happen. It applies any ambiguous situation where multiple results are possible. It is the ambiguity, regardless of tense, that requires probability. And the paradox in the particular case of Sleeping Beauty is that both tenses are required to represent her state of ambiguity, but we can't seem to get past our need to fix the ambiguity in just one tense. JeffJor (talk) 10:58, 19 April 2013 (UTC)

Can you provide evidence that this isn't just the Bayesian fringe philosophy that it sounds like? That it's a notable mainstream definition?198.228.228.169 (talk) 14:51, 2 July 2014 (UTC)Collin237

Sure. http://wiki.riteme.site/wiki/Classical_definition_of_probability. Do you see anything in that definition that requires future tense to be used? Or what if I pose the classic problem (which I truthfully can, btw): "I have two children, and at least one of them is a boy. What is the probability both are boys?" The answer you should give is not 0, or 1, even though they were born long ago. JeffJor (talk) 22:25, 22 July 2014 (UTC)

^That's just wrong. When the question is posed, the event has already happened, the time of evaluation has occurred. There is no ambiguity there. In fact, you'll find it impossible to phrase the scenario equivalently so that the coin flip had not already occurred at the time of questioning.

No, it is your response that is "just wrong." For example, if you look at the actual wording of the problem in the article, you'll see that it doesn't mention when the coin is flipped. And if you understand the experiment, you will realize that it doesn't have to be flipped until Tuesday morning, well after the only certain asking of the question. Not only is what you claim to be "impossible" possible, it is a standard facet of the problem that is explored in the literature.

Probability is a function of what you know about a situation, not the situation itself. So different people, with different knowledge sets, can assign different (and correct) probabilities to the same outcome. For example, say I draw a card from a standard deck, and tell (1) Alice nothing, (2) Bob that it is black, (3) Carla that it is a spade, (4) David that its face value is odd, and (5) Eve that it is an honor card (ten thru ace). Alice will say the probability it is the Ace of Spades is 1/52, Bob will say 1/26, Carla will say 1/13, David will say 1/28 (there are 28 "odd" cards if the values are 1-13), Eve will say 1/20, and I will say either 0 or 1, because I know what the card is. All are correct, because probability is based on the knowledge you have; or more specifically, the ambiguity in the knowledge you have. JeffJor (talk) 22:26, 9 May 2013 (UTC)

^What's your point? That probability of a card being Ace of Spades given (nothing) (black) (spade) (odd) (10-ace) are different? That's just making a strawman. — Preceding unsigned comment added by 140.180.253.155 (talk) 06:41, 13 May 2013 (UTC)

My point is that probability is not a function of the situation, it is a function of what you know about the situation. Just imagine what odds each of these people would expect to place a bet in "Ace of Spades," even after the card is drawn. Each is different, and is based on their assessment of the probability. If you can't understand this, you shouldn't comment on probability. JeffJor (talk) 21:30, 13 May 2013 (UTC)

It's not that the question is ambiguous, it's that "Heads" is not explicitly defined, and the implicit definition is changed depending on the interpretation. In the "Thirder" variant, "Heads" means "Heads upon waking", so P(Heads) = 1/3. In the "Halfer" variant, "Heads" means "Heads flipped", so P(Heads) = 1/2. This is like saying that there exists an s and t such that 5s-4t=1, and that there exists an s such that 5s=10, and therefore 4 divides 9. The "s" is different in each part, they aren't interchangeable. Aielyn (talk) 13:30, 8 September 2013 (UTC)

"Heads" means the same thing to both Halfers and Thirders: "By the end of the experiment, the single coin flip will have resulted in Heads." You just can't accept that when the Thirder changes P(Heads|Sunday)=1/2 to P(Heads|Awake)=1/3, that the change is due to the fact that "Awake" has a different information content than "Sunday" (and here "Awake" means "Awake and it is either Monday or Tuesday"). Yet you are perfectly willing to change P(Awake on Tuesday|Sunday)=1/2 to P(Awake on Tuesday|Awake)=1/4, which proves that you think there is an information change. JeffJor (talk) 11:41, 14 September 2013 (UTC)

If we're going to cut into the philosophical side of the question, and debate what kind of details and information matter for the credence update, then could it not be said that it is misleading to write "Awake on Tuesday" in both cases, while in the first it is "Will be awake on Tuesday" and in the second is "Right now it is Tuesday", which are perhaps not necessarily exactly the same thing or type of information. If such detailed information matters for the update, which I would think it can, then I think it's rather too simple to stop taking into account the difference in information at 'Awake' and not going as far as fully detailing the different situations. Think of this what you wish. In either case what we are looking for is a quantification of our uncertainty, but depending on the exact wording and interpretation, that value may differ (duh, that's the whole point of all these sorts of problems). I simply wanted to remark that the wording is either possibly misleading or incorrect, because "I am awake on Tuesday/and it is Tuesday" has probability of 0 if it is Sunday, or did you mean "given the information of the events from Sunday, meaning being informed about the experiment and the fact that a fair coin with probability 1/2 for either option will be used". These are different things, yet you never detailed that. So there must be some error or some missing clarification. It seems to also be the core of the problem that, yes, the probabilities can change, but wether to 1/4 or 1/3 (or 1/2) is the issue, disagreeing with the method of updating does not mean there is an inconsistency in either argument, as you do seem to imply to favour one side, though maybe I am too boldly assuming this based on your writing on the matter.

Missing problem statement?

The last sentence of the problem statement reads Any time Sleeping Beauty is wakened and interviewed, she is asked, "What is your belief now for the proposition that the coin landed heads?" That doesn't sound like a problem statement. Is there a missing sentence, e.g. "What should she answer?" --Vaughan Pratt (talk) 04:04, 1 June 2013 (UTC)

Equivalent problem without time or consciousness

Scenario #1: An experimenter has two baskets. The first is labeled “M” and the second is labeled ”T”. The baskets are placed near a princess but she cannot see into the baskets, nor can she see what the experimenter is taking in or out of the baskets. The experimenter flips a coin out of sight. If the coin lands on heads, the experimenter places a raw egg in basket M and a hard-boiled egg in basket T. If it lands on tails, he places a raw egg in basket M and two hard-boiled eggs in basket T. The experimenter then reaches into one of the baskets and pulls out an egg and places it in front of the princess. What is the probability that the egg in front of the princess is raw?

Scenario #2: The experimenter has three baskets. The first is labeled “M1” the second is labeled ”T1” and the third is labeled “T2. The experimenter flips a coin out of sight. If the coin lands on heads, the experimenter places a raw egg in basket M1 and a hard-boiled egg in basket T1. If the coin lands on tails, the experimenter places a raw egg in basket M1, a hard-boiled egg in basket T1 and a hard-boiled egg in basket T2. The experimenter then reaches into one of the baskets and pulls out an egg and places it in front of the princess. (Assume that the first part of the test is repeated multiple times so that there is at least one egg in every basket.) What is the probability that the egg in front of the princess is raw?

Which scenario is most like the sleeping beauty problem?192.104.67.121 (talk) 20:46, 14 January 2014 (UTC)

Neither, but a good and conclusive analogy can be made from them.

Your scenarios are confusing because the analogies are inconsistent and overlapping. Since it is what you asked about, it is your egg, not your coin, that is most analogous to the coin in the Sleeping Beauty Problem. But then you twist the analogies differently in the two scenarios.

In Scenario #1, the baskets are labeled M and T, implying an analogy to Monday and Tuesday. But choosing between them is no different from choosing which kind of egg, so again they correspond better to Sleeping Beauty’s Heads and Tails.

I suppose we are to assume each basket is equally likely to be selected, so the answer is 1/2. And the solution here is indeed the same as the halfer solution. The problem is, Sleeping Beauty (or her experimenter) never chooses between two days like your experimenter chooses between two hard-boiled eggs in the T basket. Both days happen to Sleeping Beauty, she just doesn’t know which is happening “now,” and might sleep through one. To match the problem batter, your princess should be given all of the eggs in the basket, one at a time, with an amnesia drug in between. So you captured the halfer argument, but you did not capture the Sleeping Beauty Problem.

In Scenario #2, the coin changes the number of baskets. There is a 1/2 chance (heads) that there is a 1/2 chance (two baskets) for a raw egg, and a 1/2 chance (tails) that there is a 1/3 chance (three baskets) for a raw egg. That makes the answer (1/2)*(1/2)+(1/2)*(1/3)=5/12. Simulate it, if you don’t believe me. This doesn’t compare to either the halfer or thirder argument, but it seems you were trying to make it into the thirder argument. That you wanted the distinction between the two hard-boiled eggs to reflect their argument that each Coin+Day (equivalent to Basket+Egg) combination was also distinct. But your version essentially selects the egg differently depending on the coin, which doesn’t happen in the Sleeping Beauty Problem. That’s why you get 5/12 instead of 1/3.

A better analogy is this: Four hard-boiled eggs are divided between two baskets. One of the eggs is injected with a sleeping drug. A coin is flipped to select a basket – Heads means the one with the drugged egg. The experimenter then picks an egg from the selected basket, and gives it to the princess to eat. If she remains awake, she is asked for the probability that the coin came up heads. Each of the four eggs is equally likely to be selected this way, but if the princess remains awake, she knows of one that couldn’t have been selected. This is “new information,” and allows her to update the probability, from the prior 1/2, to 1/3.

The point is that Heads+Tuesday is an outcome of the Sleeping Beauty Experiment, just like drawing the drugged egg is an outcome here. Not being able to observe it is not the same as it not happening, it is the “new information” the halfers claim does not exist.

You could try to justify several differences between my analogy and the Sleeping Beauty Problem. The first is that there is no equivalent to the two different baskets. That there are only two possible days, as opposed to four possible eggs. This point is irrelevant; we could use one basket and two eggs, but only inject an egg after Heads is flipped. I only used the two baskets because you did, and because avoids confusion with the next alleged difference.

That alleged difference is that Sleeping Beauty is awake twice after tails, but the princess only eats one egg; or equivalently, that Sleeping Beauty is guaranteed to be asked the question, but the princess is not. These are really the same point, just phrased differently. But in the analogy the princess could be wakened (if needs be), given the amnesia drug, fed the second egg, and asked again. She now eats all of the eggs, and is guaranteed to be asked the question. But because of the amnesia drug, each time the princess eats an egg, to her it looks exactly like my analogy.

And finally, you might think it is different because Sleeping Beauty only sleeps through the second day, but the princess could sleep after the first egg. Or the equivalent, that the coin doesn’t need to be flipped until after the first day. Again, my analogy can do the same thing, and each egg still looks just my original analogy.

And in fact, it is these three sets of points that obfuscates the Sleeping Beauty Problem. I don’t really think you will be confused by one vs. two baskets, or one vs. two wakings; each is easily extended. But the combination of the two, when seen as a single change, is not as easy to extend – and in fact, is what is wrong with your Scenario #1. The Sleeping Beauty Problem is confusing because it starts with them combined; my analogy, which is clearly answered 1/3, starts with them separate. JeffJor (talk) 22:27, 26 February 2014 (UTC)

Thanks. Based on your input I have tweaked the scenarios. The whole raw vs. hard-boiled issue is redundant and the scenario can be greatly simplified as followed:

Scenario #1: An experimenter has two baskets. The first is labeled “M” and the second is labeled ”T”. The baskets are placed near a princess but she cannot see into the baskets, nor can she see what the experimenter is taking in or out of the baskets. The experimenter flips a coin out of sight. If the coin lands on heads, the experimenter places an egg in basket M. If it lands on tails, he places an egg in basket M an egg in basket T. The experimenter then, after one trial, reaches into one of the baskets. If there is an egg he and pulls it out places it in front of the princess and asks, "Based on the fact that there is an egg in front of you, What is the probability that the coin landed heads?"

Scenario #2: The experimenter has three baskets. The first is labeled “MH” the second is labeled ”MT” and the third is labeled “TT". The experimenter flips a coin out of sight. If the coin lands on heads, the experimenter places an egg in basket MH. If the coin lands on tails, the experimenter places an egg in basket MT, and an egg in basket TT. The experimenter then after one trial, reaches into one of the baskets. If it contains an egg, he pulls it out and places it in front of the princess and asks, "Based on the fact that there is an egg in front of you, what is the probability that the coin landed heads?" If it is empty, he places nothing in front of the princess and says nothing.

To me, the key statement you made is "I suppose we are to assume each basket is equally likely to be selected." I purposely left this ambiguous because it is the philosophical question that is at the heart of the problem. Notice that it is the structure of how the baskets are laid out and the method of selecting the basket that at least partly drives the probability in these scenarios, the bias or outcome of the coin or the number of eggs placed are not the only relevant factor. Just like in the original problem it is the structure of time and the method of selecting "this moment" from all possible moments that drives the probability. We don't know what this structure is, therefore we can't answer the question. One school of thought points to the Self-Indication Assumption, another points to the Self-Sampling Assumption. The grander question being asked is, of all possible moments in the multiverse, what is the probability that we are experiencing the moment that we are in right here and now and not some other moment?192.104.67.122 (talk) 16:28, 7 March 2014 (UTC)

Jargon

Thirder position?

There isn't such a word as "thirder", shouldn't appropriate English be used?

And why use "P" if it is for Probability. Just use the word probability so there is a clear understanding. — Preceding unsigned comment added by 121.222.63.221 (talk) 00:17, 25 August 2015 (UTC)

I agree. Until I realised that it meant "someone who belongs in the one-third camp", I thought that "Thirder" was a person. 79.79.253.71 (talk) 15:31, 30 July 2018 (UTC)

Computer Program Analogy

How about replacing SB with a computer device that is powered up on Monday and possibly Tuesday according to the same rules as the SB problem, it will have no clock or writable storage device, but it will have a stored program to run. Ok! The computer now runs in a state of un-knowingness regarding any other possible run, what code might you right to give a prediction as to what the coin toss was? I would say that always printing out 'tails' is a strategy that would score 66.6% over time - so therefore you'd then seem to have written a computer program that 'thinks' that tails is the most likely correct answer. If the program printed out its 'guess' (tails) and the actual outcome, it would give the appearance of a situation where there is a bias towards tails and that the program exploits this by always selecting tails. The computer is no different to SB, it could have any conceivable level of AI you want to envisage (but not time based features or writable memory). It does seem that the computer has 'credence' of a 1/3 probability. — Preceding unsigned comment added by Gomez2002 (talk • contribs) 15:33, 23 December 2015 (UTC)

I think the new section on operationalization answers your question: As soon as you pin down a protocol (or a computer program), the problem becomes entirely trivial. But it's about words, not about facts: it's philosophy. -- Oisguad (talk) 12:02, 24 December 2015 (UTC)

yes it does answer it thanks! how about if sleeping beauty gets a prize afterwards only if no wrong answer is given (and can remember a strategy), there's no benefit in changing between head or tails for all answers, so it seems perfectly fair for her to say "tails because it gives me a 50% chance of winning", as a pre-decided answer Gomez2002 (talk) 18:08, 4 January 2016 (UTC)

Suggestions for rewrite

Proposals for fixing some problems with the article:

1. Change second paragraph. Something like the remainder of this bullet point, perhaps. Traceable to Arnold Zuboff originally, later named the "Sleeping Beauty problem" by Robert Stalnaker, the puzzle in its current form was popularised by Jamie Dreier in a posting to rec.puzzles in 1999. Papers by Elga and Lewis followed, after which the published literature expanded rapidly.
   • Article appears to imply the sequence is Zuboff -> Elga -> Dreier -> ...
2. Replace "belief" for the proposition that the coin landed Heads with "credence" for ...(etc)...
   • The literature talks of the specific "credence of 1/2", for example, and general "self-locating beliefs". It's the specific numerical value we want here.
3. In the solutions section, change "thirder position" to "Credence for heads = 1/3" and similarly for "halfer position". Modify the treatment in each section slightly to bring the original Elga/Lewis arguments to the fore and to be careful about how the "thirder"/"halfer" nicknames are used:
   • Credence 1/3. Argument as given. The term "thirders" is used ambiguously either for those subscribing to Elga's exact argument or, more widely, those asserting that credence(Heads) = 1/3. Note long-run argument as additional support for 1/3. Describe 1/3 as "aligned to" the Self-Indication Assumption of anthropic probability, not "implied by" (in the article as presently written, it looks like an endorsement).
   • Credence 1/2. Argument is simply that of no new information. The term "halfers" is used ambiguously either for those asserting that credence(Heads)=1/2 or, more specifically, for those subscribing to the whole of Lewis's argument in his "reply to Elga" paper. Many who agree that credence(Heads)=1/2 would disagree with Lewis on the follow-on question of credence(Heads|Monday), Sleeping Beauty's credence for Heads upon learning that today is Monday:
     • Credence(Heads|Monday) = 2/3 as originally argued by Lewis, on the basis that Pr(Heads|Mon) = Pr(Heads,Mon)/(Pr(Heads,Mon)+Pr(Tails,Mon)) = 1/2 / (1/2+1/4), where here "Pr" means "probability" as a stand-in for "credence".
     • Credence(Heads|Monday) = 1/2 (sometimes termed the "double halfer" position) in analyses aligned to the Self-Sampling Assumption of anthropic probability. Cozic's "imaging" paper was certainly not the first to offer this argument; Bostrum's "hybrid model" paper pre-dates it.
4. Variations section needs references, as already noted in the article. Saying that "days of the week are irrelevant" is unhelpful, as it is important in some treatments of the problem that there is a temporal element. The thing about computers does seem irrelevant, though, as Sleeping Beauty is a puzzle in anthropic probability in which the human element is core. — Preceding unsigned comment added by 109.152.206.9 (talk) 08:54, 29 August 2015 (UTC)

Paradox Resolved (It's 1/2): Conditional Probability Formula Does Not Apply

The answer is 1/2, as common sense suggests. The only reason for the apparent paradox is the misapplication of Bayes' Theorem/ The "Conditional Probability Formula". Taking P(A | B) = P(A&B)/P(B) as the definition of conditional probability is a common practice but one that misses the point of probability: conditional probability already has a definition as the degree of belief one may assign to B given the truth of A. Math does not get to define probability as it defines most of the structures it deals with; it studies it. Probability, like truth and space, can be modeled with mathematics but is not itself a purely mathematical concept.

Suppose we have a probability measure defined on an algebra (set) of events (things, not necessarily exclusive, that can occur); each event has a certain probability associated with it. If the events can be partitioned (via disjunction) into a family of mutually exclusive, collectively exhaustive events, we shall call these "outcomes." If we are now given the further piece of information (and only this piece of information) that some of our outcomes are in fact not any longer possibilities, the relative weights (probabilities) of the outcomes still in play do not change, as we have stipulated that we are doing nothing more than discarding certain outcomes. For example, if 20 contestants are participating in a Olympic event, and by virtue of her skills, Sally is twice as likely to win as Bob, if some other contestants drop out, Sally is still twice as likely to win as Bob, because nothing has occurred to change their relative potential (probability) to win. But we need the total probability, which would be the combined probability of all the non-excluded outcomes, to be 1, by convention. So, we divide the probabilities of all the outcomes by the total probability of the non-excluded outcomes to keep the relative probabilities the same. On the other hand, the probabilities of the excluded outcomes are now, of course, 0. Since the probability of an event is the sum of the probabilities of its constituent outcomes (their being mutually exclusive), the probability of an event A given B (that only certain outcomes are eligible) is P(A&B)/P(B), as only the outcomes in A that also fall within the allowed outcomes B contribute non-zero probabilities. This is the conditional probability formula. What did we assume? We assumed that in being given B we were simply limiting our scope of allowed outcomes. We did not repudiate any previously accepted information (which corresponds to the sample space), for example.

Now to the SB problem. It is true that P(H| Monday) = P(T| Monday)=1/2, the most straightforward argument being that the coin toss could be not yet performed. Yet, the argument goes, if P(H)=1/2, then P(H) = P(H&Monday)= P(H | Monday)*P(Monday) = 1/2*P(Monday), so P(Monday)=1, a contradiction. The fallacy is in the conditional probability formula. In calculating P(H | Monday), we exclude the possibility of its being Tuesday. But we aren't simply eliminating the "weight" of Tuesday from the match. What we're being told is that the current day is Monday; this changes the whole sample space, so that probabilities of non-excluded events (e.g., heads and tails) intersected with the new whole (Monday) are no longer in the same ratio. 2:1 turns into 1:1. There is no contradiction. The conditional probability formula can be applied only when the outcomes have no relation to one another: one is simply being removed from the race, or perhaps in another type of situation that hasn't occurred to me. In any case, it does not apply here.

That was a bit too vague. Another, better way to think about it is that the outcome (toss, day) is a chimera of two different processes, a coin flip that occurs on Tuesday morning before the potential waking (say) and whatever day the interview occurs. Thus, the first part of the outcome makes sense as a variable throughout the whole experiment, while the second part changes, obviously. The most intuitive way (and the correct way, incidentally) to get the probabilities is to say that P(H)=P(T)=1/2, as no new information has been obtained, and by symmetry, P(H%M) = 1/2, P(T%M)= P(T&Tu)=1/4. We can diagram this: make one half of a circle heads, the other half tails. Monday fills the heads half and half of the tails half; the last quarter is Tuesday. Our three outcomes are H&M, T&M, T&Tu. But the issue with applying the conditional probability formula is that they are not simply separate events vying for actuality based on their probabilities. Heads and tails fit this description, as do Monday and Tuesday, considered as outcomes of the toss and day processes, respectively. But the two groups are different sorts of outcomes: the toss outcomes are outcomes of an actual, physical process that produces a single output. The day outcomes, on the other hand, are not the outcomes of a process but simply the location of the interview; if the toss is tails, both the outcomes Monday and Tuesday are produced; if the toss is heads, only Monday is produced. Thus, the outcomes H&M and T&M don't compete for occurrence in the same way that H and T do. Exactly one of them occurs at some point. We can't use the conditional probability formula on the whole mess (that is, assume that when one outcome is removed, the ratios of the remaining outcomes' probabilities remain unchanged) because not all outcomes are created equally: while T&M and T&Tu occur either together or not at all at some point, H&M is completely separate. In fact, we can deal with this situation using conditional probability, but only if we separate Heads and Tails. We begin with the assumption (by intuition) that P(H)=P(T) but now show that this leads to a consistent system. Applying the conditional probability formula just to tails works fine, as all outcomes there are competing as members of the same class. We see that P(M | T | M) = 1, a somewhat obvious assertion. This amounts to enlarging the T&M quadrant to fill half of the circle diagram after eliminating Tuesday. Since Tuesday is not involved with the interviews in the case of heads, nothing changes there. We thus get that P(T | M) = P(M | T | M)* P(T) = 1/2, as expected. In essence, knowing that it is not (tails, Tuesday) does not get rid of any tails area from our circle, as if tails occurs at all, it occurs with Monday and Tuesday. Tails remains rigid, with P=1/2, and the distribution of day probabilities changes inside of it. Basically, apple outcomes can't be compared to orange outcomes.

If anyone reads this, let me know what you think of the argument. I hope I've convinced you that the answer is probability 1/2 with probability at least 1/3. David815 (talk) 05:15, 5 March 2016 (UTC)

Have you fully read the article? Which of the cited original papers have you consulted? As the article says: the SB problem is not about probability theory, but about certain philosophical notions. -- Oisguad (talk) 19:56, 5 March 2016 (UTC)

Yeah, I read the article. I feel that my above analysis does touch upon philosophical concepts; pure math would not cut it, as I noted. I don't agree with the use of gambling analogies in the Operationalization section, though. It mixes the two different types of outcomes (outcomes relating to the whole week and outcomes relating only to the day in question) too freely. I didn't realize that Mikal Cozic's paper was available online, so I went back and read that. Our arguments seem to be mostly different, except that we both use fruit analogies and object to conditionalization, as any double-halfer obviously must. I don't quite understand the imaging process, and though it is certainly interesting and important, I think that this particular problem can be solved without it. My main point is just that T&M and T&Tu do not comprise T in the usual sense: they must both occur at some point, or else neither occurs. That it is Monday does not provide evidence against its being tails; tails ensures that there will be a waking on Monday. The conditional probability of heads given Monday (which Cozic would not term "conditional" at all) is not given by the usual formula, as T&M and H&M are not directly comparable events. I thought of a better analogy just now than I was able to earlier. Suppose that there are two types of countries in the world: type A consists of those with a population of 100, and type B consists of those with a population of 10000. Suppose that citizens can be totally ordered by wealth in any given country. And suppose that you reside in country X. Now, given nothing else, the probability that X is in A, P(A), = P(B) = 1/2. However, suppose that you know that you are not among the top 10 wealthiest citizens, as they have a special club (call this fact C), to which you do not belong. Naive application of the conditional probability formula would then yield revised probabilities (noting that P(C) = 1/2*90/100 + 1/2*9990/10000 = .9495) P(A | C) = 1/2*90/100/.9495=.4739, and of course P(B | C) = .5261. Suppose you are friends with someone who is one of the top 10 wealthiest citizens. Applying the same method, they will revise their probability for P(A) to .99. Then the two of you discuss your calculations. Since you now have all of the same information, you should come to a common conclusion, but this does not seem likely. How can one person be mostly ambivalent on which type of country you both live in while the other is so convinced that it is type A? The problem is that in situations like this and the SB problem conditional probability does not apply: knowing that you are in the top ten tells you only that you are in a country with at least ten people; the illusion that you are gaining useful information comes from the illusion that you are a random sample from the population. But you are not: no selection process has been applied to select you. You are simply a member of the population. The same type of reasoning applies in SB. Your day is not randomly selected by any means; it is simply one of the days on which a waking occurs. The analogy is essentially the Doomsday argument; the fallacy, there too, lies with conditional probability. David815 (talk) 01:27, 6 March 2016 (UTC)

As stated, asking SB to define the probability that the coin was heads, her answer can only be 1/2. The key aspect is that SB has no knowledge of the coin toss, any previous wakings, or which day she is woken on. Therefore which day she is asked on is completely irrelevant to her answer, as she is unaware of any difference. The only information she has is the rules and that a coin will be (or has been) flipped exactly once, and thus her only possible answer is 1/2. Now, if she could tell which day she were woken on, or that she had been woken twice, things would be different (as *this* is when conditional probability can come into play -- obviously if she can tell that it is Tuesday or that it is the second time she is asked, then her answer must be zero). Thinking of thirds is attractive because as an outside observer (and an irrational actor) you are aware that there are three possible states in which she could be asked the question, but these states do not have equal probability and she cannot distinguish between them, so they are of no relevance to the problem as stated. Similarly, phrasing it instead as a bet makes it seem like the third is correct, but this presupposes that she is instead asked to state which side came up, and that she "wins" if she is correct on any query (in which case "tails" makes the most sense, as it's a guaranteed win on Tuesday and a 50:50 chance on Monday; even absent the knowledge of which day it is, it's a reasonable bet). Increasing the number of times she is asked doesn't really make any difference; either it was heads and she is only asked once, or it is tails and she is asked N times; and for any N > 1, "tails" is the most rational answer. 118.92.29.76 (talk) 22:41, 15 July 2016 (UTC)

Halfer position needs clarity

I would say for the original problem that the chance of it being Monday and the coin having landed on heads is 1/2, the chance of it being Monday and the coin having landed on tails is 1/4, and the chance of it being Tuesday and the coin having landed on tails is also 1/4. Obviously this means the chance of the day being Monday is 3/4. Are these the probabilities of the halfer position? It doesn't seem to say what the probabilities are of each case. 2602:306:802B:1520:EC42:13A9:99F0:DFC4 (talk) 16:05, 2 July 2016 (UTC) It seems like this might be a bit better explained later in the article, but I think the reasoning should be more detailed in the "Halfer position" section. It sounds like this is a question of whether the probability is subjective to the experimentee (in which case the probability is 1/2 heads & Monday, 1/4 tails and Tuesday, and 1/4 tails and Monday) or objective (in which case the probability is 1/3 for each case). Subjectively, the chance of tails and Monday is only half the chance of heads and Monday; objectively, the expected result of repeating this experiment would be that each case happens equally often. 2602:306:802B:1520:EC42:13A9:99F0:DFC4 (talk) 16:31, 2 July 2016 (UTC)

Lack of definition

There was not stated whether the coin will be tossed in certain time or is it varying, and if so, what are probabilities that coin will be tossed before Monday or after Monday. Only information available is that coin must be tossed before Tuesday morning by definition.

Now the question is: "What is your credence now for the proposition that the coin landed heads?" Answer for this question requires information on probabilities of two conditions; that the coin has already been tossed and it turned out to be heads.

Maybe did not understand the article correctly, but I cannot see any other reason for a claimed controversy than the lack of definition on what are conditions. For example, is half of he cases are where coin is tossed in Monday morning and in another half it is tossed in Tuesday morning, then we are ending up with following probabilities. Let's say X1=Finding oneself from Monday morning where coin has already been tossed and turned out to be tails; X2=Finding oneself from Tuesday morning (and by definiton, coin has been tossed and turned out to be tails); Y1=Finding oneself from Monday morning where coin has already been tossed and turned out to be heads; and Z1=Finding oneself from the morning where coin has not been yet tossed. Now, we know that X1+X2=50% and Y1=50% after the coin has been tossed, thus X1=25% and X2=25%. And we know that for Monday, probability of Z1=50% and for Tuesday it is 0% by definition. Thus probability for position of being in Z1 = 50% x (100%-25%) = 37,5%. Question was what is your credence now for the proposition that you are in the position of Y1 and the answer in above case is 100%-(X1+X2+Z1). To solve this, one needs to know expected value of X1 and as we know that Z1+X1=25% and as both have equal value, X1=12,5%. Thus Y1=100%-(12,5%+25%+37.5%)=25%.

Moreover, if coin is tossed always in Tuesday morning, answer will never be yes, because it requires that coin has been tossed when question is asked and if it is heads, question will not be asked in Tuesday. Thus in this case, credence for proposition must be 0%. First proposed solution in the article implied that the time when coin is tossed can change, because at first it has been tossed already before Monday when you learn it's tail and while at next the coin might be tossed in Tuesday morning when you learn it's Monday. 2001:14BB:70:308E:2CFD:6EE9:8748:E116 (talk) 05:40, 18 January 2017 (UTC)

The problem as stated is a false dilemma.

Are there any reliable sources that explain why the 1/2 vs 1/3 question is a false dilemma?

It seems obvious to me: The initial coin toss has a 50/50 chance of being either heads or tails, but from Beauty's perspective, there is a 50% chance she has woken up on Monday and the coin is heads, a 25% chance she has woken up on Monday and the coin is tails, and a 25% chance she has woken up on Tuesday and the coin is tails.

There is no 1/3 chance of anything. The 50% chance of tails is split in half.

Chinagreenelvis (talk) 05:39, 19 July 2017 (UTC)

There are four possible outcomes of the coin-tossing, each with identical probability. Beauty know that she will never observe exactly one of them, so for her in two out of only three cases, the coin will have come up tails: 2/3rds

To me, David Lewis and others of the 1/2 position are simply giving a nice example of Cargo Cult Science: They repeat an otherwise undubious rule without questioning the applicability, and draw a wilful , and obviously wrong, conclusion. Rigourless (=quack) philosophy. --129.13.72.197 (talk) 11:39, 9 August 2017 (UTC)

Bad information in "Thirder"

Section The Problem states:

A fair coin will be tossed to determine which experimental procedure to undertake:

If the coin comes up heads, Beauty will be awakened and interviewed on Monday only.

If the coin comes up tails, she will be awakened and interviewed on Monday and Tuesday.

Section Thirder position states:

Beauty knows the experimental procedure doesn't require the coin to actually be tossed until Tuesday morning, as the result only affects what happens after the Monday interview.

What Beauty "knows", according to the Thirder position, is incorrect. While it's true that the coin toss isn't apparently effective until Tuesday, because the two procedural paths overlap up to that point, the statement of the problem says that the coin is tossed at the start of the experiment and the procedural path determined at that time. For example, for heads, Beauty will be (future tense) woken on Monday. The coin toss only incidentally determines what happens on Tuesday; it's an outcome of what coin does select - the procedural path which is to be taken - and that must occur at the start. 79.79.253.71 (talk) 14:45, 30 July 2018 (UTC)

On further reflection, I have decided to remove the sentence given above from the Thirder section. It provides no additional information that's useful to Beauty. On the contrary, if the idea is acted upon then it's bad information. If Beauty assumes that the coin is not tossed until Tuesday ("because it doesn't need to be tossed until then") then the Monday awakening will have an untossed coin that has no value. She will only be awoken on a Tuesday if a tail has been tossed therefore the probability of her being woken after a heads is zero. That makes a nonsense of the problem. 79.79.253.71 (talk) 15:18, 30 July 2018 (UTC)

Simple probability problem, if correctly defined

To an outside observer, each day will occur with equal probability, whether SB is awake or asleep:

P(Mon)= P(Tue) = 0.5

-

For a fair coin:

P(heads) = P(tails) = 0.5

-

There are four possible states, and each state corresponds to a state of SB

-

Mon and heads (MH) -awake

Mon and tails (MT) -awake

Tue and heads (TH) -asleep

Tue and tails (TT) -awake

-

To an ouside observer:

P(MH) = 0.5*0.5 = 0.25

P(MT) = 0.5*0.5 = 0.25

P(TH) = 0.5*0.5 = 0.25

P(TT) = 0.5*0.5 = 0.25

-

From this we can calulate the probability that SB is awake:

P(awake) = P(MH)+P(MT)+P(TT) = 0.25+0.25+0.25 = 0.75

-

SB is being asked to state the probabilty of heads, given she is awake.

P(heads|awake) = P(heads and awake) / P(awake)

-

The only state where the coin is heads and SB is awake is MH therefore

P(heads and awake) = P(MH)

-

P(heads|awake) = P(MH)/ P(awake) = 0.25/0.75 = 1/3 — Preceding unsigned comment added by 174.136.133.132 (talk) 16:56, 14 March 2019 (UTC)

In my opinion, people has forgotten the fact that some of the 4 events are mutually exclusive and some are "simultaneous". To avoid confusion when we do the maths, it is easier for people to look at P(no interview) and then use P(at least one interview ) = 1 - P(no interview). Then calculate P(Head | at least one interview) as the solution Ehung606631 (talk) 17:03, 5 June 2023 (UTC)

You are perfectly right. This problem is entirely trivial, and our text said so, before an entire section was deleted for lack of external sources: https://wiki.riteme.site/w/index.php?title=Sleeping_Beauty_problem&diff=845602153&oldid=837230593. Feel free to restore that section, and let's see what happens. -- Oisguad (talk) 21:47, 15 March 2019 (UTC)

Sailor’s Problem is Non-Equivalent

The Sailor’s Problem is equivalent to the conditional probability of the coin flip given that an interview is occurring (tails leads to both more interviews and more children). However, this is not what the original problem is asking. It’s asking the belief of sleeping beauty. The fact there is twice as many interviews given tails is irrelevant because they are coupled together: the same sleeping beauty will wake up on both Monday and Tuesday, but she gains no additional information on Tuesday. The way I interpret this is that the Tuesday results can be ignored when it comes to calculating belief, resulting in the Double Halfer solution. This is not the case in the Sailor Problem though, as each child only questions once, and they question sepperately. Thus each questioning gives additional information for the inquirer and it should be given full weight. I believe this distinction is intentional as the paper it’s from was using Nick Bostrom’s terms, and thus was probably assuming his solution.

Ganondox (talk) 03:00, 15 June 2019 (UTC)

Solutions

SB problem solution: Four equally likely events:

  Coin toss   Day       Result
  H           Mon       SB awake
  H           Tues      SB asleep
  T           Mon       SB awake
  T           Tues      SB awake

P(H|awake) = 1/3

SC problem solution Four equally likely events:

  Coin toss   First port in SG    Result
  H		   port 1              1 child in port 1, you are an only child.
  H		   port 2              1 child in port 2, you are an only child.
  T		   port 1              2 children, you are not an only child.
  T		   port 2              2 children, you are not an only child.

P(one child) = 2/4 = 1/2

Remarks: Posting a 'solution' to SB does not seem appropriate here, not in this manner, as the problem is not generally considered resolved. Additionally, it is contested wether it is appropriate to phrase the problem and go about solving it in this manner. Some would argue that we are not considering four equally likely events at all, because SB asleep is not a relevant event and redistributing the probabilities, updating the credence, and how this is supposed to happen is exactly the subject of the debate, which is nicely brushed aside and assumed in this solution. It may be a fitting solution to a particular interpretation, but posting it here as being 'solutions' to the given problems is misleading and inappropriate.

---

My intent was not to brush aside the subject of the debate, but to make a statement about it. The four equally likely events are how an observer would view the problem. This does not change from SB's point of view. From her view all 4 events are still equally likely, the difference is that the act of waking her and posing the question gives her additional information. This tells her that it is not heads and Tuesday, therefore she knows she is is one of three of the four equally possible events so her creedence should be 1/3. 97.70.180.226 (talk) 16:29, 17 February 2022 (UTC)

How come the problem does not state "now that it is Wednesday ..."?

""The Sleeping Beauty problem is a puzzle in decision theory in which whenever an ideally rational epistemic agent is awoken from sleep, she has no memory of whether she has been awoken before. Upon being told that she has been woken once or twice according to the toss of a coin, once if heads and twice if tails, she is asked her degree of belief for the coin having come up heads.

Sleeping Beauty volunteers to undergo the following experiment and is told all of the following details: On Sunday she will be put to sleep. Once or twice, during the experiment, Sleeping Beauty will be awakened, interviewed, and put back to sleep with an amnesia-inducing drug that makes her forget that awakening. A fair coin will be tossed to determine which experimental procedure to undertake:

If the coin comes up heads, Sleeping Beauty will be awakened and interviewed on Monday only. If the coin comes up tails, she will be awakened and interviewed on Monday and Tuesday. In either case, she will be awakened on Wednesday without interview and the experiment ends.

Any time Sleeping Beauty is awakened and interviewed she will not be able to tell which day it is or whether she has been awakened before. During the interview Sleeping Beauty is asked: "What is your credence now for the proposition that the coin landed heads?"""

The way the introduction is worded, related to the problem being described in the subsequent paragraph, reads to me, that Sleeping Beauty could be told on Tuesday "Upon being told that she has been woken once or twice according to the toss of a coin, once if heads and twice if tails, she is asked her degree of belief for the coin having come up heads." and be under the belief that she is about to be sedated and asked the same question tomorrow.

Is it just me? Or does the wording imply, "it is Wednesday for Sleeping Beauty" is a fact, and cannot be unproven?

If it was Tuesday, the criteria would still hold true (I think, but, I am far from right about these kinds of logic puzzles) in my belief, it seems that due to the wording, it could only be Tuesday, because on Wednesday the patient is not asked any question, according to the phrase "In either case, she will be awakened on Wednesday without interview and the experiment ends.", and, since the patient is told as a truth "you have been woken up before", and asked to give an evaluation on credence, that it could not be Wednesday. Therefore, to me, the chance of it being a heads result is 0 since the patient would not be told on Monday that they have been woken up before, and on Wednesday they are not given a query on whether they believe that there is any credence to the coin flip, as the wording states, "In either case, she will be awakened on Wednesday without interview and the experiment ends." therefore, I wholeheartedly believe the wording is wrong, or the probability is 0.

Signed - Sleeping Beauty, on the day of Tuesday, in the year of our Lord, God knows? 49.185.130.1 (talk) 14:52, 24 April 2022 (UTC)

Regulating her subjectivity

Immediately after my explanation, she said, "I had a revelation. If I am not woken up on Tuesday, I will dream twice that I am awakened and questioned. Therefore, the credence that the coin will show the head is 3/5". Would she be mathematically wrong? --Dummy index (talk) 13:52, 27 May 2022 (UTC)

Sounds like a homework question. This isn't a forum for general discussion of the topic. MrOllie (talk) 14:31, 27 May 2022 (UTC)

Editing needed

The article mentions Nick Bostrom three times as arguing for the thirder position. However from what I understood from his works he is actually a double halfer. He actually brings the "Extreme sleeping beauty" scenario to argue against both the halfer and thirder positions. Also, he argues against the Self indication assumption and argues that the halfer position is implied by the Self sampling assumption.

Here is a quote from his paper Sleeping Beauty and Self-Location: A Hybrid Model page 17:

"If the hybrid model is correct, it might explain the fact that both the 1/3- and the 1/2-views have some intuitive appeal. According to the hybrid model, both these views get something right. The 1/3-view is right that Beauty’s posterior credence in HEADS after being informed that it is Monday should be one-half. The 1/2-view is right that Beauty’s prior credence in HEADS, after awakening but before learning that it is Monday, should be one-half." 108.30.23.32 (talk) 21:04, 9 October 2022 (UTC)

I will attempt editing. 108.30.23.32 (talk) 16:59, 14 October 2022 (UTC)

Er... what ?

"the credence of HEADS after being informed it is monday is 50%" is obviously right.

But how on earth does it imply that the credence of HEADS before learning what day it is has to be the same ? SB has two different relevant pieces of information in one case ("I'm awake" and "it is monday") only one in the other ("I'm awake"), why should the credence be the same before and after learning the day ?

The math clearly shows it isn't at all.

It feels like debating the plausibility of different origins of the golden tooth in Fontenelle's story (mais on commença par faire des livres, et puis on consulta l'orfèvre.) 2A01:E0A:BED:6AA0:44C1:2CAB:37F2:B5EF (talk) 22:05, 21 October 2024 (UTC)

STOP DEBATING WHAT THE ANSWER IS

Sheesh, people, that's not what this page is for. -- Jibal (talk) 23:53, 15 August 2023 (UTC)

Alternative problem in "Ambiguous question position" is not analogous

From the text in this section: "Imagine tossing a coin, if the coin comes up heads, a green ball is placed into a box, if, instead, the coin comes up tails, two red balls are placed into a box; then, a single ball is then drawn from the box" (emphasis mine). So long as only one ball is ever drawn, the probability that a green ball is drawn from the box under these conditions is 50%. For the new problem to be analogous to the original, the number of drawings would need to depend on the outcome of the coin toss as in the original problem, but there's no indication in the text that it does. In my opinion after adding this correction it's unclear that the analogy would still support the conclusion that the question is ambiguous. Baha Mk5 (talk) 18:21, 24 August 2023 (UTC)

The two statements of 'The Problem' are not equivalent.

The original statement of the problem is that Sleeping Beauty is having the experiment explained to her. At that point, before the experiment is even started, she is asked this: "When you are first awakened, to what degree ought you believe that the outcome of the coin toss is Heads?"

She asked a forward looking question about what her degree of belief should be on being first awakened, which is going to be Monday. What her degree of belief should be on Tuesday, if she is woken up on Tuesday, is not asked.

The schedule Elga created to implement his solution then substantially changes the problem description. It introduces actual questions on both Monday and Tuesday after the coin toss.

My point is that these two statements of the problem are not equivalent, and I think this is the source of a lot of the disagreement on the problem. Simon.hibbs (talk) 14:49, 14 December 2023 (UTC)

Well, it just proves that the wording is as bad as the reasoning here. The expression "when you are first awaken" makes no sense as we are explained before that there is a drug that makes you forget that you were awaken before. You don't know whether you were awake before, therefore you don't know whether you are "first awaken" or if it is your second time waking up. The only way that it could be construed as a consistent description of an experiment, is if "when you are first awaken" means "you are just awake but don't know what day it is, nor if you were awaken before. what is the likelihood of..."

If the question is to be understood as "you are first awaken, therefore it has to be monday, what is the likelihood of..." then the question is trivial and the dainty description of an elaborate experiment was for nothing, wasn't it ? 2A01:E0A:BED:6AA0:44C1:2CAB:37F2:B5EF (talk) 21:51, 21 October 2024 (UTC)

Looking for Reliable Source to this

Someone help me find the Reliable Source of this put-down to the so-called paradox. In my words but the reasoning is not originally mine:

The coin is fair. Sleeping Beauty is never awakened if Heads is tossed, and is wakened once if Tails is tossed. As before there is no hidden info and SB knows this. After being awoken SB is asked what probability is it that Heads was tossed. 0%, she says, correctly, as she flounces off in search of kinder people.

Paul Beardsell (talk) 11:32, 2 January 2024 (UTC)

It's just bad math and bad reasoning

Ok, first of all, sorry for the spelling and grammar mistakes: not a native English speaker. This sleeping beauty paradox thing seems to be a pointless discussion around faulty bayesianish demonstrations. Here is why and how. The experiment : there is one fair coin toss, resulting in H or T, and a sleeping subject of experiment, called B. If T, B is awaken on monday and left to sleep on wednesday. if H, B is awaken on monday and wednesday (there is no tuesday in this story, because both tuesday and tails begin with a t, which would throw off my notations). B knows all about the experiment, but forgets she has been awake as soon as she gets back to sleep, therefore does not know if she has been awaken before, does never know what day it is, nor if the result of the coin toss was H or T. Every time she is awake, she is asked what is the likelihood of H and the likelihood of T, knowing what she knows. What should she answer ?

Let's go for notations : uppercase are for parameters (M for monday, W for wednesday), states (A for being awake), results (H or T) ; lowercase are for the info B has on this objective reality (m if she knows it's monday and so on ; not m if she does not know that it is monday and so on) ; C is the probability function for the coin toss : C(T)=1/2 ; C(H)=1/2 ; c is for the knowledge of this function ; e is for the knowledge of the details of the experiment ; let's avoid frivolous distraction by assuming both "if A then a" and "if a then A" ; as we are talking to B, who is awake when we talk to her, there is no such thing as not a ; knowing e, not m and not w are obviously the same thing : B does not know what day it is ; L(X/y,z...) is the likelihood of the state/result/parameter X knowing the information y, z... It goes without saying that when B is awake, she has no context other than a, e, c, therefore should give the same answer in the three possible instances when she is awake : they are strictly identical from her point of view.

What we are looking for is L(H/a,e,c,not m) and L(T/a,e,c,not m). The 1/2 vs 1/2 intuition is obviously false : L(H/info)=C(H) only if the info is not relevant or linked to the result or the coin toss at all, but B knows she is awake (a) and that the experiment (e) provides a strong link between the result of the coin toss and the waking state (if H, B is awaken more often). In the frame of e, a is a crucial piece of information. There is therefore no reason to suppose L(./a,e,c)=C(.), quite the opposite. There should be a strong suspicion that L(T/a,e,c)<C(T) and L(H/a,e,c)>C(H). The 2/3 vs 1/3 intuition is quite lazy, does not use the available info and has no chance to be true : when B is awake, even if there are 3 situations when she will be, she has no reason to go for the L(M/a,c,not e) and L(W/a,c, not e) likelihoods with each of the three possibilities being equally plausible, because she knows e, and the experiment provides a strong link between the day it should be when she is awake and the result of the coin toss. Elga's "demonstration" in his 2000 article hinges solely on lousy and improper notations, where L(M and H), L(M/h) and L(H/m) are all written P(M,H) and used interchangeably, while they are different things with potentially different values.

Now for the calculus (we'll drop the notations e, c and not m from now on, for the sake of simplicity).

-first of all, we have to determine L(M/a) and L(W/a). if B knows the result of the toss was Tails, she also knows that if she is awake, it must be monday : L(M/a,t)=1 and L(W/a,t)=0 ; if B knows the result of the toss was Heads, she knows it's either monday or wednesday but does not know which, both being deemed equally likely for lack of relevant information : L(M/a,h)=L(W/a,h)=1/2 ; hence L(M/a)=C(H).L(M/a,h)+C(T).L(M/a,t)=1/2 . 1/2 + 1/2 . 1 =3/4 and L(W/a)=C(H).L(W/a,h)+C(T).L(M/a,t)=1/2 . 1/2 + 1/2 . 0 =1/4 (there we have it, we thirdists are definitely wrong).

-then we must calculate L(H/a,m), L(T/a,m), L(H/a,w) and L(T/a,w). obviously, if B knows it is wednesday, she knows the result of the toss must have been Heads, otherwise she would be sleeping : L(H/a,w)=1 and L(T/a,w)=0 ; if B knows it is monday, then she has no relevant info on the result of the coin toss : when it is monday, she is awaken anyway, whatever the result of the toss (in the context of e, m makes a irrelevant). Therefore : L(H/a,m)=C(H)=1/2 and L(T/a,m)=C(T)=1/2.

-this being done, we just have to calculate : L(H/a,not m)=L(H/a,m).L(M/a)+L(H/a,w).L(W/a)=1/2 . 3/4 + 1 . 1/4 =5/8 L(T/a,not m)=L(T/a,m).L(M/a)+L(T/a,w).L(W/a)=1/2 . 3/4 + 0 . 1/4 =3/8

So, there we are : neither 1/2 vs 1/2 nor 1/3 vs 2/3 and no paradox but an only mathematically rigourous solution 3/8 vs 5/8. If my reasoning is false, thank you for telling me where. If it isn't, I cannot fathom how philosophers have spent more than 20 years writing about this. 2A01:E0A:BED:6AA0:A5:3C67:B2E6:9DBB (talk) 19:29, 21 October 2024 (UTC)

Zuboff's answer = 1/2 or 1/3?

So what's Zuboff's opinion? Is it 1/2 or 1/3? I feel like he never gave his answer, but I might be wrong. He just says that we're all the same person, but what's his answer then?

(the objective answer, I believe, is that it depends on the question you ask. but what's Zuboff's stance?) Niepodkoloryzowany (talk) 09:51, 29 October 2024 (UTC)