Talk:Sleeping Beauty problem

message: 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)
thread_title:  Error in Introduction 

message: 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)
thread_title: Anthropomorphic Principle

message: 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)
thread_title: 100:1 probability

message: The sleeping beauty paradox was originally formulated in unpublished work by Arnold Zuboff (a lecturer at my university)as Elga acknowledges in his paper - which is available online, the reference is at the bottom of the entry.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)
thread_title: Zuboff

message: 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)
replies: text: "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)
thread_title: re-formulation

message: 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)
thread_title: by the Pseudonymous 81.152.235.34 - now registered as [[User:BernardSumption|BernardSumption]]

message: 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)
thread_title:  Is it a frivolous article? 

message: 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)
replies: text: 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)
thread_title:  What is ''credence''? 

message: 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)
replies: text: Make a redirect. - Rainwarrior 04:19, 21 June 2006 (UTC)
replies: text: Thanks PJTraill 00:55, 6 November 2006 (UTC)
thread_title:  Duplicate 

message: 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)
replies: text: 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)
thread_title:  Example 

message: 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)
replies: text: 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)
thread_title:  New paragraph 

message: "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)
thread_title:  Dispute 1/3 answer 

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

message: 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)
thread_title:  False consensus 

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

text: 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)
thread_title:  Bernoulli trial 

message: 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)
thread_title:  Feb. 2013 corrections 

message: 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)
thread_title:  Missing problem statement? 

message: 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)
replies: text: 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)
thread_title:  Jargon 

message: 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)
replies: text: 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)
replies: text: 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)
thread_title:  Computer Program Analogy 

message: Proposals for fixing some problems with the article:
#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 -> ...
#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.
#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.
#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)
thread_title:  Suggestions for rewrite 

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

message: 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)
thread_title:  Halfer position needs clarity 

message: 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)
thread_title:  Lack of definition 

message: 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)
replies: text: 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)
thread_title:  Bad information in "Thirder" 

message: :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)
replies: text: 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)
thread_title:  Simple probability problem, if correctly defined 

message: 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)
thread_title:  Sailor’s Problem is Non-Equivalent 

message: 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.
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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)
thread_title:  Solutions 

message: ""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)
thread_title:  How come the problem does not state "now that it is Wednesday ..."? 

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

message: 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)
replies: text: I will attempt editing. 108.30.23.32 (talk) 16:59, 14 October 2022 (UTC)
thread_title:  Editing needed  

message: Sheesh, people, that's not what this page is for. -- Jibal (talk) 23:53, 15 August 2023 (UTC)
thread_title:  STOP DEBATING WHAT THE ANSWER IS 

message: 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)
thread_title:  Alternative problem in "Ambiguous question position" is not analogous 

message: 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)
thread_title:  The two statements of 'The Problem' are not equivalent. 

message: 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)
thread_title:  Looking for Reliable Source to this