Beauty quips, "I'd shut up and multiply!"

As Wei Dai has said, arguing about which probability is "right" is futile until you have fixed your decision theory and goals that will actually make use of use those probabilities to act. In most use-cases of probability theory, such issues don't come up.

In sleeping beauty, you are in a situation where such considerations do matter.

If we further specify that sleeping beauty can make a bet and (if she wins) will get the money straight away on Monday, and be allowed to spend it straight away on eating a chocolate bar, and will then be put to sleep again (if the coin came up tails), woken up on Tuesday and be given the same money again and allowed to eat another chocolate bar, then she will do best by saying that the probability of tails is 2/3.

But if we specify that the money will be put into an account (and she will only be paid one winning) that she can spend after the experiment is over, which is next week, then she will find that 1/2 is the "right" answer

In the sleeping beauty problem, whether the 2/3 or 1/2 is "right" is just a debate about words. The real issue is what kind of many-instance decision algorithm you are running.

EDIT: Another way of putting this would be to simply abandon the concept of probability altogether and use something like UDT. Probability theory doesn't work in cases where you have multiple instances of your decision algorithm running.

Add a payoff and the answer becomes clear, and it also becomes clear that the answer depends entirely on how the payoff works.

Without a payoff, this is a semantics problem revolving around the ill-defined concept of expectation and will continue to circle it endlessly.

This is one of those cases where we need to disentangle the dispute over definitions (1), forget about the notion of subjective anticipation (2), list the well-defined questions and ask which we mean.

If by the probability we mean the fraction of waking moments, the answer is 1/3.

If by the probability we mean the fraction of branches, the answer is 1/2.

1. http://lesswrong.com/lw/np/disputing_definitions/

2. http://lesswrong.com/lw/208/the_iless_eye/

Odds(monday and heads : monday and tails : tuesday and heads : tuesday and tails | wakes up)
= Odds(monday and heads : monday and tails : tuesday and heads : tuesday and tails) * Likelihood(wakes up | monday and heads : monday and tails : tuesday and heads : tuesday and tails)
= (1 : 1 : 1 : 1) * (1 : 1 : 0 : 1)
= (1 : 1 : 0 : 1)
= (1/3 : 1/3 : 0 : 1/3)

The coverage on http://en.wikipedia.org/wiki/Sleeping_Beauty_problem seems much less confused than this post.

Beauty just knows that she'll win the bet twice if tails landed. We double count for tails...

That doesn't mean your credence for heads is 1 -- it just means I added a greater penalty to the other option.

You don't need a monetary reward for this reasoning to work. It's a funny ambiguity, I think, in what 'credence' means. Intuitively, a well-calibrated person A should assign a probability of P% to X iff X happens on P% of the occasions where A assigned a P% probability to X.

If we accept this, then clearly 1/3 is correct. If we run this experiment multiple times and Beauty guessed 1/3 for heads, then we'd find heads actually came up 1/3 of the times she said "1/3". Therefore, a well-calibrated Beauty guesses "1/3".

A reasonable an idea for this and other problems that don't' seem to suffer from ugly asymptotics would simply to mechanically test it.

That is to say that it may be more efficient, requiring less brain power, to believe the results of repeated simulations. After going through the Monty Hall tree and statistics with people who can't really understand either, then end up believing the results of a simulation whose code is straightforward to read, I advocate this method--empirical verification over intuition or mathematics that are fallible (because you yourself are fallible in your understanding, not because they contain a contradiction).

For my own benefit, i'll try to explain my thinking on this problem, in my own words, because the discussions here are making my head spin. Then the rest of you can tell me whether i understand. The following is what i reasoned out before looking at neq1's explanations.

Firstly, before the experiment begins, i'd expect a 50% chance of heads and a 50% chance of tails. Simple enough.

If it lands on heads, then i wake up only once, on Monday. If it lands on tails, then i wake up once on Monday, and a second time on Tuesday.

So, upon waking with amnesia, i'd expect a 50% chance of it being my first-and-only interview on Monday. I'd expect a 25% chance of it being my first-of-two interviews on Monday, and a 25% chance of it being my second-of-two interviews on Tuesday.

And due to the amnesia, and my having no indication of what day it is, i'd basically have no new information to act on after i wake up. So my probability estimates would remain the same after waking as they were before.

So, upon waking, i'd say:

• 50% chance that the coin landed on heads, and it's currently Monday.
• 25% chance that the coin landed on tails, and it's currently Monday.
• 25% chance that the coin landed on tails, and it's currently Tuesday.

In other words, neq1's probability-tree picture turned out to most clearly match my own reasoning on the problem. Does this make sense?

* 33% chance that the coin landed on heads, and it's currently Monday.
* 33% chance that the coin landed on tails, and it's currently Monday.
* 33% chance that the coin landed on tails, and it's currently Tuesday.

Please insert a section break near the start of this post, so the whole thing doesn't show up on "NEW".

The 1/3 argument says with heads there is 1 interview, with tails there are 2 interviews, and therefore the probability of heads is 1/3. However, the argument would only hold if all 3 interview days were equally likely. That's not the case here. (on a wake up day, heads&Monday is more likely than tails&Monday, for example).

Um... why? There are the same number of heads&Monday as tails&Monday; why would heads&Monday be more likely?

So, I'm still working on this in my plodding, newbie-at-probability-math fashion.

What I took away from my exchanges with AlephNeil is that I get the clearest picture if I think in terms of a joint probability distribution, and attempt to justify mathematically each step of my building the table, as well as the operations of conditioning and marginalizing.

In the original Sleeping Beauty problem, we have three variables: x is how the coin came up {heads, tails}, y is the day of the week {monday, tuesday}, and z is whether I am asked for my credence (i.e. woken) {wake, sleep}.

P(x,y,z)=P(x)P(y|x)P(z|x,y) and unlike in the "revival" case x and y aren't clearly independent. In fact the answer very much seems to hinge on what we take the probability of it being tuesday, given that the coin came up heads.

The relevant possible outcomes are: (H,M,W) (H,T,W) (T,M,W) (T,T,W) (H,M,S) (H,T,S) (T,M,S) (T,T,S) - eight in all.

Conditioning on z=W consists of deleting the part of the table that has z=S, summing up all the remaining values, and renormalizing by dividing every cell in cell in the table by the total.

The rules for filling the table are: the values must add up to 1; the "heads" and "tails" branches must receive equal probability mass from P(x); and P(z|x,y) must reflect the experimental rules. So we must have the following:

• P(H,M,W) - see below
• P(H,T,W)=0
• P(T,M,W)=1/4
• P(T,T,W)=1/4
• P(H,M,S)=0
• P(H,T,S) - see below
• P(T,M,S)=0
• P(T,T,S)=0

The ambiguity seems to arise in allocating probability mass to the outcomes: "the coin comes up heads; it is Monday; I get woken up", and "the coin comes up heads; it is Tuesday; I do no get woken up". That is, I'm not sure what the correct conditional distribution P(y|x) should be.

The 1/2 answer corresponds to allocating all of the available 1/2 probability mass to the first of these outcomes in the joint table, saying P(y=M|x=H)=1 and P(y=T|x=H)=0. Or verbally, "it's certain that I get woken up on Monday if the coin comes up heads, and after that the experiment is over". The "not woken up" half of the table receives no probability mass at all.

The 1/3 answer corresponds to distributing that probability mass among the two outcomes, saying P(y=M|x)=P(y=T|x)=1/2. Verbally: "however the coin comes up, it could be either Monday or Tuesday". Here 1/4 of the total probability mass is in the "not woken up" half of the table and gets deleted when we condition on being woken.

(ETA: Where does the amnesia appear in this formalization? It doesn't, but neither does it need to. Its only practical consequence is to outlaw conditioning on the day, so working out the distribution P(x|z) conforms to the amnesia.)

Variation Alpha:

10 people. If heads, one of the ten is randomly selected to be revived. If tails, all ten are revived. (If you like, suppose that the ten are revived one at a time on consecutive days - but it doesn't make any difference.)

Variation Beta:

Same as Alpha except the 10 people are clones of yours, with mental state identical to your own.

Variation Gamma:

Same as Beta except the cloning is done after you fall asleep.

Variation Delta:

Same as Gamma except that the way the clones are not created all at once. Rather, successive clones are created on subsequent days by erasing one days' worth of memory of the previous clone.

It seems clear to me that in variation Alpha, 1/11 is the answer and not 1/2. And clearly variation Delta is isomorphic to the Sleeping Beauty problem (except with 10 days rather than 2). And clearly each step from Alpha to Delta doesn't change anything essential.

Right?

I have a question for those more familiar with the discussions surrounding this problem: is there anything really relevant about the sleeping/waking/amnesia story here? What if instead the experimenter just went out and asked the next random passerby on the street each time?

It seems to me that the problem could be formulated less confusingly that way. Am I missing something?

I agree with the others about worrying about the decision theory before talking about probability theory that includes indexical uncertainty, but separately I think there's an issue with your calculation.

"P(Beauty woken up at least once| heads)=P(Beauty woken up at least once | tails)=1"

Consider the case where a biased quantum coin is flipped and the people in 'heads' branches are awoken in green rooms while the 'tails' branches are awoken in red rooms.

Upon awakening, you should figure that the coin was probably biased to put you there. However, P(at least one version of you seeing this color room |heads) = P(at least one version of you seeing this color room |tails) = 1. The problem is that "at least 1" throws away information. p(I see this color|heads) != p(I see this color tails). The fact that you're there can be evidence that the 'measure' is bigger. The problem lies with this 'measure' thing, and seeing what counts for what kinds of decision problems.

The blue eyes problem is similar. Everyone knows that someone has blue eyes, and everyone knows that everyone knows that someone has blue eyes, yet "they gained no knew information because he only told them that at least one person has blue eyes!" doesn't hold.

Just as a better intuition pump, we can imagine the "really extreme" sleeping beauty problem.

Omega rolls a d20

If it comes up a "1", then beauty is woken 400 times, otherwise she is woken once only. Questions:

• Should beauty bet her $100K house in a treble-or-nothing bet against omega that the d20 came up "1"? (She will have to go and live in the house again after the experiment is over) (treble or nothing means that if the coin did come up "1", then she gets 3* her stake, otherwise she loses it).

but what if:

• Omega gives beauty one small chocolate bar, which she can either eat now, or risk in a treble-or-nothing bet against omega. Should she take the bet? Chocolate bars have to be consumed immediately, before she is sent back to sleep, and obviously each time she is woken, another set of chocolate bars is used.

The OP is correct. There are actually all the same issues here as with the Self Indication Assumption; it is wrong for the same reasons as the 1/3 probability. I predict that a great majority of those who accept SIA will also favor the probability of 1/3.

Sleeping Beauty does not sleep well. She has three dreams before awakening. The Ghost of Mathematicians Past warns her that there are two models of probability, and that adherents to each have little that is good to say about adherents to the other. The Ghost of Mathematicians Present shows her volumes of papers and articles where both 1/2 and 1/3 are "proven" to be the correct answer based on intuitive arguments. The Ghost of Mathematicians Future doesn't speak, but shows her how reliance on intuition alone leads to misery. Only strict adherence to theory can provide an answer.

Illuminated by these spirits, once she is fully awake she reasons: "I have no idea whether today is Monday or Tuesday; but it seems that if I did know, I would have no problem answering the question. For example, if I knew it was Monday, my credence that the coin landed heads could only be 1/2. On the other hand, if I knew it was Tuesday, my credence would have to be 0. But on the gripping hand, these two incontrovertible truths can help me answer as my night visitors suggested. There is a theorem in probability, called the Theorem of Total Probability, that says the probability for event A is equal to the probability of the sum of the events (A intersect B(i)), where B(i) partitions the entire event space.

"Today has to be either Monday or Tuesday, and it can't be both, so these two days represent such a partition. Since I want to avoid making any assumptions as long as I can, let me say that the probability that today is Monday is X, and the probability that it is Tuesday is (1-X). Now I can use this Theorem to state, unequivocally, that my credence that the coin landed heads is P(heads)=(1/2)X+0(1-X)=X/2.

"But I know that it is possible that today is Tuesday; even a Bayesian has to admit that X<1. So I know that 1/2 cannot be correct; the answer has to be less than that. A Frequentist would say that X=2/3 because, if this experiment were repeated many times, two out of every three interviews would take place on Monday. And while a Bayesian could, in theory, choose any value that is less than 1, it is a violation of Occam's Razor to assume there is a factor present that would make X different than 2/3. So, it seems my answer must be 1/3.

Proof that neq1 is wrong:

Let H be the event that heads was flipped in this experiment instance. We're going to let Beauty experience a waking now. Let M be the event that the waking is on Monday. Let B be the information that Beauty (knowing the experiment design) has upon waking. Let h=P(H|B), and let m=P(M|B).

We wish to discover the true values of h and m. Clearly in the context of someone being asked about the expected outcome of the experiment, P(H)=1/2, but h may (or may not) differ from 1/2.

• Fact 1: P(H|M,B)=P(H)=1/2

• Fact 2: P(H|~M,B)=0 (by ~M I mean the complement of M, i.e. that it's not Monday)

Given the above two facts, we know enough to solve for h and m.

lemma 1:

• P(~H|B)=P(M,B)P(~H|M,B)+P(~M)P(~H|~M,B) ; probability axiom
• (1-h)=m(1/2)+(1-m)(1) ; by facts 1-2 and above axiom
• 1-(h)=1-(m/2) ; above simplified
• h=m/2

lemma 2:

• P(H|B)=P(M)P(H|M,B)+P(~M)(P(H|~M,B) ; probability axiom
• h=m(1/2)+(1-m)(0) ; facts 1-2 and above
• h=m/2 ; simplified
• m=2h

(oops, that turned out to be redundant; not surprising since I'm using in lemma 2 the variants p(~X)=1-P(X) from the same facts 1+2).

P(H|B) is a weighted average of the probability for heads given Monday (1/2) and given Tuesday (0). It turns out that, according to thirders, it's more likely that it's Monday (m=2h=2/3).

The thirder argument is that m=2/3 (that is, 2 out of 3 wakings on average are on Monday). The halfer argument that h=1/2 implies that m=1; that is, that Beauty is certain that it's Monday (but this is obviously stupid of her).

I was originally sympathetic to neq1's argument that B is merely "1 or more wakings occur" and that P(H|1 or more wakings occur)=P(H)=1/2, since 1 or more wakings always occur, no matter whether H or ~H. But B is better characterized as "Beauty has just been woken, not knowing whether it's the first or second waking, but knowing the experiment design".

I would like to strengthen this argument to prove that m=2/3.

Just an observation: I've mostly ignored this discussion, but it appears to have generated a lot of meaningful debate about the very fundamental epistemic issues at play (though a lot of unproductive debate as well). No consensus on which position is idiotic has apparently arisen.

With that in mind, surely this article should be rated above 1? Are the upvotes being canceled by downvotes, or are people just not voting it either way? Why isn't this rated higher?

After tinkering with a solution, and debating with myself how or whether to try it again here, I decided to post a definitive counter-argument to neq1's article as a comment. It starts with the correct probability tree, which has (at least) five outcomes, not three. But I'll use the unknown Q for one probability in it:

••••••• Monday---1---Waken; Pr(observe Heads and Monday)=Q/2 ••••••••••/ ••••••••Q •••••••/ ••• Heads •••••/••\••••••••••••1---Sleep; Pr(sleep thru Heads and Tuesday)=(1-Q)/2 ••••/•••1-Q•••••••/ ••1/2••••\••••••••/ ••/•••• Tuesday--0---Waken; Pr(observe Heads and Tuesday)=0 •/ + •\ ••\•••• Monday---1---Waken; Pr(observe Tails and Monday)=1/4 ••1/2••••/ ••••\••1/2 •••••\••/ ••• Tails •••••••\ •••••••1/2 ••••••••••\ ••••••• Tuesday--1---Waken; Pr(observe Tails and Tuesday)=1/4

What halfers refuse to recognize, is that whether Beauty is awakened in any specific circumstance is a decision that is part of the process. It is based on the other two random variables, after both – repeat, both – have been determined. The event “Heads and Tuesday” is an event that exists in the sample space, and the decision to not awaken her is made only after that event has occurred. Halfers think they have to force that event into non-existence by making Q=1, when all the experiment requires is that the probability Beauty will observe it is zero. This is the point one thirder argument utilizes, that of Radford Neal’s companion Prince who is always awakened but only asked if Beauty is awakened.

In fact, there is no reason why the probability that it is Monday, given Heads, should be any different than the probability it is Monday, given Tails. So, with Q=1/2, we get that Pr(observe heads)=1/4, Pr(observe anything)=3/4, so Pr(Heads|observe anything)=1/3. QED.

Neq1’s arguments that the thirder positions are wrong are all examples of circular reasoning. He makes some assumption equivalent to saying the answer is 1/2, and from that proves the answer is 1/2. For example, when he uses “Beauty woken up at least once” as a condition, all his terms are also conditioned on the fact that the rules of the experiment were followed. So when he inserts the completely unconditional “Pr(Heads)=1/2” on the right-hand side of the equation, he really should use Pr(heads|rules followed), which is the unknown we are trying to find. It is then unsurprising that he gets the number he inserted, especially if you consider what using a probability-one event as a condition in Bayes’ Rule means.

Where neq1 claims that Nick Bostrom’s argument is wrong in “Disclosure Process 1,” I suggest he go back and use the values from his probability tree. Her credence of heads is (1/2)/(1/2+1/2/1,000,000). In the second process, it is either (1/2)/(1/2+1/2/7,000,000) of (1/2)/(1/2+1/2/1,000,000,000,000), depending on what “specific day” means.

The whole anthropics debate is over things that you have taken as assumptions e.g. whether waking up is identical evidence to merely knowing that you wake at least once, whether the three days are equally likely

Your update doesn't solve the problem. It's a semantic issue about what credence we are being asked. If we are being asked about the probability of our coin flip associated with this iteration of the experiment, then the answer is 1/2. If we are being asked about the probability of the coin flip associated with this particular awakening, then it must be 1/3.

You say that you must use cell counts of 500,250,250, but the fact is that if you repeat the experiment 1000 times, sleeping beauty will be awoken 1500 times, not 1000. So what are you doing with the other 500 awakenings? I would say you are implicitly ignoring them, as you do when you say "we only accept her last decision" in the bet scenario. The reformulations of this using different people, rather than the same person being awoken multiple times don't seem to cause as much trouble.

The semantic issue here is reminiscent of arguments I've seen over the Monty Hall problem when it is misstated so that Monty's algorithm is not clear. People who assume what he usually does on the show come up with 2/3, and people who don't make any assumptions come up with 1/2 (as do most of the people who simply don't understand restricted choice).

It doesn't make sense to assert that probability of Tuesday is 1/4 (in the sense that it'd take a really bad model to give this answer). Monday and Tuesday of the "tails" case shouldn't be distinct elements of the sample space. What happens when you've observed that "it's not Tuesday", and the next day it's Tuesday? Have you encountered an event of zero probability? This is exactly the same reason why the solution of 1/3 can't be backed up by a reasonable model.

In the classical possible worlds model, you've got two worlds for each outcome of the coin flip, with probabilities 1/2 apiece, and so (Tuesday, tails) is the same event as (Monday, tails), weighing probability of 1/2. Thus, for example, probability that we are in the possible world where Monday can be observed, given that Tuesday can be observed, is 1, but it doesn't make sense to ask "What is probability of it being Tuesday?", unless this question is interpreted as "What is probability of us being in the possible world where it's possible to observe Tuesday?", in which case the question "What is the probability of it being Monday, given that it's Tuesday?", interpreted the same way, has "100%" as the answer.

By the way, you may have noticed that the wiki has an article on the Sleeping Beauty problem. Also, it's been referenced before Sleeping beauty gets counterfacutally mugged in a top-level post, and it was mentioned in the context of a general solution in The I-Less Eye. And the comments on How many LHC failures is too many are relevant to the problem too.

Did you even search to see if someone had done a post on this topic before?

There is a difference between P("Heads came up") and P("Heads came up" given that "I was just woken up"). Since you will be woken up (memory-less) multiple times if tails came up, the fact that you are just getting woken up gives you information and increases the probability that tails came up.

Let's consider P(H | JustWoken) = P(H and Monday | JustWoken) + P(H and Tuesday | JustWoken) Because I have no information about the scientist's behavior (when he chooses to ask the question), I have to assign equal probabilities (one third) to P(H and Monday | JustWoken), P(T and Monday | JustWoken) and P(T and Tuesday | JustWoken). And it's impossible to be woken up on Tuesday if Heads came up, so P(H and Tuesday | JustWoken) = 0. In result, P(H | JustWoken) = 1/3.

If anyone doubts that, we could set up a computer simulation (you write the scientist and coin code and I write code for the beauty answering the question) and we bet. But I would require an experimental condition, stating that the scientist will ask the beauty the question every time she wakes up. Under those conditions, a beauty which always bets that "tails came up" any time she gets woken up will win 2/3 of the time. If we could not agree to those conditions (getting interviewed by the scientist on every occasion), the bet would be broken because you know what answer I will give and you have information that I don't have (strategy for when to interview).

Robert Wiblin - Thoughts on the sleeping beauty problem

http://robertwiblin.wordpress.com/2010/03/26/news-flash-multiverse-theory-proven-right/

this is a probability tree corresponding to an arbitrary wake up day

Huh? If tails, then Beauty is (always) woken on Monday. Why do you have probability=1/2 there?

(likewise for Tuesday)

If we want to replicate the situation 1000 times, we shouldn't end up with 1500 observations. The correct way to replicate the awakening decision is to use the probability tree I included above. You'd end up with expected cell counts of 500, 250, 250, instead of 500, 500, 500.

Beauty ends up with 1500 observations on average (maybe as few as 1000 or as many as 2000). Imagine a sequence of Beauty-observations in (H|TT)^1000 , where by r^1000 I mean 1000 repetitions of r. This string is from 1000-2000 letters long.

If you consider the scenario from a non-amnesiac perspective, then you can consider the TT - the two forgetful-Beauty observations in the tails case, as a single event, which is indeed equally likely to the alternative, H. In fact, the shortest possible coding to describe one of the beauty-observation strings is just a 1000-bit string where the nth bit indicates the result of the nth coin flip.

But what are you thinking when you say there are two "cells" each with p=1/4 (count 250 out of 1000)? What, exactly, would happen to 250 times on average? Certainly we expect Beauty waking on Mon. with it being tails 500 times (and also 500 times on Tue.).

PhilGoetz writes:

I had expected that people would read posts and comments by other people, and take special note of comments by people who had a prior history of being right, and thereby improve their own accuracy.

I would like to do this. However, it's time consuming to sort through people's posts and see what they think. (You have to read carefully, because they may be critiquing a particular argument rather than the value 1/2 or 1/3 presented in the parent.) Would people mind stating their position on the Sleeping Beauty problem with a single sentence explaining the core detail of the argument that persuades them?

I don't follow your latest argument against thirders. You claim that the denominator

#(heads & monday) + #(tails & monday) + #(tails & tuesday)

counts events that are not mutually exclusive. I don't see this. They look mutually exclusive to me-- heads is exclusive of tails, and monday is exclusive of tuesday, Could you elaborate this argument? Where does exclusivity fail? Are you saying tails&monday is not distinct from tails&tuesday, or all three overlap, or something else?

You also assert that the denominator is not determined by n. (I assume by n you mean replications of the SB experiment, where each replication has a randomly varying number of awakenings. That's true in a way-- particular values that you will see in particular replications will vary, because the denominator is a random variable with a definite distribution (Bernoulli, in fact). But that's not a problem when computing expected values for random processes in general; they often have perfectly definite and easily computed expected values. Are you arguing that this makes that ratio undefined, or problematic in some way? I can tell easily what this ratio converges to, but you won't like it.

Disclosure process 1: regardless of the result of the coin toss she will be informed it's Monday on Monday with probability 1

Under disclosure process 1, her credence of heads on Monday is still 1/2.

SB would start out with P(tails) = 1,000,001/1,000,002 and on being informed that it is monday would update:

P(tails | told monday) / P(heads | told monday)
= P(tails) / P(heads) * P(told monday | tails) / P(told monday | tails)
= (1,000,001/1,000,002)/(1/1,000,002) * (1 / 1,000,001) /(1) 
= 1

The initial strong belief in tails is cancelled by the strong evidence of being told that it is monday, which only happens in one of many wakings if the coin landed tails.

Disclosure process 2: if heads she'll be woken up and informed that it's Monday. If tails, she'll be woken up on Monday and one million subsequent days, and only be told the specific day on one randomly selected day.

Under disclosure process 2, if she's informed it's Monday, her credence of heads is 1,000,001/1,000,002. However, this is not implausible at all. It's correct. This statement is misleading: "It is, after all, rather gutsy to have credence 0.999999% in the proposition that an unobserved fair coin will fall heads." Beauty isn't predicting what will happen on the flip of a coin, she's predicting what did happen after receiving strong evidence that it's heads.

P(tails | told what day AND it is monday) / P(heads | told what day AND it is monday)
= P(tails) / P(heads) * P(told what day | tails) * P(it is monday | tails) / (P(told what day | heads) * P(it is monday | heads))
= (1,000,001/1,000,002)/(1/1,000,002) * ((1 / 1,000,001)*(1/1,000,001)) /(1 * 1) 
= 1 / 1,000,001

The initial strong belief in tails is canceled by the strong evidence of being told what day it is, and then updated further to strong belief in heads by the strong evidence of it being Monday.

The next program works well:

R=Random(0,1) If R=0 SAY "P(R=0)=1/2" Elseif SAY "P(R=0)=1/2": SAY "P(R=0)=1/2" Endif

The next doesn't:

R=Random(0,1) If R=0 SAY "P(R=0)=1/3" Elseif SAY "P(R=0)=1/3": SAY "P(R=0)=1/3" Endif

Run it many times and you will clearly see, that the first program will be right, since it will be about the same number of cases when R will be 0 and the other cases when R will be 1.

Just what the first program keep saying.

I'm not convinced that 1/2 is the right answer. I actually started out thinking it was obviously 1/2, and then switched to 1/3 after thinking about it for a while (I had thought of Bostrom's variant (without the disclosure bit) before I got to that part).

Let's say we're doing the Extreme version, no disclosure. You're Sleeping Beauty, you just woke up, that's all the new information you have. You know that there are 1,000,001 different ways this could have happened. It seems clear that you should assign tails a probability of 1,000,000/1,000,001.

Now I'll go think about this some more and probably change my mind a few more times.

I agree with the author of this article. After having done a lot of research on the Sleeping Beauty Problem as it was the topic of my bachelor's thesis (philosophy), I came to the conclusion that anthropic reasoning is wrong in the Sleeping Beauty Problem. I will explain my argument (shortly) below:

The principle that Elga uses in his first paper to validate his argument for 1/3 is an anthropic principle he calls the Principle of Indifference:

"Equal probabilities should be assigned to any collection of indistinguishable, mutually exclusive and exhaustive events."

The Principle of Indifference is in fact a more restricted version of the Self-Indication Assumption:

"All other things equal, an observer should reason as if they are randomly selected from the set of all possible observers."

Both principles are to be accepted a priori as they can not be attributed to empirical considerations. They are therefore vulnerable to counterarguments...

The counterargument:

Suppose that the original experiment is modified a little:

If the outcome of the coin flip is Heads, they wake Beauty up at exactly 8:00. If the outcome of the first coin flip is Tails, the reasearchers flip another coin. If it lands Heads they wake Beauty at 7:00, if Tails at 9:00. That means that when Beauty wakes up she can be in one of 5 situations:

Heads and Monday 8:00

Tails and Monday 7:00

Tails and Monday 9:00

Tails and Tuesday 7:00

Tails and Tuesday 9:00

Again, these situations are mutually exclusive, indistinguishable and exhaustive. Hence thirders are forced to conclude that P(Heads) = 1/5.

Thirders might object that Beauty's total credence in the Tails-world would still have to equal 2/3, as Beauty is awakened twice as many times in the Tails-world as in the Heads-world. They are then forced to explain why temporal uncertainty regarding an awakening (Monday or Tuesday) is different from temporal uncertainty regarding the time (7:00 or 9:00 o’clock). Both classify as temporal uncertainties within the same possible world, what could possibly set them apart?

An explanation could be that Beauty is only is asked for her credence in Heads during an awakening event, regardless of the time, and that such an event occurs twice in the Tails-world. That is, out of the 4 possible observer-moments in the Tails-world there are only two in which she is interviewed. That means that simply the fact that she is asked the same question twice is reason enough for thirders to distribute their credence, and it is no longer about the number of observer moments. So if she would be asked the same question a million times then her credence in Heads would drop to 1/1000001!

We can magnify the absurdity of this reasoning by imagining a modified version of the Sleeping Beauty Problem in which a coin is tossed that always lands on Tails. Again, she is awakened one million times and given an amnesia-inducing potion after each awakening. Thirder logic would lead to Beauty’s credence in Tails being 1/1000000, as there are one million observer-moments where she is asked for her credence within the only possible world; the Tails-world. To recapitulate: Beauty is certain that she lives in a world where a coin lands Tails, but due to the fact that she knows that she will answer the same question a million times her answer is 1/1000000. This would be tantamount to saying that Mt. Everest is only 1m high when knowing it will be asked 8848 times! It is very hard to see how amnesia could have such an effect on rationality.

Conclusion:

The thirder argument is false. The fact that there are multiple possible observer-moments within a possible world does not justify dividing your credences equally among these observer-moments, as this leads to absurd consequences. The anthropic reasoning exhibited by the Principle of Indifference and the Self-Indication Assumption cannot be applied to the Sleeping Beauty Problem and I seriously doubt if it can be applied to other cases...

If Sleeping Beauty doesn't know what day it is, what could possibly motivate her to say that the probability of heads is something other than 50%? I mean, she knows nothing about the coin except that it's round and shiny, and the metal costs more than the coin does.

Unless I misunderstood, this problem is smoke and mirrors.

I updated the post one more time. I think this time I more effectively explain where the thirder logic fails. Correct me if I'm wrong...

I updated the post. Thanks to the many interesting comments, I think I am now better able to describe why the 1/3 solution is wrong.

And to be clear, the main point of the post isn't to show that 1/2 is right, but to make the observation about how easy it is to be confident in the wrong answer when it comes to probability problems.