The Sleeping Beauty Problem: Any halfers here?

[Dale]

Why is the ratio of number of days awake the same as the probability of being awake?

It isn't. The ratio of the number of days awake is the same as the ratio of the probabilities of being awake.

You said you're assuming P(A|H)/P(A|T) = 1/2. That sounds plausible, but I think it requires proof.

It is given in the problem!

because P(A|T) = 1. (If the coin toss is tails, then she's going to be awake every morning).

That depends on the sample space you are considering. If the "now" in "what is your credence now" is taken to be a day then yes the frequency is 2 out of 2 days for probability 1. If "now" is one hour then the frequency is 2 out of 48 hours for probability 1/24. If "now" is one minute then the frequency is 2 out of 48*60 ...

In all of those cases, to actually calculate P(A|T) requires assuming information that is not given in the problem. But the ratio P(A|H)/P(A|T) is the same in all those cases and is what is directly given in the problem.

 

[JeffJo]

If the question said "The experiment is run. From those days on which Sleeping Beauty is awakened, a day is picked at random..." this would favor the "halfer" viewpoint.

I disagree that you should include "was awakened."

Say that, on Sunday night, Beauty takes an imaginary coin out of her purse. Please note that this is consistent with the OP, since Beauty is free to imagine anything she wants to.

This imaginary coin says "Monday" on one side, and "Tuesday" on the other. She imaginary-flips it, and sets it on her nightstand, thereby picking a random day. She can't see what side is face up because, after all, it is just an imaginary coin. But nobody can move it, or change it, for the same reason.

On any of the two days, at 8AM (when I'll assume Beauty is to be wakened), there is a 1/4 chance that the coin landed heads, and the face-up side of the imaginary coin correctly names the current day. There is a 1/4 chance that the coin landed heads, and the face-up side of the imaginary coin does not correctly name the current day. And there are two similar outcomes, and probabilities, for Tails. These probabilities apply whether or not Beauty is actually wakened, so they are prior probabilities.

But if Beauty is actually wakened, one of the first two possibilities is eliminated. It doesn't matter if Beauty knows which, since they are symmetric. This constitutes "new information" that allows an update, and the posterior probability of the set {(H,Correct),(H,Incorrect)} is 1/3.

Now, it shouldn't be necessary to resort to the ploy of using the imaginary coin, because the selection of a random day is not dependent Beauty's ability to observe that day. Tuesday still happens after heads. Tuesday happens in conjunction with Heads with the same probability it happens in conjunction with Tails, and that Monday happens in conjunction with either single result.

You need to define what probability space you are using in your reasoning and what events that space contains.

I have. You just refuse to recognize the outcomes for what they are, because you confuse an occurrence with Beauty's ability to observe it when it happens.

At 8AM, the sample space is {(H,Mon),(H,Tue),(T,Mon),(T,Tue)}. This is different from the sample space on Sunday, since it refers to one day and Sunday's refers to two. Each has a prior probability of 1/4. Waking Beauty is a method of sampling from this space, and it eliminates (H,Tue).

The term "frame of reference" has a meaning in physics, but it doesn't have a specific meaning in the theory of probability.

Neither does "outcome," unless you apply a frame of reference. By that I mean that the only definition "outcome" has is "a possible result of your experiment." Your frame of reference defines how you quantify the random variables, and there is no definition of how one should do that. But it is not the same thing as observing a result.

A different model is the probability space T consisting of the set of events {A1,B,C} with corresponding probabilites {1/2, 1/4, 1/4}.

In what warped world is the prior probability of waking up on Monday, after Heads, twice as likely as waking up on Monday, after tails? If you identify a day in your probability space, the set of possibilities has to be all that can occur. Again, if you want to debate this, address the problem where she is wakened on (H,Tue) but taken to Disneyworld instead of being interviewed. Then, her confidence of Heads, in an interview, is clearly 1/3. But you do need to use my probability space.

That model uses the concept that the experiment is run and Beauty's situation is selected only from those situations where Beauty was awake in the particular run of the experiment.

Since I have [proven that this model gets the wrong answer, without having to resort to either of these debated models, we can conclude that this one is incorrect. Whether or not you agree with the models I described above, or even we we used them, I have shown that the answer is 1/3.

The "thirder model" It is a reasonable opinion, but it can't be proven that the information given in the problem implies that it is the only model for the process "Sleeping Beauty is awakened in some situation".

There is no single correct model, so this is a non sequitur. But the correct answer can be, and has been, proven. And there can only be one correct answer.

 

[stevendaryl]

It isn't. The ratio of the number of days awake is the same as the ratio of the probabilities of being awake.

That's what I meant.

It is given in the problem!

No, it isn't. The probabilities in the problem can be summarized as:

P(H) = P(T) = 1/2 (it's a fair coin)
P(Awake|H & Monday) = 1 (She's awake Monday regardless of the coin toss)
P(Awake|H & Tuesday) = 0 (She's asleep Tuesday if the coin toss was heads)
P(Awake|T) = 1 (She's awake if the coin toss is heads, regardless of the day)
P(Monday) + P(Tuesday) = 1 (We're only considering those two possible days)

The claim that P(Awake | H)/P(Awake | T) = 1/2 may or may not be derivable from those facts, but it certainly was not stated explicitly.

 

[Dale]

So the assumption that P(A|H)/P(A|T) = 1/2 isn't sufficient to determine P(H|A).

The only additional thing you need is that the coin is fair, which is also given.

 

[stevendaryl]

No, it isn't. The probabilities in the problem can be summarized as:

P(H) = P(T) = 1/2 (it's a fair coin)
P(Awake|H & Monday) = 1 (She's awake Monday regardless of the coin toss)
P(Awake|H & Tuesday) = 0 (She's asleep Tuesday if the coin toss was heads)
P(Awake|T) = 1 (She's awake if the coin toss is heads, regardless of the day)
P(Monday) + P(Tuesday) = 1 (We're only considering those two possible days)

The claim that P(Awake | H)/P(Awake | T) = 1/2 may or may not be derivable from those facts, but it certainly was not stated explicitly.

It's clear that the claim P(Awake|H)/P(Awake|T) = 1/2 is not derivable from the five probabilities above. So that means that if it's implicit in the problem statement, it must be an additional assumption. But what is the nature of that assumption? What is left out from the 5 claims above, and what general principle is being invoked?

I think what's left out is the fact that Monday inevitably is followed by Tuesday, and Tuesday is inevitably preceded by Monday. This implies that the relative frequencies of Monday and Tuesday are equal, and if we assume relative frequency is the same as subjective probability, then

P(Monday)= P(Tuesday) = 1/2

This would allow us to derive Dale's claim. From that and the fair coin assumption, we can uniquely determine all the probabilities:

• Let x = P(H & Monday)
• Let y = P(H & Tuesday)
• Let z = P(T & Monday)
• Let w = P(T & Tuesday)

Awake is definable in terms of the day and the coin result: Awake = Monday or T.

Constraints:

1. P(H) = 1/2, so x+y = 1/2
2. P(Monday) = 1/2, so x+z = 1/2
3. P(T) = 1/2, so z+w = 1/2
4. P(Tuesday) = 1/2, so z+w = 1/2

These 4 constraints uniquely determine that x = y = z = w = 1/4.

From this, it follows that P(Awake) = P(Monday & H) + P(Monday & T) + P(Tuesday & T) = 3/4
P(H | Awake) = P(H & Awake)/P(Awake) = P(H & Monday)/P(Awake) = (1/4)/(3/4) = 1/3

 

[Stephen Tashi]

The only additional thing you need is that the coin is fair, which is also given.

If that were true the "halfer" method for selecting which situation applies today (described in post #412) wouldn't give the answer P(H|A) = 1/2.

 

[Dale]

If that were true the "halfer" method for selecting which situation applies today (described in post #412) wouldn't give the answer P(H|A) = 1/2.

A wrong method gives a wrong answer. If the prior odds are 1:1 and the Bayes factor is 1/2, then the posterior probability is 1/3. A method that gives a different answer is flat out wrong.

 

[Stephen Tashi]

This would allow us to derive Dale's claim. From that and the fair coin assumption, we can uniquely determine all the probabilities.

A method that gives a different answer is flat out wrong.

"Thirders" consider the Sleeping Beauty problem to be completely analogous to a particular "balls-in-urns" problem. (I think this is why many "thirders" are absolutely positive that they have obtained the correct answer using only information in the problem.) Urn H contains two balls, one amber and one sienna. Urn T contains two balls, both amber. A fair coin is flipped and if lands heads, urn H is selected. If it lands tails then urn T is selected. A ball is drawn at random from the selected urn. Given the color of the ball is amber, what is the probability that urn H was selected.? The answer 1/3 is not controversial.

The distinction between the Sleeping Beauty problem and the above balls-in-urns problem is that the information that "a ball is drawn from the selected urn" tells us that this is the first time a ball is drawn from the selected urn. In the Sleeping Beauty "Sleeping Beauty awakens" does not tell us that this is the first time she has awakened. One may take the philosophical view that since we don't know whether she has awakened before, we may assume the awakening is the first time she has awakened, but this is not given in the problem.

A version of the balls-in-urns problem analogous to the Sleeping Beauty Problem would say that after the urn is selected, both balls are drawn from the urn. If one of the balls is amber, what is the probability that urn H was selected? The information "one of the balls is amber" doesn't tell us anything because one of the balls is always amber.

I'd like to see those "thirders" who can solve the first balls-in-urns problem try their hand at solving the second one. Or perhaps they can pose a different balls-in-urns version of the Sleeping Beauty Problem that is more to their liking.

 

[Dale]

I'd like to see those "thirders" who can solve the first balls-in-urns problem try their hand at solving the second one.

I would like to see people stick to the problem at hand.

This is a straightforward Bayesian probability problem. You have prior odds and a Bayes factor, both of which we agree on according your previous posts. So the posterior odds are given by their product. Anything else is wrong, regardless of urns or any other irrelevant distraction.

 

[Stephen Tashi]

I would like to see people stick to the problem at hand.

I'd like to see people define the problem at hand to match the statement of the Sleeping Beauty Problem.

 

[Dale]

I'd like to see people define the problem at hand to match the statement of the Sleeping Beauty Problem.

Done. See above.

The problem is that you want the specification to include lots of other things that are not given and that are not necessary. With those two pieces of information (prior odds and Bayes factor) the answer is determined. You want all sorts of other things specified that don't matter and are not given.

 

[Stephen Tashi]

Done. See above.

The problem is that you want the specification to include lots of other things that are not given and that are not necessary. With those two pieces of information (prior odds and Bayes factor) the answer is determined. You want all sorts of other things specified that don't matter and are not given.

You are solving your own interpretation of the Sleeping Beauty problem and I agree that your solution to your own interpretation is correct. However, your interpretation of the Sleeping Beauty Problem is controversial - it's controversial among academics who are familiar with Bayesian analysis. If you could demonstrate that your interpretation of the problem is the only correct interpretation, you would have a publishable result.

 

[Dale]

However, your interpretation of the Sleeping Beauty Problem is controversial

Whether or not the Earth is flat is also controversial. The mere fact of a controversy doesn't mean much.

What does not seem controversial between you and me is the prior odds and the Bayes factor. Given agreement on those two pieces, then the posterior probability is unambiguously 1/3 by Bayes rule.

 

[JeffJo]

I think what's left out is the fact that Monday inevitably is followed by Tuesday, and Tuesday is inevitably preceded by Monday.

And I think (make that "know") that this ignores the facts that Monday is still a different day than Tuesday, Tuesday is still a different day than Monday, and Beauty's probability space for the question she is asked can only deal with one day.

Each time Beauty is awake, she is concerned with only one day - "today." The probability space she needs is not the same as the one she may have formed on Sunday, that includes both days as a single outcome. To her, "today" is the only thing that matters.

"Thirders" consider the Sleeping Beauty problem to be completely analogous to a particular "balls-in-urns" problem.

And I believe I have shown, in three different ways, that it is:

Four Beauties: Each is undergoing an experiment that is either identical to, or functionally equivalent to, the OP. But now, each can consider the card of the missing Beauty (reading (H,Mon), (H,Tue), (T,Mon), or (T,Tue)) to be the ball from the urn. There is a 1/3 chance that is has the same coin result as her card. Always.

Taken to Disneyworld: the bogus point (that it matters how, or if, she would know it is (H,Tue) when she has positive evidence that it isn't) that Beauty cannot observe (H,Tue) is removed. Please, PLEASE, PLEASE note that I am not asking for her confidence that (H,Tue) will/did happen. Just that it isn't happening now. You are confusing these two very different occurrences.

Imaginary coin (or you could use a real one): The extra coin identifies one day as a "match" and one as a "mismatch." Without caring which comes first. Whan you also consider the OP coin, this makes for an exact analogy to four balls from an urn.

A version of the balls-in-urns problem analogous to the Sleeping Beauty Problem would say that after the urn is selected, both balls are drawn from the urn. If one of the balls is amber, what is the probability that urn H was selected?

No, this isn't analogous.

To see why, imagine (well, this is equivalent to the "imaginary coins" version) that there are two different lab techs running the experiment. Maude works on Mondays, and Tony works on Tuesdays. Both remain hidden from Beauty during the experiment, so she can't deduce the day. Neither knows the result of the coin flip until arriving at work. Your version here is equivalent to an absurd scenario where the calendar miraculously changes from saying "Tuesday" to saying "Monday" if Tony looks at the coin and sees "Heads."

Tuesday still happens, whether the coin lands on Heads or Tails.

The analogous version would still look at only the one ball, and not ask the question if the ball is sienna. How this "not asking " is accomplished - Beauty isn't wakened, or the interview is skipped - is completely irrelevant.

 

[PeterDonis]

To me, it seems that we can reason:

• If the coin toss is tails, then there is nothing observably different about Monday and Tuesday for Sleeping Beauty. So P(Tails & Monday) = P(Tails & Tuesday)
• If the day is Monday, then there is nothing observably different between heads and tails (because the difference only shows up on Tuesday). So P(Tails & Monday) = P(Heads & Monday)

The above assumptions imply that all three are equally likely.

A similar argument appeared in the Wikipedia article linked to in the OP (IIRC), and I responded to it many, many posts ago, but I'll briefly recap my response.

To me, the first bullet of yours above is an argument for P(Monday|Tails) = P(Tuesday|Tails). Note that these are conditional probabilities, not probabilities of conjunctions. I don't think the premise of the bullet justifies a claim about conjunctions, only about conditionals.

Similarly, the second bullet of yours is an argument for P(Tails|Monday) = P(Heads|Monday); once again, conditionals, not conjunctions.

But if all we have is the above conditionals, we can't get to the conclusion, because that would require P(Monday|Tails) = P(Tails|Monday), and we don't have an argument for that. The only argument I can see for it would be to numerically evaluate the conditionals; since we are assuming them to be mutually exclusive and exhaustive in each case, we have P(Monday|Tails) = P(Tuesday|Tails) = 1/2, and P(Tails|Monday) = P(Heads|Monday) = 1/2. But the argument purports to show that we have P(Heads & Monday) = 1/3. So something appears to be wrong with this argument.

 

[Stephen Tashi]

Whether or not the Earth is flat is also controversial. The mere fact of a controversy doesn't mean much.

There is a distinction between the events "a controversy" and "a controversy given it is a controversy among many experts".

What does not seem controversial between you and me is the prior odds and the Bayes factor. Given agreement on those two pieces, then the posterior probability is unambiguously 1/3 by Bayes rule.

That's essentially correct.

I infer that the events in your prior probability space are the 4 situations that can arise in the experiment and that you assign each a probability 1/4.

I don't agree that your prior probability space is derived from information in the problem. I do agree that it is a plausible Bayesian prior. The simplest way to justify that prior probability space is to assume the Principle of Indifference as applied to those 4 events.

I definitely agree with your answer P(H|A) = 1/3 if we assume your prior probability space and if we assume the question in the Sleeping Beauty Problem is asking for P(H|A) as those events are defined in you prior probability space.

Controversies about the problem, which you may not care to entertain, involve the question of whether there can be other plausible prior probability spaces and whether the event "H|A" in your prior probability space is the only possible interpretation of what the Sleeping Beauty Problem is asking for.

 

[Dale]

I infer that the events in your prior probability space are the 4 situations that can arise in the experiment and that you assign each a probability 1/4.

I explicitly and repeatedly do not assign probabilities to your 4 situations. You keep going back to your misstatements of my position and your demands to specify things that do not need to be specified. You are setting up a strawman to knock down, and regardless of how many times you set it up and knock it down I am not going to accept any ownership of the strawman. I do not assign probabilities to any of your 4 situations, and I do not apply the principle of indifference to anything. Please read my actual analysis and not your strawman.

I don't agree that your prior probability space is derived from information in the problem.

The prior is that P(H) = 1/2 which is explicitly given in the problem when they state that the coin is fair. This is the ONLY probability that I assign.

Controversies about the problem, which you may not care to entertain,

I am happy to entertain arguments, but the mere fact of a controversy is uninformative.

 

[Stephen Tashi]

And I believe I have shown, in three different ways, that it is:

Four Beauties:

If you want your examples to be taken seriously as mathematical arguments, you'll have to explain exactly how the events in the problems you make up correspond to the events in the Sleeping Beauty Problem. I don't mind working such details out for myself in examples like the balls-in-urns problem where it's simple to map events like "amber" and "Awake". But when you create problems that involve multiple Beauties, cards, additional coins etc. you're need to explain how events in probabilty space for your problem correspond to events in the probability space for Sleeping Beauty Problem. (That would imply that the probability space for the Sleeping Beauty problem is well defined in the first place.)

If you make up N problems and do bother to show they are equivalent to the Sleeping Beauty Problem and all give the same answer, then we must ask if you used The Principle of Indifference in solving any of those N problems. If you did use such assumptions then your N examples show that with certain assumptions, we can get the same answer to the Sleeping Beauty Problem. This does not show that the Sleeping Beauty Problem has one and only one correct answer.

One theoretical question raised by the Sleeping Beauty Problem is whether Bayesian analysis is consistent. If a person is allowed free choice of picking sets of "indistinguishable" events and assigning them equal probability then do people always get the same answer to problems by using Bayesian analysis?

No, this isn't analogous.

To see why, imagine (well, this is equivalent to the "imaginary coins" version) that there are two different lab techs running the experiment.

Again, you leave it to the reader to make a correspondence between the events in your problem and events in whatever probability space your are using for the Sleeping Beauty Problem.

 

[Stephen Tashi]

I explicitly and repeatedly do not assign probabilities to your 4 situations. You keep going back to your misstatements of my position and your demands to specify things that do not need to be specified. You are setting up a strawman to knock down,

I apologize, but I think my strawman is more convincing that your assertion that P(A|H)/P(A|T) = 1/2 based soley on the number of events in A|H and A|T. You seem to be assuming each event in those sets has the same probability - otherwise it makes no sense to compute a probability as a ratio of numbers of events.

 

[stevendaryl]

A similar argument appeared in the Wikipedia article linked to in the OP (IIRC), and I responded to it many, many posts ago, but I'll briefly recap my response.

To me, the first bullet of yours above is an argument for P(Monday|Tails) = P(Tuesday|Tails). Note that these are conditional probabilities, not probabilities of conjunctions. I don't think the premise of the bullet justifies a claim about conjunctions, only about conditionals.

Well, the rules of conditional logic say that P(Monday & Tails) = P(Monday | Tails) P(Tails) and P(Tuesday & Tails) = P(Tuesday|Tails) P(Tails). So the conjunctions would have to have equal probability, as well, right?

Similarly, the second bullet of yours is an argument for P(Tails|Monday) = P(Heads|Monday); once again, conditionals, not conjunctions.

Once again, the laws of conditional logic imply that the conjunctions have to have equal probability.

But if all we have is the above conditionals, we can't get to the conclusion, because that would require P(Monday|Tails) = P(Tails|Monday), and we don't have an argument for that.

Maybe I'm not understanding your objection--it seems that if you grant the two premises:

P(Monday | Tails) = P(Tuesday | Tails)

and

P(Heads | Monday) = P(Tails | Monday)

Then it immediately follows that:

P(Monday & Tails) = P(Tuesday & Tails) = P(Heads & Monday)

 

[Dale]

I debated with myself whether or not to post this. I want to make it clear from the beginning that this is NOT what I am claiming. I am explicitly NOT calculating any of the probabilities below. The purpose of this post is to show why the problem specification does indeed give P(A|H)/P(A|T)=1/2 without calculating P(A|H) or P(A|T).

P(Awake|H & Monday) = 1 (She's awake Monday regardless of the coin toss)
P(Awake|H & Tuesday) = 0 (She's asleep Tuesday if the coin toss was heads)
P(Awake|T) = 1 (She's awake if the coin toss is heads, regardless of the day)

If you take your sample space for P(A) to mean "awakened and interviewed sometime during a given day" then this is correct.

However, suppose (consistent with the problem description) that you assume that an awaken-and-interview lasts for about 14 min, so that there are 100 of them in a day, and suppose that you take P(A) to mean "awakened and interviewed in a particular 14 min slot", then P(A|H&Mon) = 1/100, not 1. Meaning that she is asleep for 0.99 of the day and only awake and interviewed for 0.01 of the day. So, in this case there are 200 possible 14 min slots over the two days so P(A|H) = 1/200 and P(A|T) = 2/200. The ratio P(A|H)/P(A|T)=1/2. The same thing holds for any length that you might choose.

Now, suppose that the probabilities change from day to day (i.e. principle of indifference does not apply). Specifically, say that the staff is hung over on Monday so they are slower, and it takes them twice as long to do an awake-and-interview as normal. Now there are only 50 slots on Monday and 100 on Tuesday. So then P(A|H) = 1/150 and P(A|T) = 2/150. The ratio P(A|H)/P(A|T)=1/2.

You can even change the probability of different slots, so say that the staff only wakes subjects between 8:00 am and noon, and you can even make it a continuous probability distribution (I leave the demonstration to the interested reader). It doesn't matter, regardless of how you partition the space and assign probabilities, regardless of the actual value of P(A|H), what the problem actually gives us is P(A|H)/P(A|T)=1/2. The details of the partitioning always come down to a factor (the factor of 2 or 200 or 150 above) which gets divided out in the ratio.

This is why I explicitly and repeatedly insist that I do not know and do not need to know P(A|H) or any other specific similar probability. They are not given by the problem, and they are not necessary to calculate in the problem whereas the ratio P(A|H)/P(A|T)=1/2 is given and is necessary.

 

[Stephen Tashi]

• If the coin toss is tails, then there is nothing observably different about Monday and Tuesday for Sleeping Beauty. So P(Tails & Monday) = P(Tails & Tuesday)

Those interested in complexities of interpreting the Sleeping Beauty problem can consider whether there is anything observably different about Monday and Tuesday for Sleeping Beauty if the coin lands heads.

If the coin lands head, Sleeping Beauty doesn't know that. Sleeping Beauty can reason "if the coin landed heads, today is not Tuesday". In what sense does an if...then statement count as an observable difference when the if... part of the statement cannot be observed?

 

[JeffJo]

If you want your examples to be taken seriously as mathematical arguments, you'll have to explain exactly how the events in the problems you make up correspond to the events in the Sleeping Beauty Problem.

?

I have. Several times. Here it is again:

Four Beauties: Each Beauty is participating in an experiment where she will be woken once, and maybe twice, depending on the result of a coin flip.Drugs will be used to prevent her from remembering any possible wakings in the past. When awake, she is asked for her confidence in the proposition that the coin result was the one where she was to be only woken once. With the exception of naming the sleep day, and what that coin result is, this is the OP. Whatever probability space can be called correct in the OP can be used for each of these four women, by simply re-arranging the names on the outcomes. That is, switching "Heads" and "Tails" everywhere they occur, and/or switching "Monday" and "Tuesday" everywhere they occur.

But by using four Beauties, one with each possible combination, it is possible to treat it as a ball-in-urn problem, with the (strong) PoI applying to which combination belongs to the sleeping Beauty. This probability space is not ambiguous, as you feel the OP's is. I'm not claiming that it is the same probability space that you can use in the OP, only that it is the same problem. So if there is a correct answer to either, it is the correct answer for both. And this one has an unambiguous answer.

Taken to Disneyworld: Any probability space is the same as that in the OP, with the event "asleep" replaced with "taken to Disneyworld." You seem to think that Beauty's ability to observe the event changes the probability space that contains the event. Why? Beauty's information corresponds exactly to "that other outcome is not the current outcome" regardless of what happens in it. What if Beauty has a subconscious that is aware of everything when she is asleep, but she can't recall it when she is awake?

Imaginary Coin: This occurs in the OP itself. She just picks a random day somehow, and it doesn't matter if she knows what it is, or tells anybody what it is. Any probability space that applies to the OP is merely augmented by adding this selection to it. So, let C be the event where the chosen day correctly names today, and W the state where it is wrong. When the lab tech determines whether she is to be wakened or not on either morning (see note below), a probability space is {(C,H),(C,T),(W,H),(W,T)} and the strong PoI applies.When she finds herself awake, either (C,H) or (W,H) is eliminated, and it doesn't matter which since her confidence in H is 1/3 either way. Note that I don't have to prove that this is the same as the OP, because it is happening in the OP.

Note: This is really the point. "Monday will happen" is not a random occurrence, it is a certainty. It cannot be part of a valid probability space, which is why you find it ambiguous to try to include it. But when the lab tech comes to either wake beauty up, or extend her sleep, "Today" can be either "Monday" or "Tuesday," and that is a random occurrence to Beauty since she is prevented from knowing what day it is.

I don't mind working such details out for myself in examples like the balls-in-urns problem where it's simple to map events like "amber" and "Awake".

Itr is equally simple to map "Today".

But when you create problems that involve multiple Beauties, cards, additional coins etc. you're need to explain how events in probability space for your problem correspond to events in the probability space for Sleeping Beauty Problem. (That would imply that the probability space for the Sleeping Beauty problem is well defined in the first place.)

There is a well-defined one, you just aren't using it. It can't be defined on Sunday, since it requires the random variable TODAY to take on the values Monday, or Tuesday.

The prior probability space, before the information-attaining act of waking Beauty up occurs, is that (COIN,TODAY) is in the set {(H,Mon),(H,Tue),(T,Mon),(T,Tue)}. Unlike your ambiguous, determined-on-Sunday probability space, the strong PoI applies to this one. Given that Beauty is woken and interviewed instead of [left asleep, taken to Disneyworld, subconscience, whatever], she knows she can eliminate (H,Tue) from this non-ambiguous probability space. So the conditional probability that Heads was flipped is 1/3.

+++++

One last variation:

1) It doesn't matter if Beauty sleeps after Heads, or Tails, as long as we ask her for her confidence in that result. Agreed?
2) It can't matter if Beauty sleeps through Monday, or Tuesday, since she does not know the day. Agreed?

So, flip three coins on Sunday: a quarter, a dime, and a nickel. Call them Q, D, and N. Replace "Heads" in the OP, and any valid probability space, with Q=D. Replace the Monday/Tuesday mechanic by flipping the nickel over after Monday's activities. Wake Beauty on either day, unless Q=D=N on that day.

Using Halfer logic, there is a new, correct probability space that applies to this new problem. In the prior, the PoI applies to the eight possible combinations of (Q,D,N). When Beauty is awake, she knows that two are eliminated, leaving six. Of those six, two have Q=D. And since symmetry exists, this solution applies when you invert N.

 

[stevendaryl]

I debated with myself whether or not to post this. I want to make it clear from the beginning that this is NOT what I am claiming. I am explicitly NOT calculating any of the probabilities below.

I know that you're not. I'm saying that without something like the following, I consider your claim that P(A|H)/P(A|T) = 1/2 is not derived, but simply asserted. On what basis?

However, suppose (consistent with the problem description) that you assume that an awaken-and-interview lasts for about 14 min, so that there are 100 of them in a day, and suppose that you take P(A) to mean "awakened and interviewed in a particular 14 min slot", then P(A|H&Mon) = 1/100, not 1.

Okay, that's fine. But the memory happens once per day, not 100 times per day, so she knows that this is the first interview, right? So we can define A to be "Sleeping Beauty is interviewed during the first 14 minutes". Then P(A | H & Monday) = 1.

Meaning that she is asleep for 0.99 of the day and only awake and interviewed for 0.01 of the day. So, in this case there are 200 possible 14 min slots over the two days so P(A|H) = 1/200 and P(A|T) = 2/200. The ratio P(A|H)/P(A|T)=1/2. The same thing holds for any length that you might choose.

I don't understand how that logic is any different from saying that the subjective a priori probability of each time slot is equal. That's what I don't understand. You're assuming something that is equivalent to something that you're denying assuming. To use a counting argument as an estimate for probability is mathematically equivalent to assuming equal prior likelihood for different time slots. Whether the number of time slots is 2 (Monday or Tuesday) or 200 (divide each day into 14 minute chunks) doesn't make any difference.

 

[Stephen Tashi]

If you take your sample space for P(A) to mean "awakened and interviewed sometime during a given day" then this is correct.

You (I think) assume a prior distribution for selecting a time from a uniform distribution defined over some time intervals.

That transforms the question of how to slect the situation "Sleeping Beauty is awakened" to the question of how to select the time interval containing "Sleeping Beauty is awakened right now".

Being a sticker, I'll point out that there is no information about selecting time intervals given in the problem. In fact, the problem doesn't exclude the real life possibility that interviews vary in duration, even within one run of the experiment.

Considering time points out that the Sleeping Beauty problem is ill-posed as probability problem. (It's status as a problem about "credence" may be a different story.) The statement of the problem does not rule out the case that interviews have varying lengths and does not define any particular distribution for the time selected to be the time that Sleeping Beauty is interviewed. Problems in mathematical probability don't have a tacit rule that says "Two things whose difference is not mentioned may be assumed to be equal". If two things are not explicity given as equiprobable, the additional assumption that they are equiprobable needs to be stated by the problem solver.

One advantage of your method is that it avoids assigning any probability to the event (heads, Tuesday, asleep). In glancing at other treatments of the Sleeping Beauty problem on the web, it appears to me that many people shy away from talking about the probability of that event. A flaming Bayesian has no qualms about hypothesizing a prior distrbution over all 4 situations, including (heads, Tuesday, asleep). However, one can ask what (heads,Tuesday,asleep) has to do with modeling the situation when Sleeping Beauty is awake.

 

[stevendaryl]

If you want your examples to be taken seriously as mathematical arguments, you'll have to explain exactly how the events in the problems you make up correspond to the events in the Sleeping Beauty Problem.

The point of @JeffJo's variant is that in his twist on the story, the situation for Sleeping Beauty herself is the same as in original problem, but in the new variant, there is an additional symmetry so that symmetry arguments can allow Sleeping Beauty to uniquely determine her probability.

 

[Stephen Tashi]

The point of @JeffJo's variant is that in his twist on the story, the situation for Sleeping Beauty herself is the same as in original problem, but in the new variant, there is an additional symmetry so that symmetry arguments can allow Sleeping Beauty to uniquely determine her probability.

Which variant ? - of the many variants!

What, if anything, is the distinction between a "symmetry argument" and the Principle of Indifference?

If the Principle of Indifference is use in solving a mathematical probability problem, it is considered an assumption added to the information given in the problem.

Perhaps among those whose discuss problems in "credence" there is a ground rule that says one is always permitted to apply The Principle Of Indifference. (And perhaps a discussion of "credence" is more on-topic in this thread that a discussion of mathematical probability, but at the moment, I am only discussing whether the Sleeping Beauty problem is well-posed as a problem in mathematical probability. )

 

[Dale]

So we can define A to be "Sleeping Beauty is interviewed during the first 14 minutes"

Yes, you certainly can, but you don't have to, and the problem specification doesn't tell you that.

The point is that the problem specification does not give P(A|H&Mon)=1. By choosing a different sample space for A you could have P(A|H&Mon)=0.01

I don't understand how that logic is any different from saying that the subjective a priori probability of each time slot is equal.

It is not different. That was the next paragraph.

 

[stevendaryl]

Which variant ? - of the many variants!

Here's a quick recap, leaving out a bunch of details: Suppose we have four test subjects. We randomly pick two to be "one-wakers" (only awakened on one day) and two to be "two-wakers" (awakened both days). Of the one-wakers, we wake each on a different day.

Now, on each day, we explain to the subjects the situation, and we ask of each of those who are awake: What is the likelihood that you are a one-waker?

They can see that there are three people awake, and they know from the rules that exactly one of them is a one-waker. It's equally likely that it's any of the three. So each should conclude that the probability she is the one one-waker among those who are awake is 1/3.

The similarities and differences with the original Sleeping Beauty problem are that:

1. In both cases, she is randomly (50/50 chance) assigned to be a one-waker or a two-waker. In the original, it's decided by a coin flip.
2. In both cases, if she's a two-waker, then she is awakened on both days.
   
3. In both cases, if she's a one-waker, then she is awakened one of the two days. In the original, that day is Monday.

Maybe you would argue that it makes a difference that she's awakened on Monday, rather than Tuesday?

 

[JeffJo]

If the Principle of Indifference is use in solving a mathematical probability problem, it is considered an assumption added to the information given in the problem.

If no probability value is given in a mathematical probability problem, the PoI is the only way to supply one. For example, you are applying it when you say the probability that the coin will land with heads showing is 50%.

Kolmogorov didn't tell us how to assign values to probabilities, he just laid out the properties the values need to have.

 

[Dale]

In fact, the problem doesn't exclude the real life possibility that interviews vary in duration, even within one run of the experiment.

I covered that possibility too.

Considering time points out that the Sleeping Beauty problem is ill-posed as probability problem. (It's status as a problem about "credence" may be a different story.)

Credence is a probability. It satisfies all of the axioms of probability. The sleeping beauty problem is well posed, the point of the consideration of time is to show that it doesn't matter. It drops out of the ratio.

If two things are not explicity given as equiprobable, the additional assumption that they are equiprobable needs to be stated by the problem solver.

Alternatively, a method may be used which does not depend on whether or not they are equiprobable!

 

[stevendaryl]

Perhaps among those whose discuss problems in "credence" there is a ground rule that says one is always permitted to apply The Principle Of Indifference. (And perhaps a discussion of "credence" is more on-topic in this thread that a discussion of mathematical probability, but at the moment, I am only discussing whether the Sleeping Beauty problem is well-posed as a problem in mathematical probability. )

The procedure---what actually is done--is completely well-specified, so I don't understand this business about selection procedures. There is no selection procedure, other than the initial coin toss. The question is whether is a non-arbitrary way for Sleeping Beauty to quantify her uncertainty about her situation when awakened. I would say that the principle of indifference is definitely a non-arbitrary means of solving the problem. If you want to say that it's an additional assumption, I guess I would concede that point. Although historically, probability theory was originally developed by exploiting the principle of indifference, and the more general notion of probability was developed to generalize to cases where the principle was too weak to give an answer.

 

[Stephen Tashi]

If no probability value is given in a mathematical probability problem, the PoI is the only way to supply one. For example, you are applying it when you say the probability that the coin will land with heads showing is 50%.

The fact that making an assumption is the only way to solve mathematical probability problem demonstrates that the problem is ill-posed.

One does not automatically assume a coin in a mathematical probability problem has an equal probability of landing heads or tails. It should be explicitly stated that it is a "fair" coin.

Kolmogorov didn't tell us how to assign values to probabilities, he just laid out the properties the values need to have.

Nevertheless, for a specific probability space to be explicitly described in problem, the details of assigning probabilities must be given.

 

[stevendaryl]

If no probability value is given in a mathematical probability problem, the PoI is the only way to supply one. For example, you are applying it when you say the probability that the coin will land with heads showing is 50%.

Kolmogorov didn't tell us how to assign values to probabilities, he just laid out the properties the values need to have.

I would have to agree with @Stephen Tashi that the POI is an additional assumption, although in most informal word problems involving probability, it's implicit that you're supposed to use it whenever applicable.

 

[Dale]

The fact that making an assumption is the only way to solve mathematical probability problem demonstrates that the problem is ill-posed

But making that assumption is not the only way to solve it.