The Sleeping Beauty Problem: Any halfers here?

[PeroK]

@Marana If you were the sleeper, please tell me what you would answer to these three questions:

You wake up:

What is the probability that the coin is heads?

If it's Monday, what would be your answer?

If it's Tuesday, what would be your answer?

 

[PeroK]

I may be different than a typical halfer, because I don't think we can use anything like this.

You are treating it as though MH, MT, and TT are the possible outcomes of a random experiment, each with probability 1/3. In that case, it is certainly true that the probability of heads is 1/3, and the probability of heads after conditioning on Monday is 1/2. But this is not an accurate description of sleeping beauty: when she wakes up it is not a random experiment. Tuesday always follows monday, tuesday tails always follows monday tails. Sleeping beauty doesn't lose her memory of this fact, nor does she lose her memory of what week is coming. The usual probability methods you are using need to be justified somehow for this situation which is not a random experiment, and I don't see how.

A typical halfer may also treat it like a random experiment, except they would break it into two stages. First the coin toss, then the waking up. In other words, the typical halfer is saying that waking up can only be viewed as a selection of the possible days to wake up. So they would say that MH has probability 1/2, MT has probability 1/4, and TT has probability 1/4. Probability of heads is 1/2, and probability of heads after conditioning on monday is 2/3. I disagree with this for the same reason I disagree with thirders: it isn't a random experiment.

So to me, the question is how do we define probability for such a weird situation which isn't a random experiment at all? It may be that there is currently no definition for that situation, and it may be that we don't need one. Notice that our strategies are already set in stone on sunday, before the coin flip. Whether or not there is new, relevant information on monday (which I can't imagine what it could be) it definitely won't change our strategy, calling into question if there is any importance to it. Halfers and thirders will always agree on how to act for any betting setup, even if they disagree on how probability should be defined. A halfer may argue "I prefer to bet on tails because if it is tails the bet is offered twice", a thirder may say "I prefer to bet on tails because I am defining the probability as 2/3 for tails", but they will both bet the same thing.

But to me, the most natural way to define it is to start with something solid, a genuine random experiment. The coin flip, which has sample space of H and T, each with probability 1/2. Then if you believe there is no new, relevant information on monday, you could hold onto that 1/2 probability when waking up. Or you could use the principle of reflection, knowing that you would believe 1/2 at noon on wednesday. I admit, neither of these are fully convincing.

According to the way I calculate it, it is not possible to condition on it being monday or tuesday. Which I think makes sense, because when using conditioning you can't have impossible events become possible. "It is monday" can't be followed by "it is tuesday." The probability for "it is tuesday" became 0 when you learned "it is monday". Losing memory of monday does not fix this problem, so conditioning is not justified.

So, in effect, in your solution Tuesday and everything except the coin toss is irrelevant? Nothing that happens in the experiment can change the probability of 1/2?

You might as well just wake the sleeper on the Monday, tell her it's Monday and ask her. She says 1/2. Nothing else in the experiment makes any difference to this?

The coin is tossed. It's 50-50 heads or.tails. End of.

PS this is now the psychological issue I mentioned in an earlier post. Your a priori conviction that the answer of 1/2 must be correct is so strong that simple questions such as "is it Monday?" become invalid. Anything that disproves the a priori answer must be invalid.

 

[Demystifier]

when she wakes up it is not a random experiment.

Probability is not only about randomness, but also about absence of knowledge. Suppose that I pick one of the letters A or B, by will. Then I ask you, what is the probability that I picked A? What is your answer?

 

[PeroK]

Probability is not only about randomness, but also about absence of knowledge. Suppose that I pick one of the letters A or B, by will. Then I ask you, what is the probability that I picked A? What is your answer?

Or, someone has two children. The probability of two boys is 1/4.

If they have two girls then they come to see you on a Monday; otherwise, they come to see you on a Tuesday. Nothing random. Yet, if they come to see you on a Tuesday, the probability of two boys has increased to 1/3.

 

[stevendaryl]

But to me, the most natural way to define it is to start with something solid, a genuine random experiment. The coin flip, which has sample space of H and T, each with probability 1/2. Then if you believe there is no new, relevant information on monday, you could hold onto that 1/2 probability when waking up. Or you could use the principle of reflection, knowing that you would believe 1/2 at noon on wednesday. I admit, neither of these are fully convincing.

Here's the way that I became a thirder, which I think is convincing (even if it is much more work than the original, one-line argument for 2/3 or 1/2).

Imagine that experimenters are doing this experimenter over and over, with lots of different test subjects (sleeping beauties). At any given moment, there will be some group of people who are either experiencing Day 1 or Day2 (rather than Monday and Tuesday, because the subjects may start the experiment on different days). Each such person has an associated coin flip result. So each subject has an associated label (known to the researcher, but not the subject): $(D,C)$ where $D$ is the day number, either 1 or 2, and $C$ is the coin flip result, either $H$ or $T$.

The researchers provide you with a list of all current subjects, and you pick one at random.

Here, there is a technical question about what it means to be in Day 2, and what it means to pick a subject at random.

First interpretation: You can only be experiencing Day 2 if your coin flip result was "heads". So your sample should include only those whose labels are:

• $(D=1, C=H)$
• $(D=1, C=T)$
• $(D=2, C=H)$

So there are no subjects with label $(D=2, C=T)$.

According to this interpretation, you are twice as likely to pick $C=H$ subject. Supposing that there are N experiments started each day, then let $N(D,C)$ be the number of subjects with label $(D,C)$. Then typically, you would expect that:

$N(1, C=H) = N(1,C=T) = N/2$ (of the new Day-1s, half typically are associated with a result of heads, and half are associated with a result of tails).
$N(2, C=H) = N/2$ (if a subject has label $(D=2, C=H)$ today, then she had label $(D=1, C=H)$ yesterday)
$N(2, C=T) = 0$ (no Day-2 for tails)

So on a typical day, there are $3N/2$ subjects, with the following statistics:

• $N$ have $C=H$
• $N/2$ have $C=T$
• $N$ have $D=1$
• $N/2$ have $D=2$

So for a randomly selected subject, the odds are

• $2/3$ that $C=H$
• $2/3$ that $D=1$.

Second interpretation: We can allow the possibility of $(D=2, C=T)$, but these subjects will know that that label applies to them, because their memories are not wiped after the first day. We'll call these subjects "informed".

With this interpretation, on a typical day, there are $2N$ subjects, with the following statistics:

• $N$ have $D=1$
• $N$ have $D=2$
  
• $N$ have $C=H$ (as before)
• $N$ have $C=T$ (including the $N/2$ "informed" subjects)

So for a randomly selected subject, the odds are:

• $1/2$ that $C=H$
• $1/2$ that $D=1$
• $1/4$ that the subject is "informed" (that is, $(D=2, C=T)$)
  
• $3/4$ that the subject is "uninformed"

So this interpretation restores the intuitive idea that there should be equal likelihood of heads and tails. However, we can compute conditional probabilities:

$P(H| uninformed) = P(H)/P(uninformed) = \frac{1/2}{3/4} = 2/3$

So under either interpretation, if the random subject is uninformed (doesn't know his coin result), then there is a 2/3 likelihood that her result was "heads".

 

[DrClaude]

Now seriously. I am sorry that I didn't open a poll, I guess it's too late now. The possible poll answers would be:
- 1/2
- 1/3
- It depends on the precise formulation of the problem.

It is never too late! I have added the poll.

 

[Demystifier]

It is never too late! I have added the poll.

That's great, thanks!

 

[PeroK]

One more thought. If the sleeper doesn't know about the drug, then every time she wakes she will think it is Monday and her answer will be 1/2.

The halfer's position is that knowing about the drug doesn't change this. Effectively, for reasons that I admit I don't follow, knowing about the drug and knowing that it might be Tuesday doesn't change the answer.

 

[Charles Link]

I side with the halfers. And here is an argument for why I side with the halfers: (I don't think it has been presented yet in this form=I didn't read every single post...) Suppose we weight the coin so that it has a $p= .99$ chance for heads. But suppose we also change the rules(as previously mentioned) so that she will be woken up, let's say 1000 times, to be interviewed if it comes up tails. I do think if she says .99 as the probability that it was heads that she has calculated it correctly. In all likelihood, (with 99% probability), it is the first Monday.

 

[PeroK]

I side with the halfers. And here is an argument for why I side with the halfers: (I don't think it has been presented yet in this form=I didn't read every single post...) Suppose we weight the coin so that it has a $p= .99$ chance for heads. But suppose we also change the rules(as previously mentioned) so that she will be woken up, let's say 1000 times, to be interviewed if it comes up tails. I do think if she says .99 as the probability that it was heads that she has calculated it correctly. In all likelihood, (with 99% probability), it is the first Monday.

Before you vote for 1/2, you may wish to consider this post:

Okay, here's my last idea to disprove the answer of 1/2:

You wake sleeper and she gives the answer of 1/2. Then:

a) You ask her: if it's Monday, what would be your answer? She says 1/2.

b) You ask her: if it's Tuesday, what would be your answer? She says 0.

The only combination of those probabilities that gives 1/2 is that it must be Monday. Therefore, if the sleeper is consistent she must conclude that it is definitely Monday.

 

[Charles Link]

Before you vote for 1/2, you may wish to consider this post:

The probability for Monday is 3/4 and for Tuesday 1/4.

 

[PeroK]

The probability for Monday is 3/4 and for Tuesday 1/4.

If that is true, then the probability of heads is:

$(3/4)(1/2) + (1/4)(0) = 3/8$

 

[Charles Link]

I think the alternative problem I posed (in post #99) answers the question: .99+.01(.001) it is Monday, .01(.001) it is the second day, .01(.001) that it is the 3rd day, etc.

 

[PeroK]

I think the alternative problem I posed answers the question: .99+.01(.001) it is Monday, .01(.001) it is the second day, .01(.001) that it is the 3rd day, etc.

That makes no sense to me. The problem is about a coin with 50-50 heads and tails. Please explain your solution for that, especially in light of post #100 and the fact that if it's Monday it's 50-50 heads/tails and if it's Tuesday it's 100% tails.

 

[Charles Link]

That makes no sense to me. The problem is about a coin with 50-50 heads and tails. Please explain your solution for that, especially in light of post #100 and the fact that if it's Monday it's 50-50 heads/tails and if it's Tuesday it's 100% tails.

It is somewhat of a puzzle. My logic of post #99 was somewhat of a response to the logic previously posted of adding extra days if it came up tails. I'm trying to demonstrate a possibility using a weighted coin, and extrapolating it to a non-weighted coin. $\\$ In some ways, the calculation that Sleeping Beauty does here is similar to what happens in science when we try to compute a probability under the assumption that the state we are considering is completely random. $\\$ I think my example in post #99 is worth consideration, because it illustrates the type of assessment that ultimately results: Is Sleeping Beauty likely to be awake on the Monday (with a weighted coin) of a very likely event, or did the unlikely occur, so that she is now part of a long chain of what would occur if the unlikely took place? I can also follow the logic of assigning a 1/3 to the heads condition, but, in some ways, this problem defies logic.

 

[Marana]

@Marana If you were the sleeper, please tell me what you would answer to these three questions:

You wake up:

What is the probability that the coin is heads?

If it's Monday, what would be your answer?

If it's Tuesday, what would be your answer?

Probability that the coin is heads: 1/2

If it's Monday: 1/2

If it's Tuesday: 0

Or, someone has two children. The probability of two boys is 1/4.

If they have two girls then they come to see you on a Monday; otherwise, they come to see you on a Tuesday. Nothing random. Yet, if they come to see you on a Tuesday, the probability of two boys has increased to 1/3.

Probability is not only about randomness, but also about absence of knowledge. Suppose that I pick one of the letters A or B, by will. Then I ask you, what is the probability that I picked A? What is your answer?

Here's the way that I became a thirder, which I think is convincing (even if it is much more work than the original, one-line argument for 2/3 or 1/2).

Imagine that experimenters are doing this experimenter over and over, with lots of different test subjects (sleeping beauties).

It's not just the randomness that concerns me, it is whether it is an experiment. "In probability theory, an experiment or trial is any procedure that can be infinitely repeated and has a well-defined set of possible outcomes, known as the sample space."

I can select a new two-child family each week. The experimenters can select a random beauty each day. Both are easily repeatable and have a clear probability. As for picking a letter, I don't believe I could put an exact number on the probability (principle of indifference would say 1/2, but I'm not totally indifferent as I'd guess A is more popular, similar to how certain numbers show up more than others).

The difficulty with the sleeping beauty problem is that waking up isn't an experiment at all. It is, by definition, impossible to repeat. Tuesday follows Monday by the laws of the universe, TT follows MT by the laws of the study. A single waking is insufficient to model the situation, the rules of which are known to sleeping beauty.

"It is Monday" and "it is Tuesday" can both be learned for a single coin flip. That is not consistent with conditioning. So if we are asked about the result of the coin flip, it isn't justified to condition on the day of the week. It only seems reasonable at first because of the memory loss.

So I'd begin with the coin flip, a random experiment with sample space {H, T} and probability 1/2 for each. When I wake up I would maintain probability 1/2 for various reasons (lack of new relevant info, principle of reflection, intuition due to thirders being able to all believe they won billion dollar lottery) while admitting none of those reasons are fully convincing, just more convincing than the alternative. Then if I learn "it is Monday" I will recall that I may also learn "it is Tuesday", so that this is not the kind of thing I can use conditioning on. Time marches on, and if it is Monday, that means the probability is 1/2. Either because the coin doesn't need to be flipped yet, or because Monday tails is the precursor to Tuesday tails (really "MT followed by TT" as a whole is an outcome of the coin flip experiment).

 

[PeroK]

Tuesday follows Monday by the laws of the universe.

Not if you've been given an amnesia drug. Then "I don't know what day it is" is followed by "I don't know what day it is". Only those with a good memory, or who can look up the information, know what day it is.

 

[Demystifier]

but I'm not totally indifferent as I'd guess A is more popular

And I know that, so to deceive you I will be more prone to choose B. But I know that you know that too, so I will deceive you at a higher level by being more prone to choose A. But I know that you know that I know that you know that, so perhaps I should deceive you at an even higher level be being more prone to choose B ... When one takes all this into account, A and B seem about equally likely.

 

[stevendaryl]

It's not just the randomness that concerns me, it is whether it is an experiment. "In probability theory, an experiment or trial is any procedure that can be infinitely repeated and has a well-defined set of possible outcomes, known as the sample space."

You quoted one line of my post, but I don't see that you responded to it. Do you agree that in the setup I described, it makes sense for an observer to assign a 2/3 probability that a randomly selected sleeping beauty has an associated coin flip result of heads? The way I described that thought-experiment seems perfectly amenable to usual probabilistic reasoning. Right?

The next step would be for each sleeping beauty herself to consider herself a random choice. She knows that there are $3N/2$ people in the same situation she is in--not knowing whether they have been awakened one time, or two. Of those, she knows that

• $N/2$ had a coin toss result of heads, and are awakening for the first time.
• $N/2$ had a coin toss result of heads, and are awakening for the second time.
• $N/2$ had a coin toss result of tails, and are awakening for the first time.

So it seems completely straight-forward that she would assume that

• Her probability of having a result of heads is 2/3
• The probability that she is being awoken the first time is 2/3
• The probability that she is being awoken the second time is 1/3.

The final logical leap is to assume that the probabilities that apply in a repeated experiment also apply in a single experiment.

 

[stevendaryl]

And I know that, so to deceive you I will be more prone to choose B. But I know that you know that too, so I will deceive you at a higher level by being more prone to choose A. But I know that you know that I know that you know that, so perhaps I should deceive you at an even higher level be being more prone to choose B ... When one takes all this into account, A and B seem about equally likely.

That's a standard result in game theory, called a "mixed strategy". In a game like chess, there is always a "best move" (or perhaps a number of equally good moves), and there is no reason not to make it. But in certain types of games, your best move is to be random. An example is "Battleship".

 

[stevendaryl]

The probability for Monday is 3/4 and for Tuesday 1/4.

Let me try another variant of the experiment that I think will convince you that you're wrong.

Have you seen the movie "Memento"? The main character has a form of amnesia where he wakes up every morning having no idea what happened the previous day, unless he left notes for himself beside his bed (or pinned to his pajamas, or whatever). So we can redo the Sleeping Beauty problem using such an amnesiac. There is no need to wipe memories, but instead, we just control what notes she has waiting for her on the two mornings, Monday and Tuesday.

We prepare two envelopes. The first envelope says "Read me first" on the outside. Inside is a note explaining the rules of the experiment. Regardless of the coin toss, she gets this note on both Monday and Tuesday. The second envelope says "Read me second" on the outside. Inside is a note saying either "Today is Tuesday" or "Today is either Monday or Tuesday". She only gets the note saying "Today is Tuesday" if it actually is Tuesday, and the coin toss result was tails.

She wakes up and reads her first envelope, and is asked her subjective probabilities for various things. I believe everyone would agree that her answers would be:

1. There is a 1/4 chance that it's Monday and the coin toss result was Heads.
2. There is a 1/4 chance that it's Monday and the coin toss result was Tails.
3. There is a 1/4 chance that it's Tuesday and the coin toss result was Heads.
4. There is a 1/4 chance that it's Tuesday and the coin toss result was Tails

All four situations are equally likely.

Now, she reads the second note and finds out that she is not in situation 4. The other three situations are still equally likely, though.

 

[Charles Link]

@stevendaryl Thank you. :) Very interesting, but this is one of those that I think I could study for years and not be convinced that there is one answer that is completely correct. It's like the riddle of "Who's on first. (baseball=first base). Is "Who" on first, etc.? It comes up in the Dustin Hoffman, Tom Cruise movie "Rain Man", and Tom Cruise tells his brother that it is a riddle that has no correct answer. :) :)

 

[PeroK]

@stevendaryl Thank you. Very interesting, but this is one of those that I think I could study for years and not be convinced that there is one answer that is completely correct. It's like the riddle of "Who's on first. (baseball=first base). Is "Who" on first, etc.? It comes up in the Dustin Hoffman, Tom Cruise movie "Rain Man", and Tom Cruise tells his brother that it is a riddle that has no correct answer. :) :)

Hmm. I'm not prepared to give up on mathematics just because a problem is not intuitively obvious. If we do that, then we are left with nothing. If there is a compelling argument that probability theory cannot be used in this case, then I'm happy to listen and accept that the problem has no solution. I've not seen any such argument yet, I have to say.

 

[PeterDonis]

It looks like I'm the only vote in the poll for "it depends on the precise formulation of the problem", which seems strange, since the fact of this thread going on for 6 pages would seem to be evidence in favor of that choice. (A better basis for it, though, IMO is Demystifier's post #67.)

 

[PeroK]

It looks like I'm the only vote in the poll for "it depends on the precise formulation of the problem", which seems strange, since the fact of this thread going on for 6 pages would seem to be evidence in favor of that choice. (A better basis for it, though, IMO is Demystifier's post #67.)

I looked again at post #67. I cannot see any reason that in this experiment we would only count one of the instances in the case of tails. Yes, we can change the problem to say that we will deselect one of the tails instances. But that reduces the problem to something really quite trivial. I cannot see that in the case of rule B there is anything worth analysing or discussing.

In other words, the answer of 1/2 only applies in a case where the problem is so trivial as to merit no attention.

But, what makes the problem interesting is to explain the apparent paradox that the answer of 1/3 arises after "no new information". It is a seductive argument but we have clearly identified a change to the sleeper's information:

She no longer knows what day it is. And, in fact her information precisely coincides with a "random observer" who randomly stumbles over the experiment and doesn't know what stage the experiment has reached.

Finally, we could easily have six pages of argument about whether $0.999 \dots \ne 1$ - or whether simultaneity is absolute. But, it doesn't matter how long and hard someone argues either of those points or that the answer to the sleeping beauty problem is 1/2. Pertinancity alone doesn't make their case.

 

[PeterDonis]

I cannot see any reason that in this experiment we would only count one of the instances in the case of tails.

Um, because the experimenter decided to define the experiment that way? The scenario described in post #67 is about betting, and a bet can be whatever the bettors want it to be. Unless you're claiming that it's somehow physically impossible for an experimenter to offer Beauty the bet described by rule B in post #67.

Basically, you're trying to go from "I can't see any reason..." to "obviously my answer is the only right answer". But that's not a valid argument, logically speaking; it's just a statement of your opinion. There is no unique right answer until you've specified a question that's precise enough to have a unique right answer. Post #67 simply illustrates one of the things that has to be part of that precise specification in order for your answer to be the unique right answer.

 

[PeroK]

Um, because the experimenter decided to define the experiment that way? The scenario described in post #67 is about betting, and a bet can be whatever the bettors want it to be.

Then that is a very different problem from the one (in my opinion) precisely described in the original problem statement. The original problem statement (from the experimenters' point of view) is clear. If it's heads, they wake her on the Monday and if it's tails on the Monday and the Tuesday.

It says nothing in the original problem statement about possibly waking her on the Tuesday if it's tails or possibly not waking her on the Monday if it's tails.

I don't really see how that could be any clearer. Post #67, I believe, only introduced this variation to highlight that the interpretation inferred by the 1/2 argument was a different problem altogether.

In fact, if you start with the trivial problem that they wake her only on the Monday and hence the answer is 1/2, then the halfer position is, effectively, that whatever happens on subsequent days is irrelevant.

You can see this more clearly in the $n, k$ variation, where she is woken on $n$ consecutive days (heads) and $k$ consecutive days (tails). The 1/2 answer remains 1/2 in all cases. The 1/3 answer is $n/(n+k)$.

In other words, the halfer position in general only considers the first day. Or changes the problem so that only one of the $n$ days and one of the $k$ days "counts". That is a different problem altogether.

Furthermore, the only thing that makes this problem difficult is the amnesia drug. The 1/2 position can only arise in the case of an amnesia drug. Without the drug, the problem is trivial and the answer is 1/3. And, I suggest, that without the drug, no one would be claiming that the problem is not precise.

Finally, I see the fundamental difference in this thread as the 1/3 position has been backed up by analysis; whereas, it is the 1/2 position that is largely opinion: we can't use relative frequencies; we can't use a Bayesian approach; it's not a random experiment; there is no sample space; conditional probabilities don't apply; etc.

In each case, an analysis of why these objections are invalid has been presented, although I guess now the core analysis has been lost in 6 pages of claim and counterclaim.

 

[PeroK]

I think it's interesting to quote Wikipedia on this:

"The Sleeping Beauty puzzle reduces to an easy and uncontroversial probability theory problem as soon as we agree on an objective procedure how to assess whether Beauty's subjective credence is correct. Such an operationalization can be done in different ways: By offering Beauty a bet; more elaborately by setting up a Dutch book; or by repeating the experiment many times and collecting statistics. For any such protocol, the outcome depends on how Beauty's Monday responses and her Tuesday responses are combined.

Consider long-run average outcomes. Suppose the experiment were repeated 1,000 times. It is expected that there would be about 500 heads and 500 tails. So Beauty would be awoken 500 times after heads on Monday, 500 times after tails on Monday, and 500 times after tails on Tuesday.

• If Beauty herself collects statistics about the coin tosses (in a way that is not obstructed by memory erasure when she is put back to sleep), she would register one-third of heads. If this long-run average should equal her credence, then she should answer P(Heads) = 1/3.
• However, being fully aware about the experimental protocol and its implications, Beauty may reason that she is not requested to estimate statistics of the circumstances of her awakenings, but statistics of coin tosses that precede all awakenings. She would therefore answer P(Heads) = 1/2.

It's even simpler with bets: If Beauty and the experimenter agree that bets from her different awakenings are cumulative, then a heads quota of 1/3 would be fair. If on the other hand Tuesday bets are to be discarded (being dummy bets, undertaken only to keep Monday and Tuesday awakenings indistinguishable for Beauty), then the fair quota would be 1/2."

That is, then, fairly explicit on the circumstances under which 1/2 is a valid answer. I've highlighted one sentence. My question is this: why would one calculate a probability on that basis? Without the drug, the second operationalisation is not logical at all. It's Tuesday, you are asked to estimate the probability that the coin was heads. You know it's tails, but, nevertheless, you answer 1/2 because you are still thinking about the coin before any awakenings?

With the drug, you have some information (you are not sure what day it is), but for some reason you don't or can't use that information. You simply revert to the pre-experiment answer?

So, it seems the second operationalisation (halfer position) works on the basis of: if your knowldege is imperfect then yoiu ignore that knowldege and the answer is 1/2. Only if you definitely know the result can you answer 1 or 0. But, nothing in between.

(But, this only applies in the case of drug-induced memory loss. Outside of this special case, halfers revert (hopefully) to normal probability theory where you use all the information at your disposal. So, for halfers this is very much a one-off special case where the normal rules of probability theory do not apply.)

This isn't probability theory as I understand it. I believe you are free to use all the information you have to calculate a probability. And, if you deliberately or unwittingly don't use information at your disposal, then you are not calculating the probability as I would understand it.

And, by using all the information you have about the experiment, that you have been woken, that you don't know what day it is (it could be Tuesday and it could definitely be tails), then you get an answer of 1/3.

If you do not use this information, then you can get an answer of 1/2, but that is not the probability of its being heads, given everything you know.

PS And, if the answer is 1/2, then you can deduce with certainty that it is Monday. Proof:

Suppose the probability that it is heads is 1/2 and the probability it is Monday is $p$. Then the probability it is heads is:

$\frac{p}{2} + 0 = \frac{p}{2} = 1/2$

Hence, $p = 1$.

Therefore, whatever the answer is, it cannot be 1/2, unless the problem is changed so that Tuesday is excluded.

 

[Marana]

Furthermore, the only thing that makes this problem difficult is the amnesia drug. The 1/2 position can only arise in the case of an amnesia drug. Without the drug, the problem is trivial and the answer is 1/3. And, I suggest, that without the drug, no one would be claiming that the problem is not precise.

Finally, I see the fundamental difference in this thread as the 1/3 position has been backed up by analysis; whereas, it is the 1/2 position that is largely opinion: we can't use relative frequencies; we can't use a Bayesian approach; it's not a random experiment; there is no sample space; conditional probabilities don't apply; etc.

In each case, an analysis of why these objections are invalid has been presented, although I guess now the core analysis has been lost in 6 pages of claim and counterclaim.

Nobody would be answering 1/3 without the drug. Without the drug the answer is 1/2 on monday and 0 on tuesday (because you keep your memory you always know what day it is).

Indeed, that illustrates why the frequency argument fails. Without the drug, the frequencies are the same, but not a single person would answer 1/3. The fact that it is not a repeatable random experiment, and not solvable by frequency, would be crystal clear.

That leaves us with the conditioning method, but I can demonstrate that it clearly fails as well. All I have to do is mention that you can learn both "it is monday" and "it is tuesday" for a single coin flip, totally impossible for conditioning. Therefore you can't condition on the day when trying to decide on the result of the coin flip.

We need a different way to compute probabilities with information like "it is monday" and "it is tuesday" that can't be used with conditioning. My idea so far is that if you have a correspondence of non-temporal information between "starting time" and "some specified later time", and you learn only that "some specified later time" is true, then your probabilities in things that don't change with time (like the result of a coin flip) must be equal to what they were at the starting time. So in sleeping beauty, there is a correspondence between yourself on sunday and yourself on monday (because you are always awake monday). If you learn that it is monday your probability should be 1/2, because that is what it was on sunday (the starting time) and the only thing that changed is time (as it inevitably does) marched on.

For a slightly more complicated example, if you wake up every day for a year on tails, and only prime days for heads, and you learn it is day 137, then your probability should be 1/2, because the only change from the starting time is the passage of time, which has no relevance to the result of the coin flip. This method can probably be improved to deal with more complicated situations, the main thing is that conditioning does not apply.

PS And, if the answer is 1/2, then you can deduce with certainty that it is Monday. Proof:

Suppose the probability that it is heads is 1/2 and the probability it is Monday is $p$. Then the probability it is heads is:

$\frac{p}{2} + 0 = \frac{p}{2} = 1/2$

Hence, $p = 1$.

Therefore, whatever the answer is, it cannot be 1/2, unless the problem is changed so that Tuesday is excluded.

I don't think I am being all that radical by saying things like this are not allowed, for the reason I mentioned above. "It is Monday" and "it is Tuesday" are a different kind of information which can't use the usual probability techniques. But I'm not just making that up: it is fact that you may learn both "it is Monday" and "it is Tuesday" for a single flip, and it is a fact that that is absolutely impossible with normal probability techniques.

It's like saying you "learned" it was 2pm. Then you "learned" it was 2:01. Then you "learned" it was 2:02. Well, no, you're not learning all that in the usual sense. Time is just going forward.

 

[PeroK]

Nobody would be answering 1/3 without the drug. Without the drug the answer is 1/2 on monday and 0 on tuesday (because you keep your memory you always know what day it is).

The answer is 1/3 as follows:

$P(H) = P(H|Mon)P(Mon) + P(H|Tue)P(Tue) = (1/2)(1/3) + 0(2/3) = 1/3$

That's the conditional probability of its being Heads given a random awakening.

The specific answers given are 1/2 and 0, but this equates to a conditional probability of 1/3.

The point is that the overall conditional probability in this case is not 1/2.

That's the way conditional probabilities work. In the same way that an average can be a number that it not iself attainable in any experiment, it is also the case with a conditional probability.

I don't think I am being all that radical by saying things like this are not allowed, for the reason I mentioned above. "It is Monday" and "it is Tuesday" are a different kind of information which can't use the usual probability techniques.

You are being totally radical. There is no reason not to use probability theory in this case.

Yours is an extreme position: adopted in support of your a priori requirement that the answer to the sleeping beauty problem is 1/2.