I The Sleeping Beauty Problem: Any halfers here?

[Demystifier]

The selective amnesia problem

I want to make the 1/3 solution more interesting (and perhaps more intuitive) through an analogy with selective amnesia.

We have all noticed selective amnesia in others. Let me explain what that means. We all know that other people have many prejudices. When they see evidence which confirms their prejudices, they tend to remember this evidence for a long time. But when they see evidence against their prejudices, they tend to forget this evidence quickly. This psychological effect is called selective amnesia.

We cannot observe the selective amnesia in ourselves. But if others suffer from it, it is reasonable to assume we ourselves are not immune. So let us assume that we ourselves also suffer from selective amnesia. What can we conclude from that?

Suppose that I believe (by means of a vague intuitive feeling) that some statement $A$ is true. And suppose that, at the moment, I cannot recall any actual evidence that it is true. Then I can argue at a meta-level as follows. A priori, without any other information, the probability that $A$ is true is equal to the probability that $A$ is not true. But I do have some additional information. First, I know that I have some vague feeling that it is true. Second, I know that I cannot recall any actual evidence that it is true. But if I saw evidence (that $A$ is true) in the past, I could probably recall it now. And I cannot say the same for counter-evidence, because even if I saw some counter-evidence in the past, I would probably forget it by now by the selective amnesia. So the fact that I cannot recall any evidence for $A$ and the fact that I still feel that $A$ is true implies that $A$ is probably not true. This seemingly paradoxical conclusion follows from the assumption that I suffer from selective amnesia.

Now how is it related to the Sleeping Beauty problem? The Sleeping Beauty also suffers from a selective amnesia problem, although due to a different reason. She has an induced amnesia only when she is awaken twice, i.e. only in the case of tails. And this fact alone (according to thirders) is sufficient to conclude that from her perspective tails is more probable than heads. The thing for which you have selective amnesia about evidence is more probable than the thing for which you don't have selective amnesia about evidence.

 

[stevendaryl]

Aha! I think I have found the X that restores equivalence between the relative frequency and Bayesian calculations.

Let's not say that we only wake up Sleeping Beauty once or twice. She wakes up every day, whatever the coin flip was. But the difference is this:

1. If the coin flip is heads, we don't tell her the result until after she has made a second guess about probabilities.
2. If the coin flip is tails, we tell her the result after she has made just one guess.

So there still is a second wakening in the tails case, it's just that there is no suspense in that case, because she already knows the result.

Now, let X be the statement: "Sleeping Beauty has not yet been told the result".
Let MT be the statement: "Today is either Monday or Tuesday"

Then we can compute:
P(X|H \wedge MT) = 1
P(X|T \wedge MT) = 1/2
P(H) = P(T) = 1/2

So P(X| MT) = P(X | H \wedge MT) P(H) + P(X | T \wedge MT) P(T) = 1/2 + 1/4 = 3/4

Now we can do ordinary Bayesian updating:

P(H| X \wedge MT) = P(H \wedge X | MT)/P(X|MT) = P(H) P(X| H \wedge MT)/P(X|MT) = \frac{1/2 \cdot 1}{3/4} = 2/3
P(T| X \wedge MT) = P(T \wedge X | MT)/P(X|MT) = P(T) P(X| T \wedge MT)/P(X|MT) = \frac{1/2 \cdot 1/2}{3/4} = 1/3

 

[Demystifier]

Aha! I think I have found the X that restores equivalence between the relative frequency and Bayesian calculations.

Would you agree that the frequentist approach is superior over Bayesian one, in the sense that with the frequentist approach it is much simpler to get the correct result 1/3?

 

[Marana]

Now I see why halfers think that 1/2 is correct. But let me challenge halfers with another variation of the problem (inspired by the quantum many-world version in the papers above). Now the coin is not randomly flipped, but deterministically set according to certain rules. More precisely, the experimenters perform the following fully deterministic 3-step procedure:
Step 1: The Beauty is awaken for the 1st time and the coin is set to heads.
Step 2: The Beauty is awaken for the 2nd time and the coin is set to tails.
Step 3: The Beauty is awaken for the 3rd time and the coin is set to tails.
The Beauty knows the procedure, but when she is awaken she does not know whether she is awaken for the 1st, 2nd or 3rd time. So when the Beauty is awaken, what is her probability that the coin is set to heads?

I think everybody agrees that in this case $p=1/3$.

But if probability is defined in terms of frequencies, it seems to me that this version of the problem is equivalent to the original version. In other words, for a frequentist definition of probability, it seems to me that everybody should agree that the solution of the original problem is 1/3. It is only with the Bayesian definition of probability that a potential ambiguity remains. The question "What new information does she receive when she is awaken?" may be relevant in the Bayesian approach, but not in the frequentist approach. And this looks like an argument that the frequentist definition of probability is superior (in the sense of being well defined) over the Bayesian one.

Any comments?

I would argue that we can't use the frequentist approach or Bayesian approach (at least with the way the problem is usually set up). Consider that if you didn't lose your memory the frequencies would be the same but you definitely wouldn't answer 1/3, so we all agree that frequency alone is not enough to answer 1/3. We must explain why this specific type of memory loss makes the answer 1/3. But the issue is that even with memory loss it is not a random experiment: tuesday inevitably follows monday, tuesday tails inevitably follows monday tails.

In your simplified example, the memory loss makes it possible to use the principle of indifference to fix the problem. But that doesn't work in the sleeping beauty problem. Indifference can only be used when comparing monday tails and tuesday tails.

One approach is to base the answer on a solid random experiment. On wednesday at noon we know we will believe in 1/2. By the principle of reflection we should believe 1/2 now. Another approach is to argue that there is no relevant information change. Halfers would say that waking up only tells you "I wake up at least once", which you already knew. It is true that there are other changes, like not knowing what day it is, but it isn't clear why that is relevant, or why it would lead to 1/3. Another option is to say the probability is not currently well defined.

 

[PeroK]

I would argue that we can't use the frequentist approach

What about the following:

Someone else wanders into the experiment, which is explained to them. Let's assume they don't know what day it is (or, perhaps more sensibly, don't know the stage that the experiment has reached). All they see is the sleeper about to be woken.

The frequentist approach would would routinely be used. I see an event. I know that event can happen equally likely under three circumstances:

Monday/ first awakening (coin is heads)
Monday/first awakening (coin is tails)
Tuesday/second awakening (coin is tails)

With no further information and knowing these events to be equally likely, they would assign a probability of 1/3 to each.

Note that if you deny the frequentist approach here, then you must deny it in almost all cases. This is standard probability theory.

The sleeper is in precisley the same situation as the random observer. They have precisely the same information. You could write down everything they know about the experiment and their knowledge would be identical.

Therefore: if you deny the use of relative frequencies in this case, you must deny it across all of probability theory.

PS And, again, changing the numbers exposes the problem with the answer of 1/2. If the experiment lasts 365 days, then the random observer cannot possibly conclude that with 50% probability, they have randomly walked in on day 1. That is absurd. They must conclude that, if they just happen to see an awenkening on the random day they enter, then the coin was almost certainly tails. If it was heads, they would have seen nothing as the sleeper would be left alone for 364 days.

 

[Marana]

What about the following:

Someone else wanders into the experiment, which is explained to them. Let's assume they don't know what day it is (or, perhaps more sensibly, don't know the stage that the experiment has reached). All they see is the sleeper about to be woken.

The person wandering into the experiment can use the frequentist approach because from their perspective it is a random experiment. It would be possible for them to observe monday tails their first wander, monday heads their second wander. But that is impossible for sleeping beauty.

 

[PeroK]

The person wandering into the experiment can use the frequentist approach because from their perspective it is a random experiment. It would be possible for them to observe monday tails their first wander, monday heads their second wander. But that is impossible for sleeping beauty.

The random observer can't witness both, because those two events take place at the same time. In effect, they are the same physical event. The experimenters don't even have to look at the coin until Tuesday. They just wake the sleeper in any case on Monday.

What information does the observer have that the sleeper does not or vice versa?

The simple fact is that the Monday awakening is twice as likely as the Tuesday awakening. And, this applies equally to experimenters, random observers and the sleeper. The amnesia drug does nothing except effectively turn the sleeper into a random observer. There's no more to it than that.

 

[Marana]

The random observer can't witness both, because those two events take place at the same time. In effect, they are the same physical event. The experimenters don't even have to look at the coin until Tuesday. They just wake the sleeper in any case on Monday.

What information does the observer have that the sleeper does not or vice versa?

What I mean is the observer can repeat their wandering, so they could wander into one experiment and see monday with a tails coin flip, then wander into a totally different one and see monday with a heads coin flip. The observer can go around and observe many sleeping beauties with random results.

But each individual sleeping beauty has a different perspective. From their perspective tuesday tails will most definitely follow monday tails. Waking up is not a random experiment for them.

 

[PeroK]

What I mean is the observer can repeat their wandering, so they could wander into one experiment and see monday with a tails coin flip, then wander into a totally different one and see monday with a heads coin flip. The observer can go around and observe many sleeping beauties with random results.

But each individual sleeping beauty has a different perspective. From their perspective tuesday tails will most definitely follow monday tails. Waking up is not a random experiment for them.

This makes no sense. First, you can consider the relative frequencies hypothetically, even in the case of a single experiment. That's standard probability theory.

Otherwise, your answer of 1/2 is invalid for a single experiment as 1/2 is the relative frequency of many coin tosses. A single coin toss is either 100% heads or 100% tails. It's never 50-50.

Second, you could have different (individual) random observers each time: it doesn't have to be the same person each time. The relative frequency approach works equally well in this case.

There is no difference. There is no reason to ditch probability theory. You are clutching at straws to sustain your a priori intuitive conclusion.

 

[PeroK]

The person wandering into the experiment can use the frequentist approach because from their perspective it is a random experiment. It would be possible for them to observe monday tails their first wander, monday heads their second wander. But that is impossible for sleeping beauty.

I looked again at the Wikepedia article. In particular:

"Therefore, the Sleeping Beauty problem is not about mathematical probability theory. Rather, the question is whether subjective probability or credence are well-defined concepts, and how they must be operationalized."

What I believe is that if you analyse the problem using probability theory (in particular relative frequencies), then you get a completely self-consistnet answer of 1/3. If you try to answer 1/2, then the probability of heads is inconsistent with the probability of 2/3 that it is Monday. And, if the probability is not 2/3 that it is Monday, then there is something very badly wrong.

An answer of 1/2, therefore, in my opinion takes you out of the realms of self-consistent probability theory.

You can, of course, take a philosophical position that the answer is 1/2 and probability theory does not apply. And, that is, of course, your philosophical prerogative. Although, quite what an answer of 1/2 actually means if it's not a probability I'll have to leave to the philosophers.

But, as far as mathematics is concerned, the answer is 1/3.

 

[Marana]

This makes no sense. First, you can consider the relative frequencies hypothetically, even in the case of a single experiment. That's standard probability theory.

https://en.wikipedia.org/wiki/Frequentist_probability
https://en.wikipedia.org/wiki/Experiment_(probability_theory)

"In the frequentist interpretation, probabilities are discussed only when dealing with well-defined random experiments"
"In probability theory, an experiment or trial (see below) is any procedure that can be infinitely repeated and has a well-defined set of possible outcomes, known as the sample space"

Going directly from the definition, it does not apply to sleeping beauty. Even in a hypothetical sense, it can't be repeated. Tuesday tails must follow monday tails.

This is not the same as a coin flip, which hypothetically could be repeated as much as you want.

 

[stevendaryl]

One approach is to base the answer on a solid random experiment. On wednesday at noon we know we will believe in 1/2. By the principle of reflection we should believe 1/2 now. Another approach is to argue that there is no relevant information change. Halfers would say that waking up only tells you "I wake up at least once", which you already knew. It is true that there are other changes, like not knowing what day it is, but it isn't clear why that is relevant, or why it would lead to 1/3. Another option is to say the probability is not currently well defined.

I already mentioned this, but it's possibly worth repeating: The halfer answer can be thrown into doubt by considering a similar thought-experiment that sounds like the probabilities should be the same, but which justifies the thirder answer.

Instead of considering just three possibilities--two wakenings for heads, one for tails--let's make it a little more symmetric by throwing in a second wakening for the tails case as well. But in the tails case, the subject is told the coin flip result on his second wakening.

Then letting X be the statement "the subject has not been told the result", we have:

P(X) = 3/4 (because there are 4 awakenings, and X is true for 3 of them)

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

I think both the Bayesian and frequentist accounts would agree with this answer. And it seems that the way in which it differs from the original Sleeping Beauty problem is irrelevant to the probabilities.

 

[stevendaryl]

What I believe is that if you analyse the problem using probability theory (in particular relative frequencies), then you get a completely self-consistnet answer of 1/3. If you try to answer 1/2, then the probability of heads is inconsistent with the probability of 2/3 that it is Monday. And, if the probability is not 2/3 that it is Monday, then there is something very badly wrong.

I'm about 98% in agreement with the thirder position, but it seems strange to me to base it on the fact that there is a 2/3 chance that today is Monday. Isn't that number just as contentious as the 2/3 versus 1/2 number?

 

[PeroK]

https://en.wikipedia.org/wiki/Frequentist_probability
https://en.wikipedia.org/wiki/Experiment_(probability_theory)

"In the frequentist interpretation, probabilities are discussed only when dealing with well-defined random experiments"
"In probability theory, an experiment or trial (see below) is any procedure that can be infinitely repeated and has a well-defined set of possible outcomes, known as the sample space"

As I said, you can certainly state that probability theory does not apply in this case. And, the fact that 1/3 is a self-consistent answer is then irrelevant.

But, you can't then take another tack and use some wooly intuitive alternative to probability theory to give an answer of 1/2, which is not even self consistent.

Finally, as is covered in the Wikipedia analysis. If the sleeper procedes on the basis of 1/3, she will make the right decision in terms of prizes and fates. But, if she procedes on the the basis of 1/2 she will make less favourable decisions. And, since no one has actualy lied to her: i.e. she has not been given any false information, why does she get things wrong?

The random observer could win more prizes by using 1/3 heads than she could be using 1/2. So, why is she disadvantaged if no one has given her false information and she has as much information as the random observer?

Finally, I see no reason why you couldn't repeat the experiment many times, awarding her a prize every time she guesses heads/tails correctly.

The sleeper who uses 1/3 and always guesses tails will win more than the sleeper who uses 1/2 and guesses equally heads/tails. And that, IMHO, is as good a measure of whether you have the right calculation or not (again, with the proviso that no one has actually lied to you).

 

[stevendaryl]

https://en.wikipedia.org/wiki/Frequentist_probability
https://en.wikipedia.org/wiki/Experiment_(probability_theory)

"In the frequentist interpretation, probabilities are discussed only when dealing with well-defined random experiments"
"In probability theory, an experiment or trial (see below) is any procedure that can be infinitely repeated and has a well-defined set of possible outcomes, known as the sample space"

Going directly from the definition, it does not apply to sleeping beauty. Even in a hypothetical sense, it can't be repeated. Tuesday tails must follow monday tails.

This is not the same as a coin flip, which hypothetically could be repeated as much as you want.

Right. There is something a little weird about a question like "What is the probability that today is Monday?" You can't randomly pick what day it is.

Your point about this being neither purely Bayesian nor purely frequentist is right. We're really being asked to compute a subjective probability, which makes it sound Bayesian. But the most straightforward way of computing it is to use a frequentist definition of probability.

 

[PeroK]

I'm about 98% in agreement with the thirder position, but it seems strange to me to base it on the fact that there is a 2/3 chance that today is Monday. Isn't that number just as contentious as the 2/3 versus 1/2 number?

If I were the sleeper I would bet on it being Monday with 2/3 probability. Or, more simply, bet on its being tails.

I would be happy to take bets with anyone on that. There's no way I could (on average) lose. Every time I get woken I bet tails on a 50-50 bet and I win 2/3 of the time. After being a sleeper for a year or two, I retire on my winnings. I could even give odds 55-45 and still come out ahead.

The halfer who bets equally on heads/tails can only break even.

And, the confident halfer who bets heads all the time will lose!

PS I don't even begin to understand why relative frequency wouldn't apply to that situation. Gee, I'd be happy to find a philosopher who was willing to lose all his money that way!

PPS Although maybe the amnesia drug would have side-effects!

 

[Demystifier]

A Decision Theory Approach

A part of the problem is that, in the usual formulation of the Sleeping Beauty problem, the probability and degree of belief are abstract ambiguous concepts the meaning of which is not completely clear. To overcome this ambiguity let me reformulate the problem as a decision problem. With a given rules of game, what is wise for the Beauty to do? I will present two versions of the game rules, one corresponding to 1/3 and the other to 1/2.

Let me explain the rules through a dialog:
Experimenter: Hi Beauty, do you want to play a game with me?
Beauty: First I need to know the rules.
Experimenter: I shall flip a coin. If you guess correctly what the coin has shown, I will give you 100$. If you guess incorrectly what the coin has shown, you will give me 100$.
Beauty: Sounds boring. Is there a catch?
Experimenter: Yes. I will flip the coin only ones, but you will make two guesses in the case of tails and one guess in the case of heads. For each guess you will be awaken from a sleep and you will not remember anything about previous (if any) awaking.
Beauty: Sounds interesting, but something is still not crystal clear to me. In the case of tails, will the 100$ be payed for each guess?
Experimenter: Hmm, you are a smart beauty, I didn't think of it.
Beauty: Well, I need to know, my strategy will depend on it.
Experimenter: OK, suppose that I propose the rule A: The 100$ is payed for each guess.
Beauty: Then I would be very happy to play this game, because I have a strategy which puts me in an advantage over you.
Experimenter: And what if, instead, I propose the rule B: The 100$ is payed for the first guess only.
Beauty: Then it is a fair game, where nobody has any advantage over the other one.

So what is the best Beauty's strategy in the case of rule A? What about the case of rule B? The answers should be obvious.

In the case of rule A the Beauty's strategy is always to say "tails". (It corresponds to the 1/3 answer in the usual formulation of the problem.)

In the case of rule B, there is no special strategy for Beauty because any guess is as good as any other. (It corresponds to the 1/2 answer in the usual formulation of the problem.)

 

[PeroK]

A Decision Theory Approach

A part of the problem is that, in the usual formulation of the Sleeping Beauty problem, the probability and degree of belief are abstract ambiguous concepts the meaning of which is not completely clear. To overcome this ambiguity let me reformulate the problem as a decision problem. With a given rules of game, what is wise for the Beauty to do? I will present two versions of the game rules, one corresponding to 1/3 and the other to 1/2...

As I understand it, everyone is supposed to agree on this, because this formalises things.

In my opinion, with rule B there is really no point in the second awakening. Unless we have rule A, then it's not the problem as stated.

What I can't grasp is the halfer objection to this as a logical formulation of the problem. It's exactly the way I think of probabilities: calculations based on the available information.

For me, the halfers must fall into two categories: those that disagree philosophically with your formulation; and, those who simply miscalculate on the basis of rule A.

 

[Demystifier]

As I understand it, everyone is supposed to agree on this, because this formalises things.

Yes.

In my opinion, with rule B there is really no point in the second awakening.

Well, in that case the rule B can be modified, e.g. such that another coin is flipped to decide whether the payment will be according to the first or second guess (but not both). It doesn't affect the conclusions and strategies.

 

[Demystifier]

For me, the halfers must fall into two categories: those that disagree philosophically with your formulation; and, those who simply miscalculate on the basis of rule A.

I want to believe that only the first category is represented in this thread.

 

[PeroK]

I'm about 98% in agreement with the thirder position, but it seems strange to me to base it on the fact that there is a 2/3 chance that today is Monday. Isn't that number just as contentious as the 2/3 versus 1/2 number?

Suppose the sleeper is told that she will always be woken on the Monday and there is a 50-50 probability of being woken on the Tuesday. Let's assume the decision mechanism is not specified.

Now we have no coin that must already be heads or tails to distract us.

In being woken, she must calculate it's 2/3 Monday and 1/3 Tuesday.

If not, please justify another answer.

If the decision is made via the outcome of a pre tossed coin, then this is just one of many possible mechanisms.

How could the specific mechanism affect the probability in this case?

And, if the mechanism does affect the outcome, how do you calculate in more complex cases where there is a probability $p_n$ of being woken on day $n$, where there is no intuitive a priori answer such as 1/2?

 

[stevendaryl]

Finally, as is covered in the Wikipedia analysis. If the sleeper procedes on the basis of 1/3, she will make the right decision in terms of prizes and fates.

The definition in terms of bets is dependent on how you set up the bets. If you say that upon each awakening, the sleeper is given the opportunity to bet, then of course she should bet on heads, because if it's heads, she gets two opportunities to bet, while if it's tails, she gets one opportunity. But the dependence on opportunity is kind of strange. It's sort of like the following game:

• A player is allowed to place two bets (each with a payoff of $1) on the outcome of a coin flip.
• If the result is heads, both bets are honored (win or lose).
• If the result is tails, only the first bet counts.

Given this game, it makes sense to bet on heads, because then you have an opportunity to win $2 or to lose only $1. But I wouldn't say that the odds of the coin landing heads are affected by the number of bets that are honored.

The Sleeping Beauty problem is equivalent to this game, as far as betting is concerned.

 

[Demystifier]

The definition in terms of bets is dependent on how you set up the bets. If you say that upon each awakening, the sleeper is given the opportunity to bet, then of course she should bet on heads, because if it's heads, she gets two opportunities to bet, while if it's tails, she gets one opportunity. But the dependence on opportunity is kind of strange. It's sort of like the following game:

• A player is allowed to place two bets (each with a payoff of $1) on the outcome of a coin flip.
• If the result is heads, both bets are honored (win or lose).
• If the result is tails, only the first bet counts.

Given this game, it makes sense to bet on heads, because then you have an opportunity to win $2 or to lose only $1. But I wouldn't say that the odds of the coin landing heads are affected by the number of bets that are honored.

The Sleeping Beauty problem is equivalent to this game, as far as betting is concerned.

Isn't this equivalent to my post #67?

 

[PeroK]

The definition in terms of bets is dependent on how you set up the bets. If you say that upon each awakening, the sleeper is given the opportunity to bet, then of course she should bet on heads, because if it's heads, she gets two opportunities to bet, while if it's tails, she gets one opportunity. But the dependence on opportunity is kind of strange. It's sort of like the following game:

• A player is allowed to place two bets (each with a payoff of $1) on the outcome of a coin flip.
• If the result is heads, both bets are honored (win or lose).
• If the result is tails, only the first bet counts.

Given this game, it makes sense to bet on heads, because then you have an opportunity to win $2 or to lose only $1. But I wouldn't say that the odds of the coin landing heads are affected by the number of bets that are honored.

The Sleeping Beauty problem is equivalent to this game, as far as betting is concerned.

PS I'm not sure I understand this. My reply below was based on misreading this post, I think. In any case, my reply simply restates how I believe the sleeper can make money given the scenario. As I understand it, everyone is supposed to agree with this is any case.

I'll keep my money until it's time to bet. So that I don't guess what day it is by looking at how much money I've got, I'll have an automatic system that simply gives me $1 any time I want (although, with this game, I'm on for more than a $1 a time!).

I get woken, I get my $1, I put down my bet and I say "tails". That is it. The rest is obfuscation!

And, I'd have a side bet on the day of the week as well, if I could.

The whole point of my argument is that if the coin is initially heads, you only get one bet. The counterargument says you must lose $2 on heads, because you only get to bet once if it's heads. Umm ... isn't that just about the definition of the conditional probability that if I get woken it's only half the chance it's heads than tails? That is absolutely the definition of "probability that it's heads".

This is really about putting your money where your philosophy is!

 

[PeroK]

For me, it is quite obvious that the correct answer is 1/2.
There are 3 cases: Head-Monday, Tail-Monday, Tail-Tuesday
Head vs Tail are 50% 50%
so we have: Head-Monday 50%, Tail-Monday 25%, Tail-Tuesday 25%
So probability it was head is 50% or 1/2.

Is that the case without the amnesia drug as well?

Isn't it strange that if it's Monday, it's twice as likely to be a Head as a Tail?

 

[stevendaryl]

Isn't this equivalent to my post #67?

Yes. I posted before reading your post.

 

[Khashishi]

Rather than calculating conditional probabilities, it's simpler if we just calculate all the total probabilities from the experimenter's point of view.
From the experimenter's point of view, there's a 50/50 chance of heads/tails.
So, there's a 50% chance of waking Sleeping Beauty once, and 50% chance of waking them twice.

So, Sleeping Beauty will be asked to answer once or twice. If they answer heads (or tails), there's a 50% chance that they are right.
The issue is that the trials are not independent. If Sleeping Beauty answers heads, they're either right once (50% chance) or wrong twice (50% chance). If they answer tails, they are right twice (50% chance) or wrong once (50%).

If they want to make a fair wager, they'd better bet more on tails, because even though the probability is the same, they stand to lose twice on the same flip on heads. Therefore, level of belief is NOT equal to level of fair wager.

 

[PeroK]

Rather than calculating conditional probabilities, it's simpler if we just calculate all the total probabilities from the experimenter's point of view.
From the experimenter's point of view, there's a 50/50 chance of heads/tails.
So, there's a 50% chance of waking Sleeping Beauty once, and 50% chance of waking them twice.

So, Sleeping Beauty will be asked to answer once or twice. If they answer heads (or tails), there's a 50% chance that they are right.
The issue is that the trials are not independent. If Sleeping Beauty answers heads, they're either right once (50% chance) or wrong twice (50% chance). If they answer tails, they are right twice (50% chance) or wrong once (50%).

If they want to make a fair wager, they'd better bet more on tails, because even though the probability is the same, they stand to lose twice on the same flip on heads. Therefore, level of belief is NOT equal to level of fair wager.

If you take the idea of a random observer who happens on the experiment. Why is their level of belief different from the sleeper?

Also, unless someone has lied to you, why would your level of belief be different from the probabilities you can calculate.

If I understand your position, it is that that the sleeper would say her belief is that heads is 50-50, even though she can calculate she would lose money by betting in this?

Personally, I can't see how belief can be mathematically different from the odds I can calculate. For example, I can't imagine a situation where I believe that a coin is definitely heads but am unwilling to bet on it - unless I know or suspect I have been lied to.

 

[PeroK]

The sleeping beauty problem is a well known problem in probability theory, see e.g.
https://en.wikipedia.org/wiki/Sleeping_Beauty_problem
http://allendowney.blogspot.hr/2015/06/the-sleeping-beauty-problem.html
https://www.quantamagazine.org/solution-sleeping-beautys-dilemma-20160129
or just google.

Allegedly, there are many "thirders" who think that the correct answer is 1/3, but also many "halfers" who think that the correct answer is 1/2. For me, it is quite obvious that the correct answer is 1/3. Is there anybody here who is convinced that the correct answer is 1/2? If you are one of them, what is your argument for 1/2?

Sadly, dispiritingly, shamefully(?) it appears that on the whole of PF, in terms of thirders, it's only thee and me.

I thought this was a clear-thinking scientific forum, but, in this case, we appear to be outnumbered by woolly philosophical thinking.

In the spirit of some of the posts above I believe that everyone on PF is a thirder, even though I can calculate that halfers are in the majority.

 

[stevendaryl]

Sadly, dispiritingly, shamefully(?) it appears that on the whole of PF, in terms of thirders, it's only thee and me.

You don't think 98% agreement is good enough to join your club?

 

[forcefield]

The coin flip creates two equally likely scenarios. Hence the probability is 1/2.

 

[PeroK]

You don't think 98% agreement is good enough to join your club?

Given the meagre membership, I think we can lower the entrance criteria in your case.

 

[forcefield]

The coin flip creates two equally likely scenarios. Hence the probability is 1/2.

I was only a part-time halfer: I should add that she has information about the relative probabilities of her being awake in the two scenarios. Hence 1/3.

 

[lewando]

I like to think the SB is the inquisitive type and would be happy to submit to a large series of trials. SB will precede each trial with a simple wager to probe her experience. The wager (claim, really) is this: “the coin flip will be heads”. She will always make the same wager before each trial. At the end of each trial (Wednesday), she will be informed of, and record, the result of the actual coin flip and will have access to this information at all times throughout future trials. As the number of completed trials becomes large enough to be statistically significant, her answer during the interview phase will become more correct.

 

[Demystifier]

You don't think 98% agreement is good enough to join your club?

It is always healthy to question one owns convictions. For that purpose, in post #67 I have given a case in which 1/2 is correct. I hope it will not me banish from the club.

 

[PeroK]

It is always healthy to question one owns convictions. For that purpose, in post #67 I have given a case in which 1/2 is correct. I hope it will not me banish from the club.

You were the founder member!

 

[Demystifier]

You were the founder member!

The best way to become the club president is to banish the founder.

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.

 

[PeroK]

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.

 

[Demystifier]

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.

The best disproof of 1/2 so far.

 

[Marana]

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.

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.

 

[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.