The Sleeping Beauty Problem: Any halfers here?

[PeroK]

PS just to confirm one point. The sleeper knows she will be given the amnesia drug and if she is woken on a Tuesday she knows she might have been given it.

If she doesn't know about the drug but knows about the experiment otherwise, then that is a different problem. In that case, whenever she is woken she is sure that it is Monday (but may be wrong on account of the drug) and will say 1/2 on heads/tails.

If the drug erases all her memory of the experiment, then if she gets woken on the Tuesday, she will have no idea why she is being asked and would guess 1/2 having list all the information about the experiment.

Those are different problems. In the main problem she remembers everything except the first awakening.

 

[PeroK]

PPS I've distilled the issue down to this.

The sleeper is asked the following question:

How many times have you been awoken?

At the start of the experiment, she answers "none".

When she is awoken she answers " I don't know".

Therefore, she does have different information in the two cases. QED

 

[Marana]

This post contains ubsubstantiated opinions with no hard analysis.

None of the claims you attribute to "thirders" can be attributed to my posts.

The thirder position does not depend on intuition but on solid analysis of probabilities.

Moreover, my post #28 shows the two reasons that the 1/2 answer is actually wrong.

Which of those things I attributed to thirders do you disagree with? I think they are all standard for thirders.

 

[stevendaryl]

This post contains unsubstantiated opinions with no hard analysis.

Well, an analogous lottery thought-experiment is this: 1000 people play the lottery. The winner is kept a secret for the sake of the thought experiment. A researcher finds out who the real winner is, and does the following:

1. Each loser is woken on day 1 and asked what the odds are that he won.
2. The one winner is woken up 1 billion days in a row and asked what the odds are that he won (and memory is erased afterward)

So what answer should one of the lottery players give, when asked what are the odds that he won?

By analogy with the thirder reasoning, there are on the average 1 billion wakenings in which the person being woken was the winner, and only 999 wakenings in which the person was the loser. So using relative frequency, you conclude that there is about a million to 1 odds of being the winner. So each of the 1000 should be jumping for joy, since he is almost certainly the winner (in his own mind).

 

[stevendaryl]

PPS I've distilled the issue down to this.

The sleeper is asked the following question:

How many times have you been awoken?

At the start of the experiment, she answers "none".

When she is awoken she answers " I don't know".

Therefore, she does have different information in the two cases. QED

Ah! That's interesting. So let's change the experiment slightly. You flip the coin, and if heads, you wake up the sleeper on three consecutive days. If tails, only on two consecutive days. Then by analogy with the thirder position, you would say that when woken, the sleeper should say that the odds are 3/5 that the answer was "heads", and 2/5 that the answer was "tails". Now, the twist: On day 1, we tell the sleeper "This is day 1". On the other days, we do not.

Now, the relative frequency of day 1 is 2/5. The relative frequency of day 2 is 2/5. the relative frequency of day 3 is 1/5.

Now, the usual conditional probability formulas would tell us:

1. P(H| D=1) = P(H \wedge D=1)/P(D=1) = \frac{1}{5}/ \frac{2}{5} = \frac{1}{2}
2. P(H| D \neq 1) = P(H \wedge D \neq 1)/P(D \neq 1) = \frac{2}{5} / \frac{3}{5} = \frac{2}{3}

This scenario is practically the same as the original Sleeping Beauty scenario, and it shows that it is self-consistent to think the probability of heads is 1/2 on day 1 and 2/3 on subsequent days.

So I'm convinced that it's all self-consistent.

 

[Demystifier]

It seems that nobody payed attention to my $k=0$ argument in post #4. Let me explain this argument in more detail. Consider the generalized Sleeping Beauty problem defined as follows:
In the case of tails, the Beauty will be awaken $n$ times.
In the case heads, the Beauty will be awaken $k$ times.
What is the probability $p(heads;n,k)$?
The original Sleeping Beauty problem corresponds to $n=2$, $k=1$.

According to halfers, the solution of the original problem is $p(heads;2,1)=1/2$. But this means that halfers think that $p(heads;n,k)=1/2$ is independent of $n$ and $k$. On the other hand this cannot be correct because it is certainly wrong for $k=0$. It is quite obvious that $p(heads;n,0)=0$.

The halfer might argue that $p(heads;n,k)=1/2$ is correct only for $n\neq 0$ and $k\neq 0$. But then he/she must explain why $n=0$ or $k=0$ is an exception.

EDIT: If someone wonders, the general solution (consistent with a thirder's way of reasoning) is
$$p(heads;n,k)=\frac{k}{n+k}$$
It is ill defined only for $n=k=0$, which is perfectly sensible because in that case the Beauty is never awaken so the question "what is her probability when she is awaken" does not make sense. It must be assumed from the beginning that at least one of the numbers $n$ and $k$ is non-zero.

 

[PeroK]

It seems that nobody payed attention to my $k=0$ argument in post #4. Let me explain this argument in more detail. Consider the generalized Sleeping Beauty problem defined as follows:
In the case of tails, the Beauty will be awaken $n$ times.
In the case heads, the Beauty will be awaken $k$ times.
What is the probability $p(heads;n,k)$?
The original Sleeping Beauty problem corresponds to $n=2$, $k=1$.

According to halfers, the result of the original problem is $p(heads;2,1)=1/2$. But this means that halfers think that $p(heads;n,k)=1/2$ is independent of $n$ and $k$. On the other hand this cannot be correct because it is certainly wrong for $k=0$. It is quite obvious that $p(heads;n,0)=0$.

The halfer might argue that $p(heads;n,k)=1/2$ is correct only for $n\neq 0$ and $k\neq 0$. But then he/she must explain why $n=0$ or $k=0$ is an exception.

Your $n, k$ argument is sound and should be yet another nail in the 1/2 coffin.

For me, this has become more of a psychological question about how far one is prepared to go in denying mathematical principles and calculations in order to defend an a priori intuitive conclusion.

In this respect, it is a bit like the question of whether $0.999 \dots = 1$. Although rigorous calculations show this to be true, those that oppose it do so on "intuitive" grounds and - to some extent - no amount of rigorous calculation can dissuade them. Eventually, they demand that basic mathemtical principles are overturned in order to preserve this particular inequality.

One argument evinced against your $n, k$ solution on this thread is that "you are changing the problem". The problem must be left precisely as it is and any attempt to illuminate the flaw in the 1/2 intuitive logic by altering the numbers is not allowed.

 

[Demystifier]

One argument evinced against your $n, k$ solution on this thread is that "you are changing the problem". The problem must be left precisely as it is and any attempt to illuminate the flaw in the 1/2 intuitive logic by altering the numbers is not allowed.

Yes, but my generalized problem contains the original problem as a special case. It is not rare in mathematics that a specific problem is easier to solve by considering it as a special case of a more general problem.

 

[PeroK]

Yes, but my generalized problem contains the original problem as a special case. It is not rare in mathematics that a specific problem is easier to solve by considering it as a special case of a more general problem.

That is exactly the sort of mathematical principle that must be abandoned in order to preserve an answer of 1/2 in this particular case.

 

[Marana]

The halfer might argue that $p(heads;n,k)=1/2$ is correct only for $n\neq 0$ and $k\neq 0$. But then he/she must explain why $n=0$ or $k=0$ is an exception.

To a halfer, waking up gives you the information that "I wake up at least once." So it is self evident why 0 is an exception: you don't always wake up at least once in that case.

We should also remember that waking up is not a random experiment. Waking up is not independent of other times you wake up. For example, it is impossible to wake up in the order HTH. The T is forced to have another T next to it. I believe that this disqualifies the typical thirder analysis, which just uses the basic stuff we do for random experiments. Since waking up is not a random experiment it isn't totally clear how, or even if, we should define the probability.

Halfers can call upon the principle of reflection, saying that at noon on wednesday they will believe in 1/2 (because wednesday is a random experiment), and therefore argue that they should believe in 1/2 now.

Thirders can call upon the principle of indifference, but again they have a problem, because monday and tuesday are not random experiments. I'm not saying that proves 1/2 is the answer, but 1/2 does have a more solid foundation because it is based on a genuine random experiment (wednesday), while 1/3 is not.

 

[Demystifier]

To a halfer, waking up gives you the information that "I wake up at least once." So it is self evident why 0 is an exception: you don't always wake up at least once in that case.

So your point is: When $k=0$, $n\neq 0$, then waking up gives a new information because you didn't know that you will wake up. When $k\neq 0$, $n\neq 0$, then waking up does not give a new information. Interesting!

However, the thirder can reply that waking up gives a new information even for $k\neq 0$, $n\neq 0$, with the only caveat that it is an uncertain information. (I don't know why, but it reminds me of the raven paradox https://en.wikipedia.org/wiki/Raven_paradox where observing a red apple increases probability that the hypothesis "all ravens are black" is true.)

 

[Demystifier]

If someone is interested, there is also a quantum version of the problem, related to the concept of probability in many-worlds interpretation of quantum mechanics:
http://www.tau.ac.il/~vaidman/lvhp/m123.pdf
http://philsci-archive.pitt.edu/9948/1/QSBP_pitts.pdf

 

[stevendaryl]

It seems that nobody payed attention to my $k=0$ argument in post #4. Let me explain this argument in more detail. Consider the generalized Sleeping Beauty problem defined as follows:
In the case of tails, the Beauty will be awaken $n$ times.
In the case heads, the Beauty will be awaken $k$ times.
What is the probability $p(heads;n,k)$?
The original Sleeping Beauty problem corresponds to $n=2$, $k=1$.

According to halfers, the solution of the original problem is $p(heads;2,1)=1/2$. But this means that halfers think that $p(heads;n,k)=1/2$ is independent of $n$ and $k$. On the other hand this cannot be correct because it is certainly wrong for $k=0$. It is quite obvious that $p(heads;n,0)=0$.

Well, the case k=0 is a more straight-forward application of the usual conditional probability formula.

Let A be "the coin landed heads"
Let B be "the sleeper just woke up and was asked to give her estimation of the probability"

Then the usual conditional probability formula gives:

P(A | B) = P(A \wedge B)/P(B)

But if the sleeper is never awakened unless the coin landed heads, then P(A \wedge B) = P(B) and the conditional probability yields 1.

In the case of n > 0 and k > 0, the conditional probability formula doesn't seem to help, because P(B) = 1.

So I don't think that the k=0 case tells us much about the case n=2, k=1

 

[Demystifier]

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?

 

[PeroK]

Halfers can call upon the principle of reflection, saying that at noon on wednesday they will believe in 1/2 (because wednesday is a random experiment), and therefore argue that they should believe in 1/2 now.

Thirders can call upon the principle of indifference, but again they have a problem, because monday and tuesday are not random experiments. I'm not saying that proves 1/2 is the answer, but 1/2 does have a more solid foundation because it is based on a genuine random experiment (wednesday), while 1/3 is not.

This is a good example of how a lack of precision leads to the difference of answers. You introduce Wednesday, but do not analyse what happens on Wednesday. If we do that:

a) In the problem as stated, she is woken on Wednesday and told the experiment is over (effectively telling her what day it is). Now, she reverts to the pre-experiment answer, as she has no knowledge of what happened during the experiment.

This exemplifies why the "no new information" argument is wrong:

At the beginning and end of the experiment she knows what day of the week it is.

When she is woken during the experiment, she does not know what day of the week it is. That is a significance difference in her knowldege.

b) If you do not initially tell her that it is Wednesday, then you effectively extend the problem to the 3-2 version, where she is always woken on Mon and Wed but only woken on Tuesday after a tail.

In that experiment (until she is told the day of the week), then she answers 2/5 each time she is woken.

As far as I can see, 1/2 has no solid foundation except intuition and a reluctance to analyse the number.

 

[stevendaryl]

As far as I can see, 1/2 has no solid foundation except intuition and a reluctance to analyse the number.

The fact that in this case, the conditional probability cannot be obtained by the Bayesian updating rules seems to me to make this problem very different from other applications of probability.

A priori, before the cockamamie scheme is even mentioned, someone flips a coin and asks the subject what the probabilities are, and she says 50/50 heads or tails. So the prior probability of heads is P(H) = 1/2. Then the experimenter explains the rules for wakening and memory wipes and so forth. The next morning, the sleeper is awakened, and is asked what the probability is. She now says 2/3 heads.

Bayesian probability would say that if you learn new information X, then your revised probability of heads will be given by:

P(H | X) = P(H \wedge X)/P(X) = P(H) P(X | H)/P(X) = 1/2 P(X | H)/P(X)

We can similarly compute:

P(T | X) = 1/2 P(X | T)/P(X)

So whatever X is, if P(H|X) = 2/3 and P(T|X) = 1/3, then we have:

P(X|H) = 2 P(X|T)

So what is X? On the one hand, it's twice as likely when the coin lands heads than when the coin lands tails. On the other hand, for X to even come into play in the Bayesian updating, it has to be something that the sleeper learns upon wakening. But if she is guaranteed to learn it, then I would think that P(X | H) = P(X | T) = 1. That seems like a contradiction.

So this seems to be an example where Bayesian updating fails. That's pretty significant, so it's not correct to treat this as a "Monty-Hall" or "1 = 0.999..." type confusion over mathematical principles.

 

[stevendaryl]

Here's yet another take on the paradox that perhaps sheds light on the strange change of probability from 1/2 to 2/3.

Instead of coin flips and wakenings, let's look at offspring. Suppose that in a certain country where the people reproduce asexually, the statistics are that half the people have one child, and half the people have two children. Then the a priori probabilities are: P(two-child) = P(one-child) = 1/2. But if you take a random person and ask whether their parent is a one-child parent or a two-child parent, and it's twice as likely that they will answer "two-child".

Let's suppose further that the custom in this country is to take children from the parents at birth and raise them in orphanages. So typically, nobody knows who their parents are, or who their siblings are. Now, take a random adult and find their children and ask each child: What are the odds that this person has two children? They will answer 1/2. But now tell them that this person is their parent, and they will revise the answer to 2/3.

The sleeping beauty case is sort of similar, if you consider each wakening as a different person (after all, the person at second wakening doesn't share any memories created by the person at first wakening). The coin flip is their parent. So if you ask a sleeping beauty what are the odds that the coin flip was heads, she will say 1/2. If you then explain that this coin flip was her "parent", she will revise it to 2/3.

 

[PeroK]

The fact that in this case, the conditional probability cannot be obtained by the Bayesian updating rules seems to me to make this problem very different from other applications of probability.

A priori, before the cockamamie scheme is even mentioned, someone flips a coin and asks the subject what the probabilities are, and she says 50/50 heads or tails. So the prior probability of heads is P(H) = 1/2. Then the experimenter explains the rules for wakening and memory wipes and so forth. The next morning, the sleeper is awakened, and is asked what the probability is. She now says 2/3 heads.

Bayesian probability would say that if you learn new information X, then your revised probability of heads will be given by:

P(H | X) = P(H \wedge X)/P(X) = P(H) P(X | H)/P(X) = 1/2 P(X | H)/P(X)

We can similarly compute:

P(T | X) = 1/2 P(X | T)/P(X)

So whatever X is, if P(H|X) = 2/3 and P(T|X) = 1/3, then we have:

P(X|H) = 2 P(X|T)

So what is X? On the one hand, it's twice as likely when the coin lands heads than when the coin lands tails. On the other hand, for X to even come into play in the Bayesian updating, it has to be something that the sleeper learns upon wakening. But if she is guaranteed to learn it, then I would think that P(X | H) = P(X | T) = 1. That seems like a contradiction.

So this seems to be an example where Bayesian updating fails. That's pretty significant, so it's not correct to treat this as a "Monty-Hall" or "1 = 0.999..." type confusion over mathematical principles.

I'll accept that it's trickier and perhaps there is something deeper. But, to be honest, I don't see it.

Your Bayesian analysis would seem to depend on a new piece of information $X$, which is not applicable in this case. It's not a new piece of information but the change in circumstances, scenario and knowldege caused by the amnesia drug.

One solution to the Bayesian conumdrum, which I originally suggested, is to regard the sleeper as effectively two different people (given that the amnesia drug has potentially removed information). Then the $X$ is simply "I have been selected". That seems valid to me.

And, the fundamental problem with 1/2 as an answer is that you are forced to conclude that it 3/4 probability of being Monday. And, that cannot be explained if you analyse the day of the week first. Show me the analysis that confirms that it is 3/4 that it is Monday.

In other words, instead of trying to introduce a new piece of information $X$, you ask the sleeper two questions:

a) Do you know what day of the week it is? If not, from what you remember, can you calculate the probability that it is Monday or Tuesday?

b) Do you know what the result of the coin toss was? If not, from what you remember, can you calculate the probability that it is Heads or Tails?

This exposes the difference between the sleeper at the beginning and end from the sleeper during the experiment. Trying to fit that into a specific new piece of information $X$ may fail. But, it's a clear change of scenario/information/call it what you will.

The sleeper can itemise the things she remembers, so that everyone is clear on what basis she makes her calculation.

 

[Demystifier]

Here's yet another take on the paradox that perhaps sheds light on the strange change of probability from 1/2 to 2/3.

Instead of coin flips and wakenings, let's look at offspring. Suppose that in a certain country where the people reproduce asexually, the statistics are that half the people have one child, and half the people have two children. Then the a priori probabilities are: P(two-child) = P(one-child) = 1/2. But if you take a random person and ask whether their parent is a one-child parent or a two-child parent, and it's twice as likely that they will answer "two-child".

Let's suppose further that the custom in this country is to take children from the parents at birth and raise them in orphanages. So typically, nobody knows who their parents are, or who their siblings are. Now, take a random adult and find their children and ask each child: What are the odds that this person has two children? They will answer 1/2. But now tell them that this person is their parent, and they will revise the answer to 2/3.

The sleeping beauty case is sort of similar, if you consider each wakening as a different person (after all, the person at second wakening doesn't share any memories created by the person at first wakening). The coin flip is their parent. So if you ask a sleeping beauty what are the odds that the coin flip was heads, she will say 1/2. If you then explain that this coin flip was her "parent", she will revise it to 2/3.

This is very similar to the boy-or-girl paradox
https://en.wikipedia.org/wiki/Boy_or_Girl_paradox
the most interesting version of which is the tuesday paradox
https://en.wikipedia.org/wiki/Boy_or_Girl_paradox#Information_about_the_child
https://www.jesperjuul.net/ludologist/2010/06/08/tuesday-changes-everything-a-mathematical-puzzle/

 

[PeroK]

In other words, instead of trying to introduce a new piece of information $X$, you ask the sleeper two questions:

a) Do you know what day of the week it is? If not, from what you remember, can you calculate the probability that it is Monday or Tuesday?

b) Do you know what the result of the coin toss was? If not, from what you remember, can you calculate the probability that it is Heads or Tails?

PS This, to me, exposes why the Bayesian approach fails. You cannot map this change of scenario into a single additional piece of information. It's trying to fit a cockamamie peg into a Bayesian hole, if you'll pardon that expression.

 

[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.