I The Sleeping Beauty Problem: Any halfers here?

[Demystifier]

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?

 

[fresh_42]

What is your argument of a third? She will not be interviewed on Wednesday, which leaves two possible days.
As the coin decides on which day(s) she will be awakened, her a priori probability for heads (= Monday only) is .5.
Now the question is, whether her a posteriori probability changed during the experiment. However, due to her amnesia, there is no way for her to gain additional information, except it is neither Sunday nor Wednesday which she knew already before. Consequently her guess still cannot be better (or worse) than .5.

 

[Demystifier]

Let me present an argument for 1/3, by analysing a similar problem from my actual real experience. Let us divide a day into a two equal halves, namely the range 24:00-12:00 and the range 12:00-24:00. When I awake during a sleep, at first I am not fully conscious, so at first I don't know what part of the day it is. Nevertheless, almost always I automatically assume that I am somewhere in the 24:00-12:00 range. And almost always I turn out to be right, despite the fact that, a priori, the two ranges should be equally likely. That's justified because I rarely go to sleep in the afternoon. (But sometimes I do go to sleep at afternoon hours, and in such cases, when I awake, at first I get confused when I see what time it is.)

 

[Demystifier]

What is your argument of a third? She will not be interviewed on Wednesday, which leaves two possible days.
As the coin decides on which day(s) she will be awakened, her a priori probability for heads (= Monday only) is .5.
Now the question is, whether her a posteriori probability changed during the experiment. However, due to her amnesia, there is no way for her to gain additional information, except it is neither Sunday nor Wednesday which she knew already before. Consequently her guess still cannot be better (or worse) than .5.

Are you a genuine halfer, or just a devil's advocate?

Anyway, perhaps the best way to see why 1/3 is the right answer is to go to the extreme. Instead of awaking the Beauty n=2 times, let us awake her n=1.000.000 times in the case of tails (and only k=1 times in the case of heads). So when she is awake, she can be almost certain that it was tails, so the probability for heads must be very small. And if n=1.000.000 is not convincing enough, consider the limit $n\rightarrow\infty$. If this is still not enough, vary also k and consider the limit $k\rightarrow 0$.

 

[fresh_42]

But this changes her a priori chances: she knows beforehand that heads is basically impossible. If we agree on the fact, that the two probabilities (before and after) don't differ, then the question is only what she can say on Sunday, even without the entire experiment. And these chances are even. I simply don't see how additional information should enter the system, like it does in the three doors riddle.

 

[PeroK]

It depends how you define "credence". You could extend the problem thus:

If she guesses wrong, then another coin is tossed and if it's tails she is executed. Now, if she guesses heads every time, then she has a 50-50 chance of being wrong but only once. However, if she guesses tails every time she has a 50-50 chance of being wrong, but she will be wrong twice, so has an increased chance of execution.

This leads to a bias for heads over tails.

I would compare it to a similar scenario: toss a coin and ask one or two different people at random. If the people know this rule then the fact that they have been picked favours heads. If they don't know this rule, they must go 50-50.

The princess is effectively two different people on account of the amnesia.

 

[PeroK]

PS change the rule so that she gets woken every day for a year (heads) or only once a year (tails). I think that blows the 50-50 argument.

 

[fresh_42]

PS change the rule so that she gets woken every day for a year (heads) or only once a year (tails). I think that blows the 50-50 argument.

Wouldn't this contradict

The princess is effectively two different people on account of the amnesia.

So if you have a different person every day of the year, it is still 50:50 for each single person assuming they cannot communicate.

 

[PeroK]

Wouldn't this contradict

So if you have a different person every day of the year, it is still 50:50 for each single person assuming they cannot communicate.

Not if they know the rule about two or more people getting woken if it's heads. I see I got the role of heads and tails mixed up, but if I persevere with the way I wrote it.

If you toss a coin and if it's heads you ask everyone in Belgium, but if it's tails you pick one person at random and ask only them. If everyone knows this rule, even if they cannot communicate, they know that by virtue of being asked it is almost certainly heads.

If there were a cash prize for being right, it would be foolish to guess tails.

 

[PeroK]

PS having read a bit more. One of the halfer arguments is that the princess "learns nothing new" about the coin when she is woken, so should stick with 1/2. Therefore, being woken involves learning nothing new.

Now, change the problem so that she is only woken if it is heads and not tails. She has still learned nothing new if she is woken, so sticks with 1/2, even though it's now 100% heads.

 

[TeethWhitener]

So when she is awake, she can be almost certain that it was tails

I don't think this is obvious at all. The probability that she will be woken up is 1 (regardless of whether the coin is heads or tails). She has no way of knowing whether she has previously been woken up, so I don't see how having n=a zillion, k=1 changes anything at all. I think of it this way: let's say we have a two-day period. We flip a coin, and if it's heads, sleeping beauty is awake for the entire 2-day period. If it's tails, she's awake for a random subset of that period whose measure equals 1 day. At some point while sleeping beauty is awake, we receive a note in that period saying "Sleeping beauty is awake right now." At least as far as I'm interpreting the problem, we receive the note regardless of what the outcome of the coin toss is. I can't see how we've gained any knowledge about the situation by receiving the note, and yet to me, the problem is essentially equivalent to the initial problem.

 

[fresh_42]

PS having read a bit more. One of the halfer arguments is that the princess "learns nothing new" about the coin when she is woken, so should stick with 1/2. Therefore, being woken involves learning nothing new.

Now, change the problem so that she is only woken if it is heads and not tails. She has still learned nothing new if she is woken, so sticks with 1/2, even though it's now 100% heads.

All these counter arguments against the .5 fraction involve a change on the set up and the knowledge of it. Therefore it doesn't affect the original question with a .5 chance plus the argument, that there is no additional information available during the experiment. It's exactly this lack of information that makes the situation different from the standard three doors example for conditional probabilities.

 

[PeroK]

All these counter arguments against the .5 fraction involve a change on the set up and the knowledge of it. Therefore it doesn't affect the original question with a .5 chance plus the argument, that there is no additional information available during the experiment. It's exactly this lack of information that makes the situation different from the standard three doors example for conditional probabilities.

The flaw in your argument is this: The question you imagine asking her at the start is:
If I toss a coin now, what is the probability it is heads? To which the answer is 1/2.

But, later, this is not precisely the question. The question now is: We have woken you up ...

If you asked her the same question at the beginning it would be:

If we toss a coin and wake you up twice if it's heads and once if it's tails, then at the time we wake you,what is the probability it is heads?

To which the answer is 2/3.

So, the answer to the precise question she is asked under the precise circumstances is 1/3 - 2/3. It is never 1/2 to begin with.

That's the fundamental flaw. The answer to the question is not 1/2 in the first place, so has no need to change. The information is all there at the outset about the precise circumstances under which the question will be asked.

The question is always asked under conditional circumstances. It is never asked under the unconditional circumstances of a single coin toss.

 

[PeroK]

PS and, moreover, it's never 1/2 for the experimenters. It's always known to them by the time they wake her and ask the question.

When the question is asked the coin toss is known. But, the princess has the knowledge she always had that when she is wakened the coin is more likely to already have been a head than a tail.

The experimenters know this because they know that if they are waking her it is more likely to have been a head.

 

[fresh_42]

Let's take the Wikipedia setting: heads = Monday only ; tails = Monday and Tuesday
The chances to get awoken twice equals the chances to get awoken once. That doubles the chances for a Monday.
However, in case of Monday she still don't know anything about the coin flip. But she will be asked: "What is your credence now for the proposition that the coin landed heads?" So although the chances for Monday are biased, the chances for the coin flip are not.

 

[PeroK]

PPS here's my solution:

At the outset, the experimenters and the princess have the same information about the experiment.

You ask the experimenters first: If you are waking the princess, what is the probability the coin is a head. This is clearly 1/3.

But, as the princess has the same information at the start, she must answer the same:

If you are being woken, what is the probability the coin is heads. She must answer 1/3.

When she is being woken, nothing has changed for her - no new info - so she sticks with 1/3.

It's an illusion that the probability is ever 1/2 and must somehow change to 1/3.

 

[fresh_42]

You ask the experimenters first: If you are waking the princess, what is the probability the coin is a head. This is clearly 2/3.

This is what I don't see. The chances are 50:50 on Tuesday and 50:50 on not Tuesday as rest. But I get the feeling this riddle is very similar to one I once read in a little book from Martin Gardner. A criminal has been sentenced to death within a week and the judge says, he won't know the day. His lawyer celebrated this as a victory, because it obviously can't be on Sunday, which would be the last day and in which case he knew on Saturday. But for the same reason it can't be Saturday for he would know on Friday and so on. Finally he was very surprised as it happened on Wednesday morning.

 

[PeroK]

This is what I don't see. .

That's just elementary conditional probability theory, surely. No tricks. No embellishments.

The conditional probability of a head given the instance of a waking.

You can do it with a simple probability tree.

 

[fresh_42]

That's just elementary conditional probability theory, surely. No tricks. No embellishments.

The conditional probability of a head given the instance of a waking.

You can do it with a simple probability tree.

I know. That's why the day of the week is 2:1, but that doesn't affect the coin flip, which remains 1:1. And she isn't asked for the day, only for the coin.

 

[PeroK]

I know. That's why the day of the week is 2:1, but that doesn't affect the coin flip, which remains 1:1. And she isn't asked for the day, only for the coin.

If you take away all the confusing complications you are left with - for the experiment at least - a simple conditional probability. You can't look at it as a simple coin flip. It's the conditional probability of heads given some event.

Effectively you have a random event that occurs twice after a tail and once after a head. The conditional probability of a head, given the event is 1/3.

This is basic stuff. The problem has muddled this - with its talk of sleeping beauty - and obscured this.

The coin flip isn't affected as such by the event, but the probability the event resulted from a head is not necessarily 1/2.

For the experimenters, how is this problem different from any conditional probability?

If we hadn't been introduced to the sleeping beauty, we would never have thought twice about this.

The problem with the 1/2 answer - for the experimenters at least - is that it essentially denies that the conditional probability of a coin toss can ever be different from 1/2, because the subsequent event selection process cannot affect the original coin toss.

 

[PeroK]

Here's another argument. Suppose the answer is 1/2. She gets woken. It's probability 1/2 that the coin was a head and hence Monday. And, it's probability 1/2 that the coin was a tail, hence 1/4 that it's Monday and 1/4 that it's Tuesday.

The princess, therefore, calculates that it's 3/4 Monday and 1/4 Tuesday.

In other words, out of every 4 times she gets woken, 3 are Monday and one is Tuesday.

But, she knows from the outset that if she gets woken it's 2/3 Monday and 1/3 Tuesday.

It can't be 1/2, therefore, because the calculation of how likely it is to be Monday goes wrong.

It should be clear that she is always woken on a Monday and only 1/2 of the time on a Tuesday. So, if she is woken and has no other information, it must be 2/3 that it's.Monday and 1/3 that it's Tuesday.

 

[fresh_42]

I do follow your argument, that's not the point. But there is only one coin flip, and one experiment. And as you mentioned earlier, the coin flip already took place when she wakes up. So there is no way she can improve her chances, as they are the same as they were when she fell asleep. And due to missing information on former wake-ups and current day, she can't improve on it. It's the lack of information that makes me think, that it is not a case of conditional probabilities. Being awake cannot influence the outcome of a past event which had been 50:50.
And as I mentioned earlier, and also in the discussion of it on the Wiki page, all arguments against a 50:50 call rely on different set ups like repetitions or similar. Her wake up call is a post selection event.

I guess there's something to the summary on Wiki: "All this seems to be consensual among philosophers. 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."

 

[PeroK]

I do follow your argument, that's not the point. But there is only one coin flip, and one experiment. And as you mentioned earlier, the coin flip already took place when she wakes up. So there is no way she can improve her chances, as they are the same as they were when she fell asleep. And due to missing information on former wake-ups and current day, she can't improve on it. It's the lack of information that makes me think, that it is not a case of conditional probabilities. Being awake cannot influence the outcome of a past event which had been 50-50.

This is missing the point. The experimenters looked at the coin and this resolved, with 100% probability, what the coin actually was. That has changed. What the coin landed is now known.

The experimenters then follow a course of action based on the actual outcome of the coin flip. This gives someone else the ability to deduce the coin flip from their actions. Or to adjust the likelihood that it was heads.

One of the reasons you cannot allow the game to be changed at all is that there are factors that allow 1/2 to seem like a plausible answer. It's not that different numerically from 1/3. But if you change the game so that your answer, by the same logic, remains 1/2 while the alternative reduces to 1/365, say, then it becomes harder to justify 1/2 as it's become numerically absurd.

Another change that exposes the issue is to extend the game by a day. If it's heads she gets woken on Monday and if it's tails on Tuesday and Wednesday. The game ends on Thursday.

Now, if she answers 1/2 for heads, she must also answer 1/2 for Monday. But, if all she knows is that she has been wakened, it's equally likely to be Monday, Tuesday or Wednesday. How can she tell it's (more likely to be) Monday?

How does she know it's twice as likely to be Monday as Tuesday?

Again, extend this to 365 days and the bias towards her thinking it's the first day becomes numerically absurd.

Finally, the argument that the experiment only happens once, so we can't use relative frequencies is very weak. That argument effectively scuppers the use of probability theory in any situation where there is only one or a small number of experiments.

 

[PeroK]

PS the key is simply this: The experimenters look at the coin and do a certain thing more often if it is tails than jf it is heads. Someone who experiences this thing - and knows their rule - knows it's more likely they are experiencing that thing through a tail than a head.

That's conditional probability and is the case here, once you see through the fog.

The original probability of a head or a tail never changes and never needs to change.

Relative frequencies and probability trees are applicable tools, as they always are.

 

[stevendaryl]

PS the key is simply this: The experimenters look at the coin and do a certain thing more often if it is tails than if it is heads. Someone who experiences this thing - and knows their rule - knows it's more likely they are experiencing that thing through a tail than a head.

But there hasn't been a satisfactory account of why the probability of heads changes. Let's restate the experiment slightly differently:

1. The experimenter flips a coin, and asks Sleeping Beauty what is the probability that the result is "heads".
2. Presumably, she says 1/2.
3. He then tells her that he plans to ask her the question again, either once or twice depending on the result.
4. Now, he asks her again what the probability of heads is, given all this information.
5. Presumably, she again says 1/2.
6. Now, he asks her what answer she will give when she is waken up.
7. Now she says 2/3.

So she knows with certainty that her answer in the future will be 2/3. So why isn't it 2/3 now? You may say that the 2/3 is a conditional probability, while the 1/2 is the unconditioned probability. But normally, conditioning on an event that is certain doesn't change the probability.

There is a related thought experiment with the same numbers where the conditional probability works out sensibly. Suppose you do the following:

1. Flip a coin.
2. If it's tails, you pick a random name out of the phone book, and ask him what he thinks the probability of heads is.
3. If it's heads, you pick two random names out of the phone book, and ask them both what they think the probability is.

In this case, assuming there are N names in the phone book, your test subject can reason as follows:

1. The probability that I will be picked if heads is P(me | H) = \frac{2}{N}
2. The probability that I will be picked if tails is P(me | T) = \frac{1}{N}
3. So the cumulative probability of being picked is P(me) = P(me | H) \cdot P(H) + P(me | T) \cdot P(T) = \frac{2}{N} \cdot \frac{1}{2} + \frac{1}{N} \cdot \frac{1}{2} = \frac{3N}{2}
4. The probability of heads and my being picked is: P(me \wedge H) = P(me | H) \cdot P(H) = \frac{2}{N} \cdot \frac{1}{2} = \frac{1}{N}
   
5. The conditional probability of heads given that I was picked is: P(H | me) = \frac{P(H \wedge me)}{P(me)} = \frac{\frac{1}{N}}{\frac{3N}{2}} = \frac{2}{3}

The numbers work out the same as in Sleeping Beauty. But in this case, the fact that I am picked is additional information that changes the conditional probability of heads. In the Sleeping Beauty case, the fact that she is asked the probability upon waking is no additional information, since it was a certainty that that would happen.

 

[PeroK]

I know. That's why the day of the week is 2:1, but that doesn't affect the coin flip, which remains 1:1. And she isn't asked for the day, only for the coin.

You can't deduce one without the other!

You cannot have the probability of a head of 1/2 without the probability of Monday of 3/4. Anything else is immediately self-contradictory.

 

[PeroK]

But there hasn't been a satisfactory account of why the probability of heads changes. Let's restate the experiment slightly differently:

1. The experimenter flips a coin, and asks Sleeping Beauty what is the probability that the result is "heads".
2. Presumably, she says 1/2.
3. He then tells her that he plans to ask her the question again, either once or twice depending on the result.
4. Now, he asks her again what the probability of heads is, given all this information.
5. Presumably, she again says 1/2.
6. Now, he asks her what answer she will give when she is waken up.
7. Now she says 2/3.

So she knows with certainty that her answer in the future will be 2/3. So why isn't it 2/3 now? You may say that the 2/3 is a conditional probability, while the 1/2 is the unconditioned probability. But normally, conditioning on an event that is certain doesn't change the probability.

Let me try to give my answer in two parts. We have a careful analysis using relative frequencies that gives an answer of 1/3. And we have an intuitive assumption that there is no new information, hence conditional probabilities don't apply.

I trust the careful analysis, so it remains to find the flaw in the intuitive assumption.

The assumption is that there is no new information as only something that was inevitable has happened. But, is this the case?

There are precisely three different events: Woken on a Monday after a head; woken on a Monday after a tail; and, woken on a Tuesday after a tail. The sleeper cannot tell these apart, but I suggest they are still technically different events and the sleeper can still reason on the basis that one (not inevitable) event has happened.

I agree that "normally" the new information is more obvious than this. And in this case the new information is subtlely presented and not intuitively obvious.

But, to return to my first point, I would rather trust the careful analysis than to rely on an intuitively appealing analysis which may/must prove false, even if to falsify it is not particularly easy.

 

[PeroK]

And in this case the new information is subtlely presented and not intuitively obvious.
.

I suggest the following explanation:

First, the answer of 1/2 must be wrong, as it leads to a contradiction to the probability of which day of the week it is.

The complicating factor is the selective amnesia. Without the amnesia, we will all agree on the probability calculations:

When the sleeper is first woken, it is definitely Monday and 50-50 heads/tails. If she is woken a second time it is Tuesday and definitely tails.

Note that on the Monday she has no new information as being woken on the Monday was inevitable.

With the amnesia, she doesn't know whether it's the first or second time. Whether she has new information, as such, is perhaps a moot point; but, the scenario has definitely changed, because now there is a 1/3 chance that this is the second time she has been woken and the first time has been forgotten.

So, when you ask her initially (at the start of the experiment) about the coin, her answer is 1/2. That is scenario A.

But, when she is awoken, we do not have the same scenario and, critically, the sleeper knows it is not the same scenario. She knows that there is a 1/3 chance that it's Tuesday and that she has already been woken once before.

And that, I believe, is the new information she has ( and can predict she will have).

In short: when she is woken she knows that it may not be the first time she has been awake. That definitely changes the scenario and the criteria based on which she calculates probabilities.

 

[Marana]

Since halfers and thirders never disagree on how they would act (both would make the correct bet under all conditions) the disagreement is most likely about the definition of credence, or if there even is a good definition in this scenario. Personally I am a halfer (with reservations about whether an answer exists).

Some of the intuitive reasons are:

1. If the number of days increases, thirders can become certain that they just won the lottery. You could have a group of 1000000 people, and every single one of them wakes up on monday believing they just won the lottery... even though they also know for sure only one of them actually won. But according to thirders all 1000000 people are dancing around the room when they wake up, absolutely convinced they are the one winner, rich for life. While halfers think "it probably wasn't me that won", which is more intuitive to me.

2. Both halfers and thiders agree than on wednesday, even at a specific time like noon on wednesday, they will believe the probability is 1/2. So the question is, how can they know they will believe in 1/2 at noon on wednesday, but believe in 1/3 now? What is the difference in information?

3. According to halfers, the probability changes when you become unsure of what week it is. In the original problem, you know there is only one coin and one week, and then you will regain your memory. But what if the number of coins and weeks approached infinity? Then halfers would become thirders. As the uncertainty over the week increases, the probability approaches 1/3 by the halfer view. For thirders, uncertainty over the week doesn't change anything.

 

[PeroK]

Since halfers and thirders never disagree on how they would act (both would make the correct bet under all conditions) the disagreement is most likely about the definition of credence, or if there even is a good definition in this scenario. Personally I am a halfer (with reservations about whether an answer exists).

Some of the intuitive reasons are:

1. If the number of days increases, thirders can become certain that they just won the lottery. You could have a group of 1000000 people, and every single one of them wakes up on monday believing they just won the lottery... even though they also know for sure only one of them actually won. But according to thirders all 1000000 people are dancing around the room when they wake up, absolutely convinced they are the one winner, rich for life. While halfers think "it probably wasn't me that won", which is more intuitive to me.

2. Both halfers and thiders agree than on wednesday, even at a specific time like noon on wednesday, they will believe the probability is 1/2. So the question is, how can they know they will believe in 1/2 at noon on wednesday, but believe in 1/3 now? What is the difference in information?

3. According to halfers, the probability changes when you become unsure of what week it is. In the original problem, you know there is only one coin and one week, and then you will regain your memory. But what if the number of coins and weeks approached infinity? Then halfers would become thirders. As the uncertainty over the week increases, the probability approaches 1/3 by the halfer view. For thirders, uncertainty over the week doesn't change anything.

This post contains unsubstantiated 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.

 

[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.