A variation of the sleeping beauty problem

[Ibix]

1/2 represents zero knowldege about the coin toss; and 1/3 represents slightly more knowldege about the coin toss.

Actually, I wonder if this is a sticking point. Intuitively, Sleeping Beauty gets no more information about the state of the coin when the experiment starts. However, she has information about the consequences of the coin toss that can modify her beliefs. She knows she'll be asked twice as many times if the coin came up tails, so she knows she'll be right twice as many times if she says tails. It's her knowledge of the deterministic consequences of a probabalistic outcome that lets her beat 50:50.

I wonder if it's that mix of probability and deterministic counting that confuses the issue.

 

[PeroK]

Actually, I wonder if this is a sticking point. Intuitively, Sleeping Beauty gets no more information about the state of the coin when the experiment starts. However, she has information about the consequences of the coin toss that can modify her beliefs. She knows she'll be asked twice as many times if the coin came up tails, so she knows she'll be right twice as many times if she says tails. It's her knowledge of the deterministic consequences of a probabalistic outcome that lets her beat 50:50.

I wonder if it's that mix of probability and deterministic counting that confuses the issue.

As I understand it, there are two levels of halfers. There are the "extreme halfers" who don't allow the possibility of anything other than 0, 1/2 or 1 in the first place. Then, there are the subtle halfers, who focus on the information issue. If they would accept that "the game has started" is new information, they would change their position (ha, ha!).

The real problem, as I see it, is that a halfer clearly loses money to someone who uses computer simulation and takes the thirder position. This is why, I think, all professional statisticians take the thirder position and it's mostly philosophers who take the halfer position - who are less easily swayed by concrete calculations!

 

[Dale]

Ok, everyone, the topic of this thread is specifically the new variant proposed by @Demystifier. The usual version has been discussed already.

A post on stack exchange is only a valid source here insofar as it is consistent with the professional scientific literature. I personally have never seen any peer reviewed paper that claimed 42% for two experiments. So that would not be an acceptable reference even on a thread about the original paradox.

Please refocus the thread, and make sure that the discussion is based on better sources.

 

[Dale]

I propose a variation of the problem, hoping that it will help to solve the problem in a way that everyone can agree on

I appreciate your optimism!

The question is: If you are one of the examinees, and if you are asked what is the probability that the coin landed heads, what is your answer?

I claim that this version of the problem is equivalent to the original sleeping beauty problem

I guess one difference between your scenario and the standard one is that the standard one asks about credence and you are asking about probability. I don’t think that should make much difference, but I suspect that some people would consider them to be not equivalent on that basis.

 

[Vanadium 50]

This is why we don't restart closed threads.

 

[Dale]

If the coin landed heads, one examine is asked what is the probability that the coin landed heads.

How is the one examinee chosen?

 

[Demystifier]

How is the one examinee chosen?

Good question! It can be chosen randomly, by another fair coin.

 

[PeroK]

How is the one examinee chosen?

I may be misremembering, but I think the scenario in the OP was presented in my first year university probability/stats course in 1980-81!

 

[Dale]

Good question! It can be chosen randomly, by another fair coin.

Since you mention in the OP generalizing to $n$ participants, then we could consider the experiment to be the result of both flipping a fair coin and rolling a fair $n$-sided die. If the $i$-th participant is asked, then the $i$-th participant knows that either the $n$-sided die is $i$ or the coin is tails (or both).

 

[Dale]

So we can simply build a truth table for the statement: "Participant $i$ is asked". The columns are the different rolls of the fair die, and the rows are the different flips of the fair coin. Each element in the table is equiprobable.

	1	2	...	i	...	n
Heads	False	False	False	True	False	False
Tails	True	True	True	True	True	True

The probability of the coin being Heads given that participant $i$ is asked then is simply the count of the "True" values in the top row divided by the total count of the "True" values. So it is $1/(n+1)$.

I don't see how your scenario can be analyzed any other way.

So the only objections I could see is that it is not equivalent to the usual problem in some way. That could be someone who thinks that probability and credence are different as I mentioned above, or perhaps there are some other objections. But I don't see any other probability that can be obtained from your scenario.

 

[jbergman]

TL;DR Summary: A version of the "sleeping beauty" problem, in which a single sleeping examinee is replaced by two awake examinees.

Some time ago we had a discussion of the sleeping beauty problem
https://www.physicsforums.com/threads/the-sleeping-beauty-problem-any-halfers-here.916459/
which is a well known problem in probability theory. In that thread, there was no consensus whether the probability of heads is 1/2 or 1/3. Since the thread is closed now, I open a separate thread where I propose a variation of the problem, hoping that it will help to solve the problem in a way that everyone can agree on.

The rules of the game are the following. There are two examinees that cannot communicate with each other. A fair coin is tossed, and the result is not shown to examinees. If the coin landed heads, one examine is asked what is the probability that the coin landed heads. If the coin landed tails, both examinees are asked what is the probability that the coin landed heads. All these rules of the game are known to both examinees. The question is: If you are one of the examinees, and if you are asked what is the probability that the coin landed heads, what is your answer?

I claim that this version of the problem is equivalent to the original sleeping beauty problem, and that in this version of the problem it is clear that the correct answer is 1/3 (and not 1/2). Do you agree?

My argument is the following. For the sake of intuition, I first go to the extreme. Instead of considering $n=2$ examinees, I consider $n=1.000.000$ examinees. Either only one of them is asked the question (if the coin landed heads), or all of them are asked the question (if the coin landed tails). So if I'm asked the question at all, it is very unlikely that I am the lucky one that has been asked, while all others have not been asked. Given that I was asked the question, it is much more likely that I was not particularly lucky, i.e., that I am just one of $n=1.000.000$ examinees who were all asked the question. In other words, it is much more likely that the coin landed tails, so the probability that the coin landed heads is much smaller than 1/2. With this intuition in mind it is not difficult to compute that the probability of heads is $1/(n+1)$ (rather than 1/2), which in the case of $n=2$ gives the probability $1/3$.

After thinking about this more, I think this game is not equivalent to sleeping beauty. The reason is that we have more information when being asked. Specifically each individual is different and knows they are not the other individuals.

This is equivalent to the following modified version of Sleeping Beauty. We put her to sleep. We then flip a coin. If it is heads, we flip a second coin and if heads wake her on Monday, tell her it is Monday, and ask her and if tails wake her on Tuesday and tell her it is Tuesday, and ask. If the first coin was tails, we then wake her on Monday, telling her the day, and ask and then put her back to sleep and wake her on Tuesday, again telling her the day, and ask.

For me it is clear, that this is inequivalent so the standard Sleeping Beauty Problem.

 

[jbergman]

After thinking about this more, I think this game is not equivalent to sleeping beauty. The reason is that we have more information when being asked. Specifically each individual is different and knows they are not the other individuals.

Actually maybe it is the same, but I think it is more helpful to frame as I have so we can clearly see how we have changed the structure from the original game.

 

[Dale]

For me it is clear, that this is inequivalent so the standard Sleeping Beauty Problem.

Personally, this is one reason that I don't like any modifications to the scenario. I feel that any modification will be criticized by someone as being inequivalent.

Given that you think it is inequivalent and therefore that the answer has no bearing on the original problem: do you agree that the probability for the OP scenario is indeed $1/3$? (or $1/(n+1)$ if doing $2<n$)

 

[PeroK]

It definitely isn't equivalent. It's standard elementary probability theory. The SB problem has an important twist. At the very least, you have to reassess the problem, even if ultimately it has the same numerical answer.

That said, if this problem were put into the public domain, there would be plenty of halfers!

 

[jbergman]

Personally, this is one reason that I don't like any modifications to the scenario. I feel that any modification will be criticized by someone as being inequivalent.

Given that you think it is inequivalent and therefore that the answer has no bearing on the original problem: do you agree that the probability for the OP scenario is indeed $1/3$? (or $1/(n+1)$ if doing $2<n$)

Actually, I am not sure they are inequivalent. Rosenthal, claims they are and analyzes this exact variant in A Mathematical Analysis of the Sleeping Beauty Problem. I agree with his analysis there and the consensus that the answer is 1/3.

 

[Dale]

I agree that this scenario is pretty clearly 1/3. It seems that this is just a straightforward conditional probability exercise.

Interestingly, a glance at the truth table shows that anyone that is not asked knows that the coin was heads with (conditional) probability 1. That is kind of obvious, but it hadn't struck me until I did the table. One possible inequivalence is that in the setting of repeated experiments there would be participants with this complete certainty. That never happens in the original.

On the other hand since such individuals are never asked about the probability, the fact that they exist doesn't necessarily make the scenario inequivalent with respect to probability. But I think that it at least makes the claim of equivalence require some solid justification. On what basis can equivalence be claimed and/or rejected? I really don't know.

 

[jbergman]

Actually, I am not sure they are inequivalent. Rosenthal, claims they are and analyzes this exact variant in A Mathematical Analysis of the Sleeping Beauty Problem. I agree with his analysis there and the consensus that the answer is 1/3.

Just to post a slight translation of Rosenthal's analysis here, since it is fairly convincing for this variant of the problem. He actually introduces two coins one is a nickel that we flip first and then we flip a dime if the nickel was a heads to decide whether to interview SB on Monday or Tuesday. Then we have the following:

P(Nickel = Heads and Asked On Monday) = P(Nickel = Heads and Dime = Heads) = 1/4
P(Asked On Monday) = P(Nickel = Tail or Heads and Dime = Heads) = P(Nickel = Tail) + P(Nickel = Heads and Dime = Heads) = 3/4

Therefore P(Nickel = Heads | Asked On Monday) = P(Nickel = Heads and Asked On Monday) / P(Asked On Monday) = 1/3.

 

[PeroK]

There is another "paradox" in elementary probability theory. Suppose we toss three coins. They can't all be different, so we can find two that are the same. The third is then the same or different with equal probability. Hence, the probability they are all the same must be 1/2.

But, if we run a computer simulation, we find that the probability they are all the same is 1/4.

We can draw two possible conclusions:

Probability theory is ambiguous and both 1/2 and 1/4 are valid answers.

The answer of 1/4 must be correct, by experiment. The 1/2 argument must be flawed.

The question in the OP can be resolved by computer simulation. The arguments are only important in terms of understanding the answer; not in establishing it. The answer is 1/3, whether we like it or not. And if someone produces an argument that the probability is not 1/3, then (as Feynman would have said) it doesn't matter what their name is, and it doesn't matter how clever they are, it's wrong!

 

[PeterDonis]

The arguments are only important in terms of understanding the answer; not in establishing it.

I'm not so sure. Suppose I simulate a million trials and all three simulated coin flips are the same in 243,596 of them. Is the probability of the outcome 243,596/1,000,000? Or is it 1/4? Experiment can't tell me the answer to that question; only theory can.

This experiment can of course rule out 1/2 as a valid answer, at least to a very high confidence level. But it can't tell me that the answer is exactly 1/4, instead of whatever ratio the experiment yielded. I need a theoretical argument to predict the theoretical ratio of 1/4, to which I can then compare what I see in the experiment and decide whether I accept the theory or need to modify it.

 

[fresh_42]

Before we run again in a complete subthread which would be hard to separate among those many posts I'd like to ask you to either stay on topic or create a new thread with this new topic. Thank you.

 

[PeroK]

Before we run again in a complete subthread which would be hard to separate among those many posts I'd like to ask you to either stay on topic or create a new thread with this new topic. Thank you.

Okay, but I'm not sure how much more we can squeeze out of this thread.

 

[PeroK]

This experiment can of course rule out 1/2 as a valid answer, at least to a very high confidence level.

Not if it's the sleeping beauty problem it can't!

 

[fresh_42]

Okay, but I'm not sure how much more we can squeeze out of this thread.

I think that you are very right here, so I will close the discussion.
Please contact me in case you still have something important to say and I closed it too early.

As for the latest development (posts #53 f.):

"Significance of probability measurements and distribution of error margins."​

seems to be a debatable subject among applied mathematicians and physicists.