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- 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,

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##.

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##.

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