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116: 2022/10/26(水)20:16 ID:b4wD2Jth(4/5) AAS
>>115

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懐疑派3 回答者 DR Tony Huynh氏
外部ﾘﾝｸ:mathoverflow.net
Probabilities in a riddle involving axiom of choice
asked Dec 9 '13 at 16:16 Denis

<回答者 DR Tony Huynh氏>
I also like this version of the riddle. To answer the actual question though, I would say that it is not possible to guess incorrectly with probability only 1/N, even for N=2. In order for such a question to make sense, it is necessary to put a probability measure on the space of functions f:N→R. Note that to execute your proposed strategy, we only need a uniform measure on {1,…,N}, but to make sense of the phrase it fails with probability at most 1/N, we need a measure on the space of all outcomes. The answer will be different depending on what probability space is chosen of course.

Here's a concrete choice for a probability space that shows that your proposal will fail. Suppose that for each index i we sample a real number Xi from the normal distribution so that the Xis are independent random variables. If there is only person, no matter which boxes they view, they gain no information about the un-opened boxes due to independence. Thus, their probability of guessing correctly is actually 0, not (N?1)/N, say.

If it were somehow possible to put a 'uniform' measure on the space of all outcomes, then indeed one could guess correctly with arbitrarily high precision, but such a measure doesn't exist.
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