>>573 補足 mathoverflow で、下記２の議論があるね。この解法の不成立を主張している ”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. ”だと。質問者のDenisは同意していないがね まあ、おっちゃんには読めないだろうが（＾＾； mathoverflow.net/questions/151286/probabilities-in-a-riddle-involving-axiom-of-choice Probabilities in a riddle involving axiom of choice Dec 9 '13 Denis (抜粋) 2 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 Xi
