現代数学の系譜11 ガロア理論を読む24at MATH
現代数学の系譜11 ガロア理論を読む24 - 暇つぶし2ch636:現代数学の系譜11 ガロア理論を読む
16/10/23 19:09:25.30 MjfWcywG.net
>>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(^^;
URLリンク(mathoverflow.net)
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



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