19/08/28 07:30:55.88 MajO1X6X.net
>>238
いや、だから
(引用開始)
サイコロに勝手な自然数６コを記載する
ツボの中でサイコロを転がして開ける
表から見えない真下の面に書かれてる数字が
６コ中の最大値である確率はいくらか？
(引用終り)
その論法が、時枝の場合に適用できるという厳密な数学的証明がないと言っているんだよ
６コ中の最大値である確率は、1/6
まあ、時枝で言えば、6列に並べて、6列のある列が決定番号の最大値は？
と言いたいんだろ？
そこを（数学的に厳密でないと）批判しているのが、Alexander Pruss氏だよ
URLﾘﾝｸ(mathoverflow.net)
Probabilities in a riddle involving axiom of choice Dec 9 '13
(抜粋)
asked Dec 9 '13 at 16:16 Denis
I think it is ok, because the only probability measure we need is uniform probability on {0,1,…,N-1}, but other people argue it's not ok, because we would need to define a measure on sequences, and moreover axiom of choice messes everything up.
Alexander Pruss answered
The probabilistic reasoning depends on a conglomerability assumption, namely that given a fixed sequence u￣ , the probability of guessing correctly is (n?1)/n, then for a randomly selected sequence, the probability of guessing correctly is (n?1)/n.
But we have no reason to think the event of guessing correctly is measurable with respect to the probability measure induced by the random choice of sequence and index i, and we have no reason to think that the conglomerability assumption is appropriate.
A quick way to see that the conglomerability assumption is going to be dubious is to consider the analogy of the Brown-Freiling argument against the Continuum Hypothesis (see here for a discussion).
URLﾘﾝｸ(www.mdpi.com)