>>249
>上記の関数をつかって代表元を選べると認めたならば
>決定番号がいかほど巨大であろうが決まるのだから
>その次の箱を選べばいいだけのこと　何の問題もない

「何の問題もない」と思い込ませるところが、このパズル（ mathoverflow では、”riddle”）のキモだろう
問題は、Probabilitiesに関することだから、確率計算に乗らないとまずいのだ。下記ご参照

スレ34 http://rio2016.2ch.net/test/read.cgi/math/1496568298/649
(抜粋)
下記英 mathoverflowは参考になる。要するに、時枝記事類似”Riddle”で、Alexander Pruss氏は、2013年に
”But we have no reason to think the event of guessing correctly ・・..で、非可測経由だとまずいと言っている。これ如何に？
http://mathoverflow.net/questions/151286/probabilities-in-a-riddle-involving-axiom-of-choice
Probabilities in a riddle involving axiom of choice - MathOverflow: edited Dec 9 '13 Denis
(抜粋)
answered Dec 11 '13 at 21:07 Alexander Pruss
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.
(引用終り)