>>1よりHuynh氏
https://mathoverflow.net/questions/151286/probabilities-in-a-riddle-involving-axiom-of-choice
Probabilities in a riddle involving axiom of choice
asked Dec 9 '13 at 16:16 Denis
（Huynh氏）
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.
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.
(引用終り)

さて、数学DR Huynh氏**)の批判を取り上げる

１）いま、決定番号で、上記{1,…,N}でN→∞として、自然数全体に渡るとする。自然数全体は非正則分布を成す>>4
２）1<k<Nなるkを考える。N有限で一様分布{1,…,N}の場合
　1～kの場合は k/Nで、N→∞ではk/N→0。kを幾ら大きく取っても、有限である限りk/N→0
３）だから、（上記Denis氏質問にある）有限の決定番号Mは、自然数全体からなる非正則分布中では、確率的零事*）である
　注*）cf 前スレ https://rio2016.5ch.net/test/read.cgi/math/1667737961/469

これも、時枝>>1のトリックの一つです！ｗ

（注**)数学DR Huynh氏のプロフィール）
https://mathoverflow.net/users/2233/tony-huynh
About
I am currently an Assisant Professor in the Department of Computer Science at Sapienza Universita di Roma.
I completed my PhD in the Department of Combinatorics & Optimization at the University of Waterloo. My supervisor was Jim Geelen. I am mainly interested in graphs, matroids, and combinatorial optimization,