22/11/04 08:28:38.68 sQY7VXAT.net
>>653
それについては
下記のPruss氏の全文をキチンと読んだらどうですか？
(参考)>>650より再録
URLﾘﾝｸ(mathoverflow.net)
Probabilities in a riddle involving axiom of choice
asked Dec 9 '13
（Pruss氏）
That's a fine argument assuming the function is measurable. But what if it's not?
Here is a strategy: Check if X1,X2,... fit with the relevant representative.
If so, then guess according to the representative. If not, then guess π.
(Yes, I realize that π not∈{0,1}.)
Intuitively this seems a really dumb strategy.
(引用終り)