[過去ﾛｸﾞ] スレタイ 箱入り無数目を語る部屋4 (1002ﾚｽ)
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115(1): 2022/10/26(水)20:15 ID:b4wD2Jth(3/5) AAS
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懐疑派2 DR Alexander Pruss氏
外部ﾘﾝｸ:mathoverflow.net
Probabilities in a riddle involving axiom of choice
asked Dec 9 '13 at 16:16 Denis

<回答者 DR Alexander Pruss氏>
Here's an amusing thing that may help see how measurability enters into these things. Consider a single sequence of infinitely many independent fair coin flips. Our state space is Ω={0,1}^N, corresponding to an infinite sequence (Xi)∞i=0 of i.i.d.r.v.s with P(Xi=1)=P(Xi=0)=1/2. Start with P being the completion of the natural product measure on Ω.

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.

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