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 (Denis質問) 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. (Pruss氏) The probabilistic reasoning depends on a conglomerability assumption, ・・・and we have no reason to think that the conglomerability assumption is appropriate. (Huynh氏) 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.
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0795132人目の素数さん2022/05/13(金) 08:57:07.90ID:ISbFbGqJ>>794 100人のペテン師を用意する。 任意の s∈R^N に対して、背番号kのペテン師は番号kを選び、 この k と決定番号 d(s) から箱の中身を回答する。 この回答は s と k に依存して決まるので、ans(s,k) と書くことにする。 従って、任意の s∈R^N に対して、100通りの回答 ans(s,1), ans(s,2), …, ans(s,100) が一括で得られる。