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・Pruss氏のAnswerより（冒頭部分）
　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.

・Our choice of index i is made randomly, but for this we only need the uniform distribution on {0,…,n}. It is made independently of the opponent's choice. ? Denis Dec 17 '13 at 15:21

・What we have then is this: For each fixed opponent strategy, if i is chosen uniformly independently of that strategy (where the "independently" here isn't in the probabilistic sense), we win with probability at least (n?1)/n. That's right.
　But now the question is whether we can translate this to a statement without the conditional "For each fixed opponent strategy". ? Alexander Pruss Dec 19 '13 at 15:05

・How about describing the riddle as this game, where we have to first explicit our strategy, then an opponent can choose any sequence. then it is obvious than our strategy cannot depend on the sequence. The riddle is "find how to win this game with proba (n-1)/n, for any n." ? Denis Dec 19 '13 at 19:43

・But the opponent can win by foreseeing what which value of i we're going to choose and which choice of representatives we'll make. I suppose we would ban foresight of i? ? Alexander Pruss Dec 19 '13 at 21:25
(引用終り)

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