[過去ﾛｸﾞ] 現代数学の系譜　カントル　超限集合論2 (1002ﾚｽ)
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582(1): 2020/03/22(日)01:44 ID:da5TzX47(1/2) AAS
>>579
> 「measurable」で議論できるでしょ

Denisが以下のようにコメントして
> ah ok I see where the misunderstanding comes from,

結局Prussは戦略が数列に依存しないことに納得したのでしょ

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586(3): 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2020/03/22(日)09:53 ID:TMbOZsnt(3/22) AAS
>>582
おサル、それ誤読だよ
”misunderstanding”は、下記引用の3)のとこでしょ
でも、面白いね、文献の”philosophical reason”の「 independently」の
”orthodox (Kolmogorovian) probability theory”と異なる見方（哲学だけれど）

（>>553より参考）
外部ﾘﾝｸ:mathoverflow.net
Probabilities in a riddle involving axiom of choice Denis氏 Dec 9 '13
DR Pruss氏
(抜粋)
show 6 more comments
1)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
2)I was assuming that "independently" has the meaning it does in probability theory (P(AB)=P(A)P(B) and generalizations for σ-fields). But that does require a probabilistic description of the opponent's choice.
Of course, one could mean "independently" here in some non-mathematical causal sense. (And there may be philosophical reason for doing this: fitelson.org/doi.pdf )
Still, mixing the probabilistic with nonprobabilistic concepts might lead to some difficulties, though. ? Alexander Pruss Dec 18 '13 at 15:21
3)ah ok I see where the misunderstanding comes from, it's true that "independently" is ambiguous, because only one random variable is involved here.
But I think it still has a mathematical meaning in the sense "it does not depend on the opponent's choice", namely we have ∃x∀y where x is our strategy and y is our opponent's strategy (i.e. the sequence),
and we still win this game because we can choose devise a (probabilistic) strategy that works on all sequences. ? Denis Dec 19 '13 at 11:54

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