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240: 現代数学の系譜 雑談 ◆yH25M02vWFhP  [] 2020/07/26(日) 12:22:17.29 ID:uQ4z/5zX

>>233
>箱入り無数目の同値関係をフレシェ・フィルタを用いて定義することはできると思う？

同値関係は、別に問題ない
問題は、時枝の確率計算 99/100が、測度論的に正当化されないってことだよ
（なお、確率論のIID（独立同分布）が時枝の反例になっているよ。IID（独立同分布）が理解できないようだね（＾＾；）

時枝記事の類似は、2013年12月09日にmathoverflowで、議論されている
二人の数学Dr Alexander Pruss 氏と Tony Huynh氏と、それ以外に質問者Denis氏（彼はコンピュータサインスの人）の周囲の人（"other people argue it's not ok"）
たちは、「時枝の議論は測度論的に不成立」と言っている

（参考）
https://mathoverflow.net/questions/151286/probabilities-in-a-riddle-involving-axiom-of-choice
Probabilities in a riddle involving axiom of choice Denis氏 Dec 9 '13
(抜粋)
・・・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.

answered Dec 11 '13 at 21:07 Math Dr. Alexander Pruss 氏
・・・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
・・・Intuitively this seems a really dumb strategy.

answered Dec 9 '13 at 17:37 Math Dr. Tony 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.

http://rio2016.5ch.net/test/read.cgi/math/1592654877/240

243: １３２人目の素数さん [] 2020/07/26(日) 14:00:17.78 ID:9ZaudBKU

>>240
https://mathoverflow.net/questions/151286/probabilities-in-a-riddle-involving-axiom-of-choice
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

http://rio2016.5ch.net/test/read.cgi/math/1592654877/243

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