[過去ﾛｸﾞ] 現代数学の系譜 工学物理雑談 古典ガロア理論も読む80 (1002ﾚｽ)

現代数学の系譜 工学物理雑談 古典ガロア理論も読む80 http://rio2016.5ch.net/test/read.cgi/math/1578091012/

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206: 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE [] 2020/01/10(金) 00:43:07.94 ID:KeHo+Wgs >>202 (引用開始) 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. (引用終り) おっさん、思い込み激しいな（＾＾ そんな、議論の途中をつまみ食いして、Pruss氏の結論にするなよ、おいおいｗｗ（＾＾ （>>109より） https://mathoverflow.net/questions/151286/probabilities-in-a-riddle-involving-axiom-of-choice Probabilities in a riddle involving axiom of choice Dec 9 '13 3 Answers中のanswered Dec 11 '13 at 21:07 Alexander Pruss に対する議論の中で関連を抜粋すると つづく http://rio2016.5ch.net/test/read.cgi/math/1578091012/206

207: 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE [] 2020/01/10(金) 00:43:57.52 ID:KeHo+Wgs >>206 つづき ・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) つづく http://rio2016.5ch.net/test/read.cgi/math/1578091012/207

221: 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE [] 2020/01/10(金) 10:26:57.58 ID:ebMXZTdz >>218 ＞確率のやつは完成にアウトだけど おおっ！ 同意ありがとう！！（＾＾ ＞確率でないやつというのはどんなのですか？ 下記引用が、元のmathoverflowからなのだが 冒頭のThe Riddleが、確率でない版だと思う 後のThe Modificationが、確率版でしょう なお、両者の記述の間に、The Riddleの解法の記述があるよ （>>206より） https://mathoverflow.net/questions/151286/probabilities-in-a-riddle-involving-axiom-of-choice Probabilities in a riddle involving axiom of choice Dec 9 '13 (抜粋) The question is about a modification of the following riddle (you can think about it before reading the answer if you like riddles, but that's not the point of my question): つづく http://rio2016.5ch.net/test/read.cgi/math/1578091012/221

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