[過去ﾛｸﾞ] 現代数学の系譜 カントル 超限集合論他 3 (548ﾚｽ)
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77(2): 現代数学の系譜 雑談 ◆yH25M02vWFhP 2020/07/31(金)20:57 ID:W/05pVKh(1/2)調 AAS
>>74
あなた、それ不正確引用ですよ
というか、意図してゴマカシていますね
＜正確な引用＞
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
より
(引用開始)
「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 」
(引用終り)
いいですか
あなたは、”But・・・”の前段の文だけを引用しましたね
それは全くのゴマカシです
当然、Math Dr. Alexander Pruss 氏の主張の力点は、後段の But 以下の文
But now the question is whether we can translate this to a statement without the conditional "For each fixed opponent strategy".
にあります
QED
（＾＾；
78: 現代数学の系譜 雑談 ◆yH25M02vWFhP 2020/07/31(金)21:06 ID:W/05pVKh(2/2)調 AAS
>>77 文字化け訂正
「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 」
↓
「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」
”-”が、文字化けしました（＾＾；
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