>>482
私は、名前の議論はしない。だれか、他人の第三者に迷惑をかけるとも限らないからね

それはさておき、”数学Dr. Prussだって認めたぞ？
　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.”
は違うよ

その話は、スレチだが下記な
箱入り無数目を語る部屋
https://rio2016.5ch.net/test/read.cgi/math/1609427846/52-56
https://mathoverflow.net/questions/151286/probabilities-in-a-riddle-involving-axiom-of-choice
mathoverflow
Probabilities in a riddle involving axiom of choice asked Dec 9 '13 at 16:16 Denis
＜回答12の質疑応答＞
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

まず
>That's right. But now the question
典型的な「イエスバット法」（下記）でしょ
会話の基本テクニック
ある程度相手の言い分を認めつつ、自分の主張を展開するのです
”But”以下に力点がありますよ

つづく