だったら、これな >>192 返信:132人目の素数さん[] 投稿日:2019/08/17(土) 10:52:37.17 ID:+5QXhyrz [2/27] Pruss(^^ 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. Prussは間違いを認めた。未だに認められないのはおまえ一人(^^; (引用終り)
(なぜ、mathoverflow>>465 の手法が成立たないのか? ”CONGLOMERABILITY”が成立ってないというのが、数学DR Alexander Pruss氏の指摘(2013)で、それを2018年の著書で詳しく解説している) スレ65 https://rio2016.5ch.net/test/read.cgi/math/1557142618/750-754 https://books.google.co.jp/books?id=RXBoDwAAQBAJ&pg=PA77&lpg=PA77&dq=%22conglomerability%22+assumption+math&source=bl&ots=8Ol1uFrjJQ&sig=ACfU3U1bAurNGJm5872wDblskzsSgsU0iA&hl=ja&sa=X&ved=2ahUKEwioiPyV_IPiAhXHxrwKHUeaArUQ6AEwCXoECEoQAQ#v=onepage&q=%22conglomerability%22%20assumption%20math&f=false Infinity, Causation, and Paradox 著者: Alexander R. Pruss Oxford University Press, 2018 P75 (抜粋) 2.5.3 COUNTABLE ADDITITVITY AND CONGLOMERABILITY (引用終り)
(mathoverflowの”conglomerability”関連箇所) 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) (Alexander Pruss氏) <12> (抜粋) The probabilistic reasoning depends on a conglomerability assumption・・ 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. A quick way to see that the conglomerability assumption is going to be dubious is to consider the analogy of the Brown-Freiling argument against the Continuum Hypothesis (see here for a discussion). http://www.mdpi.com/2073-8994/3/3/636 (引用終り) 以上 0264132人目の素数さん2019/08/17(土) 19:45:26.20ID:+5QXhyrz>>262 >「Prussは間違いを認めた」が”嘘デタラメ垂れ流し”(>>251-252)な え? おまえ↓の意味わからんの? バカ? we win with probability at least (n−1)/n. That's right. 0265現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2019/08/17(土) 19:46:28.67ID:sbItYGIt>>260 >だからいってるだろう >反スレ主の書き込みしてるのは少なくとも2人いるって
確かに、サルが二匹いることは分った というか、二匹いることは認識している注*)
但し、人間からは、 なかなかサルの見分けは 難しいな(^^;
注*) テンプレより (>>2より) 知能が低下してサルになっています (>>3より) 知能の低い者が、サルと呼ばれるようになり、残りました。w(^^;) 0266現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2019/08/17(土) 19:48:29.21ID:sbItYGIt>>264 >we win with probability at least (n−1)/n. That's right.
ごまかしてもムダだよ。 0270132人目の素数さん2019/08/17(土) 20:11:46.58ID:+5QXhyrz>>266 バカ丸出し 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 yes the order would be: 1)describe the probabilistic strategy 2)opponent choses a sequence 3)probabilistic variable i is instanciated – Denis Dec 19 '13 at 23:02 この流れを分かってないの、おまえだよw
要するにPrussは ”if i is chosen uniformly independently of that strategy” の"independently"を否定したいらしいが、それは無理だろうw 0274132人目の素数さん2019/08/17(土) 20:21:38.74ID:+5QXhyrz Pruss氏「ランダム選択の結果を事前予想できれば時枝は不成立」 だそうですw 哲学に転向して正解だったかもw 0275132人目の素数さん2019/08/17(土) 20:22:44.46ID:4Tqla7J5>>272
最後Denisに yes the order would be: 1)describe the probabilistic strategy 2)opponent choses a sequence 3)probabilistic variable i is instanciated. と諭され、さすがにこれ以上の抵抗は無理と諦めたw
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) (Alexander Pruss氏) <12> The probabilistic reasoning depends on a conglomerability assumption・・ 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. A quick way to see that the conglomerability assumption is going to be dubious is to consider the analogy of the Brown-Freiling argument against the Continuum Hypothesis (see here for a discussion). http://www.mdpi.com/2073-8994/3/3/636 (引用終り)
ところが、最近 >>262の下記を見つけてね。やっぱり、 ”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”だねと、分ったのだった
(なぜ、mathoverflow>>465 の手法が成立たないのか? ”CONGLOMERABILITY”が成立ってないというのが、数学DR Alexander Pruss氏の指摘(2013)で、それを2018年の著書で詳しく解説している) スレ65 https://rio2016.5ch.net/test/read.cgi/math/1557142618/750-754 https://books.google.co.jp/books?id=RXBoDwAAQBAJ&pg=PA77&lpg=PA77&dq=%22conglomerability%22+assumption+math&source=bl&ots=8Ol1uFrjJQ&sig=ACfU3U1bAurNGJm5872wDblskzsSgsU0iA&hl=ja&sa=X&ved=2ahUKEwioiPyV_IPiAhXHxrwKHUeaArUQ6AEwCXoECEoQAQ#v=onepage&q=%22conglomerability%22%20assumption%20math&f=false Infinity, Causation, and Paradox 著者: Alexander R. Pruss Oxford University Press, 2018 P75 (抜粋) 2.5.3 COUNTABLE ADDITITVITY AND CONGLOMERABILITY (引用終り) 0302132人目の素数さん2019/08/17(土) 22:14:47.86ID:+5QXhyrz バカ主は選択公理も分かってなければ、時枝解法で選択公理がどう使われてるかも分かってない サル畜生にはサル知恵しかありませんでした(^^; 0303132人目の素数さん2019/08/17(土) 22:16:49.74ID:+5QXhyrz>>301 じゃあ時枝解法の確率変数を書いてみ? おまえ雄弁に語る割に確率変数一つ書けないじゃんw サル畜生にはサル知恵しかありませんでした(^^; 0304132人目の素数さん2019/08/17(土) 22:22:36.62ID:+5QXhyrz こら、サル畜生、白状しなさい
(参考) スレ62 https://rio2016.5ch.net/test/read.cgi/math/1551963737/915- (抜粋) 915 名前:現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE [sage] 投稿日:2019/03/27(水) Sergiu Hart氏のPDF http://www.ma.huji.ac.il/hart/puzzle/choice.pdf このP2に Remark. When the number of boxes is finite Player 1 can guarantee a win with probability 1 in game1, and with probability 9/10 in game2, by choosing the xi independently and uniformly on [0, 1] and {0, 1, ・・・, 9}, respectively. とある
ここで”independently and uniformly”が、独立同分布(IID)を含意することは、知る人がみればすぐ分かること で、例えば、{0, 1, ・・・, 9}ならば、的中確率は、1/10(for Player 2)(つまり、出題者Player 1は、確率9/10で勝てる) つまり、独立同分布(IID)を仮定すれば、どの箱も同じで、例外はない
game1(選択公理を使う)→game2(選択公理を使わない)→boxes is finite (有限の場合は通常確率論通り) と並べて説明している まあ、落語の落ちですね。最後”boxes is finite (有限の場合は通常確率論通り)”ですから