0192１３２人目の素数さん2019/08/17(土) 10:52:37.17ID:+5QXhyrz>>157 おまえ（＾＾ but other people argue it's not ok, because we would need to define a measure on sequences,
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.
>>64 名前：１３２人目の素数さん[] 投稿日：2019/08/16(金) 10:56:38.15 ID:WjfkqcDK [21/40] >32 ＞スレ主にしつこく質問 ＞・３進カントール集合を[0,1]と同相にする簡単な方法は？ (引用終り) 0194１３２人目の素数さん2019/08/17(土) 12:10:07.59ID:+5QXhyrz>>159 Pruss 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? 100列からランダム選択される列を事前予測すれば出題者側の勝ちだとさｗ Pruss、論破されて発狂ｗ
∞は、｛ や ｝の数が2^∞になるぅ 可算無限個を定義するのに、 非可算無限個の｛｝を必要となんて、 滅茶苦茶な気がする。 0199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) 12:30:43.54ID:sbItYGIt>>179 補足 >Here's an amusing thing that may help see how measurability enters into these things. Consider a single sequence of infinitely many independent fair coin flips. Our state space is Ω={0,1}^N, corresponding to an infinite sequence (Xi)^∞ i=0 of i.i.d. r.v.s with P(Xi=1)=P(Xi=0)=1/2. >Can you guess the first coin flip on the basis of all the others? You might think: "Of course not! No matter what function from the values of flips X1,X2,... to {0,1} is chosen, the probability that the value of the function equals X0 is going to be 1/2."
おサル I think it is ok, because the only probability measure we need is uniform probability on {0,1,…,N-1}
おれ（＾＾ but other people argue it's not ok, because we would need to define a measure on sequences,
（参考） 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) asked Dec 9 '13 at 16:16 Denis I think it is ok, because the only probability measure we need is uniform probability on {0,1,…,N-1}, 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. 0203１３２人目の素数さん2019/08/17(土) 12:40:03.72ID:+5QXhyrz>>164 おまえは本当に頭が悪いというか固いというか(笑
４）Pruss氏はいう（>>158） ”Can you guess the first coin flip on the basis of all the others? You might think: "Of course not! No matter what function from the values of flips X1,X2,... to {0,1} is chosen, the probability that the value of the function equals X0 is going to be 1/2." この議論は、任意の有限i番目に拡張できる 即ち ”Can you guess the n-th coin flip on the basis of all the others? You might think: "Of course not! No matter what function from the values of flips X0,X1,X2,...,Xi-1,Xi+1,... to {0,1} is chosen, the probability that the value of the function equals Xi is going to be 1/2." となる 以上
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
