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.
Alexander Pruss answered 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.
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
Let's go back to the riddle. Suppose u￣ is chosen randomly. The most natural option is that it is a nontrivial i.i.d. sequence (uk), independent of the random index i which is uniformly distributed over [100]={0,...,99}. In general, Mj will be nonmeasurable (one can prove this in at least some cases). We likewise have no reason to think that M is measurable. But without measurability, we can't make sense of talk of the probability that the guess will be correct.
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."
Denis 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.
Tony Huynh In order for such a question to make sense, it is necessary to put a probability measure on the space of functions f:N→R. Note that to execute your proposed strategy, we only need a uniform measure on {1,…,N}, but to make sense of the phrase it fails with probability at most 1/N, we need a measure on the space of all outcomes. The answer will be different depending on what probability space is chosen of course.
If it were somehow possible to put a 'uniform' measure on the space of all outcomes, then indeed one could guess correctly with arbitrarily high precision, but such a measure doesn't exist. (引用終り) 以上 0160哀れな素人2019/08/17(土) 07:58:48.96ID:W8+WGsHp ID:W6QnSuYA
誰も同相だとはいってない 同相でないものを同相にするのに どういう操作が必要か？ というのが問い その答えが 「ある箇所から ０２２２・・・ ２０００・・・ となる２つの３進小数をくっつけて１つにする」 これが位相を変える操作だと気づけないスレ主って…馬鹿だろｗ 0168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) 09:31:11.40ID:sbItYGIt>>158 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."
この時点で時枝戦略成立 Prussは死んだｗ スレ主も死んだｗ 0179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) 09:56:23.42ID:sbItYGIt>>168 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."
お前はアホだから集合というのは あくまで人間の頭の中にある概念だ ということが分っていない(笑 0191１３２人目の素数さん2019/08/17(土) 10:46:27.75ID:+5QXhyrz>>156 どうぞ、 時枝解法の確率変数を書いて下さい 書けないなら、 自分の言葉通り、 数学板から出て行けよ！ 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
