[過去ﾛｸﾞ] 現代数学の系譜 工学物理雑談 古典ガロア理論も読む75 (1002ﾚｽ)
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12: 2019/08/15(木)23:09:12.68 ID:eHJsXL8K(1) AAS
スレ主、問題に答えられずコピペ連投で誤魔化すの図ｗ
18: 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2019/08/15(木)23:56:52.68 ID:brP98meI(17/17) AAS
>>17
＞X?/?〜 に商位相、すなわち、写像 m： x→ conn (x) が連続になる最も粗い位相を与える。
妙なところが文字化けしている
”?/?”で？の半角部分はスペースなんだけど、特殊文字なのか（＾＾；
まあ、原文見てください（＾＾
125(1): 2019/08/16(金)17:28:40.68 ID:W6QnSuYA(26/35) AAS
哀れ過ぎるド素人は失せろよ
バカはスレ主一人で十分だ
158(3): 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)07:32:33.68 ID:sbItYGIt(3/35) AAS
>>157
つづき
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).
外部ﾘﾝｸ:www.mdpi.com
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."
つづく
245: 哀れな素人 2019/08/17(土)17:28:59.68 ID:W8+WGsHp(29/30) AAS
ID:h0TAsPzg
ID:+5QXhyrz
ID:4Tqla7J5
実はこれは全部アホのサル石なのである(笑
スレ主は成立派が二名いると思っているが、
実はサル石だけである(笑
603: 2019/08/21(水)20:27:23.68 ID:SfXTc3qP(20/20) AAS
在阪朝鮮人はこれを毎日聞いて涙を流すｗ
動画ﾘﾝｸ[YouTube]
714(1): 2019/08/23(金)23:53:03.68 ID:WGfkBzAI(5/5) AAS
>>713
>２）決定番号は、一様分布ではない。いや、「一様分布」と言いたいなら証明してみろ（それ無理です）
「さて， 1〜100 のいずれかをランダムに選ぶ. 例えばkが選ばれたとせよ. s^kの決定番号が他の列の決定番号どれよりも大きい確率は1/100に過ぎない. 」
↑
100個の決定番号のいずれかをランダムに（一様分布で）選択しているんだがｗ
バカザルは人間の言葉がわからんのか？ｗ
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