現代数学の系譜工学物理雑談古典ガロア理論も読む74

[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74 (1002ﾚｽ)
上下前次1-新
通常表示 512ﾊﾞｲﾄ分割ﾚｽ栞
抽出解除ﾚｽ栞

このｽﾚｯﾄﾞは過去ﾛｸﾞ倉庫に格納されています｡
次ｽﾚ検索歴削→次ｽﾚ栞削→次ｽﾚ過去ﾛｸﾞﾒﾆｭｰ

ﾘﾛｰﾄﾞ規制です｡10分ほどで解除するので､他のﾌﾞﾗｳｻﾞへ避難してください｡

781: 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE  [] 2019/08/14(水) 15:27:20.20 ID:rg2Nhb+h

おサルさん、（>>692より）【必死のパッチ】やなｗｗ（＾＾；

>>775
”if i is chosen uniformly independently of that strategy”
のindex iについてｗｗ

これ下記の”Alexander Pruss Dec 19 '13 at 15:05 ”の抜粋なｗ
以下の応答を嫁めｗ（＾＾

要するに、力点は、But以下の文にあるってことと
前文の”if i is chosen uniformly independently of that strategy”の部分が未証明だってことよｗ

(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)
sked Dec 9 '13 at 16:16 Denis
The Modification
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. a

answered Dec 11 '13 at 21:07
Alexander Pruss
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.

Denis Dec 17 '13 at 15:21
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.

つづき

http://rio2016.5ch.net/test/read.cgi/math/1564659345/781

782: 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE  [] 2019/08/14(水) 15:28:16.70 ID:rg2Nhb+h

>>781
つづく

Alexander Pruss Dec 19 '13 at 15:05
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".

answered Dec 9 '13 at 17:37
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

http://rio2016.5ch.net/test/read.cgi/math/1564659345/782

上下前次1-新書関写板覧索設栞歴

ｽﾚ情報赤ﾚｽ抽出画像ﾚｽ抽出歴の未読ｽﾚ AAｻﾑﾈｲﾙ

ぬこの手ぬこTOP 0.031s