[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80 (1002ﾚｽ)
上下前次1-新
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

---

206(2): 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2020/01/10(金)00:43 ID:KeHo+Wgs(1/19) AAS
>>202
(引用開始)
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.
(引用終り)

おっさん、思い込み激しいな（＾＾
そんな、議論の途中をつまみ食いして、Pruss氏の結論にするなよ、おいおいｗｗ（＾＾
省6

---

207(2): 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2020/01/10(金)00:43 ID:KeHo+Wgs(2/19) AAS
>>206
つづき

・Pruss氏のAnswerより（冒頭部分）
　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.

・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. ? Denis Dec 17 '13 at 15:21

・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.
省5

---

221(4): 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2020/01/10(金)10:26 ID:ebMXZTdz(3/16) AAS
>>218
＞確率のやつは完成にアウトだけど

おおっ！ 同意ありがとう！！（＾＾

＞確率でないやつというのはどんなのですか？

下記引用が、元のmathoverflowからなのだが
冒頭のThe Riddleが、確率でない版だと思う
後のThe Modificationが、確率版でしょう
省7

---

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 0.038s