作者查詢 / kirimaru73
作者 kirimaru73 在 PTT [ Math ] 看板的留言(推文), 共21則
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1F推: 後面幾輪的X_i都是大型同捆包，或許不能從iid出發01/09 12:16
2F→: 另外為了減少歧義，可以追加幾點假設：01/09 12:17
3F→: (1) 該遊戲僅進行正好一輪且已結束01/09 12:17
4F→: (2) 獨裁者玩膩了，完全沒有進行下一輪的打算01/09 12:18
5F→: (代表朋友生還了就是生還了，不用擔心下一輪)01/09 12:18
6F→: (3) 已參與遊戲的生還者不會被重複選擇01/09 12:19
7F→: 這樣仍然可以產生爭議，但能簡化很多問題01/09 12:19
8F推: 直接固定R確實不合理 那如果改為計算一整場遊戲的01/09 15:17
9F→: 「生存率的期望值」 = Σ(N=1到無限)P(R=N)P(Z|R=N)01/09 15:19
10F→: 這樣第一項是 0.10 x 0.0 (衰人)01/09 15:21
11F→: 第二項是 0.09 x (1/11) 第三項是 0.081 x (11/111)01/09 15:22
12F→: 這應該就是你說的8.9%01/09 15:24
13F→: 我傾向覺得單一參賽者的生存率也是8.9% 那個90%其實01/09 15:33
14F→: 根本是不存在的，但90%又太過真實，不知道怎麼解釋01/09 15:33
17F推: 好像真的是樓上所說，如果加上"某人必定參與遊戲且01/10 00:07
18F→: 必定在第N輪被選中"的前提，並假設遊戲會進行正好01/10 00:07
19F→: 一場，則無論N是多少，他的生存率都是90%01/10 00:07
20F→: N=1時顯而易見，N=K則代表遊戲必進行K輪以上01/10 00:08
21F→: 這時K輪沒有爆掉的機率也是90%01/10 00:08
18F推:這個問題是沒有公式的，雖然可以算出來09/12 19:58
19F→:但是對於每一個不同的N都要實際運算一次09/12 19:59
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