經典概率問題(啤牌遊戲)
2010-08-18 6:40 pm
現在莊家桌上有三張啤牌A,B,C，其中一隻是joker(只有莊家知道哪一隻是)。現在X先生要猜中哪一隻為joker。他先估計A 為joker，然後莊家揭開B，而B不是joker。那麼X先生應繼續選擇A還是轉為選擇C？請從概率上分析。
回答 (3)
2010-08-18 8:35 pm
✔ 最佳答案
這是一道很經典的老題,什至曾有統計學家也失魂答錯,也引起過不多不少的爭論。結論是應轉為選擇C,猜中概率會由 1/3 提升至 2/3 。下面是一個很直觀的分析,當X先生估計A 為joker, 猜中概率 = 1/3 ,換句話說 , B , C 為 joker 之概率 = 2/3 ,如今莊家故意公然展示B(無論你中不中)不是joker( 故意很重要,如果莊家只是隨意揭開一隻,那麼這問題就沒意思,這時未揭的牌joker機會也是1/2) , 於是 2/3 的概率全落在 C 牌身上(因已人為排除了B+C 中的 B) , 所以轉為選擇 C 較明智。試想像是100 張牌,你估 1 張 , 猜中機率 = 1/100 ,
然後莊家故意展示其餘 98 張不是joker的牌給你看(無論你中不中) ,
那你信自己牌是 Joker 還是剩下的最後一張是 joker ?
快轉軚吧,剩下的牌是 JOKER 機會有 99% !
2010-08-18 18:32:45 補充：
請看 :
http://zh.wikipedia.org/zh-hk/%E8%92%99%E6%8F%90%E9%9C%8D%E7%88%BE%E5%95%8F%E9%A1%8C
2010-08-27 9:40 pm
Before the dealer reveals the card B,
P(A = Joker) = 1/3 and P(A ≠ Joker) = 2/3
P(B = Joker) = 1/3 and P(B ≠ Joker) = 2/3
P(C = Joker) = 1/3 and P(C ≠ Joker) = 2/3
The assumption on these probability is:
i) the game is fair and unbiased (the dealer puts the joker in these 3 cards randomly)
2010-08-18 14:13:54 補充：
After the dealer reveals the card B,
P(A = Joker | B ≠ Joker) = P(A = Joker & B ≠ Joker) / P(B ≠ Joker) = (1/3) / (2/3) = 1/2 because A = Joker is a subset of B ≠ Joker
2010-08-18 14:13:59 補充：
P(C = Joker | B ≠ Joker) = P(C = Joker & B ≠ Joker) / P(B ≠ Joker) = (1/3) / (2/3) = 1/2 because C = Joker is a subset of B ≠ Joker
2010-08-18 14:14:11 補充：
The chance is still equal between card A or C.
Regarding雨後陽光’s answer, it is a logical mistake. In line with his/her principle, we could consider
P(A = Joker or B = Joker) = (1/3) + (1/3) = 2/3
因已人為排除了A+B 中的 B, then P(A = Joker) = 2/3 and the Mr. X should choose A again!
You see.
2010-08-24 09:47:30 補充：
so, can you explain why:
P(A = Joker or B = Joker) = (1/3) + (1/3) = 2/3
因已人為排除了A+B 中的 B, then P(A = Joker) = 2/3 and the Mr. X should choose A again!
Why must consider P(B = Joker or C = Joker)?
2010-08-24 09:48:48 補充：
don't forget the first choice A = joker is arbitary
2010-08-27 13:40:27 補充：
so, can you explain why:P(A = Joker or B = Joker) = (1/3) + (1/3) = 2/3
因已人為排除了A+B 中的 B, then P(A = Joker) = 2/3 and the Mr. X should choose A again!Why must consider P(B = Joker or C = Joker)?
don't forget the first choice A = joker is arbitarySee my comments on 意見
P(A = Joker) = 1/3 and P(A ≠ Joker) = 2/3
P(B = Joker) = 1/3 and P(B ≠ Joker) = 2/3
P(C = Joker) = 1/3 and P(C ≠ Joker) = 2/3
The assumption on these probability is:
i) the game is fair and unbiased (the dealer puts the joker in these 3 cards randomly)
2010-08-18 14:13:54 補充：
After the dealer reveals the card B,
P(A = Joker | B ≠ Joker) = P(A = Joker & B ≠ Joker) / P(B ≠ Joker) = (1/3) / (2/3) = 1/2 because A = Joker is a subset of B ≠ Joker
2010-08-18 14:13:59 補充：
P(C = Joker | B ≠ Joker) = P(C = Joker & B ≠ Joker) / P(B ≠ Joker) = (1/3) / (2/3) = 1/2 because C = Joker is a subset of B ≠ Joker
2010-08-18 14:14:11 補充：
The chance is still equal between card A or C.
Regarding雨後陽光’s answer, it is a logical mistake. In line with his/her principle, we could consider
P(A = Joker or B = Joker) = (1/3) + (1/3) = 2/3
因已人為排除了A+B 中的 B, then P(A = Joker) = 2/3 and the Mr. X should choose A again!
You see.
2010-08-24 09:47:30 補充：
so, can you explain why:
P(A = Joker or B = Joker) = (1/3) + (1/3) = 2/3
因已人為排除了A+B 中的 B, then P(A = Joker) = 2/3 and the Mr. X should choose A again!
Why must consider P(B = Joker or C = Joker)?
2010-08-24 09:48:48 補充：
don't forget the first choice A = joker is arbitary
2010-08-27 13:40:27 補充：
so, can you explain why:P(A = Joker or B = Joker) = (1/3) + (1/3) = 2/3
因已人為排除了A+B 中的 B, then P(A = Joker) = 2/3 and the Mr. X should choose A again!Why must consider P(B = Joker or C = Joker)?
don't forget the first choice A = joker is arbitarySee my comments on 意見
2010-08-18 9:54 pm
這當中包含了心理環境因素~_~
收錄日期: 2021-04-18 18:32:03
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100818000051KK00491