回复：Fun Probability Question 来源: TCT

来源: zeal626 于 2009-11-09 23:37:50 [] [旧帖] [给我悄悄话] 阅读数 : (198 bytes)
本帖于 2009-12-19 10:42:34 时间, 由普通用户 康MM 编辑

0.5*(1 - P[having 10k after n tosses]) ?

and

P[having 10k after 30 tosses]
= (30 choose 15)*(1/2)^30

P[having 10k after 100 tosses]
= (100 choose 50)*(1/2)^100

hehe

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• hmmm -dynamic- ♂ 给 dynamic 发送悄悄话 (184 bytes) () 11/10/2009 postreply 02:44:00

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