回复：Fun Probability Question 来源: TCT

来源: 于 2009-11-09 23:37:50 [旧帖] [给我悄悄话] 本文已被阅读：次

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