回复:Fun Probability Question 来源: TCT
zeal626
2009-11-09 23:37:50
( reads)
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
dynamic
2009-11-10 02:44:00hmmm