f*****k 发帖数: 353 | 1 61. Coin toss problem #3
Suppose you have a coin in which the probability of flipping a heads is p,
where p>=0.5 . What is the expected number of flips it will take for the
number of heads to equal the number of tails, assuming the first flip is a
tails?
答案给的是:1/(2p - 1),solution给的挺含糊的
我想用conditional expectation: let E[N | -1] be the expect number if the
first flip is a tail
E[N | -1] = p*2 + (1 - p)*( E[N | -2 ] + 1)
往后就不好算了。。。
请教一下解法,谢谢 | v*******y 发帖数: 1586 | | u****4 发帖数: 5 | 3 random walk. X(head)=-1, X(tail)=1. let u be the expected number. then
u=p*1+(1-p)*(u+u+1)
where the first term on the right corresponds to the case that the next flip
is a head. the second term is for a tail, and you need 2->1, then 1->0, so
it's u+u+1 (1 is the initial tail).
【在 f*****k 的大作中提到】![](/moin_static193/solenoid/img/up.png) : 61. Coin toss problem #3 : : Suppose you have a coin in which the probability of flipping a heads is p, : where p>=0.5 . What is the expected number of flips it will take for the : number of heads to equal the number of tails, assuming the first flip is a : tails? : 答案给的是:1/(2p - 1),solution给的挺含糊的 : 我想用conditional expectation: let E[N | -1] be the expect number if the : first flip is a tail : E[N | -1] = p*2 + (1 - p)*( E[N | -2 ] + 1)
| f*****k 发帖数: 353 | 4 i c, thanks
flip
so
【在 u****4 的大作中提到】![](/moin_static193/solenoid/img/up.png) : random walk. X(head)=-1, X(tail)=1. let u be the expected number. then : u=p*1+(1-p)*(u+u+1) : where the first term on the right corresponds to the case that the next flip : is a head. the second term is for a tail, and you need 2->1, then 1->0, so : it's u+u+1 (1 is the initial tail).
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