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】
: 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】
: 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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