WXC又掷硬币了

constant · 发表于 2006-3-8 21:30:44

假设投掷n次才第一次出现连续三次正面。问n的期望值是多少？www.ddhw.com

这题以前争论过很长时间，答案是14。（Oops！说漏了）现在问更一般的问题：www.ddhw.com

难度：+++www.ddhw.com

任给一个正反面的序列，问第一次出现这个序列时投掷次数的期望值是多少？

本贴由[constant]最后编辑于：2006-3-9 12:14:36

QL · 发表于 2006-3-9 06:36:43

I think what matters is the length of the sequence, not the specific string of heads or tails. We can just assume the sequence in question is 1,1,...,1, with n 1's.

At any time, we can count the number K of consecutive 1's, going back from the lastest toss. For example, after 5 tosses, the sequcence of head/tails is 1 -1 1 1 1, then K=3. Specifically, if the lastest toss is a -1, then K = 0.

Let A_K be the expected number of more tosses needed to end the game. We have the following system of equations:

A_0 = 1/2(A_0+A_1)+1

A_1= 1/2(A_0+A_2) +1

....

A_{n-2} = 1/2(A_0+A_{n-1})+1,

A_{n-1} = 1/2(A_0+A_n) +1

A_n=0

n+1 unkowns, n+1 equations, and the final answer is 1/2(A_0+A_1)+1

本贴由[QL]最后编辑于：2006-3-8 22:45:14

大头羊 · 发表于 2006-3-9 07:38:19

为什么是14？

QL · 发表于 2006-3-9 16:11:46

"I think what matters is the length of the sequence, not the specific string of heads or tails. We can just assume the sequence in question is 1,1,...,1, with n 1's."

Oops, wrong. So the previous solution is only for 1,1,...,1 case. For general case, the transition between states can be more complicated.

本贴由[QL]最后编辑于：2006-3-9 8:27:52

constant · 发表于 2006-3-9 18:11:35

就是问你呢

QFT · 发表于 2006-3-9 18:21:24

sean gave 14 using Markov method.
Anyway it can be solved through simple reasoning similar to that of QL.

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WXC又掷硬币了

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就是问你呢[:D)]

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