• 公达
  • 2 月,3 周前
  • 2021年2月21日 16:31
  • 卡主

解决了。答案如下:

It is a problem of shift geometric distribution. If $X$ is the number of draws until you get Pikachu, $Y$ is the number of draws until you get two consecutive Pikachu.

We know that $E(X)=1/p=4$, where p = 1/4.

Then you will either get a Pikachu on the next card. There is a 3/4 chance you don’t, so we have a recursive definition for the expected number:

$E(Y)=E(X) + 1 + \frac{3}{4} E(Y)$

=> $E(Y)= 20$

也可以参考页面 https://math.stackexchange.com/questions/112726/tossing-a-fair-coin-until-two-consecutive-tosses-are-the-same

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