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:
© 行恒 2013
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