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

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:

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

=> $E(Y)= 20$