The question: What is the expected number of cards required to be drawn in order to draw 5 spades.
What I have:
Let $X=X_1+\cdots+X_{43}$ (43 because we're examining the case when 4 spades have been drawn and we're waiting for the 5th) with $X_i=1$ iff card $i$ is drawn before the fifth spade. Then $X=$ the number of cards drawn before the 5th spade. Let $Y=X+1$. Then $Y=$ the number of cards drawn before and including the 5th spade.
$P\left\{X_i=1\right\}=$ the probability that card $i$ is drawn before the 5th spade $=\frac{9!}{10!}=\frac{1}{10}$ (because there are $9!$ ways of ordering the $i$th card at the front of 9 spades and $10$ ways of ordering 10 cards).
Then $E\left[Y\right]=\frac{43}{10}+1=5.3$. But this is far too small. What went wrong?
Thanks.
Best Answer
There is a distribution to cover these waiting time problems without replacement. It is called the Negative Hypergeometric Distribution. It has an expected value of
$ \frac{N+1}{M+1} \cdot k = \frac{265}{14}$
where N ( 52 cards ) is the total number, M ( 13 spades ) is the total number of the target type and k ( 5 spades needed ) is the number wanted from M.