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The game of Kings is a drinking game played with a standard 52-card deck. The rules are irrelevant to the nature of this question; we only wish to calculate the expected length of a game of Kings. The game ends when all four kings have been drawn (without replacement of any cards).

I've tried using the definition of the hypergeometric distribution's expectation, but it has not gotten me any closer to a solution.

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    Yes, I'm$a$counselor this summer!2012-07-29

1 Answers 1

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Here's a relationship that's very useful for solving this kind of problem:

$\langle n\rangle=\sum_n np(N=n)=\sum_np(N\ge n)\;.$

Thus we don't need to calculate $p(N=n)$, only $p(N\ge n)$, which is easier. The probability that $4$ kings occur somewhere in the first $n$ out of $52$ cards is

$p(N\le n)=\binom n4\Big/\binom{52}4\;.$

Then using

$\sum_{n=m}^k\binom nm=\binom{k+1}{m+1}$

we get

$ \begin{align} \langle n\rangle &= \sum_{n=1}^{52}p(N\ge n) \\ &=52-\sum_{n=0}^{51}p(N\le n)\\ &=52-\sum_{n=4}^{51}p(N\le n)\\ &=52-\sum_{n=4}^{51}\binom n4\Big/\binom{52}4\\ &=52-\binom{52}5\Big/\binom{52}4\\ &=52-\frac{48}5\\ &=42\frac25\;. \end{align} $

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    @$N$icolásKim Could you copy and paste and add that as another answer? It's nice, but hidden in a comment!2013-06-09