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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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It's natural to try to solve the more general problem of the expected number of draws needed to get $a$ particular cards from a stack of $b$ cards, the reason being that you might observe more structure in this function as a function of both $a$ and $b$ (e.g. a recurrence). –  Qiaochu Yuan Jul 29 '12 at 6:34
    
Hey, you're at Promys right? I know Dylan and Joe, haha. I've seen you around Warren Dining. I'm here over the summer as well (I'm a BU student). And thank you for the tip—I'll work on that now. –  Nicolás Kim Jul 29 '12 at 6:35
    
Yes, I'm a counselor this summer! –  Qiaochu Yuan Jul 29 '12 at 6:37

1 Answer 1

up vote 4 down vote accepted

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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Thank you for your solution! –  Nicolás Kim Jul 29 '12 at 7:06
4  
One solution that I came up with just now is the following: Say we fix the four kings in the deck and now we have 48 cards to fill into 5 holes. As there is equally probability any card goes into any hole, each one is filled with 48/5 cards. Thus, we get the number of cards (52) minus the number of cards in the last hole (48/5), which equates to 42.4. –  Nicolás Kim Jul 29 '12 at 7:07
    
@NicolásKim Could you copy and paste and add that as another answer? It's nice, but hidden in a comment! –  Potato Jun 9 '13 at 22:24

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