# Expected number of card draws to get all 4 suits

You have a standard 52 card deck, with 13 cards of each of the 4 suits (Hearts, Diamonds, Spades, Clubs). What is the expected number of cards you have to draw from the deck until you have all 4 suits represented in your hand?

I couldn't think of how to get the negative binomial to work, since this is sampling without replacement and has 4 suits instead of just 2. I imagine a distribution that could solve this might be called the Negative Hypergeometric Multivariate. Anyone have any ideas? Thanks very much.

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Nice question! Running some simulations, the average seems to be about $7.66$, but I don't know which distribution is appropriate here. – TMM Jan 1 '13 at 2:03
Yep, you need the Multivariate Hypergeometric Distribution, see for example in wikipedia: en.wikipedia.org/wiki/… – Bitwise Jan 1 '13 at 2:11

Let $s=13$ denote the number of cards in each suit and $N$ denote the (random) number of cards drawn when all $k=4$ suits are represented for the first time. Hence, $k\leqslant N\leqslant(k-1)s+1$ with full probability.
For each $n\geqslant1$, the event $[N\gt n]$ depends on the $n$ first cards drawn only. There are ${ks\choose n}$ collections of $n$ cards from a full deck of $ks$ cards and each such collection has the same probability ${ks\choose n}^{-1}$ to be drawn. The event $[N\gt n]$ means that one avoids at least one suit. Using inclusion-exclusion principle, there are $A_n$ ways to do so, where $$A_n={k\choose1}{(k-1)s\choose n}-{k\choose2}{(k-2)s\choose n}+\cdots\pm{k\choose k-1}{s\choose n}\mp{k\choose 0}{0\choose n}.$$ This yields the expectation $$\mathbb E(N)=\sum_{n\geqslant0}\mathbb P(N\gt n)=\sum_{n\geqslant0}{ks\choose n}^{-1}A_n.$$ In the case at hand, $$\mathbb E(N)=\sum_{n\geqslant0}{52\choose n}^{-1}\left(4{39\choose n}-6{26\choose n}+4{13\choose n}-{0\choose n}\right),$$ that is, $$\mathbb E(N)=4+\sum_{n=4}^{39}{52\choose n}^{-1}\left(4{39\choose n}-6{26\choose n}+4{13\choose n}\right),$$ or, $$\mathbb E(N)=4+4B_{39}-6B_{26}+4B_{13},\qquad B_i=\sum_{n=4}^{i}{52\choose n}^{-1}{i\choose n}.$$ Numerically, $$\mathbb E(N)=\frac{4829}{630}=7+\frac23-\frac1{630}\approx7.66508.$$
Another approach giving the same answer is that in the answer to math.stackexchange.com/questions/1660538/… so $\mathbb{E}(N)= 1+{4 \choose 1} \frac{39}{14} - {4 \choose 2} \frac{26}{27} + {4 \choose 3} \frac{13}{40} = \frac{4829}{630}$ – Henry Feb 18 at 23:08