# What is the probability in any s-of-a king in n cards?

What is the probability of getting any s of a kind (let's say 3 of a kind) when dealt n cards (let's say 13) of a standar deck? I have calculated the probability of getting a specific set of 3 of a kind (i.e. 3 Kings) which is $\frac{\left( \begin{array}{c} 4\\ 3 \end{array} \right) \left( \begin{array}{c} 52-3\\ 13-3 \end{array} \right)}{\left( \begin{array}{c} 52\\ 13 \end{array} \right)}$ (please correct me if I'm wrong).

However, what I am trying to find is the probability of getting any s of a kind, and furthermore the probability of getting any s of a kind greater than one specific value.

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Should a hand with 4 of a kind be included in the count? What about 3 kings and 3 tens in the same hand (if $n\ge2s$)? Should that contŕibute one or two or none at all to the count? –  Hagen von Eitzen Sep 8 '12 at 7:58
Yes, a hand with 4 of a kind is included in the count. I should have said at least one "s of a kind". (3 kings and 3 tens should contribute one to the count). –  auitaV Sep 8 '12 at 8:11
@anitaV: The numerator of your formula is wrong. If you want exactly $3$ kings then the number is $\binom43\binom{52-4}{13-3}$, and if you also want to count the $4$ kings possibilities, then you need to add $\binom44\binom{52-4}{13-4}$ to that (see the answer by Hagen). –  Marc van Leeuwen Sep 8 '12 at 11:54
s-of-a king –  binn Sep 8 '12 at 12:19
@MarcvanLeeuwen: Are you sure about that? I mean, the $4$ King possibilities isn't it included in the number $\left( \begin{array}{c} 52-3\\ 13-3 \end{array} \right)$ ? Does not this number represent all the possible combinations with the 52-3 cards left in the deck? 4th King included? –  auitaV Sep 11 '12 at 10:34

The possibilities for (at least) three $x$'s (where $x$ is e.g. "kings") are $4{48\choose n-3}+{48\choose n-4}$ (three kings combined with $n-3$ non-kings or all kings and $n-4$ non-kings).

Let's count how many $n$-hands there are that do not have three (or four) of a kind. Then all ranks occur once or twice or not at all. For all $0\le k\le \lfloor \frac n2\rfloor$ such a hand is obtained by

• choosing $k$ out of 13 ranks to occur twice
• for each of these choose 2 out of 4
• choosing $n-2k$ of the remaining $13-k$ ranks to occur once
• for each of those choose 1 out of 4.

In total, we have $$\sum_{k=0}^{\lfloor \frac n2\rfloor} {13\choose k}{4\choose2}^k{13-k\choose n-2k}{4\choose1}^{n-2k}=4^n\sum_{k=0}^{\lfloor \frac n2\rfloor} {13\choose k}{13-k\choose n-2k}\Bigl(\frac38\Bigr)^k.$$ In case $n=13$ this simplifies to $$4^{13}\sum_{k=0}^6{13\choose k}{13-k\choose k}\Bigl(\frac38\Bigr)^k=323603221504.$$ Finally, we find $$p = 1-{323603221504\over{52\choose 13}}\approx0.49$$

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After you've chosen $k$ out of $13$ ranks, there are $13-k$ ranks that remain, not $13-2k$... This may explain the smallness of the number you found for $n=13$. And note also that for $n>13$ your summation gives $0$, which is suspect. –  Marc van Leeuwen Sep 8 '12 at 12:04
@Marc. Thanks for the hint. After editing, the result looks more trustworthy. Also, it is now $n>26$ instead of $n>13$ that makes avoiding 3 of a kind impossible - sounds much less suspect. :) –  Hagen von Eitzen Sep 8 '12 at 12:14