2
$\begingroup$

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.

  • 0
    @Marcva$n$Leeuwen: Yes, that was exactly how I thought, I now see your point! Thanks a lot!2012-09-11

1 Answers 1

1

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$

  • 0
    @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. :)2012-09-08