# What is the probability of getting FOUR OF A KIND in a $13$-card poker game?

What is the probability of getting FOUR OF A KIND in a $13$-card poker game?

Here is my attempt:

The setup for the required poker hand would either be: $$AAAABCDEFGHIJ,$$ $$AAAABBBBCDEFG,$$ or $$AAAABBBBCCCCD,$$ where $A, B, C, D, E, F, G, H, I, J$ are distinct faces.

The total number of poker hands of type $AAAABCDEFGHIJ$ is $${13 \choose 1}{4 \choose 4} = 13.$$ The total number of poker hands of type $AAAABBBBCDEFG$ is $${13 \choose 2}{4 \choose 4}\cdot{_{2} P_{1}} = 156.$$ The total number of poker hands of type $AAAABBBBCCCCD$ is $${13 \choose 3}{4 \choose 4}\cdot{_{3} P_{1}} = 858.$$

The total possible number of $13$-card poker hands from the standard deck of $52$ playing cards is $${52 \choose 13} = 635013559600.$$

Therefore, the required probability is $$\dfrac{1027}{635013559600} \approx 0.000000001617288299555\ldots.$$

My questions are:

(1) Is this probability computation correct?

(2) If my computation is not correct, where is/are the error(s) and what hint can you give towards rectifying that error(s)?

I am a bit unsure about my computation of the total number of poker hands of types $AAAABBBBCDEFG$ and $AAAABBBBCCCCD$, as logically these should be rarer than the poker hands of type $AAAABCDEFGHIJ$.

• You would also need to consider hands of the form AAAABBCDEFGHI, AAAABBBCDEFGH, and the like. In other words, just because you don't have 4 aces doesn't rule out the possibility of having 1, 2, or 3 of them, or the possibility of having 4 pairs, 4 of a kind, and one additional card that doesn't match any of these. Apr 16, 2016 at 0:54
• So hands that have at least one $4$ of a kind are to be counted? Can some of the remaining cards outside the $4$(s) of a kind form a $3$ of a kind? Or a straight? Or a royal flush? Apr 16, 2016 at 0:55
• @AndréNicolas, yes - poker hands having at least one $4$ of a kind are to be counted. There are no restrictions to the remaining cards. Apr 16, 2016 at 0:57
• The number of $4$ of a kind hands is very very much undercounted in the post. Apr 16, 2016 at 1:01

We use Inclusion/Exclusion.

Let us first count the number of hands that have $4$ Queens. The rest of the cards can be chosen in $\binom{48}{9}$ ways.

Similarly, there are $\binom{48}{9}$ hands that have $4$ Aces, and so on.

Add up. We get $\binom{13}{1}\binom{48}{9}$.

However, this double-counts, for example, the hands that have $4$ Queens and $4$ Aces. There are $\binom{44}{5}$ such hands. The $2$ kinds we have $4$ each of can be chosen in $\binom{13}{2}$ ways, so our next estimate is $\binom{13}{1}\binom{48}{9} -\binom{13}{2}\binom{44}{5}$.

However, we have taken away too much, for we have removed one too many times the $\binom{13}{3}\binom{40}{1}$ hands that have $3$ kinds of $4$ of a kind. So our final total is $$\binom{13}{1}\binom{48}{9} -\binom{13}{2}\binom{44}{5}+\binom{13}{3}\binom{40}{1}.$$

Finally, for the probability, divide as you did by $\binom{52}{13}$.