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 $AAAABCDEFGHIJ$.
Best Answer
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 $\binom{52}{13}$.