Q:

poker full house binomial overcounting

I know that the probability of getting a full house in poker is:
$$ \frac{13 \cdot 12 \cdot \binom{4}{3} \cdot \binom{4}{2}}{\binom{52}{5}}.$$
Can someone please explain why we are not overcounting here. I believe that it does not matter whether we choose the 3 of a kind first and then the 2 of kind or if we do it vice versa choosing the 2 of a kind first and then choose the 3 of a kind second. If this is true then we would need to multiply that formula by $ \frac 12 $ to end up with $$ \frac 12 \cdot \frac{13 \cdot 12 \cdot \binom{4}{3} \cdot \binom{4}{2}}{\binom{52}{5}}.$$ Why is my thinking wrong? Thanks in advance.

A:

It doesn't matter which one you choose first, but whichever you choose first, each hand is counted exactly once, so there's no overcounting.

If you choose the $3$-of-a-kind first, and then the pair, the number of hands is the value of the expression
$${\small{\binom{13}{1}}}{\small{\binom{4}{3}}}{\small{\binom{12}{1}}}{\small{\binom{4}{2}}}$$

If you choose the pair first, and then the $3$-of-a-kind, the number of hands is the value of the expression
$${\small{\binom{13}{1}}}{\small{\binom{4}{2}}}{\small{\binom{12}{1}}}{\small{\binom{4}{3}}}$$
yeilding the same count.

But with either choice, each hand is counted exactly once.

As a way to see it, consider a reduced deck with $3$ suits, and $2$ ranks; $Q\;$(Queens), and $J\;$(Jacks), where for each rank, there is one of each suit, so the deck consists of $3$ Queens, and $3$ Jacks. Since the deck has $6$ cards, there are only $6$ different $5$-card hands, and clearly, every hand is a full house. Thus, the count for the number of full house hands is $6$, which is the same count as you would get by evaluating the expression
$${\small{\binom{2}{1}}}{\small{\binom{3}{3}}}{\small{\binom{1}{1}}}{\small{\binom{3}{2}}}$$
So it would be incorrect to divide by $2$.

As an example where you would need to divide by $2$, consider the number of hands which are two-pairs . . .

Choosing the first pair, then choosing the second pair, then choosing the remaining card, would yield a doubled count, since the hand $\text{XXYYZ}$ is the same as the hand $\text{YYXXZ}$. Hence, the correct count (using a standard $52$-card deck) would be
$$\left({\small{\frac{1}{2}}}\right)\left({\small{\binom{13}{1}}}{\small{\binom{4}{2}}}{\small{\binom{12}{1}}}{\small{\binom{4}{2}}}{\small{\binom{11}{1}}}{\small{\binom{4}{1}}}\right)$$

