poker full house binomial overcounting

Question

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.

Answer 1

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 $$\left({\small{\frac{1}{2}}}\right)\left({\small{\binom{13}{1}}}{\small{\binom{4}{2}}}{\small{\binom{12}{1}}}{\small{\binom{4}{2}}}{\small{\binom{11}{1}}}{\small{\binom{4}{1}}}\right)$$

Answer 2

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.

Correct. You can do it whichever way you like. But there is only double counting if you do it both ways, and you haven't - in your answer you have (I think) done it the first way. You could do it the second way instead and you would get the same answer, but it's only if you did it the second way as well that you would have to divide by $2$.

A rough analogy: put your full house cards down on the table and count them. If you count them from left to right and from right to left you will get an answer of $10$ and you have to divide by $2$. But if you only count them from left to right or right to left and not both, then you will get the right answer straight away, whichever direction you count.

Hope this helps!