# About picking values when counting $2$ Pair and Full House

Number of hands:

Two Pair: $\binom{13}2 \binom 42^2 \binom{44}1$

Full House: $\binom{13}1 \binom 43 \binom{12}1 \binom42$

I was wondering why we don't do $\binom{13}1 \binom{12}1 \binom 42^2 \binom{44}1$ for Two Pair and $\binom{13}2 \binom 43 \binom42$ for Full House instead. I made these mistakes and would like to understand why they don't work.

In this answer, I am assuming that a 'hand' is a set of five cards (i.e., there is no associated order to the hand). In some cards games this would be a good assumption, in others (such as 5-card stud [https://en.wikipedia.org/wiki/Five-card_stud], where cards are received and bet on sequentially), this may not be a good assumption.

Let's start with a very much simplified example: a deck with three cards, 1, 2, and 3. How many two-card hands are there? Answer: 3. The 3 hands are $\{1,2\}, \{1,3\},\mbox{and } \{2,3\}.$ If we tried to count this as pick a card ($3$ choices), pick another card ($2$ choices), giving an answer of $6$ possible hands, we would have overcounted by a factor of $2$. For example the hand $\{1,2\}$ is counted twice (once as choose $1$, then $2$; and once as choose $2$, then $1$).

The same idea occurs in your question. A hand with a pair of $7$s and a pair of jacks is the same as a hand with a pair of jacks and a pair of $7$s. But if you count the rank choices as ${{13}\choose{1}}\cdot {{12}\choose{1}}$, you are double counting these two-pair hands. That's why you get the correct number of choices if you choose the two ranks simultaneously as ${13}\choose{2}$.

By contrast in a full-house rank choices are distinguished by one rank getting three-of-a-kind and one rank getting a pair. Thus three $7$s and two jacks is a different hand than three jacks and two sevens.

In your two pair counting, you counted "AAKKJ" cases twice - once when your $\binom{13}{1}$ choice was the ace and your $\binom{12}{1}$ choice was the king, and once when they were in reverse.

On the other hand, the denominations of the full house case are not symmetric - you need to pick one denomination for the three cards, and one for the two. So if you pick one of the $\binom{13}{2}$ pairs of denominations, say, $\{A,K\}$, you still don't know whether to pick three aces or three kings.

So you can think of it as $2\binom{13}{2}\binom{4}{3}\binom{4}{2}$. But $2\binom{13}{2}=\binom{13}{1}\binom{12}{1}$.