Probability of getting two pair in poker

You have to choose the two card values you want as your pairs simultaneously. Remember--multiplying the numbers ${13\choose1}{4\choose2}{12\choose1}{4\choose2}{11\choose1}{4\choose1}$ assumes an $order$, i.e. you are counting, say, QQKK2 as different from KKQQ2.

This is why you have to do ${13\choose2}{4\choose2}{4\choose2}{11\choose1}{4\choose1}$. It makes the counting not sensitive to which pair you choose first.

First choose the two (different) values of the cards that will be pairs: $13 \choose 2$.

For each of these values, pick two suits from the four suits available: ${4 \choose 2}{4 \choose 2}$.

Then, since this is only two pair and not more, choose the value of the other card, and its suit: ${11 \choose 1}{4 \choose 1}$.

Finally, divide by the total number of combinations of all hands: $52 \choose 5$. And there it is:

$$P = \frac{{13\choose2}{4\choose2}{4\choose2}{11\choose1}{4\choose1}}{52\choose5}$$

The difference between this solution and that for the full house is that there is more "symmetry" for the two pair: both pairs are groups of two. With the full house, one is a group of three, and the other is a group of two. Aces over kings is distinct from kings over aces.

Here, you choose the card for the three of a kind, then pick the three suits: ${13 \choose 1}{4 \choose 3}$. Then, you choose the card for the pair, and pick the two suits: ${12 \choose 1}{4 \choose 2}$.

Another way is for the first to choose three values out of 13: ${13 \choose 3}$, then choose 2 values out of 3 for pair: ${3 \choose 2}$, for each pair choose 2 suits out of 4: ${4 \choose 2}$ - twice, and finally choose one suit for the 5-th card, which is not in any of pairs: ${4 \choose 1}$.

Resulting formula: $$ {13 \choose 3}*{3 \choose 2}*{4 \choose 2}*{4 \choose 2}*{4 \choose 1}=123552 $$

No difference with previous answers because of $$ {13 \choose 3}*{3 \choose 2} = {13 \choose 2}*{11 \choose 1} $$