Probability of getting two pair in poker

Solution 1:

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.

Solution 2:

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: ${12 \choose 1}{4 \choose 2}$.

Solution 3:

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 $$ {13 \choose 3}*{3 \choose 2} = {13 \choose 2}*{11 \choose 1} $$

Solution 4:

I find permutation more intuitive to follow for this kind of problems. For people like me:

We have five slots to fill: - - - - - . The first slot can take all 52 cards. The second slot can take only three cards so that they can make a pair. Similarly, the third and fourth slots can take 48 and 3 cards, respectively. The last and final slot can take any of remaining 44 cars. Therefore:

52 * 3 * 48 * 3 * 44 = 988416. Please note, this is order dependent. In other words, this is the count of x x y y z. However, we should count all the possibilities (i.e., z x y x y). Therefore, we multiply 988416 with 5! and divide by 2! (order between two xs) * 2! (order between two ys) and 2! (order between the pair of xs and ys). The total count is 14826240.

This is the numerator. The denominator is 52*51*50*49*48 = 311875200. The probability is 0.0475390156062425.

Note that if you want to count how many different hands can be dealt, then you have to divide 14826240 by 5! to compute the combination.