Probability of getting two pair in poker

Question

I was looking at this website http://www.cwu.edu/~glasbys/POKER.HTM and I read the explanation for how to calculate the probability of getting a full house. To me, the logic basically looked like you figure out the number of possible ranks and multiply by the number of ways to choose the cards from that given rank.

In other words, for a full house $P=$ $$\frac{{13\choose1}{4\choose3}{12\choose1}{4\choose2}}{52\choose5}$$

Following this logic, I tried to calculate the probability of getting two pair. My (incorrect) logic was that there are 13 possible ranks for the first pair and $4\choose2$ ways to choose two cards from that rank, 12 possible ranks for the second pair and $4\choose2$ ways to choose two cards from that rank, and 11 possible ranks for last card and $4\choose1$ ways to choose a card from that rank.

So I tried $P=$ $$\frac{{13\choose1}{4\choose2}{12\choose1}{4\choose2}{11\choose1}{4\choose1}}{52\choose5}$$

Obviously my solution was incorrect. I read explanation and the correct answer is $P=$ $$\frac{{13\choose2}{4\choose2}{4\choose2}{11\choose1}{4\choose1}}{52\choose5}$$

I'm still a bit fuzzy on where I went wrong though. Can anyone help me understand this problem a little better? Thank you very much for your help.

Answer 1

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

Answer 2

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.

Answer 3

Riffing off prony's answer, which I think is a little confusing. Here are the possibilities for each card:

• Card 1: 52 cards
• Card 2: 3, since it must match the value of Card 1
• Card 3: 48, since we can't match the value of Card 1
• Card 4: 3, since we can't match the value of Card 3
• Card 5: 44, since we can't match the value of the other cards

This will give us all orderings of the form XXYYZ. We then notice two issues that we are double counting:

1. Cards 1 and 2 can be interchanged (XX). (2!)
2. Cards 3 and 4 can be interchanged (YY). (2!)
3. Cards 1 and 2 can collectively be interchanged with Cards 3 and 4 (XX with YY). (2!)

So we have distinct, unordered $(52 \times 3 \times 48 \times 3 \times 44)/(2! 2! 2!)$ ways. Dividing this by the number of combinations ${52 \choose 5}$ yields our answer.