# Probability of 2-pair in Poker

A 2-pair is when the hand is of the form $aabbc$ where $a,b,c$ are distinct denominations (2,3,4,...,10,J,Q,K,A), and order does not matter. There are ${52}\choose {5}$ poker hands. There are ${13}\choose{2}$ ways to choose the 2 denominations for the pairs, there are ${{4}\choose{2}}{{4}\choose{2}}$ ways to choose the suits for the 2 pairs, and there are ${{11}\choose{1}}{{4}\choose{1}}$ ways to choose the remaining card. Therefore, $$P(2-pair)=\frac{{{13}\choose{2}}{{4}\choose{2}}^2{{11}\choose{1}}{{4}\choose{1}}}{{52}\choose {5}}\approx.0475$$

However, instead of writing ${{13}\choose{2}}{{4}\choose{2}}^2$ for the number of ways to choose the pairs, why can't I compute the ways to find one pair and then multiply by the number of ways to find the second pair after the first has been selected to obtain $${{13}\choose{1}}{{4}\choose{2}}\cdot{{12}\choose{1}}{{4}\choose{2}}?$$ In the latter case, the computation becomes twice the correct answer: $$\frac{{{13}\choose{1}}{{4}\choose{2}}\cdot{{12}\choose{1}}{{4}\choose{2}}{{11}\choose{1}}{{4}\choose{1}}}{{52}\choose{5}}\approx .0951$$

What would be the flaw in the reasoning in the second approach?

## 1 Answer

The flaw in the second approach is that you account for the order of the denominations. I.e. you count Aces and Kings separately from Kings and Aces. Hence you get a factor of two too much.

In the formulas the difference manifests as ${{13}\choose{1}} \cdot {{12}\choose{1}} = 13\cdot 12$ versus ${{13}\choose{2}}= \frac{13\cdot 12}{2}$.

• In the first calculation, aren't I also accounting for order when I choose the 2 pairs before I choose the isolated card? – The Substitute Jul 3 '17 at 7:24
• No, when you choose the two pairs with the forumla $\binom{13}{2}$, "Aces and Kings" are counted only once. You don't count the double pair "Aces and Kings" and then the double pair "Kings and Aces", unlike in your second, flawed approach - This is the would point of binomial factors. – Evargalo Jul 3 '17 at 7:54