# 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?

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