Question: Calculate the probability of getting a two pair hand ( e.g., two 8’s, two Queens, and a Knight )
My answer: The probability of getting a two pair hand is : $$ \frac{13\cdot\binom{4}{2}\cdot12\cdot\binom{4}{2}\cdot11\cdot\binom{4}{1}}{\binom{52}{5}} = \frac{396}{4165} $$ but the correct answer is : $$ \frac{\binom{13}{2}\cdot\binom{4}{2}^2\cdot44}{\binom{52}{5}} = \frac{198}{4165} $$ which is my answer but divided by $2$. My question is why is my answer wrong, and in the correct answer where did $\binom{13}{2}$ come from?
You are double-counting as you count any two pairs twice (as (8,8,Q,Q) and as (Q,Q,8,8)). To avoid this the correct solution suggests to use $\binom{13}2$ so that the whole pair (8,Q)$\equiv$(Q,8) is chosen.