# Question about poker hand probability

So this is really simple, but can someone shed light on this tiny issue I'm having?

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I want to find out how many combinations of cards there are in $5$-hand poker where you get two pair. The obvious solution to me, was: $$P = \binom{13}{1}\cdot\binom{12}{1}\cdot\binom{4}{2}^2\cdot\binom{11}{1}\cdot\binom{4}{1}$$

However, this was wrong; from Wikipedia I gathered that it should be: $$P = \binom{13}{2}\cdot\binom{4}{2}^2\cdot\binom{11}{1}\cdot\binom{4}{1}$$

It makes sense, but I don't see why they shouldn't be equivalent. The correct statement says choose $2$ cards out of $13$, while mine says choose $1$ card out of $13\times1$ card out of $12$

Obviously they're different calculations, but can somebody tell me why mine is wrong? I'm thinking mine is somehow counting the cards twice, but I don't know why.

In your counting getting a pair of $8$s and then a pair of $7$s is different from getting a pair of $7$s and then a pair of $8$s. You need the $\binom{13}{2}$ since you're picking 2 denominations to be your pairs.