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I know the usual approach to poker hands using combinations. For example the probability for a pair in a 5 card hand is

$$\frac{{13 \choose 1} {4 \choose 2}{12\choose 3}*4^3}{52 \choose 5} $$

I have some lecture notes that line out a different permutation based approach using

$$ \frac {{5\choose2}*52*3*48*44*40}{52*51*50*49*48}$$

to derive the probability of a single pair. First the position of the pair on the hand is chosen, then cards are put into those positions. For the first card there are 52 ways to do this for the second 3 to get a pair. For the third card one rank is striken, thus 48 possiblities and so forth.

I would like to use this approach for a two-pair and came up with the following:

$$ \frac {{5\choose2}*52*3*{3 \choose 2}*48*3*44}{52*51*50*49*48}$$

However this overcounts giving me twice the true probability. I don't understand why this is the case.

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    Your very first ${13 \choose 2}$ should be ${13 \choose 1}$, I think. [This page](http://www.math.hawaii.edu/~ramsey/Probability/PokerHands.html) backs me up. – 2017-01-19
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    @TonyK: you are of course right. – 2017-01-19

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You are counting each hand twice, once with one pair chosen first and once with the other pair chosen first. The leading $5 \choose 2$ counts the ways you can locate the first pair among the five draws. The $3 \choose 2$ counts the ways you can locate the second pair among the five draws. If your pairs are aces and twos, you could locate the aces first and the twos second or the twos first and the aces second, but swap the chosen locations. That will give the same draw.

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    I see. So the permutations of the pairs have to be discounted. Thinking of this in terms of actions, is it a good intuition that the cards determining would have to be drawn simultaneously? I.e. there is no first or second pair, there is just two pairs. Thus, if we pretend there was an order to make things actionable, we have to discount the permutations of the pairs. Likewise, if we want the probability of a three-pair on a hand of seven cards, we get $ \frac {{7\choose2}*52*3*{5 \choose 2}*48*3*{3 \choose 2}*44*3*40}{3!/(3-3)!*52*51*50*49*48*47*46}$ – 2017-01-19
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    You are right that dividing by $3!$ solves the overcount because you are counting each hand $3!$ times based on the order of the pairs. – 2017-01-19