Combinatorics / Probability Question (drawing cards)?

I'd just go to a poker site and check on the odds on getting 2 pair in a game of 7 card stud.

https://en.wikipedia.org/wiki/Poker_probability#De...

Scroll down to two-pair, specifically "2 pairs with 3 kickers".

I don't know whether or not you wish to count two-pairs that also make straights or flushes. The derivation shown will allow you to decide.

Then divide by (52 C 7) for the probability.

(It will be lower than what the probability chart says, because you're not counting the combos of 3-pair which in poker only count as 2-pair.)

Let C(n,r) be n choose r

C(n,r) = n!/(r!(n-r)!)

Probability= number of 2 pair / number of 7 card hands

Number of 7 card hands = C(52,7)

Number of 2 pairs = C(13,2)*C(4,2)*C(4,2)*C(11,3)*C(4,1)*C(4,1)*C(4,1)

The above calc broken into steps

1) C(13,2) - of the 13 possible ranks, pick 2

2) C(4,2) - of the first rank chosen, pick 2 suits.

3.C(4,2) - of the second rank chosen, pick 2 suits

4. C(11,3)- now, we pick the three remaining to each have a separate rank to prevent a third pair, a triple or a 4 of a kind

5) just like #2 and #3 but only 1 card is drawn