What are the odds of getting 2 sets in texas holdem?

Question

I'm looking to create a variant of holdem allowing 6 and 7 card hands in special situations. Namely, 3 pair, 2 set, and quads over set (a full mansion).

My probability skills are rusty, but I'm still able to break down the problems, I think!

I've started trying to calculate the odds of 2-set so I can find where to rank the hand. I'm using the probability tree method. The branch at the top is clear, pocket pair vs mixed hand.

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Pocket pair (0.059):

With a pocket pair, you need to hit a set with the card in your hand, and the board needs to hit a set. There are 12 other sets the board could hit besides the cards in your hand.

Odds of a permutation of such a board:

(2/50) * (4/49) * (3/48) * (2/47) * (44/46) = 8.3067e-06

The last value of 44/46 is because you want to preclude quads from the board. Using permutation with repetition, you get (5!/3!) or 20 for the permutations possible.

8.3067e-06 * 12 * 20 = 0.001993608 for the odds of 2 set when holding a pocket pair.

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mixed hand (0.941)

When you don't have a pocket pair, the board needs to 2 pair both of your cards.

(3/50) * (2/49) * (3/48) * (2/47) * (44/46) = 6.2301e-06

Using permutation with repetition, you find there are (5! / (2! * 2!)) or 30 permutations of such a board.

6.2301e-06 * 30 = 0.000186903 of 2 set when you have a mixed hand

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Combining both sides of the tree, you get

pocket pair: 0.059 * 0.001993608 = 0.0001176229

mixed hand: 0.941 * 0.000186903 = 0.0001758757

with a combined odds of 0.0002934986 or 0.0293%, ranking just below a straight flush (0.0279%).

Answer 1

You're not rusty at probability at all. Poker is just complicated and hard to calculate. The only thing you forgot to count was the possibility of making the "2-set" with only one card in your hand. This caused you to underestimate by around 25%.

Spoilers below. Forgive the lack of LaTeX as you can now copy paste this into Wolfram Alpha to verify.

Using a simple method considering all seven cards at once

binom(13, 2) * binom(11, 1) * binom(4, 3) ^ 2 * binom(4, 1) / binom(52, 7) = 0.00041

Calculating the probability that you forgot, using your method:

2*(3/50)*(2/49)*11*(4/48)*(3/47)*(2/46)*5!/(2!*3!)*(48/51) = 0.000117 The 2 comes from the fact that the pair on the board can hit either of your cards. The 11 is considering the 3-of-a-kind on the board can be any rank other than either of your hole cards. The 48/51 is the probability of unpaired hole cards.

I am curious about the modified rules of poker that you're trying to implement.