Full House Minute Probability

[getfunky]

The probability of getting a full house by the river when you only have one pair is so small, its probably not worth it, but I am trying to learn so bear with me.

Would you first calculate the odds of tripping your pair by the river:
Check w/1 card remaining(.04255)
Check w/2 cards remaining(.08418)

Then find the odds that one of the unpaird cards pairs up:
Check w/1 card remaining(.19148)
Check w/2 cards remaining(.34968)

Then Multiply the two together.
Check w/1 card remaining(.00814)
Check w/2 cards remaining(.02943)

Seeing that I need to have both my pair hit the third card and one of the unpaird cards pairing, I am not sure weather I need to calculate the possibilities with one or two cards remaining.

Thanks for the help
-ST

[Shawsy]

You might want to tighten up the description of the situation if you want to eliminate the case where the board shows a full house, but as you have presented the problem, here is a solution. The solution below would hold for situations where you have a pocket pair, or where the board is paired, or where you have hit one of your two hole cards to make a pair.

With 47 unseen cards and 2 to come, there are C(47,2) - (47!/(45!2!)) - or 1081 possible combinations for the turn and river cards.

You can get a full house if:
You get one of your paired cards, and one of the unpaired cards on the turn and river. This can happen in C(2,1)xC(9,1) ways. (18 ways)
The turn and river could also bring two more of the previously unpaired cards. This can happen in 3xC(3,2) or 9 ways.
I will ignore the one possibility that you make quads, since you specifically referred to a full house, but you could add this one possibility if you want.

So, probability of making a full house by the river when you have one pair on the flop is 27/1081 or about 2.5% - one in about 40.