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Paxosmotic
09-07-2005, 12:18 PM
This is a problem me and another 2+2er have been trying to figure out for a few weeks (off an on, obviously, I like sleep), but haven't pieced together. I know the chance of at least one overcard on a flop is represented by p=1-(50-n choose 3)/(50 choose 3) where n is number of overcards, but what we are trying to figure out is the probability of exactly 1 overcard to 99, exactly 2 overcards to 77, and so forth. How would one represent an equation to find the chance of 0, 1, 2, or 3 overcards to a given pocket pair, from 3 overcards to AA being 0% and 1 overcards to 22 being 100% and everything inbetween? Anyone have any ideas?

LetYouDown
09-07-2005, 12:44 PM
For 9-9:

No overcards:
C(30,3)/C(50,3) = 4060/C(50,3)

One overcard:
20 * C(30,2) = 8700/C(50,3)

Two overcards:
C(20,2) * C(30,1) = 5700/C(50,3)

Three overcards:
C(20,3)/C(50,3) = 1140/C(50,3)

Formula:

No overcards:
C(#undercards,3)/C(50,3)

One overcard:
#overcards * C(#undercards,2)

Two overcards:
C(#overcards,2) * C(#undercards,1)

Three overcards:
C(#overcards,3)/C(50,3)

Paxosmotic
09-07-2005, 01:27 PM
And you mean to have C(50,3) on the edge of the 1 and 2 overcard formulas as well, correct? Adding those gave me 19600 possible combinations, which is how I've been error checking, so it looks like it's good to go.

LetYouDown
09-07-2005, 01:32 PM
They should all have the C(50,3) if you're looking for probability. If you're just looking for # of combinations, eliminate it. My apologies for not being consistent.