View Full Version : Pocket Pair preflop, P(x) flopping 3k, FH or 4k?
DoctorWard
09-14-2005, 08:37 AM
I want to determine the probability of flopping a set, full house or quads when holding a pocket pair? When holding a pocket pair obviously can't make a flush, straight or straight flush on the flop. However, can someone please show me the full workings of flopping a set, full house or quads when holding a pocket pair?
LetYouDown
09-14-2005, 08:48 AM
Off the top of my head, including the board having trips:
1 - [C(48,3) + 36]/C(50,3) = 11.57%
BruceZ
09-14-2005, 09:06 AM
[ QUOTE ]
I want to determine the probability of flopping a set, full house or quads when holding a pocket pair? When holding a pocket pair obviously can't make a flush, straight or straight flush on the flop. However, can someone please show me the full workings of flopping a set, full house or quads when holding a pocket pair?
[/ QUOTE ]
A set, full house, or quads all require flopping at least 1 of your rank, assuming that we are excluding full houses flopped by 3-of-a-kind on the flop. The probability of flopping at least one of your rank is 1 minus the probability of flopping none of your rank:
1 - (48/50)*(47/49)*(46/48)=~ 11.76%.
or 1 - C(48,3)/C(50,3) =~ 11.76%.
If you want to include flopping a full house with 3-of-a-kind on the flop, there are 12*4 flops that do this, out of C(50,3) = 50*49*48/6 = 19600 flops. 12*4/19600 adds another 0.24%, bringing the total to about 12.0%.
BruceZ
09-14-2005, 09:08 AM
[ QUOTE ]
Off the top of my head, including the board having trips:
1 - [C(48,3) + <font color="red">36</font>]/C(50,3) = 11.57%
[/ QUOTE ]
36 -> -48. We want to count these, not exclude them, and there are 4 of each rank, times 12 ranks.
LetYouDown
09-14-2005, 09:20 AM
[ QUOTE ]
36 -> -48. We want to count these, not exclude them, and there are 4 of each rank, times 12 ranks.
[/ QUOTE ]
Yeah, whoops. Thanks for that, and now I'm off for coffee.
DoctorWard
09-14-2005, 05:38 PM
Bruce, thanks for that, very good explanation!
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