PDA

View Full Version : Probabilities the blinds hold/hit a pair on the flop?


Bukem_
11-14-2004, 07:59 PM
Assuming both blinds call any time they have two cards, what percantage of time will at least one of them hold a pocket pair, or hit a pair or better on the flop?

Stork
11-14-2004, 08:47 PM
If both blinds hold 2 non pair different cards, one of them will pair roughly 2/3 of the time I believe.

BruceZ
11-15-2004, 05:09 AM
Post deleted by Mat Sklansky

BruceZ
11-15-2004, 08:25 AM
Please disregard my previous post. When the blinds have no preflop pair, they can still share 0, 1 or 2 ranks in common, and each of these has a different conditional probability of flopping a pair. We have to consider the probability of each of these cases separately, in addition to the case of a preflop pair. As a check, we can verify that these probabilities sum to 1.

preflop pair: 1/17 + (16/17)*72/C(50,2) = 0.114141657

no preflop pair cases:

0 rank in common: (16/17)*44*40/2 / C(50,2) = 0.676110444

1 rank in common: (16/17)*6*44 / C(50,2) = 0.202833133

2 ranks in common: (16/17)*3*3 / C(50,2) = 0.006914766
-------------------------------------------------------------------------------
total: 1.000000000


Now each of these get multiplied by the appropriate conditional probability of flopping a pair. We are only counting cases where a hole card pairs, not when the flop pairs.

preflop pair: 1/17 + (16/17)*72/C(50,2) = 0.114141657

no preflop pair cases:

0 ranks in common: (16/17)*44*40/2 / C(50,2) * [1 - C(36,3) / C(48,3) ] = 0.397003797

1 rank in common: (16/17)*6*44 / C(50,2) * [1 - C(40,3)/C(48,3) ] = 0.086968693

2 ranks in common: (16/17)*3*3 / C(50,2) * [1 - C(44,3)/ C(48,3) ] = 0.001619949

-----------------------------------------------------------------------------------------------------------------
59.97%.

Stork
11-15-2004, 11:33 PM
The party don't start til' BruceZ walks in /images/graemlins/cool.gif.

phixxx
11-16-2004, 10:53 PM
Jesus :/ I could barely get through highschool math. Hahaha,