Re: odds of set over set
Hey aaron. Will you explain where I went wrong? Thanks.
The probability of just exactly 2 players geting distinct PPs is:
13(4,2)/(52,2)*12(4,2)/(48,2) *100 =0.346%
At a 9 player table you can estimate the probability of 2 players out of 9 getting a distinct PP by [C(9,2)*0.00346]*100= 12.456%
The probability of a player flopping a set is 11.76%. The probability of 2 players who start with a PP flopping a set is (0.1176^2)*100= 1.38%
The probability on any give hand that 2 players will flop a set is ~ (0.12456*0.0138)*100= 0.17%
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