Re: Odds of 3 sets being flopped and quads being hit on the turn?
Well lets see.
To get the Number of at least 3 players having PP > PP > PP in 6 max is:
C(6,3)*(13(C(4,2))*12(C(4,2))*11(C(4,2)))/C(52,12) =
20 * ((13*6) * (12*6) * (11*6))/C(52,12) = .000035919862891074924 or 1:~27838
To get set > set > set on the Flop:
(6*4*2)/(46*45*44) = 5.270092226613965E-4 or 1:~1896
So to catch a quad from the set > set > set:
3/43 or 1:~13.3
So of this ever happening is:
(1:~27838) * (1:~1896) * (1:~13.3) = 1:703,744,640
And I know I've done something wrong.
-Gryph
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