Re: Odds of at least 1 player of 6 having pocket pair
here is a nice little formula i learned from phil gordons site. this is assuming you hold a pocket pair.
C = number players behind you
N = number of pairs higher than yours
(C*N)/2 = probability(%) a higher pair is behind you
so if you have 1 opponent remaning and you are dealt kk:
1/2 = 0.5%, close enough to 221-1
if you have 22 with 1 opponent behind:
12/2 = 6%, or odds to be dealt any pocket pair
similar to the ak situation; i think its pretty sweet
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