Classic Cooler
What's the probably of aa-kk in a 6 mx table? This seems to be happening a lot latly.
|
Re: Classic Cooler
google: "wiki probabilitiy poker texas hold'em" and you'll get there
|
Re: Classic Cooler
Barring a miracle straight or flush
LOL at the use of miracle |
Re: Classic Cooler
[ QUOTE ]
What's the probably of aa-kk in a 6 mx table? This seems to be happening a lot latly. [/ QUOTE ] As usual, treating the hands as independent gives a very accurate approximation of about 0.06647% or 1 in 1504, while the exact answer is about 0.06625% or 1 in 1509. The approximation is 1 - [1 - 12*6/C(52,2)/C(50,2)]^C(6,2) =~ 0.06647% or 1 in 1504 where the expression in [] is 1 minus the probability that 2 particular players have AA and KK, which is the probability that these 2 players do not have it. This is raised to the power of C(6,2), which is the number of pairs of players. This approximates the probability that no 2 players have it, and this is all subtracted from 1 to get the probability that some 2 players have it. The exact solution from the inclusion-exclusion principle is C(6,2)*12*6/C(52,2)/C(50,2) - C(6,3)*12*6*1*3/C(52,2)/C(50,2)/C(48,2) + C(6,4)*12*6*1*3*1/C(52,2)/C(50,2)/C(48,2)/C(46,2) =~ 0.06625% or 1 in 1509. Note that this is the probability that at least 1 AA and at least 1 KK are dealt at a 6 player table. If you want the probability of being up against AA when you have KK, or other similar problems, see this post. |
All times are GMT -4. The time now is 10:01 PM. |
Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2024, vBulletin Solutions Inc.