View Full Version : AK vs KJ on KJx board
MyMindIsGoing
03-13-2005, 05:31 AM
Lets say I have AK and flop is KJx, what is the chance someone flopped top two pair with KJ?
Cobra
03-13-2005, 01:17 PM
We will find out the probability of one or more of your opponents having top two pair. You would solve this by using inclusion-exclusion and determining two terms. What we know is that there is 47 unseen cards, 2k, 3j, and 42 other cards.
First term=3*2/(47 c 2) = .00555
Second term=6*2/(47 c 2)/(45 c 2)=.000011
Multiply the two terms by number of opponents than subtract the first term from the second.
9*.00555-(9c2)*.000011=.04955 or 1 in 20.2 times.
Cobra
BruceZ
03-13-2005, 01:40 PM
[ QUOTE ]
Multiply the two terms by number of opponents than subtract the first term from the second.
9*.00555-(9c2)*.000011=.04955 or 1 in 20.2 times.
[/ QUOTE ]
This assumes that all 9 opponents see the flop, (and that they all hold random hands).
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