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09-25-2005, 03:05 AM
Bruce I am hoping you can help me with this simple calculation.

If I was 100% sure that an opponents range was either one of the following 4 hands how can I figure out what % of the time it will be AK?

AA,KK,QQ, or AK (suited or unsuited).

If you could explain the math it would be greatly appreciated. I took your advice from my last post about using the seacrh function but couldn't find the answer I was looking for.

TIA... /images/graemlins/smile.gif

BruceZ
09-25-2005, 03:18 AM
[ QUOTE ]
Bruce I am hoping you can help me with this simple calculation.

If I was 100% sure that an opponents range was either one of the following 4 hands how can I figure out what % of the time it will be AK?

AA,KK,QQ, or AK (suited or unsuited).

If you could explain the math it would be greatly appreciated. I took your advice from my last post about using the seacrh function but couldn't find the answer I was looking for.

TIA... /images/graemlins/smile.gif

[/ QUOTE ]

There are 6 ways each of making AA,KK,QQ, so 18 ways all together, and 16 ways of making AK, so if each hand is equally likely to be played, and you don't hold an A, K or Q, then the probability of AK is 16/(16+18) =~ 47.1%.

09-25-2005, 03:24 AM