Card combinations question

[angry hobo]

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For future reference it's a very simple combinatorics problem. For AK (suited and off) there are 4 cards for the aces, 4 cards for the kings, so there are 16 combinations.

The number of combos of a pair is actually slightly more complicated because of removal. There are 4 options for the first card and 3 for the second, so that implies 12 combos, but AcAd is the same as AdAc so half of them are duplicates and really there are only 6 combos.

And that's as complex as holdem gets [img]/images/graemlins/wink.gif[/img]

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you got the right answer but not in the right way. when you use 4 times 3 it is when you have something like this: an A in your hand and you think he has AK, then there are 3 aces and 4 kings, thus he can have 12 combos of AK. When determining the combos for AA, it is different. the combos for any pair is 6. When you are figuring the combos of their hand and have seen 1 or more of the card they would need to have, then the combos all change. Read Poker, Gaming, and Life by David Slansky. I didnt understand this concept until i read it. I have used some of the concepts in that book alot. I am sure it would pay for itself in one session for any winning player