Looking for a formula

[M50Paul]

How do you calculate teh following:
1. How many AK combinations are there when 5 cards are seen none of which are AK.

2. How many AA combinations are there with no cards seen

Thanks

[PaultheS]

for AK: 4 * 4 = 16 combinations. (ie. Ah can go with any of the four kings, Ac can go with any of the four kings, etc.)

for AA: 4C2 = 4! / 2!2! = 6 combinations.

[M50Paul]

thanks

[LetYouDown]

[ QUOTE ]

1. How many AK combinations are there when 5 cards are seen none of which are AK.

[/ QUOTE ]
This doesn't change the number of combinations. Are you looking for the odds of having the hand?

[M50Paul]

good point, agree. I was making some tables on odds and other measures and I kind of ran two things together. This particular question was getting at an objective way to detemine how to randomly raise 75% of the time. So in the case of AK -- 16 combinations I would raise any time there was no king of clubs in my hand.

[LetYouDown]

Not to nitpick, but that's not random...that's derived /images/graemlins/grin.gif

[M50Paul]

Well...... it's a non-random, random secletion of an integral part of my digit which gives the the non bias opportunity to rasie with impunity.

So........