Number of combinations of starting hands

foldem · #1 03-30-2005, 05:58 PM

Hello, is the correct way to calculate the number of combinations as follows?

C(AKo) = 4*4 = 16
C(QQ) = 4*3 = 12
C(AKs) = 4(1*1) = 4

Thanks.

Drunken Monkey · #2 03-30-2005, 07:02 PM

C(AKo)=12
4*3

Lost Wages · #3 03-30-2005, 07:32 PM

C(QQ) = (4*3)/2 = 6

Q[img]/images/graemlins/heart.gif[/img]Q[img]/images/graemlins/spade.gif[/img] = Q[img]/images/graemlins/spade.gif[/img]Q[img]/images/graemlins/heart.gif[/img] etc.

Lost Wages

PokerProdigy · #4 03-30-2005, 07:43 PM

Sklansky's "Getting the Best of It" has much more on this subject. Also look for counting techniques (factorials, combinations, and permutations) online or in a math book. Combinations will be of the most help.

Stipe_fan · #5 03-31-2005, 10:01 AM

you divide the total by two if you don't care if you have
Q [img]/images/graemlins/spade.gif[/img]Q [img]/images/graemlins/heart.gif[/img] OR Q [img]/images/graemlins/heart.gif[/img]Q [img]/images/graemlins/spade.gif[/img]

thus the AKo combination is actually:

4*3*2*1/2= 12

you can look, like mentioned, permutations and combinations

stipe