Ok, i feel dumb just asking this, but it's bugging me. Earlier today I had some time to kill and decided to do some poker math which involved counting possible hand combinations. For example, I tried to figure out how many possible flush draws (including SFDs, pair + fds, etc) there are on a board of 8h7h2s. I initially assumed that (since there's 11 hearts out there) there would be 11x10 combos, but quickly realized that that method is counting a lot of hands twice. Now, I could certainly work this out by hand- there's going to be 10 AhXh combos, 9 KhXh combos (X != Ah) , 8 QhXh combos (X !=Ah or Kh), etc. But adding up 10 + 9 + 8... gets arduous pretty quickly. I know there's a faster way to do this that I learned in high school, but totally forget it- someone refresh my memory?

OK, so you were on the right right. 11*10 is a good place to start, but as you said, it over-counts. This is because it will give you both
AhKh and KhAh
But you can see there are only 2 ways to express each hand, so it counts each combination exactly twice, so the answer is
11*10/2 = 55
which is the same thing you get if you add up the numbers.

I recommend looking into Combinatorics - wikipedia probably has some good entries on it.

If I recall from school (15 years ago)... such counting is called permutations and combinations... in this case (after googling)...

the formula for finding the number of combinations of k objects you can choose from a set of n objects is:


...................n!
n_C_k = --------------
..............k! (n - k)!

where n! = n x (n-1) x (n-2) x (n-3) ... x 2 x 1


.....................11!....................11 x 10
11_C_2 = ---------------- = ---------------- = 55
..................2! x 9!.....................2

Damn can't get format it so had to had dots... blah

Usually people just do it ike
n!/( k! * (n-k)! )
but if you want it nicely formatted, try this



(Go to http://hausheer.osola.com/latex2png, enter in a formula, and you get a link you can put in the forums. I've seen some more useful ones that let you put the formula in the url but I can't find one now.)

From Science, Math, and Philosophy Sticky:

FWIW, 1+2+3+4+5+...+n = n*(n+1) / 2, or using LaTeX: