Combinations
 

Sorry if this seems like such a basic question, but how would you find the possible number of combinations a holding can be dealt? For example, if you believe a player has JJ, there are 6 ways that can be dealt? So in short, whats the equation for finding this?

Re: Combinations
 

This is basic probability, pick up a high school probability maths book will explain everything.

But in essence, for JJ:

#combinations = 4 choose 2 (which is labelled 4C2 in math-speak - with the 4 a superscript and the 2 a subscript).

In fact, p choose r = #no of ways to choose R objects from a set of P elements = pCr

= p! / (p-r)!r!

where p! = p.(p-1).(p-2)....3.2.1

e.g. 4! = 4.3.2.1 = 12

So for the JJ scenario, p = 4, r = 2

4C2 = 4! / (4-2)!.2! = 12/2x2 = 12/4 = 6

Re: Combinations
 

So basically, out of four jacks, you can choose any of the four for the first card. For the second whole card, you are left with three, so that is 4x3, i got that, and you just divide by 2 always? and how would you find AK?

Re: Combinations
 

You divide by two because you would be double counting otherwise.

J [img]/images/graemlins/club.gif[/img] J [img]/images/graemlins/heart.gif[/img] is the same as J [img]/images/graemlins/heart.gif[/img] J [img]/images/graemlins/club.gif[/img]. But 4x3 counts them both individually.

When order matters, it's called a permutation. When it doesn't, it's a combination.

With AK, it's (4x4)/2 = 8.

Permutations and combinations explained.

Re: Combinations
 

Wrong.

AK is (8 X 4) / 2 = 16 combinations. Thanks for playing.