Combinations

[BloodyRomance]

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?

[Rotating Rabbit]

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

[BloodyRomance]

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?

[Huskiez]

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.

[Big_Jim]

Wrong.

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