Starting hands, hold em

[aron]

Okay, before you come with your sarcastic comments, I know this is totally pointless.

The other day I was trying to calculate how many different starting hands there were.
I'm thinking about all the different combinations A/images/graemlins/heart.gif A/images/graemlins/club.gif and A /images/graemlins/heart.gif A /images/graemlins/diamond.gif for instance would be two different hands.

Before I thought there would be 2704 different ones (52*52), but it seems I was mistaken.

I haven't studied math since highschool, and I did it on the back of a receipt, so I'm not sure I got it right.

Anyway my results were 1326 different combinations, which seems right since there are 6 different combinations in suits with pocket pairs (/images/graemlins/spade.gif /images/graemlins/diamond.gif, /images/graemlins/spade.gif /images/graemlins/heart.gif, /images/graemlins/spade.gif /images/graemlins/club.gif, /images/graemlins/diamond.gif /images/graemlins/heart.gif, /images/graemlins/diamond.gif /images/graemlins/club.gif and /images/graemlins/heart.gif /images/graemlins/club.gif) And we know that we are dealt AA 1/220. 6*220 gives 1320 which is close to the number I came up with.

Is this right?

-aron

[gaming_mouse]

1326 is correct. This is "n choose r" where n = 52 and r =2.

If you'd like to learn more about this, do a search on google for "n choose r" (with the quotes) or "combinatorics". You could also try searching the archives here for those things.

gm

[TomCollins]

If you think about it, there are 52 different Card A you can get, and 52 different Card B you could get. But you actually count twice as many this way, because Ad As is the same hand as As Ad. So we divide by 2, and come up with 52*51/2.