PDA

View Full Version : Omaha Starting Hand Probability

Popinjay
12-18-2004, 03:57 AM
What is the probability that AAxx (exactly two aces) will be dealt out to you as a starting hand?

I got close to 19:1. Here's my math:
52C4 = 270725 total starting hands
4C2 = 6 ways to get AA
6 * 48 (assuming AA dealt out to you already and no other A is wanted) * 47 = 13536 AAxx hands
13536/270725 = ~.05 = ~19:1 odds

Did I do this right?

gaming_mouse
12-18-2004, 05:30 AM
looks good.

gm

Phat Mack
12-21-2004, 07:48 PM
52C4 = 270725 total starting hands

OK

4C2 = 6 ways to get AA

OK

6 * 48 (assuming AA dealt out to you already and no other A is wanted) * 47 = 13536 AAxx hands

I've got a problem with this one. When counting total hands and number of AA's, you are not counting equivalent hands in different orders. In other words, you aren't counting both AsAh and AhAs as different AA's.

In your final calculation, you are counting AsAh3d2c and AsAh2c3d as different hands. You have to perform the entire calculation one way or the other.

gaming_mouse
12-21-2004, 08:24 PM
[ QUOTE ]
6 * 48 (assuming AA dealt out to you already and no other A is wanted) * 47 = 13536 AAxx hands

I've got a problem with this one. When counting total hands and number of AA's, you are not counting equivalent hands in different orders. In other words, you aren't counting both AsAh and AhAs as different AA's.

In your final calculation, you are counting AsAh3d2c and AsAh2c3d as different hands. You have to perform the entire calculation one way or the other.

[/ QUOTE ]

Nice catch. I missed that.

Instead of 6 * 48 * 47 it should be 6 * (48 choose 2).

gm