Basic omaha probability question

[BlueBear]

Hi,

I would like to calculate the probability of being dealt an omaha containing at least 2 aces or more.

The number of possible starting omaha hands = 52*51*50*49 = 52,4 C = 6497400
The number of possible hands containing at least 2 aces = 6 * 50 * 49 = 14700 (6 acccounts for all possible combinations of a pair of aces).

Probability = 14700/6497400 = 0.0021 = 0.2%

This looks too low and I'm sure I have made a mistake here. Could anybody take a look at this?

BlueBear

[Gonzoman]

[ QUOTE ]
Hi,

I would like to calculate the probability of being dealt an omaha containing at least 2 aces or more.

The number of possible starting omaha hands = 52*51*50*49 = 52,4 C = 6497400
The number of possible hands containing at least 2 aces = 6 * 50 * 49 = 14700 (6 acccounts for all possible combinations of a pair of aces).

Probability = 14700/6497400 = 0.0021 = 0.2%

This looks too low and I'm sure I have made a mistake here. Could anybody take a look at this?

BlueBear

[/ QUOTE ]

52 choose 4 isn't 52 * 51 * 50 * 49. It's 52!/(48!*4!), or in your case 270,725.

[BruceZ]

The number of possible starting omaha hands = 52*51*50*49 = 52,4 C = 6497400
The number of possible hands containing at least 2 aces = 6 * 50 * 49 = 14700 (6 acccounts for all possible combinations of a pair of aces).

You have to divide 52*51*50*49 by 4!=4*3*2*1 to get the number of starting hands C(52,4). You have computed P(52,4) which counts every order or permutation of the 4 cards. Also, you have to add separate terms for 2,3 and 4 aces, otherwise you will count these more than once. The correct solution is:

[ C(4,2)*C(48,2) + C(4,3)*48 + C(4,4) ] / C(52,4) = 2.6%

where C(4,2) = 4*3/2! = ways to choose 2 aces
C(48,2) = 48*47/2! = ways to choose 2 non-aces
C(4,3) = 4*3*2/3! = ways to choose 3 aces
C(4,4) = 4*3*2*1/4! = 1 (all aces)
C(52,4) = 52*51*50*49/4! = total hands.

Note that no aces is C(48,4)/C(52,4) and 1 ace is 4*C(48,3)/C(52,4), and these added to the above number sum to 1 as they should.