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gergery
03-04-2005, 10:24 PM
If have A2xx in Omaha, then

what are the odds at least one other player has A2xx?

more difficult to figure, what are the odds at least two other players have A2xx?

I'm more interested in how to calculate this, than in what the exact answer is, tho i'm interested in both.

--Greg

MickeyHoldem
03-05-2005, 01:08 PM
I'll give it a shot....

You have A2xx, so the possible number of deals for the 9 opponents
D = 48c36 * (36! / ((4!^9) * 9!))

opp. deals that contain 3 A2xx
Q3 = 9*42c2 * 4*40c2 * 38c2 / 3! * 36c24 * (24! / (4!^6 * 6!))

opp. deals that contain 2 A2xx
Q2 = 9*44c2 * 4*42c2 / 2! * 40c28 * (28! / (4!^7 * 7!)) - 3(Q3)

opp. deals that contain 1 A2xx
Q1 = 9*46c2 * 44c32 * (32! / (4!^8 * 8!) - 2(Q2) - 3(Q3)

So odds atleast 1 opp. with A2xx
= (Q1+Q2+Q3) / D = 0.391479

odds atleast 2 opp. have A2xx
= (Q2+Q3) / D = 0.038780

Cobra
03-05-2005, 08:27 PM
I will try to give you an answer using inclusion/exclusion which is often used in determining what your opponents might hold.

Given that you have A2xx with xx not being an A or 2. There will be up to three terms for our inclusion/exclusion because up to four people(including you) could have A2xx.

First term: There is 3 Aces, 3 two's, and 42 other cards.
Your opponent could have an x will always indicate a non ace or two.

A2xx = (3c1)*(3c1)*(42c2)=7749 combo's
AA2x = (3c2)*(3c1)*42=378 combo's
A22x = (3c1)*(3c2)*42=378 combo's
AAA2 = (3c3)*(3c1)=3 combo's
A222 = (3c1)*(3c3)=3 combo's

This gives a total of 8511 combinations
We divide this by the total possible combo's = (48c4)=194580

So term one ends up being 4.37%

Now we have to determine term two or two people getting an A2xx. We know that if the first person got an AAA2 or A222 we can't have a second person so we do not include those hands.

If the first peron had an A2xx we now have 2 aces, 2 two's and 40 other cards for the xx's. So the second person could have an A2xx, AA2x, or and 22Ax. We add these combo's together and multiply by what the first person had.

(2c1)*(2c1)*(40c2)+ 2*(2c2)*(2c1)*40=3280 hands

If the first person had an AA2x or A22x there would now be two of one card remaining one of the other and 41 cards for the xx. The second person could now have A2xx or either AA2x or A22x depending on what the first person. We will now add these combo's together.

(2c1)*(1c1)*(41c2)+ (2c2)*(1c1)*41=1681 hands

To find term two we take 7749*3280=25416720 which is A2xx times the sum of hands if first guy had A2xx. Then we take 378*1681=635418 which is sum of hands if first guy had AA2x. Then we take 378*1681=635418 which is sum of hands if first guy had A22x. We now take the sum of the above numbers and divide by (48c4) and (44c4) this gives us the second term.

(25416720+635418+635418)/(48c4)/(44c4)=.101%

The third term can also be figured but it is so small it will not change the answer. The only way to get to the third term is if player one gets A2xx and player two gets A2xx.

Here's what we now
Term one = 4.37%
Term two = .101%

With nine opponents we take term one *9= 39.36 %
Take term two times (9c2)= 3.64%
Term three times (9c3) opponents is less than .00%

To find probability of one or more having a hand take term one and subtract term two = 35.73%

To find probability of two people it is term two = 3.64%

Cobra

MickeyHoldem
03-06-2005, 01:55 PM
[ QUOTE ]
A2xx = (3c1)*(3c1)*(42c2)=7749 combo's
AA2x = (3c2)*(3c1)*42=378 combo's
A22x = (3c1)*(3c2)*42=378 combo's
AAA2 = (3c3)*(3c1)=3 combo's
A222 = (3c1)*(3c3)=3 combo's

[/ QUOTE ]

AA22 = 9 combo's

My solutions over counts hands... similar to the way I screwed up the other post... I'm working on fixing it!

MickeyHoldem
03-06-2005, 02:15 PM
The first term of Q1 should be (9*42c2 + 2*9*42 + 2*3 + 3*3) instead of (9c46).... there are similar mistakes for Q2 and Q3 that are more difficult to correct...

Cobra's method is much better!

This is why I hate Omaha!!!

Cobra
03-06-2005, 04:32 PM
Your right Mickeyholdem I missed that hand.

Cobra

Cobra
03-06-2005, 05:47 PM
After MickeyHoldem pointed out a hand that I left out and went back and recomputed terms one and term two.

Term one = 4.38%
Term two = 3.64%

One or more players with A2xx = term one - term two=35.77%

Two or more players = 3.64%

Cobra

gergery
03-09-2005, 02:47 AM
thanks guys.

it is unbelievably complicated to get a good # here. jesus.

EasilyFound
03-17-2005, 09:44 AM
I do not question the answer, but I've played in many home Omaha games and seems like I've seen two people have A2xx more than the percentages dictate.