PDA

View Full Version : Omaha math - chance AA is out

binions
08-09-2005, 06:44 PM
Posted this on the O8 board too.

In a 10 handed game, if you don't hold an ace, then there are 1128 possible 2 card holdem combinations left (48*47/2), and 6 AA combinations, so it's 187:1 against a single AA combination, or 187:54 that the group wont have an AA among them (since 9 players * 6 combos = 54). In other words, ~3.5:1 someone has AA when you have KK.

If you have an lone A, then it's 1128:3, or 375:1 or 375:54 that an AA is out. So, ~7:1 that an AKK hand runs into AA.

Have I got the math right?

binions
08-09-2005, 09:32 PM
Did I totally screw it up?

AaronBrown
08-09-2005, 09:41 PM
Not quite.

Your first computation is approximately correct if you're the only one at the table playing Omaha, everyone else is playing hold'em. But since they each have four cards, there's a much better chance that one has AA.

For a single hand, there are C(48,4) = 194,580 possible hands. There are 6*C(44,2) = 5,676 ways to hold AA and two non-Aces (so this does not count AAA7 but does count AA77). 5,676/194,580 = 2.92% or about 1 in 34.

The exact number with 9 players is a little more complicated, but multiplying by 9 is approximately correct.

With a lone Ace it's approximately correct that the chance is cut in half, since there are now only 3 pairs of Aces available, not 6. However, there are more available combinatinos of cards to fill out the hand, 990 instead of 946. So the probability really goes down by 48% instead of 50%.

binions
08-09-2005, 10:48 PM
[ QUOTE ]
Not quite.

Your first computation is approximately correct if you're the only one at the table playing Omaha, everyone else is playing hold'em. But since they each have four cards, there's a much better chance that one has AA.

For a single hand, there are C(48,4) = 194,580 possible hands. There are 6*C(44,2) = 5,676 ways to hold AA and two non-Aces (so this does not count AAA7 but does count AA77). 5,676/194,580 = 2.92% or about 1 in 34.

The exact number with 9 players is a little more complicated, but multiplying by 9 is approximately correct.

With a lone Ace it's approximately correct that the chance is cut in half, since there are now only 3 pairs of Aces available, not 6. However, there are more available combinatinos of cards to fill out the hand, 990 instead of 946. So the probability really goes down by 48% instead of 50%.

[/ QUOTE ]

Thanks. Interesting that you came up with 33:1, and I came up with 31:1.

I am not very familiar with the formulas, but does C(48,4) = 194,580 mean before any hands are dealt, that's how many possible Omaha hands there are. In other words, is 194,580 in Omaha equivalent to 1326 in holdem?

I wouldn't think so, since 52*51*50*49/4*3*2*1 = 270,725

LetYouDown
08-10-2005, 08:44 AM
C(48,4) is the number of hands someone can have after you have a hand that's defined. There are 48 cards left in the deck. C(52,4) is the number of hands you can possibly have if you were to deal a random hand from a full deck.