In Hold'em, the odds of being dealt 2 aces is 1-in-221 (4/52 * 3/51).

What are the odds of being dealt 2 aces in Omaha? And, just as importantly, what is the formula for calculating those odds?

A local poker room runs a cracked aces promo, and Omaha games are eligible. Problem is, Omaha games rarely run. I'm trying to convice a couple of die-hard holdem players that it will be worth their while to try Omaha for a few hours during the promotion hours. I figure I can bolster my argument if I come armed with statistics proving how they'll get the promo bucks more often if they play Omaha.

Thanks for the help.

Possible omaha starting hands: 52C4 = 52*51*50*49 = 6497400
Possible omaha starting hands with at least two aces: 6*4C2*50C2 = 6*4*3*50*51 = 183600

(the 6 in the last formula is for the six different permutations of where the aces may show up in your hand, eg the first two cards, the second and last cart, and so on)

P(at least two aces) = 183600 / 6497400 ~ 0.0281 (2.81%)

Someone, please check my math. Did this real quickly, may be wrong.

/Klyka

That's not 52C4, it is 52P4 (permutations with order, not combinations without order). Divide by 4! for 52C4 You can use either one as long as you are consistent.

You can't just multiply by 50C2 as this counts the hands with 3 aces C(3,2) = 3 times, and it counts the hands with 4 aces C(4,2) = 6 times. You must use separate terms to count these, or you can use what you computed as the first step to the inclusion-exclusion principle where the over counting is desirable.

Using what the OP already knows, we can get P(at least 2) like this:

P(at least 2) =

P(exactly 2) + P(exactly 3) + P(exactly 4) =

4/52 * 3/51 * 48/50 * 47/49 *6 +
4/52 * 3/51 * 2/50 * 48/49 *4 +
4/52 * 3/51 * 2/50 * 1/49

or more compactly

4*3*(48*47*6 + 2*48*4 + 2*1) / (52*51*50*49)

=~ 2.57%

Note that we multiply by 6 for P(exactly 2) since there are 4*3/2 = 6 pairs of cards that could be aces. For P(exactly 3) we multiply by 4 since there are 4 ways to choose the 3 cards that are aces.

Using combinations this is:

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

=~ 2.57%

Thanks!

Also, <- Klyka

For this application, C(4,2)*C(48,2) / C(52,4) is the more interesting number. 2.5%

I was just dealt aces five time in a row in omaha.

What is the chances of that and will i ever see that again in my lifetime.

I have the handhistory saved to my computer.

Better chance at having Jesus Christ walk threw your front door.

How much can you win if you get aces cracked? That happens in Omaha all the time. If it is a lot, relative to the stakes, you need to get that game going.

If you just sat down and played five hands, and got two or more Aces in each one, that's 1 in 89 million. But if 10,000 players play 10,000 hands each, you expect one of them to get five Ace pairs in a row. So it's one of those things that happens, but rarely to you.

Bump

That seems like an insightful enough post to justify a 3 year necrobump.

We get some good ninja bumps in this forum. However, this one might worth the bump, because people actually may want to know how frequently AAxx in Omaha shows up. Considering how much poker math I have bothered to do on this site, I actually never considered this question before, but glad I read it now.