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.

What seemed to me to be a more interesting but related question would be:

-What are the odds that *someone* at a 6-handed table will be dealt 2 aces? (I've approximated this as 1 in 7)

And the related questions:

-If I am dealt a hand with one ace in it, what are the odds *someone else* has been dealt 2 aces? (I've approximated this as 1 in 13.5)

-If I am dealt a hand with no aces in it, what are the odds *someone else* has been dealt 2 aces? (I've approximated this as 1 in 7.3)

These are merely approximations, but to get a more precise answer seemed complicated, and I wasn't sure it was a line worth pursuing until asking whether there is a simpler approach.

For example, I approximated 1-in-7 chance that someone would get dealt 2 aces as:

1 - (chance that no one gets 2 aces) =
1 - (0.975)^6 = 1 - 0.859 = approximately 14% = rather approximately 1 in 7

(actually 1 in 7.143)

This seems *fair* as an approximation, but I wonder if it's worth figuring the true odds and whether it would yield a significantly different answer.

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I'm not sure of a better way to describe why I think my answer is approximate, other than by way of a simplified example. Imagine a different problem. A "deck" of cards has only 4 cards in it: 2 aces and 2 twos. I deal out two cards each to two players. What are the chances that *someone* will have a pair of aces?

My crude method that I used above is analogous to this:

-The chances that player Albert gets 2 aces = 1/6
-The chances that *someone* gets 2 aces = 1 - (5/6)^2 = 0.30555

But the more correct answer would be:

there are six ways to deal out the cards:

AA 22
A2 A2
A2 2A
2A A2
2A 2A
22 AA

two of those ways result in *someone* having a pair of aces. Thus the true odds are 0.3333

This same answer would be arrived at by calculating chances of each of Albert's possible outcomes, and the odds for each outcome that the other player would get the aces, and adding them up...

1/6
+ 4/6 * 0
+ 1/6 * 1

= 0.3333

simple in *this* problem, but it seems messy in the 6-player Omaha scenario!


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The point? Obviously which cards are out in one player's hands will affect probabilities of the players that follow, and as the deck gets exhausted, this can have a measurable effect. My approximations may be marginally low... but how much so? My guess is that the difference is negligible, but unless someone else has already done the calculation, I'm not sure.

P.S. I realize that this is mainly an exercise in probability, and that it's not going to have an appreciable influence on strategy, other than to give a sense of how often opponents will have aces... and aces aren't monsters, and, and, and.... I'm still curious to know.

These are for hold'em. I will add Omaha probabilities in another post.

6*6/C(52,2) - C(6,2)*1/C(52,4)

≈ 1 in 36.9

That's 6 times the probability that a single player has it, minus the probability that 2 players have it which was double counted by the first term. This is exact.

Note that C(52,2) = 52*51/2 = 1326, the total number of hands, and C(50,2) = 50*49/2 = 1225, the number of hands out of 50 cards.


That's an easy one since only 1 other player can have aces, so they are mutually exclusive, and the probability is simply 5 times the probability that a single player has it. That is

5*3/C(50,2) ≈ 1 in 81.7


Similar to first one only now there are only 50 cards and 5 opponents.

5*6/C(50,2) - C(5,2)*1/C(50,4)

≈ 1 in 40.9


The chance that a player doesn't get 2 aces isn't 0.975, it is

1 - 6/1326 ≈ 0.995475

Then if you take 1 - (.995475)^6 you get 1 in 37.3, close to what we got above. The reason this is an approximation is that the hands are not independent due to card removal, so raising this number to the 6th power is not exact. When one player fails to get 2 aces, it is less likely that the next player won't, so the number you should subtract from 1 should be a little smaller, and the probability a little greater which it is. This approximation is almost always good enough. The exact method was easy here, and can be extended to more complicated problems which you can compute exactly or as accurately as you need. The method is called inclusion-exclusion. See this post.

Edited: I think that what I posted below is still relevant, but after posting I realized one big source of confusion: Omaha vs. Holdem! With that in mind consider the rest of the post as to why the odds shold be significantly lower than 6 times the odds of one player getting the hand in question.

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Wait, I'm sorry, but the above cannot be right. First of all, the chance that a specific individual is dealt AAxx is 2.5%, which is 1 in 40. So the chances that one of six players get it is going to be much more likely than 1 in 36.9. Also, it cannot be as simple as multiplying the odds of one player getting it by six, not even if you're discounting the odds of two players getting it. (For a simple example as to why it can't work that way, consider a different, simple problem. If I deal you one card from a 52-card shuffled deck, you have a 25% of getting a heart. But I can't say that if I deal one card each to four players that there is a 100% chance of someone getting a heart. In fact the chances are less than 100% even if you deal to 39 players!)

I'm pretty confident that my estimate of 1 in 7 is close for the odds of one in six players getting AAxx in Omaha, but I just wonder how close, and whether there's a simplified approach to getting a precise figure.

P.S. and now, with refection, it seems that your suggestion of taking the odds of one player getting AAxx, multiplying by six, and subtracting the odds of two getting it may well be the solution I was looking for after all... hmmm

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There's a deck of four cards, two red and two black. deal a card to each of two players. what are the chances that *someone* gets a red?

1/2 * 2 - (1/2 * 1/3) = 5/6

That is the same thing as asking what are the chances that the undealt cards are not both red, i.e. 1 - (1/2 * 1/3)... yup, same....

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So, if this still holds as it appears to, then in original problem, odds of one player getting AAxx = 0.025

odds of two players getting it = 0.025 * (0.00534) * 5 = 0.00067

So odds of someone getting it 0.025 * 6 - 0.00067 = 0.14933 = approximately 1 in 6.7. Well, that's significantly more likely than my earlier estimate, but then I'm burnt out at the moment and will need to review when I wake up in a few hours.

I did it for hold'em, not noticing that this ancient thread was about Omaha. The Omaha probabilities are done with the same method. When only 1 player can get AA, you simply multiply the probability for 1 player by the number of players since the players are mutually exclusive and the probabilities add. That's fundamental. If more than one player can get it, then that will double count the cases where more than one can get it, as in your heart example, so we use inclusion-exclusion to account for this. That's the standard method of doing these problems as I have demonstrated on this forum constantly for for well over a decade. I've done some Omaha problems before too, but here are yours:


6*[C(4,2)*C(48,2) + C(4,3)*48 + C(4,4)] / C(52,4) -
C(6,2)*C(4,2)*C(48,2)*C(2,2)*C(46,2)/ C(52,4)/C(48,4)

≈ 1 in 6.6

Note we must now take into account getting 2,3,or 4 aces.


That's an easy one since only 1 other player can have aces, so they are mutually exclusive, and the probability is simply 5 times the probability that a single player has it. That is

5*[C(3,2)*C(45,2) + C(3,3)*45] / C(48,4)

≈ 1 in 12.9


Similar to first one only now there are only 48 cards and 5 opponents.

5*[C(4,2)*C(44,2) + C(4,3)*44 + C(4,4)] / C(48,4) -
C(5,2)*C(4,2)*C(44,2)*C(2,2)*C(42,2)/ C(48,4)/C(44,4)

≈ 1 in 6.7


You get 1 in 6.9 if you use the exact value above for the probability that 1 player doesn't get it of 0.974287561178317. See discussion above for why this is approximate, and a link to more information on the exact method.

I had to work on this a bit on my own, with a different simple example for a while, to see if I could make sense of it.

[A deck consists of AAA222, three players are dealt one card each, what are the chances that someone gets an A. The simplicity allowed an easy solution that I could use as a "check" against the different approach you were suggesting.]

Anyway, I got to where you were leading, didn't quite see why it worked at first, but then I finally followed the link to read up on inclusion-exclusion and examine it against my own understanding of the problem, and now I see that that is exactly the solution to my dilemma. Thanks.

In my initial estimate of 1 in 7, I was a bit surprised to see that I was further away from the correct 1 in 6.6 than I supposed.

Meanwhile, in my estimation I did discount the likelihood of someone holding AAAx or AAAA, based on the fact that AAAx is 1 in 1410 for any particular player, and AAAA is 1 in 270,725. So I think the inaccuracy of my method mainly derives from the fact that I didn't yet know how to deal with the likelihood that several players could hold one-or-two aces at the same time.

Okey-doke, time to play with more numbers... or just relax and play razz... heh.

3 terms in inclusion-exclusion since up to 3 players can get them. Then

3*P(1 player) - C(3,2)*P(2 players) + C(3,3)*P(3 players)

= 3*1/2 - C(3,2)*1/2 *2/5 + C(3,3)*1/2*2/5*1/4

= 0.95


Note that this is easy to do as

1 - 1/2*2/5*1/4 = 0.95

But notice that we don't do 1 - (1/2)^6 because these aren't independent.


That's not the main inaccuracy. The main inaccuracy is that the hands aren't independent, so multiplying the probabilities by raising a single probability to a power isn't exact. With each player that does not get it, the probability that the next player doesn't get it changes.

Instead of

1- P(1')^6

where P(1') means player 1 didn't get it, you want

1 - P(1')*P(2'|1')*P(3'|1',2')*...

where P(2'|1') means the probability that 2 didn't get it given that 1 didn't get it, etc.

Bump (renecrobump).

Why though

I think a point is missing here:
2 aces in an Omaha hand is good.
3 aces is borderline bad (chance to improve to a set is cut in half and you're also at most single suited. Your chances of making a straight is also taking a big dive. Even if you make your set there's likely no one to pay you off because no one else will have paired the ace so if they have a pair or two-pair hand they have to consider that you have at least an ace when you start betting. Any two pair hand you have will be better than theirs. )
4 aces is downright terrible. Chances of winning with just a pair (and that's practically guaranteed the only thing you'll have) are low in Omaha.

So if you want the odds of something "good" happening to you (which is what poker odds calculations are all about) the correct question should be:
"What is my likelyhoood of getting dealt exactly two aces"

Get a calculator handy and see post #3