The other day I was playing some limit Omaha/8 in one of my local card rooms and then after the guys took down the pot, he made a comment that he has been playing Omaha for 25 years and referring to something that he said he had never happen for him before.

So, he showed us his had which was AATT (suit does not matter in this story). What happened was that he flopped two sets. He said that was the first time in 25 years of playing the game. At first, I thought, “No way!”. I’m pretty sure that has to have happened before. But, the more I thought about it, I don’t think it’s ever happened to me before. However, I haven’t played Omaha for as long as that guy.

Which brings me to my question. If you are dealt two pair in Omaha, what are the odd of flopping two sets? Can someone tell me and show me the math? I can find odds of being dealt two pair, but it’s the other part I’m interested in, and that is the flopping two sets part. I’m guessing it must be pretty low. Thanks in advance, guys.

Let's use AAKK as a prototype. From our perspective, there are 48 missing cards including two aces and two kings. In order to flop two sets, the flop needs exactly one ace, exactly one king, and exactly one other card.

The number of ways we can flop three cards that are anything is
C(48,3)=48*47*46/1/2/3=17,296

The number of ways we can flop three cards that don't contain an ace is
C(46,3)=46*45*44/1/2/3=15,180.
Thus the probability of flopping three cards containing an ace when we hold
AAKK is
1-15,180/17,296=0.12234

When we hold AAKK,
the probability of a three card flop having an ace is
1-C(46,3)/C(48,3) =
1-15,180/17,296=0.12234.

When we hold AAKK,
the probability of a three card flop having a king is also
1-C(46,3)/C(48,3) =
1-15,180/17,296=0.12234.

And when we hold AAKK, the probability of the flop containing an ace or a king is
1-C(44,3)/C(48,3)=
1-13,244/17,296 = 0.23427.

Neither
AAK,
AAK,
AKK, nor
AKK works because then we'd have a set and quads.
The probability of one of these flops is 4/17,296=0.00023

I'll try to use what mathematicians call the inclusion/exclusion principle.
0.12234+0.12234-0.23427-0.00023=0.01018

In other words, the probability is about 1% of flopping two sets when your hand contains two pairs .

Since I’m not a mathematician I won’t try prove it, and I don't think I've explained it very well. When you have probability questions, there are some fantastic math denizens who inhabit the probability forum.

Buzz

Oops. I didn't finish. You asked for odds, not probability.

If the probability is about 1%, then the odds are 99 to 1 against or roughly about 100 to 1 against.

But this isn't going to happen once every 100 hands. It's just going to happen once every hundred hands when you get dealt two pairs.

You get dealt two pairs
6*6*13*12/2=2808 times in 52*51*50*49/1/2/3/4=270,725 deals.
P=2808/270,725=0.01037
(That's about one time per hundred too).

Finally, (1/100)(1/100)=1/10,000

So combined, you'll be dealt two pairs and flop two sets about once every ten thousand hands... something like that... don't hold your breath.

Buzz

Oh, wow. Thanks Buzz. Your reply was quick. I know you said you're not a mathematician, but I went through all your explanations and it seems to look like it makes sense. I remember learning a little about Combinations and Permutations in high school when I was younger. Which was a loooooong time ago. Forgot most of it now. Anyway, I knew this was the right place to ask this. Lots of smart people on here.

--patho

Funny coincidence:

Ten days ago or so I was playing Omaha Hi Lo in my local poker room. I was dealt the same thing AATT. The flop was AAT. Flopped quads and a set. The turn and river didn't change anything.

It also a bad beat in this game, quads 10's beaten. A player at the table said jokingly:

"If a Ten was on the turn or the river, you were going to beat yourself". She said it happened to her (beating herself) only once in her lifetime.

You're welcome.

Some who post here are a lot smarter than I am.

Maybe I can explain the inclusion/exclusion principle, as I applied it here, better.

Using the inclusion/exclusion principle is usually a way to avoid overcounting. (But that’s not why I’m using it here).

The basic logic is the probability a flop contains an A or K is = to the probability a flop contains an ace plus the probability a flop contains a king minus the probability a flop contains them both. (You may need to think about that for a while before it makes perfect sense).
Think of the probability a flop contains both as the “overlap.” And that overlap is what really interests us here.

I used:
The probability a flop contains A or K =
1-the probability a flop doesn’t contain an ace +
1-the probability a flop doesn’t contain a king -
the probability a flop contains both.

Thus P a flop contains both is

P(AorK)= PA+PK -P(A+K)

Thus P(A+K)=PA+PK-P(AorK)

But that doesn’t exclude AAK or AKK (either of which would give Hero flopped quads). So we have to subtract that too.

Finally P(1A+1K)=PA+PK-P(AorK)-P(AAK or AKK)

I hope that makes more sense. Inclusion/exclusion is really a nifty method.

Buzz

it is possible that it was his first time seeing at, as 2 pair is very rarely played in omaha 8, it is often just mucked preflop.

Happened to me last week in a 1-2 PLO half and half PLO8, it was during the high only session and I was in the BB and had 2299 and flopped top and bottom. I have been playing Omaha and all its variants for fifteen years and I believe its happened a handful of times to me. The guy probably just runs bad lol.

If the guy plays well (or just is fairly tight), he probably folds most of the double paired hands he is dealt. I don't play many of them myself.