Hello

Can someone help me figure out the odds of anyone getting quads in 5 card omaha, if 8 people are dealt in and all go to the river. I have no idea how to calculate this.

Thanks!

Miz

My answer is 5.38% and my work is below. I put it in a spoiler in case someone else wants to try this.

I did a back of the envelope calculation and got around 3.83%.

I then kicked off a simulation of 10,000,000 trials and got 3.82%.

Weird, even excluding the two somewhat tricky cases, my subtotal is higher than that, which suggests I even got the easy cases wrong.

I did mislabel my n's (e.g n4 should have been labeled n1), but not when typing into my calculator.

I'm not seeing any obvious mistakes at the moment. But it would be a crazy coincidence if both your calculation and code were wrong in ways that led to the same answer, so right now I have to believe your answer more than mine.

You did it for 8 players, right? Mind sharing your envelope calculation? It sounds like maybe you found a more efficient solution (whereas mine required nine calculations).

My back of the envelope calcs follow.

For someone in Omaha to make quads board has to be either:

Case 1: 2 pair & 1 singleton

or

Case 2: 1 pair & 3 singletons

Using combinatorics, I find that there are 123,552 case 1 boards and 1,098,240 case 2 boards.

-----

Now consider just one player (Hero).

For Hero to make quads with a case 1 board, he must have a pocket pair of one (or both) of the board pairs.

This is approximately twice the times he has one of the pairs. Which I calculate to be 14,190 * 2 = 28,380.

For Hero to make quads with a case 2 board, he must have a pocket pair of the board pair.

I calculate this to be 14,190.

Putting this altogether, I find the (approx) prob that Hero makes quads in 5-card Omaha to be:

= 0.004788595

-----

Now multiplying this by 8 since there are 8 players at the table (over-counting the rare times more than one player makes quads) gives:

= 0.03830876

Of course, is possible that my logic is faulty or that I totally botched these calcs or that I over-looked a case, etc. But the simulation is fairly straightforward and seems to confirm this answer.


eta: Now that I see this in writing, I guess I did over-look the case of a board with a "full house". But I am guessing that this doesn't boost the overall prob very much.

Case 3: trips on board
Case 4: boat on board

Edit - oh you edited right before I posted. For me, the two cases you forgot comprised the majority a good chunk of my percentage (particularly the trips-and-singletons case).

Wait a minute!

If board has trips, and Hero has the remaining card of that rank, that give Hero quads, right?

Egads. So, kindly disregard everything I have posted above (and I programmed my simulation to not tally the trips board cases).

Mea culpa.

Heh np. I think not long ago when someone asked an omaha question, you were the one correcting a goofy rules mistake of mine, so now it's even

There is no excuse for my screwing up. Of all the poker I have played in my lifetime, Omaha probably comprises around 0.01% (but that is still no excuse since I do know the rules for Omaha, at least most days I do).

Thanks so much for the responses. I could not even begin to understand any of this.

So 3.82% is incorrect? So what is the correct answer?

Miz

Now that a simulation doesn't disagree with me, I'm again confident in 5.38%

Man, it's rare for me to go a post without brainfarting.

The 80/1081 should be multiplied by 2 because we're taking P(quad a's) + P(quad b's) and then subtracting P(both).

I think the C(8,2) should be (8P2).

These corrections change the result to 5.72736 %

However, I tried the problem a different way and got a slightly different answer:

13*8*[5 + C(5,2)]*C(5,2)/C(52,4) -
C(13,2)*{
C(8,2)*[5*C(5,2)²C(4,2)2 + 2*5*2*C(5,2)³]/C(8,4) + 8[5²C(4,2)4/C(8,3) + 2*C(5,2)²C(4,2)3/C(7,3)]
} / C(52,8)

= 5.6451 %

I don't know which to believe at the moment, but I'm confident they're close to the truth.

After double-checking that I have included all the cases (this time), I ran a simulation of 100 Million deals.

Sim Prob = 5.727071%

Thanks, that allowed me to focus on finding the mistakes in my 2nd calculation.

The subtracted part should be:

C(13,2)*{
C(8,2)*[5*C(5,2)²C(4,2)*2 + 2²*5*C(5,2)²]/C(8,4) + 8[2*C(5,2)*C(4,2)*4/C(8,3) + 5²C(4,2)3/C(7,3)]
} / C(52,8)

And now both calculations get 5.72736 %. Unfortunately I didn't get to learn anything from the mistakes because they were of the silly variety. I had calculated some terms separately and mixed up which terms to paste them next to. I also had some extra C(5,2)'s as a result of losing track of what I was counting.

Now that it's right, I'll explain it.

Picture all the cards lined up like BBBBB|hhhhh|hhhhh|...|ddddddd
where B=board card, h=hole card and d=deck card.

There are 13 ranks that can be the quads.
There are 8 players who can have it.

They can have one of the cards in their hand and trips on the board, or a pair that makes quads. If one in their hand, there are 5 cards it can be and there are C(5,3) ways the board trips can appear. If quads from a pair, there are C(5,2) ways the pair can be in their hand and C(5,2) ways the pair can be on the board.

All of that is out of C(52,4) ways to distribute the four cards of that rank to places in the line.

So we have 13*8*[5+C(5,2)]*C(5,2) / C(52,4)

However, that double-counts the chance of two quads.

There are C(13,2) ways to pick the two quads. One player can have both quads, or C(8,2) selections of two people can have quads.

If two people have quads via two-pair on the board, that means 4 of the quad-making cards must be within the BBBBB, 2 must be within an hhhhh region, and the remaining to must be in another hhhhh region. There are 5*C(5,2)² ways that can happen.

However, that doesn't account for the distinction between the two ranks involved. If it's A's and K's then we want to specify that the players have AA/KK as opposed to AK/AK, and that the board must be AAKK. There's more than one way to account for that.

I showed one: multiply by 2*C(4,2) / C(8,4)

Even though we're talking about 15 cards, only 8 are important and we can ignore the extras. There are C(8,4) ways to arrange four A's and four K's relative to one another, whether there are other cards in between or not.

Take an arrangement like AAKKAKAK and insert the dividers: AA|KK|AKAK

One player has AA and another has KK. They can be swapped, so that's 2 arrangements. Compounding those are C(4,2) board arrangements with two A's and two K's.

So of the 5*C(5,2)² valid ways to distribute the 8 cards to the board and two players, 6/35 of them have the right coordination of the two ranks.

Another way to get the 6/35 is to frame it as assigning the 8 cards into three groups, thinking only about which partners they have rather than how they're lined up. It will be one group of 4 and two groups of 2. If we don't distinguish the two smaller groups, there are C(8,4)*3 possible assignments. For our numerator, there are C(4,2) ways to pick which two A's are in the group of 4 (to be part of the board cards), and likewise for the K's. So the fraction is 36/210 = 6/35

I used the first approach twice and the 2nd approach twice when it allowed smaller denominators of C(8,3) and C(7,3).

The rest of the calculation is more of the same. For the scenarios with boat boards, I had to multiply by 2 to distinguish between AAAKK and AAKKK.