So I was playing at Global Poker today and this happened in a PLO sweeps sit n go. Big stack hits a one outer on the river to have quad Aces beat quad Threes. Three hands later I go all in pre-flop (AA88 suited hearts) and the same person makes a straight flush (KQJJ 3 clubs J of spades). What are the odds a player makes quads over quads and then a straight flush 3 hands later? Board ran out 9c2c3hTc8c.

These are independent events

Fine, what are the odds of quads over quads in PLO? Then what are the odds of hitting a straight flush in PLO? If the odds of A are say 1 in 10,000 hands and the odds of B are 1 in 1,200. Then the combined probability that both would happen in 4 total hands is astronomical.

P(board double-paired) = 3*C(13,3) * 4*6*6 / C(52,5)
P(two quads in 6max | board double-paired) = 15 * 6/35 * C(43,4) / C(47,8)

Combined: P(quads over quads in 6max) ≈ 1 in 20844

Explanation for the 6/35 term:

1. There are C(7,3) = 35 ways to split 8 cards into two unlabeled groups of 4. How to see this: if you start with an arbitrary card, you have to choose 3 partners for it out of 7 possible. Once those partners are chosen, the remaining 4 cards are automatically the other group.

2. We're interested in groupings where the two X's are together in one pile and the two Y's are together in the other. To count how many groupings meet that criterion, start by assuming the two X's are together and choose two of the four non-Y cards to go with the X's. There are C(4,2)=6 possible choices. Once that choice is made, the remaining two non-Y's are automatically grouped with the Y's, so we're done counting.

Within how large of a sample?

I know it drives some people nuts, but here is my "C-heavy" derivation of the Prob of observing two different players having quads in 6-max PLO on the same board (assuming random deals as per usual):

= C(13,3)*C(3,2)*C(4,2)*C(4,2)*C(4,1)*C(6,2)*C(2,1)* C(2,2)*C(43,2)*C(2,2)*C(41,2)*C(39,4)*C(35,4)*C(31 ,4)*C(27,4) / C(52,5)*C(47,4)*C(43,4)*C(39,4)*C(35,4)*C(31,4)*C( 27,4)

which simplifies to:

= 216 / 4,502,365

which is approx once in every 20,844.28 deals (matching what heehaww derived above).


Edit to add: wait a minute, there is another way to have quads over quads in PLO, isn't there (board could be a "full house")??


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You're right, that would use two hole cards if the player had a kicker higher than the pair on board. The boat couldn't be XXXAA or AAAKK. The kicker can't tie the pair because the villain needs quads of that rank.

This is a tedious case to calculate because there are many (110?) different probabilities to add. The board is xxxrr (where r is a rank from 1 to 12) and there is a different probability for each r and for x>r vs x<r.

The player with quad x's needs a playable kicker. The # of playable kickers (k) varies with r and r's relation to x.

If x<r then k=4(13-r) else k=4(12-r)

The # of valid ways to group the cards (p) depends on the # of playable kickers dealt (j).

P(quads>quads | xxxrr) = sum of 15 * C(k,j) * C(52-k, 5-j) * (p/35) / C(47,8) from j=1 to 5

j=1 → p = 6
j=2 → p = 9
j>2 → p = 10

P(board xxxrr such that x>r) = 24(13-r) / C(52,5)
P(board xxxrr such that x<r) = 24(r-1) / C(52,5)
For r=12 we only consider x<r.

I'll compute it another time.

You are right that could happen. In this case both players had pairs in the hole cards. There was not a full house on the board.

I just got up and haven't had my morning coffee yet, so this may be the dumbest post ever, but ...

AAAKK

A432 vs KK74

is quads over quads, isn't it??

I don't understand the "kicker" issue that heehaww is worried about. PLO just requires a player use exactly two cards from his hand and exactly 3 cards from the board.

What am I missing? (Be gentle if I am being a total idiot)

D'oh, as you can see I don't play much Omaha. I was thinking your kicker would be one of the K's on the board and thus wouldn't count, but no, your hand would be AAAA4.

After I have some lunch I'll code a loop to sum the probabilities.

What do you think the odds are that it happens somewhere this year?

Isn't the "full house on board" case just another straightforward combinatoric exercise? I will put my derivation into a spoiler so you won't be influenced by my answer (right or wrong).

Since the kicker doesn't matter, yes it's a much easier problem than I made it out to be. No double-summations are needed.

P(boat on board) = 13*12 * 4*6 / C(52,5)

P(quads>quads | boat on board) = 15 * 10/35 * C(44,5) / C(47,8)

P(quads > quads) = 1/46899.635 + 216/4502365 ≈ 1 in 14431

By the way, I think we (collectively) have encountered the double-summation issue on a prior inquiry. Maybe something to do with a NLHE high-hand jackpot where both cards have to play.

In most case doesn't the double-summation collapse to a "simple" expression via Pascal's triangle or combinatorics or something like that?

IIRC there were two times I unnecessarily used a double-summation and I think they were multinomial distribution problems. But multinomial distribution can require a multi-summation (and when it doesn't, I don't even know how I would have used one). For instance if you roll a die 10 times and want the chance of at least three 6's and three 5's, I think that's a double-summation (it would get the right answer) but maybe I'm repeating my mistake of over-complication.

The false problem I was answering earlier ITT isn't Multinomial and I don't see how it can collapse either.