PLO double board bomb pots. What are the odds of a player flopping a straight flush on one board, while a different player flops an Ace high flush on that same board?

My approach will be: P(straight flush flop) * P(SF > nut flush | SF flop).

When doing it that way, we have to distinguish between flops allowing 1/2/3 possible SF's.

The flops with 3 possible SF's are connected ones ranging from 345 to TJQ, which is 8. But actually, we can't count SF's that block the nut flush, so we'll instead categorize 345 and TJQ as only allowing two SF's. Therefore, N(flops w/ 3 possible SF's) = 4*6

The flops with 2 possible SF's are {2x5 to KxT} & {234, JQK} & {345, TJQ} but we must exclude {234, 235, 245, TJK, TQK, JQK}. This amounts to 16, then multiply by 4 suits = 64.

The flops with 3 possible are {2x6 to 9xK} & Axx & [the 6 subtracted from above], so


Oh JFC first I accidentally pressed a hotkey that submitted this in the middle fo typing it, and then while editing that I accidentally clicked the Back button on my mouse and lost what I was typing next. I'll finish this later lol...

So the other day I must have been possessed by a boomer, sorry about that. Continuing where I left off:

so 4(8*3 + 2*6 + 6) = 168

The answer will be [24*P(success | 1 sf flop) + 64*P(success | 2 sf flop) + 168*P(success | 3 sf flop)] / C(52,3)

If we're 9-handed:

Let p = [7*C(39,2) + C(7,2)*39 + C(7,3)]/C(47,4)
Let q = [6*C(39,2) + C(6,2)*39 + C(6,3)]/C(46,4)
Let r = [5*C(39,2) + C(5,2)*39 + C(5,3)]/C(45,4)

P(sf>nf | 1 possible) = 9*8 * C(4,2)/C(49,2) * p ≈ .01244

Explanation: there are 9*8 two-player permutations. The chance that the first one flopped a SF is C(4,2)/C(49,2) because there are 4 spots for the 2 SF cards out of a possible C(49,2) places those two cards can be. I multiplied that by P(2nd player NF | 1st player SF) which is p.

P(sf>nf | 2 possible) = 9*8*[2*C(4,2)*p/C(49,2) - 4*q/C(49,3)] ≈ .0244

Explanation: the factor of 2 before the C(4,2) is because there are 2 possible SF's. However, 2*C(4,2)/C(49,2) double-counts the six-card SF's. The chance of a six-card SF is 4/C(49,3). That's multiplied by a separate conditional probability "q" because in that case, 3 cards are known.

P(sf>nf | 3 possible) = 9*8*[3*C(4,2)*p/C(49,2) - 2*4*q/C(49,3)] - C(9,3)*C(4,2)²r/C(49,4) ≈ .036

Explanation: 3*C(4,2)/C(49,2) double-counts the 6-card SF's, of which there are 2, so we subtract them. After that, we don't need to add the 7-card SF because it's still counted (now only once). Straight formations aren't the same as combinations in that not all the sets intersect, which is why inclusion-exclusion behaves differently with them. Anyway, when 3 SF's are possible, it becomes possible for two players to flop a SF, and this possibility was double-counted and therefore gets subtracted. A third conditional probability "r" is necessary for that scenario. There are 3*C(9,3) ways to pick a player to have a NF and two to have SF's; the chance of the 4 SF cards being sorted correctly is 1/3, so I omit the 3 and 1/3.

answer ≈ 1 in 2797

Most likely you'd get a very similar result using only "p" throughout, but my OCD wouldn't let me submit an answer whose nth decimal place is wrong

Next time I'll tweak a few #'s to get the 6-max answer.

I'll also investigate whether another method (such as permutations) is more efficient. What I did above already hides some of the inner workings by calculating the players separately, which I think might negate the advantage of perms, but we'll see.

Thank you!!