Double board PLO bomb pot odds help

I doubt I can solve this with any confidence but am pretty sure I can simulate it, probably in less than an hour but almost certainly before year end.

But the WNBA game is just starting so if no response by tmw mid day, I’ll do it. However, guys like heehaw and whosnext can do it analytically if they happen to see it.

Wow, this is incredible. Thank you very much.

The 99 and KK seem higher than I would expect though. You are the expert but perhaps there’s a misunderstanding. There’s only two nines left and so an opponent has to have them both. So I would have thought the numerator would have been 2C2?

Go Vegas Aces…

Yeah there are many nice and helpful people here

I did a simulation for estimating the probability of at least one of five players having 2 kings or 2 nines. I got 13.27% with a run of 10 million hands. This result is consistent with an estimate I made assuming independence.

The fact that the simulation confirms the independence assumption range indicates I got both right or both wrong.

Someone verifying these results would be a good thing.

OOPs I see heehaw did an analysis. I think we agree other than a relatively minor adjustment to be made.

You can figure that out using odds oracle. I arrive at the same numbers that heehaww and statmanhal posted and it also confirms the 73,2% for a 6 to be out there.

Yeah one way to do it is (2C2)(40C2)/(42C4) for the same result, but I took a different approach: let the denominator be the number of ways to distribute the two 9's, 42C2. Then the numerator is the number of ways for them to be somewhere within the player's 4 cards, 4C2. Another way to think of it is, if the player were only dealt two cards it would be 1/42C2, so with 4 cards it's 4C2 times that because there are 4C2 mutually exclusive 1/42C2 chances at it.

For the KK I showed both approaches: either a denominator of 42C3 or 42C4. If the denominator is 42C3, the numerator is the #ways 2 or 3 kings can be within a specific 4 cards. For 2 kings it's (4C2)*38 because to distribute the 3 kings we're choosing 2 of the player's 4 spots and then one of the 38 other spots.

Now I'll show the harder P(KK and 99) calculation. It's more efficient to split it into two cases: one villain gets KK99, or two villains get them.

P(KK99) = 5*3/(42C4)

For P(two villains KK and 99) I'll show both approaches.

In the framework of choosing spots for the desired cards: (5C2)(4C2)[4C3 + (4C2)36] / (42C2)(40C3)

Explanation: there are 5C2 choices of hands for the 99 and KK to be put in. Starting with the nines, there are 4C2 ways to place them in one of the hands, which I pulled in front of the [ ]. Then for the kings, we have to deal with KK and KKK separately. 4C3 is the KKK case. To count the KK case: 4C2 placements in the non-99 hand, then the 3rd king can be anywhere other than that same villain's hand. It can be in the same hand as the 99, so that's 36 possible places. As for the denominator, we're choosing 2 places for the nines and then 3 places for the kings. The other way is fine too ofc: (42C3)(39C2). Or, my preference is to pick 5 spots for all of them and then choose which of the 5 spots are the nines: (42C5)(5C2)

With the more common approach of choosing cards for the players: (5C2)[3*37(36C2) + (37C2)*3(35C2) + (37C2)35] / (42C8)(8C4)

Those give the same result which, when added to our P(KK99) above, gives us P(KK and 99) ≈ 1/593

Since P(KK)=29/287 and P(99)=10/287, by inclusion-exclusion we have P(KK or 99) = .1342027677

Thanks for all of this heehaw!