I have a math problem I would love some help on. I've searched and studied for 8+ hours and still am unsure the correct way to approach this (seemingly simple) problem. Maybe I'm overthinking and have gone too deep...

BACKGROUND: I was dealt Ax. Board reads J, J, X, X, X (two jacks, three unconnected trash cards). No flush, straight draws. I'm sure I have the top hand with my Ace/Kicker. But to be sure...what are the odds that any of the 3 players remaining were dealt one of the 9 cards that can give them two pair? Lets leave all other possibilities (three jacks, etc.) off the table.

I have given myself a crash course in probability theory but cannot figure out the appropriate way to tackle this question. I apologize if its obvious, but any help would be appreciated. Thanks!

Would you feel comfortable posting what you have done so far?

I feel that may be a good way to proceed.


Edit to add: If it helps, we can consider the three non-jack board cards to be K82 (king, eight, deuce) for the purposes of the math or XYZ.

I don't know if you are really trying to figure out a math problem, or a poker problem. I'll look at it as a poker problem.

As a poker problem, the information is incomplete. If it was limped preflop, and checked around on the flop, turn and river, then there is a chance your ace is good. If it was raised preflop, and there were bets on the flop and turn that were called by everyone, the probability your ace is good is pretty close to zero. If you are thinking of betting the river you are certainly going to be betting as a bluff, because there is no chance that worse hands are going to call. If someone else bets the river you have to assume they have you beat.

Calculated in a total void: There are 45 unknown cards, 9 of which are threatening

The chance that these three players (6 unknown cards) do not have any of the 9 threatening cards is:
36/45 * 35/44 * 34/43 * 33/42* 32/41* 31/40 = 0.24 (rounded)

So the chance that someone does have one (or more) of the 9 threatening cards is 1 minus the above probability (since all probabilities for all independent cases that cover all possibilities must add up to 1..The two cases are in this..erm...case: Either "they don't have any of the threatening cards" or "One (or more) of them has one (or more) of the threatening cards")

So chances are 76% that at least someone has at least one matching card.

But that really doesn't help since people aren't playing random cards. People play high cards more often than low cards and connected cards more often than totally unconnected cards. With any preflop, flop and turn action you can further narrow down holdings.

(Actually you should consider anyone holding a better Ax also...or someone holding a small pocket pair and trying to get it to showdown...but since you didnt I won't overcomplicate things here)