Odds question

[kingstalker]

In a 9 handed ring game, I have a jack and one of my opponets have a jack, what are the odds (assumeing both other jacks are live) that two jacks will appear on the flop?

[BruceZ]

In a 9 handed ring game, I have a jack and one of my opponets have a jack, what are the odds (assumeing both other jacks are live) that two jacks will appear on the flop?

If by "live" you mean that you somehow know that none of your opponents holds the 2 remaining jacks, then there are 32 possible flops containing 2 jacks and one of the 32 remaining cards out of C(34,3) = 5984 total flops, so the probability is 32/5984 = 0.53% = 185-1.

If you don't know if the remaining two jacks are in your opponent's hands or not, including the guy with the jack who might have JJ, then there are 47 possible flops with JJ and one of the 47 remaining cards, since you hold number 48, out of C(49,3) = 18424 total flops, so the probability is 47/18424 = 0.24% = 391-1.

Beginners are advised to study these two examples and thoroughly understand why there are only 32 possible cards in the first case and 47 in the second case, even though we don't actually know the exact cards in your opponent's hands in either case; however, in the first case we do know that they are not jacks.

[kingstalker]

Thanks for the reply, I was playing Hold them when I had J-10 and the board was Q-J-J, my Opponet went all in and I started thinking " The odds gotta be low that he has a jack as well so i am gonna call and even if he does have a jack I prob have a better kicker" so he must have a hand like A-Q" so I called and he had Q-J.

[BruceZ]

That's a different question entirely. Once you know the flop has 2 jacks, the probability that your one opponent has a jack is much higher than the probability of the flop having 2 jacks given that you and your opponent hold a jack. That probability he holds a jack will depend on what hands he will go all-in with, but even if he holds a completely random hand, he has a 4.3% chance of holding a jack. 1 - 46/47 * 45/46 = 4.3%.

As a more realistic example, since we have QJJ JT out, suppose he could hold AQ (12 ways), <font color="red">AJ (4 ways)</font>, AA (6 ways), KK (6 ways), <font color="red">QQ (3 ways)</font>, TT (3 ways), KQs (3 ways), <font color="red">KJs (1 way)</font>, <font color="red">QJs (1 way)</font>, and JTs 1 way, then the probability that he has a better J is 6/40 = 15%. He could also have QQ with probability 3/40 = 7.5%. You can modify this to suit your particular opponent profile.