Player 1 holds KsKd
Player 2 holds KhJs

That's all we know.

What is the probability that the case King (Kc) AND any remaining Jack will come on the flop?

I think I know how to calculate this type of problem, but I would appreciate confirmation from my 2+2 mentors.

TIA!

[1*3*44 + 1*C(3,2)] / C(48,3)

=~ 0.78% or about 1 in 128

The first term is for 1 jack. That is 1 king, times 3 possible jacks, times 44 other cards. The second term is for KJJ. That is 1 king times C(3,2) = 3*2/2 = 3 possible pairs of jacks; all divided by C(48,3) = 48*47*46/(3*2*1) possible flops.


Without combinations:

(1/48 * 3/47 * 44/46)*6 + (1/48 * 3/47 * 2/46)*3

=~ 0.78% or about 1 in 128

The first summed term is for 1 jack, while the second is for KJJ. The first term is multiplied by 3! = 6 ways to order the 3 cards, while the second is multiplied by 3 possible positions for the king (since 3/47 * 2/46 already consider both possible orderings for the jacks).

Perfect explanation. Thank you.

Turns out I was not doing this correctly after all, so I'm glad I asked.

Thanks again for taking the time to explain.

A word of warning, suppose you have the KJ and the flop comes KJx. It's not correct to think "Say he had KK, there was only a 1/128 chance of this happening. So the chance he has KK is 1/128". It's easy to make this mistake subconsciously, even if you do spot the problem when I write it out like this!

True. I've heard a lot of players make the same backward application of odds when losing to flopped flushes, saying, "less than 1% chance he had a flush..." when that's clearly incorrect. Just because the odds of flopping a flush are less than 1% when you have two suited cards doesn't mean that a player is less than 1% to have a flush when a suited flop comes. The hard part's already happened at that point.