Probability of Five Card Hands with one exposed card

That's right, and it's the best approach here.

Much appreciated heehaww

What about either having three nines in your hand or two nines?

Denominator is the same

Numerator is sum of C(3, 3)*C(48, 2) + C(3, 2)*C(48, 3)

Turns out to be around 2.3% total

Right again. If we wanna be a little more efficient we can do it this way:

[C(5, 3) + C(5, 2)*46] / C(51, 3)

C(51, 3) is the # of ways to place the 3 nines among the 51 unknown cards, not caring about which suit is where.
C(5, 3) is the # of ways all the nines can be within your 5 cards.
C(5, 2)*46 is the # of ways to put 2 nines within your 5 cards and the other nine elsewhere.

That's the approach I took for OmahaDonk's AA problem, where it was a bigger shortcut.

Can you please help me with probability of getting dealt exactly one other nine and a pair? The pair can actually be trips or quads as well.

In other words 9KKA2, 9KKKA, or 9KKKK

My best guess of numerator would be: C(3, 1) * C(12, 2) * C(43, 2)

One of three nines * one of pairs * two of any cards (except for nines which we already calculated)

I’m mediocre at this so it might not be right, thanks

3[C(48, 4)−C(12, 4)4⁴] = 203, 580

C(12, 4)4⁴ is the same as 48*44*40*36/4!

If two-pair isn't allowed: 203580 − 3*C(12, 2)*C(4, 2)² = 196, 452

Or by addition: 3[C(12, 3)*C(4, 2)4² + 12*11*4² + 12] = 196, 452

Wow thanks, yeah as expected I was wrong, way wrong.

But two pair certainly allowed assuming you meant 922KK

Yeah 922KK etc, so if that counts then 203580 is the numerator.

Breaking down what I did: first I chose one of 3 nines like you did. Then for the other four cards: 48C4 is all combos of non-9's, from which I subtracted the no-pair combos. For the no-pairs: there are 12C4 choices of ranks, times 4^4 choices of suits for them.