Another question for you guys, this should be a lot easier than the last.

What are the chances of being dealt say specifically 27 playing triple draw where the other three cards can't be 3,4,5,6, or 8? So the 5 cards must contain at least one Deuce and one Seven but no 3's-6s, or 8s.

This was my quick way of doing it that may be incorrect. I calculated all the possibilities of getting hands with two or more Deuces or two or more Sevens. Then added the sum of all those instances to the chance of being dealt specifically 27xxx where x's are 9 or higher (there are 24 cards 9 and up).


Odds of getting dealt 27

22227 12 3*4
77772 12 3*4
22277 36 6*6
77722 36 6*6
2277H 864 6*6*24
2227H 288 3*4*24
7772H 288 3*4*24
277HH 6624 4*6*(24*23)/(2*1)
227HH 6624 6*4*(24*23)/(2*1)
27HHH 33856 4*4*(24*23*22)/(3*2*1)

Sum: 48640 combos

There are 2,598,260 5 card hands. So probability of specifically 27xxx is 48640/2598260

I used Pro Poker Tools 'count' to arrive at the following values


Odds of getting dealt 27

22227 4
77772 4
22277 24
77722 24
2277H 864
2227H 384
7772H 384
277HH 6624
227HH 6624
27HHH 32384

Sum: 47320 combos

Without enumerating all the possible combo types, we can use the inclusion-exclusion principle and make a little use of boolean algebra. Since we don't want 3s,4s,5s,6s and 8s, we can count the combos assuming a 32-card deck. We start by counting how many hands don't have both a 2 and a 7:

N(NOT(2 AND 7)) = N(NOT(2) OR NOT(7)) =
= N(NOT(2)) + N(NOT(7)) - N(NOT(2) AND NOT(7))
= 2*C(28,5) - C(24,5)

Therefore, the number we want is:

C(32,5) - 2*C(28,5) + C(24,5) = 47320

Perfect, thanks guys