Join Date: Mar 2009
Posts: 6,729
This can also be done using logic and combinatorics (again, doing this as a "pure" probability question). Using your example of K994 upcards.
Quad 9's (two Nines down):
= C(2,2)*C(46,1)
= 46
Quad 4's (three Fours down):
= C(3,3)
= 1
Quad K's (three Kings down):
= C(3,3)
= 1
Full House with exactly one Nine and one King and no Fours down:
= C(2,1)*C(3,1)*C(40,1)
= 240
Full House with exactly one Nine and one Four and no Kings down:
= C(2,1)*C(3,1)*C(40,1)
= 240
Full House with exactly one Nine and two Kings down:
= C(2,1)*C(3,2)
= 6
Full House with exactly one Nine and two Fours down:
= C(2,1)*C(3,2)
= 6
Full House with exactly one Nine and one King and one Four down:
= C(2,1)*C(3,1)*C(3,1)
= 18
Full House with exactly one Nine and a pair of some other rank (not Kings or Fours) down:
= C(2,1)*C(10,1)*C(4,2)
= 120
Full House with No Nines, 2 Kings down:
= C(3,2)*C(43,1)
= 129
Full House with No Nines, 2 Fours down:
= C(3,2)*C(43,1)
= 129
Full House with No Nines, trips of some other rank (not Kings or Fours) down:
= C(10,1)*C(4,3)
= 40
TOTAL = 976.
So the probability:
= 976 / 17,296
= 5.642923219% or around once every 18 times.
Last edited by whosnext; 03-06-2019 at 06:19 PM.