C(n,r) = n!/(r!(n-r)!)
You take the number of possible 13 card hands with no 9 in them and subtract that from the total number of possible 13 card hands 9's included.
so C(52,13) - C(48,13)
The number of all possible 13 card hands is:
52!/13!(52-13)! or 52!/13!*39! which is 635,013,559,600
The number of all possible 13 card hands with no 9s is:
48!/13!(48-13)! or 48!/13!*35! = 192,928,249,296
The difference is 635,013,559,600 - 192,928,249,296 = 442,085,310,304
So 442,085,310,304 out of 635,013,559,600 hands will have at least 1 nine. The ratio of 442,085,310,304 to 635,013,559,600 is 0.696182472... So roughly 70% of the time
27,630,331,894/39,688,347,475 is the best I could do for a whole number ratio.
