Probability of 5 Card Poker Hand with Only Two Suits:
Pick a card from any suit: 52
Pick a card from a different suit: 39
Pick three more cards from either of the first two suits: $24*23*22$
So, $52*39*24*23*22$ should be the number of ordered 5 card poker hands consisting of only two suits.
To get unordered hands, divide by the number of ways to organize 5 things, that is, 5!.
Why is $\frac{52*39*24*23*22}{5!}$ not the right answer?
I understand one can get the answer by doing the following:
Pick two suits: 6
Pick 5 cards from the two suits: $26 ~C~ 5$
Subtract the sets of the above that have only one of each suit: $2(13 ~ C ~ 5)$
Thus, a correct answer would be 6*(26 C 5 – 2(13 C 5))
Which could also be written $6*((26*25*24*23*22/5!) – 2*(13*12*11*10*9/5!))$
I get why that works, I just need help understanding why my first method does not, or how I could modify it so it would work? Any thoughts?
Best Answer
You are forcing the first two cards to be different suits and missing SSHHS for example.