Suppose I draw 4 cards from a fair deck. What is the probability that I will end up with a hand with 4 different suits? What about 2 distinct suits or 3 distinct suits or all the same suit?
For 4 different suits, my solution is (13^4)∕52C4 which is about 0.105
[13 possible cards from each suit]
For 3 distinct suits, my solution is (4C3)(13C2)(13^2)/(52C4) which is about 0.195
[4C3 is me choosing 3 suits out of the 4, 13C2 is me choosing 2 cards from a specific suit]
For 2, my solution is (4C2)(13C3)(13)/(52C4)+ (4C2)(13C2)(13C2)/(52C4) which is about 0.217
[Case 1: 3 of the same suit and 1 of a different suit, Case 2: 2 of the same suit then another of the same suit]
For all the same suit, my solution is 4(13C4)/(52C4) which is about 0.011
[4 suits to pick from, then choose 4 cards from the suit]
As far as I know, the total probability is supposed to add up to 1 (in this situation I can only draw either all of the same suit, 2 distinct suits, 3 distinct suits, or all different suits), but when I add up my solutions, it doesn't even reach 1. Obviously, I did something wrong, but I can't figure out where. Help?