Demi: I think there's a factor 5 missing in your first term.

Anyway I think that the best way to go about this problem is to use inclusion-exclusion style arguments. Let PX be the probabilities that X particular cards are all of the required flush suit. In the first scenario 2 cards of the suit and 2 offsuit cards are accounted for, so P3 = 11*10*9/48*47*46, ... In the second 1 card of the suit and 3 offsuit are known, so P4 = 12*11*10*9/48*47*46*45, ...

A first approximation of the probability of having at least 4 of the suit on the board is clearly given by 5*P4. The inaccuracy comes from the fact that this counts the instances of all 5 cards having the same suit 5 times, so to get the exact answer we subtract 4*P5. The exact probability is therefore 5*P4 - 4*P5.

A first approximation of the probability of having at least 3 of the same suit is given by 10*P3. This counts each occurance of 4 of the same suit 4 times, so we subtract 3*5*P4 to compensate. Then, hoewver, we count each occurance of 5 of the same suit -5 times, so to get the exact answer we fianlly need to add 6*P5. The exact probability of getting at least 3 of the specified suit is therefore 10*P3 - 15*P4 + 6*P5.