C(n,k)=n!/k! (n-k)!
Left = 13! 39!/ 52! \sum_k (52-k)! / (39-k)!13! = 1/C(52, 13) \sum_k C(52-k, 13).
Using Hockey Stick Identity (https://en.wikipedia.org/wiki/Hockey-stick_identity), the sum is C(53, 14).
So the answer is C(53, 14)/C(52, 12) =53/14.