If we do $$\frac{{^7C_2} + {^6C_2}}{{^{13}C_2}}$$
it won't be correct as by doing this we ensure that in choosing 2 elements, they both will necessarily be distinct.
But to be independent, we should be able to select an item n number of times in n trials. Because choosing the same item before shouldn't matter in the present, as the trials are independent. We need to allow repetitions.
So,
$$\frac{7}{13}*\frac{7}{13}+\frac{6}{13}*\frac{6}{13}$$
will be correct.
0.503 is the answer.