jmc

There are two cases that we have to consider.

$\bullet$ Case 1: The first card is one of 2, 3, 4, 5, 7, 8, 9, 10.

There are 32 such cards, so this occurs with probability $\frac{32}{52}$. For any of these cards, there are 4 cards left in the deck such that the cards sum to 12, so the probability of drawing one is $\frac{4}{51}$. Thus, the probability that this case occurs is $\frac{32}{52}\times \frac{4}{51}=\frac{32}{663}$.

$\bullet$ Case 2: The first card is a 6.

There are 4 of these, so this occurs with probability $\frac{4}{52}$. Now we need to draw another 6. There are only 3 left in the deck, so the probability of drawing one is $\frac{3}{51}$. Thus, the probability that this case occurs is $\frac{4}{52}\times \frac{3}{51}=\frac{3}{663}$.

Therefore the overall probability is $\frac{32}{663}+\frac{3}{663}=\boxed{\frac{35}{663}}.$