jmc

We clearly can't have both two kings and at least 1 ace, so we have two exclusive cases to evaluate separately.

Case 1: Two kings. The probability that two kings are drawn is $\frac{4}{52}\cdot \frac{3}{51}=\frac{1}{221}$.

Case 2: At least 1 ace. We can break this into two cases:

Subcase 2A: Exactly 1 ace. We can choose the ace first with probability $\frac{4}{52}\cdot \frac{48}{51}$, and we can choose the ace last with probablity $\frac{48}{52}\cdot \frac{4}{51}$. So, the total probability of getting exactly one ace is $2\cdot \frac{48}{52}\cdot \frac{4}{51}=\frac{32}{221}$.

Subcase 2B: 2 aces. The probability of this occurring is the same as that of two kings, $\frac{1}{221}$.

So, the total probability for Case 2 is $\frac{33}{221}$.

Adding this to our probability for Case 1, we have $\frac{34}{221}=\boxed{\frac{2}{13}}$.