jmc

We have two cases because if the first card is a $♢$, it could be an ace or not be an ace.

There is a $\frac{1}{52}$ chance that the ace of $♢$ is drawn first, and a $\frac{3}{51}=\frac{1}{17}$ chance that the second card drawn is one of the three remaining aces, which gives a probability of $\frac{1}{52}\cdot \frac{1}{17}=\frac{1}{884}$ chance that this occurs.

There is a $\frac{12}{52}=\frac{3}{13}$ chance that a $♢$ other than the ace is drawn first, and a $\frac{4}{51}$ chance that an ace is drawn second, giving a $\frac{3}{13}\cdot \frac{4}{51}=\frac{4}{221}$ chance that this occurs.

So the probability that one of these two cases happens is $\frac{1}{884}+\frac{4}{221}=\boxed{\frac{1}{52}}$.

Notice that we can avoid some of the large denominators above by organizing this computation as follows: $$\frac{1}{52}\cdot \frac{3}{51}+\frac{12}{52}\cdot \frac{4}{51}=\frac{1\cdot 3+12\cdot 4}{52\cdot 51}=\frac{51}{52\cdot 51}=\boxed{\frac{1}{52}}.$$