How many hashes for high probability of finding a collision (specific case)? Suppose Bob managed to obtain 220 different digests that were generated
by a hash function employed by a target system. The hash function outputs 8-byte
digest of a message. Bob now wants to find a message that hashes into 1 (one) of the
obtained digests. How many different messages should Bob approximately hash until
there is a good probability that a generated digest will match 1 of the obtained
digests?

My answer is $\sqrt{2^{64}}$ ($= 2^{32}$) messages for a probability of 0.5. Is this correct?
Let's assume that there are no prior collisions (two different messages that generate an identical hash) for the first $2^{64}$ messages.

If you want to find a message for one hash then you would have to try all these $2^{64}$ messages to reach the probability of 1 (100%). Since you mentioned good probability we can say that we would have to try half of that for a probability of 0.5 (50%). That means that we would have to try $2^{64} / \space 2$ possibilities which would be $2^{63}$.

Now you also already have a list of 220 hashes. That means that you can reduce it because we previously only calculated it for one hash, so the solution is:

$$\frac{2^{63}}{220} \approx 4.2 \times 10^{16}$$ for a probability of $0.5$ to find a message that hashes to one hash in your list.