Edit: Skip to 4gotunameagain's reply, this probably doesn't contain anything helpful to read.

The math is correct, but it's the math for the question:

  blockHashes.has(hash(domainInBlockList1) == hash(domainInBlockList2)

While the thing they probably want to check the collision math for is:

  blockHashes.has(hash(domainNotInList))

As that's the risk you do the wrong action due to a hash collision. Importantly, the 2nd calculation is not going to be bounded by the size of your block list, it's going to be bounded by the number of domains overall.

The former question isn't really useful - you can have a 100 entry blocklist with 100 collisions and it still works the same as if it had 0 collision since the action on all 100 is the same. That's more a problem for generating unique IDs per hash whereas this is flipped around because it's about classification.