This is about cryptographic hashing functions (to be fair, could have spelled that out in my prior comment, but in general when someone talks about anything security relevant in conjunction with hashing, they always mean cryptographic hashing functions).

MD5 is not a cryptographic hashing function for exactly these reasons.

Also, the example you gave in your original comment wasn't actually about distribution but about symbol space.

By multiplying by four (and I guess you implicitly meant that the bit length of the hash stays the same thus dropping two bits of information) you are reducing the number of possible hashes by a factor of four, because now 3/4 of all potential hashes can't happen any more. So sure, if your 64bit hash is actually only a 62bit hash that just includes two constant 0 bits, then of course you have to calculate the collision chance for 62bits and not 64bits.

But if all hashes are still possible, and only the distribution isn't perfectly even (like is the case with MD5), then the average chance for collisions doesn't change at all. You have some hashes where collisions are more likely, but they are perfectly balanced with hashes where collisions are less likely.