I want to truncate an md5 hash to about half size. How much does that increase the odds of collisions? if I'm dealing with around 500 000 generations, should I be worried about a collision? what about 1m generations.
The math you're looking for is on Wikipedia's birthday attack page.
We consider the following experiment. From a set of H values we choose n values uniformly at random thereby allowing repetitions. Let p(n; H) be the probability that during this experiment at least one value is chosen more than once. This probability can be approximated as
With 128 bits the chance of a collision among 500,000 hash values is around 10-28. If you halve the size of the collision space then the chance of collision is around 10-9. That is, even though the chance is vastly greater it's still very, very low. It depends on how critical it is that there be no collisions. 10-9 is on the order of one in a billion, so while extremely unlikely it's within the realm of possibility.
For reference:
1028 = 10 octillion = 10 billion billion billion
109 = 1 billion
There's an interesting mathematical problem called the birthday problem that deals with that kind of situation. The fact is that the more entries you push in, the higher the chances to have a collision.
Following the table posted on the above link, assuming your digests are 64 bits each (since a single MD5 hash is 128 bits) and that MD5 have a uniform distribution, there is a very low chance that two hashes will collide. It becomes significant (1% chance or more) at 610,000,000 entries.
来源：https://stackoverflow.com/questions/2256423/truncating-an-md5-hash-how-do-i-calculate-the-odds-of-a-collision-occurring