Birthday Attack with probability of 1
To get a collision with probability 1, you need to hash $H+1$ distinct values.  You can see that because it is possible that if you hash only $H$ values, each one might happen to hash to a unique hash output; if any reasonable sized hash function, this is extremely unlikely, but it does have a nonzero probability to happen.  On the other hand, if we hash $H+1$ distinct values, we know (by the pigeon hole principle (because there are only $H$ hash outputs), that two of the inputs must hash to the same output.