The most useful formula I’ve seen related to the birthday problem is the approximation p = (n^2)/(2m), where n is the number of random choices made, m is the nu

The most useful formula I’ve seen related to the birthday problem is the approximation p = (n^2)/(2m), where n is the number of random choices made, m is the number of choices available, and p is the probability of at least one collision. You can derive the usual formula for the number of hash bits necessary to avoid a collision with probability 1/2 (given a set of cardinality n) by just substituting 1/2 for p and solving for m. But this formula is much more useful than that special case since it lets you set the collision probability to be arbitrarily small.