birthday paradox doesn't apply
this is because the birthday paradox presupposes 23 randomly chosen birthdays. this is equivalent to choosing them all at once with a random function.
in this case we have n *distinct* identifiers and are choosing one more. we can't take advantage of a collision in the pre-allocated numbers because there isn't one. given n randomly distributed allocated numbers in a field of m size, the probability of a collisionis n/m.
otherwise of course, the previously chosen values would pre-dispose the next choice. but we know numbers aren't calvinist.
