Hacker News Reader

That doesn't matter to an individual developer. Why should they care if the set of all in use USB PIDs has collisions? It's only becomes a problem for them if their PID suffers a collision.

The Birthday Problem is only relevant to find collisions between any 2 strings, but in this case the developer is only concerned about the probability of his own string clashing with someone else’s, thus the probability of a collision is only Number of taken strings / Number of total strings.

A good rough rule is to halve the power of two: so 2^16 = 65536, which is close enough to 77163.

That's really cool. I tried it with the birthday problem[1] (real answer 23) and got 16 if I used whole exponents[2] and 19 with fractional exponents[3]. They're a little off but it gives you an idea of the value.

[1] https://en.wikipedia.org/wiki/Birthday_problem

[2] 256 days=2^8, 16=2^4

[3] 365 days=2^8.51, 19=2^4.255