[danielmarkbruce]

This seems off. A few billion seconds to have a 50:50 chance? Why wouldn't it be a billion seconds at a billion per second (2^60 total requests) would give a 1 in 2^68 chance (or 1 in 2^62 if its really only 122 bits)?

[Dylan16807]

Birthday paradox. The number of opportunities to collide is the number of items squared. (Divided by two and a smidge)

[danielmarkbruce]

Lol. I must be brain dead. Yes.

[penteract]

Because we're talking about collisions, as opposed to comparing 2^64 independent pairs. With 2^128 possible values, if you've picked 2^63 distinct ones, the chance that a randomly selected value collides with one of those is 1 in 2^65. If none of your second batch of 2^63 collide with each other, that gives a 2^63/2^65 = 1/4 chance of one of them colliding with the first batch. Considering the possibility of collisions within each batch of 2^63 brings it closer to 1 in 2.