4 ms·
For those curious on how to find that ~1200 values are needed before a 50% probability of collision, this is a generalized form of the birthday problem [0]. Wi
by yulle 2y ago
For those curious on how to find that ~1200 values are needed before a 50% probability of collision, this is a generalized form of the birthday problem [0].
With a calculator [1], we can find the exact amount. plugging in D=10^6 (# of unique codes) with P=0.5 (odds of collision) gives you 1178 values.
[0] https://en.wikipedia.org/wiki/Birthday_problem https://en.wikipedia.org/wiki/Birthday_problem
[1] https://www.bdayprob.com/ https://www.bdayprob.com/
- codeflo 2y agoThe very rough rule of thumb that I think one should remember is: You get a 50% chance of collision with sqrt(total) samples. Sqrt(1 000 000) = 1 000, close enough to the exact number you calculated.
- ColinWright 2y agoMore accurately, for 50% it's \sqrt(\ln(4))\sqrt(N) which is \sqrt( ln(4) N ). That's about 1.17741 times \sqrt(N). The reasons are explained here: https://news.ycombinator.com/item?id=40886363 https://news.ycombinator.com/item?id=40886363