2 ms·
Given they’ve already collided on the 19 digit numbers, then it’s just the probability of having the same birthday: 1/365, smoothing over the rough edges of the
by travisjungroth 4y ago
Given they’ve already collided on the 19 digit numbers, then it’s just the probability of having the same birthday: 1/365, smoothing over the rough edges of the real world.
If we’re not taking them as already collided, you multiply the two probabilities since they’re independent.
Edit: I realize you’re probably asking a more complex question I don’t yet know the answer to. Like a double birthday paradox.
- asow92 4y agoTo your edit, yes, and there would certainly still be a lot of collisions.
- travisjungroth 4y agoMy second answer is a lower bound on the correct answer and extremely close to correct. Think of finding the ID collision first. The probability of having at least one ID collision where they also share the same birthday is pr(2 matching ids) * pr(birthday match in 2 people) + pr(3 matching ids) * pr(birthday match in 3 people)... The probability of having 3 matching IDs is extremely low. Assume you already have a match of size 2. Now you're looking for 1/10^19 match with ~1/10^10 draws. Not good odds. So we can ignore everything after the ID match of size 2, and get very close with 96% * 1/365 or 0.26%.