Dropout

But suppose they just gave each individual person advantage, and then picked the most popular roll out of all those advantage'd rolls. (That is, if I rolled an 8 and a 15 on my phone, only the 15 would get compared with everyone else's advantaged roll.)

Out of 400 ordered pairs of 2d20 rolls, the number of combinations which occur for each roll are 2n-1. For example, to roll a 15 with advantage there are 2(15)-1 = 29 possible ways to do so: a 15 on the first die, and anything from 1-15 on the second, or a 15 on the second die and anything from 1-15 on the first for what seems like 30 combinations, but we've double counted the case where we rolled 15 on both tries, so we subtract one to get 29.

Notice that the sum of (2n-1) for n=1..20 (that is 1 + 3 + 5 + ... + 37 + 39) is exactly 400, the total number of combinations we expect.

So the probability of a nat 20 with advantage is 39/400. So what we're actually asking is when 20,000 people roll with advantage, what are the odds that any of the other numbers, like 19 which has a 37/400 chance each time, still somehow get rolled more often than a nat 20?

I do not know how to actually compute that - it's complicated beyond my skill - but the law of large numbers suggests that as the crowd gets bigger that number should get very, very small indeed, and in fact the probability of not rolling a nat 20 would converge to 0 as crowd size went to infinity.

I tried tossing together some python to experiment: out of 10,000 attempts using this method, and presuming a crowd of 20,000 who all rolled. Out of 10,000 trials, a nat 20 was rolled 9454 times, a 19 was rolled 545 times, and an 18 was rolled 1 time. It never rolled below 18 using this method in 10k attempts. So, I'm guessing that's not what they went with.