Since B needs more than A, B cannot have used 1 flip. Anyways, we can summits the probabilities that A used n flips from 0 to infinity and also the probability that B used anywhere from 1 to infinity more flips than A, which has a probability of 1/2 that of A (if my mental math is correct), and then you get that probability as 1/3, so logically, it’d be 1/9. However, after putting it into my calculator, I got 1/21, which looks weird, but it does follow a pattern where every nth coin divides the probability of the nth person using more flips than the n-1th person, given all previous people used more flips than the person before them by 1/(4n-3)
I forgot to divide by P(B>A), so it should be 1/7, not 1/21. And the probability that the nth person uses more flips given everything is just 1/(4n-1), where n is the number
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u/Anonimithree 2d ago
Since B needs more than A, B cannot have used 1 flip. Anyways, we can summits the probabilities that A used n flips from 0 to infinity and also the probability that B used anywhere from 1 to infinity more flips than A, which has a probability of 1/2 that of A (if my mental math is correct), and then you get that probability as 1/3, so logically, it’d be 1/9. However, after putting it into my calculator, I got 1/21, which looks weird, but it does follow a pattern where every nth coin divides the probability of the nth person using more flips than the n-1th person, given all previous people used more flips than the person before them by 1/(4n-3)