Post #2555816

2026-03-18 10:34 UTC

@LiseAndreasen@mastodon.world let a and b be the two random numbers on the Cantor set and assume without loss of generality that a <= b. Divide into cases according to which ternary digit is the first to differ. In 1/2^n cases, that digit is the nth (there is a 1/2 chance that the first pair is the same, 1/2 that the second is, and so on, and finally we have a 1/2 chance that the nth pair differs). In that case, the difference truncated at the nth digit is 2/3^n,

Replies (1)

@aleph_omega_plus_four@mastodon.world 2026-03-18 10:34

@LiseAndreasen@mastodon.world then there are some more digits which may also differ. But at each n+i, there are 4 possibilities (with equal probability): 0, 0; 0, 2; 2, 0; 2, 2. The identical pairs add nothing to the overall difference, and pair 0,2 adds 2/3^(n+i), whilst the pair 2,0 subtracts the same amount. Thus everything beyond n on average contributes 0 difference. So the probability is the sum of 1/[2^(n-1) * 3^n], which is a geometric series with limit 2/5 = 0.4 :)

Open ##2555817