Re:When U.S. air force discovered the flaw of averages (2016)
by spappal on Friday April 24, 2020 @07:38AM
(#22966893)
P(simultaneously within middle 50 % of n independent distributions) = 0.5^n = 2^(-n), like my parent comment says.
But interestingly, every relevant comment here either got something wrong about the final frequency or percentage, or corrected the wrong thing. (Parent comment, grandparent, 2 aunts and 1 cousin.)
0.5^10 = 2^-10 = 1/(2^10) = 1/1024 [i.e. exactly 1 person per 1024] ~= 0.00097656 ~= 0.09766 %.
Just to add to add to the mathematical intuition here (please correct me if I'm wrong): if you're thinking of it as a unit line/square/cube then total n-dim area is 1^n, and the portion in the 50% range is (1/2)^n, where n = number of dimensions. So that should simplify to 2^(-n).
Note this works out to 10th dimension as 0.09765% or 0.1 person per 1000.
ETA: As one of the comments below points out, you can also model it as a binomial distribution.
Probability of getting all heads, given p=0.5 is (n!)/((n-k)!k!) / 2^n. Since n=k since we're looking to get all heads at all times, this also simplifies to 2^(-n).