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 Don't Be a Square (Posted on 2003-05-19)
Given n points drawn randomly on the circumference of a circle, what is the probability they will all be within any common semicircle?

 See The Solution Submitted by DJ Rating: 4.4667 (15 votes)

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 re: Solution | Comment 6 of 21 |
(In reply to Solution by Bryan)

It occurred to me after I posted that, since Theta is less than or equal to pi, the integral should be from 0 to pi as well, not from 0 to 2pi. Considering n=3, if the first two points are nearly 180 degrees apart, the third point needs to fall between them, the odds of which are 1/2, but if the first two points are nearly on top of each other, the third point can be almost anywhere, the odds of which approach 1/1. The average of these is 3/4 and, not coincidentally,

P = 1/pi * integral[1 - Theta/2pi] from 0 to pi
P = (pi - .75pi)/pi
P = .75

and the overall probability is

P = .75^(n-2)
 Posted by Bryan on 2003-05-19 12:52:22

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