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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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 Solution | Comment 4 of 21 |
Start by considering two points on the circle: one at zero degrees from some reference point and the other at 30 degrees. A third point will work (i.e. share some semicircle with the first two) if it falls between 0 and 180 degrees or between 210 and 360 degrees. There is only a 30 degree area in which the third point wont work. In other words, for n=3 points, with one point at zero and a second point at Theta, the odds that the third point works is (2pi-Theta)/2pi, or 1  Theta/2pi. Note that, for this equation to be valid, the first point must be selected such that the angle Theta to the second point is less than or equal to pi.Since Theta can have any value between zero and pi, the probability of the third point working (for all values of Theta) is

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

If we have a condition that works for n=3, again there will be an angle Theta that is less than or equal to pi that the three points fall within, and the odds of finding a fourth point that works is again 1  Theta/2pi, or 1/2 for all values of Theta. Since a working set of 3 points was a prerequisite to finding a fourth point that works, the probability for n=4 is .5^2. From this logic, one sees that the probability for n points to work is

P = .5^(n-2)
 Posted by Bryan on 2003-05-19 10:14:41

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