what is the probability that n points drawn randomly on the circumference of a circle is in a semicircle?

Question

The answer seems to be n/[2^(n-1)]. Why?

Answer 1

This is how I would tackle the problem.

For n points, choose any given point and evaluate the probability that the other n-1 lie within a semicircle going clockwise. This probability is 1/2^(n-1).
Given that there are n points to start with the overall probability is n/2^n-1. This may seem like an abuse of taking the sum of probabilities, but in this case only zero or one of the events may be true, which eliminates the problem of joint probabilities.

Answer 2

I sometimes tackle old Putnam exams for fun, and I'm pretty sure I ran across that one. So one of the sites with old Putnams should help you out.

It's tricky. For example, you can't just say that if n-1 points are known to be in a semicircle, the probability than an nth point joins them there is 1/2. It's actually higher than 1 - (greatest distance between any two of the prior n-1 points).

Maybe it's easier to compute the whole function for each n that is the probability of n points making an arc of angle theta given that the prior n-1 made an arc of angle phi, as well as the outright probability of lying in an arc of angle theta.

Answer 3

I'm not going to spend time working out this problem -- but let me tell you how I would do it.

First -- note that this problem is equivalent to picking n points in the interval [0,1] and asking if the difference between the largest number picked and the smallest number picked is less than 1/2.

Second -- I would prove that if n=2 then it is true.

Third -- I would use an induction proof to prove it in general. The assumption is that if this is true for (n-1) points -- then I need to show that it is true for n points.

Good luck.