Points on a Sphere Part 3

[jay_shark]

I posted this problem a while back but I found it very interesting that I wanted to spend a bit more time on it .

The problem is as follows :

What is the probability that n randomly chosen points on a sphere will all lie on a single hemisphere ? Although there are several solutions to this problem , I have yet to encounter a combinatorial argument . In other words , show that the probability in question is (n^2-n+2)/2^n .

Before we solve this problem , it may be a good idea to solve the analogous case in R^2 ; that is , what is the probability that n points on a circle lie on a semicircle ?

It turns out that the answer in R^2 is n/2^(n-1) . There is a straightforward combinatorial argument to this question which goes as follows :

There are 2^n pairs of antipodes . There are also 2n semi-circles that have the property that all the points lies on that side . Therefore , the probability is simply 2n/2^n or n/2^(n-1) .

In R^3 , we know that there are 2^n pairs of antipodes , but it isn't so obvious that there are n^2-n+2 ways of having n points on a hemisphere .

I'll let others think about it for a bit and then I'll post my answer . Try to be as thorough as possible with your solution .

Good luck !!