As it turns out this problem is extremely difficult...
After lots of googling I came across a very old journal article.
The Annals of Mathematics, Vol. 2, No. 6 (Dec., 1886), pp. 133-143
A solution is given there (probability of not being in same hemisphere),
"Let three points be taken at random on the surface of a sphere; they will all lie in the same hemisphere, and will form the vertices of a spherical triangle. If right lines be drawn from these points to the centre of the sphere, and ex- tended to the surface, the points of intersection will be the vertices of a triangle opposite the first one. If the fourth random point is not in the same hemisphere as the first three points, it will be on the surface of the opposite triangle. Now the average area of this triangle is 1/2pi*r^2, or one-eighth the surface of the sphere. Hence the required probability is 1/8."
Anyone have a Phd in mathematics?
Im still unsure if the solution found by Olly is actually correct. It gives the right probability..... but..... I just dont buy it.
