I have three vectors on the unit sphere, with angles between them x, y, and z. I'm trying to find, in terms of these angles, the probability that a random plane through the origin divides any of these points.
At first, I thought that this would be related to the area of the corresponding spherical triangle defined by the vectors. However it's not as simple as the 2D case, since given a single vector on the sphere's surface, there are infinitely many planes which pass through it.
Note that this is analogous to the 2D case, where I have two unit vectors with angle theta between them. The probability that a random line through the origin divides them is theta / pi. I'm hoping to get something similar for the 3 dimensional case.
We are interested in vectors for which the normal plane to that vector passing through the origin separates all three points on the same side of the plane. This will be twice the probability that all three points lie on the same side of this plane as our chosen vector.
The probability that our random vector lies in all three hemispheres is just the area of their intersection divided by $4\pi$, so we just need to figure out what the size of this intersection of hemispheres is.
But the intersection of three hemispheres is just another spherical triangle, whose edges are given by the borders of the hemispheres and whose vertices are given by their intersection. It's easy to see that given two points $p,q$ with angle $\theta$ between them, the angle at which the corresponding hemispheres are tilted to each other will be $\pi-\theta$.
So we want the area of a spherical triangle with angles $\pi-x,\pi-y,\pi-z$, which is just $2\pi-(x+y+z)$.