What is the probability for 2 points on a sphere to be in the same hemisphere, given the angle between them? The sphere has predefined hemispheres, and 2 points are randomly selected.
If the 2 points are 180 degrees apart, we know they are not in the same hemisphere. if it's only 1 degree it probably is in the same half.
How do I calculate the probability for angle X?
 A: These spherical trigonometry problems can be hard on the brain. But in this case the answer is surprisingly simple. You ask:

Given a partition of the sphere into two hemispheres, what is the
  probability that two random points separated by an angle $X$ lie in the same
  hemisphere?

We can turn this problem inside out, and ask:

Given two points separated by an angle $X$, what is the probability
  that a random partition of the sphere into two hemispheres will result
  in the two points lying in the same hemisphere?

So let us take two points on the equator, with coordinates $A=(0,0)$ (see Null Island) and $(0,X)$, where $X\le 180$. Any point $P$ on the sphere defines a partition into two hemispheres, by considering that point as a pole. We may assume that the longitude $\theta$ of $P$ is in the range $[90^\circ,270^\circ]$, because if not, we can replace $P$ by its antipodal point. And for such a point $P=(\theta,\phi)$ to separate the points $A$ and $B$ into distinct hemispheres requires $90^\circ\le\theta\le X+90^\circ$. The probability of this is clearly $X/180$.
Hence the answer to your question is $1-X/180$.
