Spherical probability problem

Question

Planet Zog is a sphere with centre $O$. A number $N$ of spaceships land at random on its surface, their positions being independent, each uniformly distributed over the surface. A spaceship at $A$ is in direct radio contact with another point $B$ on the surface if $\angle AOB<π/2$. Calculate the probability that every point on the surface of the planet is in direct radio contact with at least one of the $N$ spaceships.

I get $1 - \dfrac{N(N-1)+2}{2^N}$. My method was to consider $N$ great circles which split the sphere up into $N(N-1)+2$ regions, and that the probability that one of these regions is not in contact with any spaceship is $1/2^N$. Is that right? It seems too simple. I would be interested in seeing other solutions in any case.

Answer 1

Your solution can be extended to a complete solution, but there's some gaps to be filled in, which I don't know if you had in mind when writing it.

Let's suppose that the spaceships choose where to land in two steps:

1. First, a spaceship chooses a uniformly random axis through the center of the planet, which determines two opposite points on the planet's surface.
2. Second, the spaceship chooses one of the two points independently at random.

We can then prove that no matter which axes all the spaceships choose in step 1, the probability of making their step-2 choices in a way that covers the sphere with radio broadcasts is $1 - \frac{N(N-1)+2}{2^N}$. (We exclude a measure-$0$ subset of degenerate step-1 outcomes, such as choosing three axes which are all collinear.)

Each axis that a spaceship chooses cuts the planet's surface into two hemispheres: the hemisphere in radio contact with one possible landing site, and the hemisphere in radio contact with the opposite landing site. These are separated by the great circle of points at an angle of $\frac\pi2$ from both of these landing sites.

The great circles for the $N$ spaceships separate the plane into $N(N-1)+2$ regions. (We can prove this by Euler's formula, if we assume that all $N(N-1)$ intersection points are distinct, so that there are $2N(N-1)$ segments joining them and therefore 2+E-V = N(N-1)+2$ faces.)

For each region, each spaceship has a $\frac12$ chance of landing in the right hemisphere to cover that region by radio broadcasts. These are all independent, so the probability is $\frac1{2^N}$ that no spaceship covers the region. Moreover, if one region is left uncovered, the spaceships nearly covering its borders cover all the remaining regions - so these probabilities are disjoint, and we can add them to get a total $\frac{N(N-1)+2}{2^N}$ probability that any region is uncovered.