Given a collection of points on the sphere, we can consider their spherical convex hull: add all points on the shortest path between two points in the set, repeat until the resulting set does not expand. (For specificity, say that the shortest path between two diametrically opposite points is the whole sphere, but it doesn't matter for our purposes.)
I am interested in the distribution of outcomes from taking the convex hull of $n$ random points on the sphere.
When $n=3$, the result is a triangle with probability 1.
When $n=4$, the result is a triangle with probability $\frac12$, a quadrilateral with probability $\frac38$, and the whole sphere with probability $\frac18$.
When $n=5$, the result is a triangle with probability $\frac5{16}$, a quadrilateral with probability $\frac5{16}$, a pentagon with probability $\frac1{16}$, and the whole sphere with probability $\frac5{16}$.
The above results can be derived given the moments of a random triangle's area and some cute counting arguments; see this question and answer for an instance of the sort of reasoning that produces these conclusions. Happy to elaborate on the proofs of these results in the comments if desired.
However, the cute geometric arguments for $n\le 5$ start falling apart at $n=6$. The probability that the convex hull of $6$ points will be a triangle is $\frac{15}{32}-\frac{15\ln(2)}{4\pi^2}\approx0.2054$; again, see the linked thread above for an explanation of why this is (basically, the issue is that the third moment of the area of a random spherical triangle gets messy in a way that earlier moments do not).
On the other hand, the odds that the convex hull of $6$ random points is the entire sphere is exactly $\frac12$; see this answer for a citation and general formula.
What are the chances that the convex hull is a quadrilateral, pentagon, or hexagon? Given the above results, it seems possible that one of these three probabilities may yet be rational (and thus possibly amenable to a natural geometric argument), even though of course not all three can be.
Empirically, in a Monte Carlo simulation with $100000$ trials, there were $20369$ triangles, $22127$ quadrilaterals, $6880$ pentagons, $591$ hexagons, and $50033$ spheres.
