Average area of choosing three points on a surface? Assume I choose three random points on the surface of a sphere. What is the average area? (Each point is independently chosen relative to a uniform distribution on the sphere) Also, what would be the average area if I choose three points on other types of surfaces? Such as 2-dimensional square/circle
 or 1-dimensional line. 
 A: An other way to see the result is by a symmetry argument.
For any given spherical triangle, you may toggle each point with their diametrical opposite to obtain a total of $2^3 = 8$ triangles. The key property to see here is that the union of all these triangles is equal to the sphere, and that the intersection of two of these triangles has an area equal to $0$ (it is at most a common side). So they basically form a partition of the sphere (area-wise, at least).
This means that the expected area of a random spherical triangle has to be an eighth that of the total sphere, ie $\pi R^2/2$.
Edit:
To clarify the last step, since these $8$ triangles have equal probability, their average area is the sum of the areas ($4\pi R^2$) divided by $8$. Then you need to see that each triangle is part of such a set of $8$ triangles, with no set more probable than any other.
The way I see it: choosing a triangle at random is the same as choosing a set of $8$ triangles (ie, choosing three great circles at random, which relates to the angle solution), and then choosing one of the $8$ triangles uniformly at random.
