Average distance of two points on circle vs sphere

The average distance of two random points on a circle of radius $1$ is $4/\pi \approx 1.27$ (you can find this online easily).

The average distance of two points on a sphere is $4/3 \approx 1.33$.

The two problems can be stated in terms of the average distance from a particular point on the circle or sphere equivalently.

I am trying to find the error in the following line of reasoning that gives a wrong result in the second question:

Say we are looking for the average distance to the north pole. Condition on a particular ring for the second point at a given longitude $\phi=\phi_1$. Then the average distance conditioned on the point being on that longitude is $4/\pi$ due to the first problem. Therefore $E[distance\ on\ sphere]=E[E[distance\ on\ sphere|\phi]]=E[4/\pi]=4/\pi$.

Can you help me identify the fallacy?

Solution 1:

What is intuitively wrong with your north pole analogy is that points nearer the equator of the sphere are more likely than your meridian ring argument would suggest.

You would be conditioning on the event that longitude $\phi=\phi_1$ of probability $0$, and that can lead to Borel–Kolmogorov paradoxes and worse: your conditional density is essentially $\frac00$.

Your ring argument led you to suppose latitude $\theta$ is uniformly distributed on $[-\frac\pi2,\frac\pi2]$ , when on a sphere it is in fact $\sin \theta$ which is uniformly distributed. [More usually latitude is described as $\phi$ and longitude as $\lambda$ but that does not matter.]

This would not affect the answer if you were measuring distances on the circumference of the circle and surface of the sphere (the errors would cancel out by symmetry and you would get $\frac{\pi}{2}$ in both cases), but it does matter when measuring distances through the interior of the circle and sphere.