Expected value of interesting random variable on sphere.

Question

Imagine we have a sphere. We will fix the orthogonal coordinate system $xyz$. First, we paint the lower hemisphere in some color(for lower I mean we place the center of sphere in the origin, and hemisphere with negative $y$-coordinate will be the lower). After that we rotate the sphere (randomly) and again paint the lower hemisphere. What is expected value for the number of such rotations for which sphere will be fully painted. First, I consider simpler case. Just circle. So, there is no information for distribution of random choose of rotation, so it will be $U[0;\pi]$ (uniform). So let $\xi \in U[0;\pi]$ the rotation angle (also, we have a freedom of choosing the clockwise rotation or not, but I couldn’t think of how I can handle it). Now, let $N_{\pi} =\{ min n : \sum_{i=0}^n \xi_i > \pi\}$. We can easily find the expected value of such random variable by recursion formula (I won’t provide it, since it is quite easy, and also we didn’t take into account choice of rotation direction). Even if I could solve the easier case. I don’t know how to apply it to original problem.

Answer 1

For a circle, the probability you need to paint exactly $n$ halves is $\frac{n-2}{2^{n-1}}$ for $n\ge 3$ and the expected number of painted halves is $5$. See the related questions Probability that n points on a circle are in one semicircle and Spinning a disk and painting one half - probability whole disk is painted after $n$ steps?

For a sphere, this was considered by Kevin Brown, where he gave the probability of not succeeding after $n$ halves (i.e. of all the points being in the same hemisphere, as pointed out by Greg Martin) of $\frac{n^2-n+2}{2^n}$. This would imply the probability of needing to paint exactly $n$ halves of $\frac{(n-3)(n-2)}{2^n}$ for $n\ge 4$ and an expected number of painted halves of $7$