What is the probability that the signs of $u \cdot g$ and $v \cdot g$ are the same?
My attempt:
Denote $\theta(x, y)$ as an angle between $a, b \in \mathbb R^n$ (between $0$ and $pi$).
Then, $$P(sign(u \cdot g) = sign(v \cdot g)) \\= P(sign(\cos(\theta(u, g))) = sign(\cos(\theta(v, g)))) \\= P(\theta(u, g) \in [0, \frac{\pi}{2}], \theta(v, g) \in [0, \frac{\pi}{2}]) + P(\theta(u, g) \in (\frac{\pi}{2}, \pi], \theta(v, g) \in (\frac{\pi}{2}, \pi])$$
$|\theta(u, g) - \theta(v, g)| \leq \theta(u, v) \leq \theta(u, g) + \theta(v, g)$, so $$P(\theta(u, g) \in [0, \frac{\pi}{2}], \theta(v, g) \in [0, \frac{\pi}{2}]) = E[1_{[\theta(u, g) \in [0, \frac{\pi}{2}], \theta(v, g) \in [0, \frac{\pi}{2}]]}] \\= E[E[1_{[\theta(u, g) \in [0, \frac{\pi}{2}], \theta(v, g) \in [0, \frac{\pi}{2}]]} | \theta(u, g)]]$$
but not sure how to proceed.