In the $n$-dimensional Euclidean space, let $H$ be a random hyperplane selected in way similar to the so called "random radial point" (used to propose a solution for the Bertrand paradox): we select a point $\mathbf{z}$ uniformly at random on the unit $(n-1)$-sphere and we select uniformly at random a hyperplane $H=\{\mathbf{x}\in\mathbb{R}^n:\langle\mathbf{x},\mathbf{z}\rangle=b\}$, for a uniformly random $b\in[-1,1]$.
Question: What is the probability
$\Pr_H$
that $H$ separates point $\mathbf{x}:=(1, 0, 0, \ldots, 0)$ from point $\mathbf{y}=(-1, 0, 0, \ldots, 0)$?
I conjecture that $\Pr_H$ approaches $0$ as $n$ increases, but I do not know how to prove it. I am especially interested in how $\Pr_H$ varies as a function of $n$.
Best Answer
Assume instead that $H$ is the hyperplane $x_1=b$, where $b$ is uniformly distributed on $[0,1]$ (as Misha Lavrov has proposed), and that $x=(x_1,\ldots,x_n)$ is uniformly distributed on the unit sphere $S^{n-1}$. We have a success iff $|x_1|>b$. This means that we have to begin with calculating the portion of the area of $S^{n-1}$ corresponding to $|x_1|\in[b,1]$, then integrate over $b\in[0,1]$.
In fact you don't need to compute this area. It is sufficient to have a formula for the density $f_{X_1}(x_1)$ $(-1\leq x_1\leq 1)$ of the first coordinate under the assumed distribution. This formula is around, but I have not found a quotable source.