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We are given a unit origin-centered sphere $S$ in $\mathbb{R}^3$, and three points $\mathbf{x},\mathbf{y},\mathbf{z}\in S$. Let $\mathbf{h}$ be a point selected uniformly at random from $S$ and let $H$ be a hyperplane that passes through the center $\mathbf{0}$ of $S$ and has $\mathbf{h}$ as normal vector. Let $E_x$ be the event that $H$ separates $\mathbf{x}$ from the other two points $\mathbf{y}$ and $\mathbf{z}$.

Questions: What is the probability of $E_x$? How can we extend this result to express the probability of separating $\mathbf{x}$ from other $n-1$ points when there are $n$ points instead of only three?

Note: I conjecture that the answer to the first question is $\frac12\left(\arccos\left(\mathbf{x}^{\top}\mathbf{y}\right)+\arccos\left(\mathbf{x}^{\top}\mathbf{z}\right)-\arccos\left(\mathbf{y}^{\top}\mathbf{z}\right)\right)$.

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    $\begingroup$ Note that $H$ separates $x$ from $y$ and $z$ if and only if $h^\perp x \cdot h^\perp y<0$ and $h^\perp x\cdot h^\perp z<0$. You didn't say what is the law of $h$, but if you know the marginal probability distributions, then you can compute everything. $\endgroup$
    – M. Dus
    Commented Mar 15, 2021 at 10:17
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    $\begingroup$ Might as well assume that $x$ is the point $(1,0,0)$. There is a half-circle of planes through $0$ and $x$. Denote by $\alpha$ the angle between the planes $oxy$ and $oxz$. The probability is $\alpha/\pi$. $\endgroup$
    – Liviu Nicolaescu
    Commented Mar 15, 2021 at 10:55
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    $\begingroup$ You probably mean $\frac12\left(\arccos\left(\mathbf{x}^{\top}\mathbf{y}\right)+\arccos\left(\mathbf{x}^{\top}\mathbf{z}\right)-\arccos\left(\mathbf{y}^{\top}\mathbf{z}\right)\right)$ $\endgroup$
    – Yaakov Baruch
    Commented Mar 15, 2021 at 11:35
  • $\begingroup$ Thank you @YaakovBaruch, there was a typo. $\endgroup$
    – Penelope Benenati
    Commented Mar 15, 2021 at 12:13
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    $\begingroup$ @PenelopeBenenati The problem is equivalent to the following. Suppose that you have $n-1$ points on a circle. What is the probability that a uniformly chosen random diameter leaves $k$ of them on one side and $n-1-k$ on the other side. The special case $n-1=4$ and the points are the vertices of a square shows that for $k=2$ the probability is $1$ while for other $k$țs the answer is zero. The answer is dependent on the configuration of those $(n-1)$ points. At one extreme they can be clustered and at the other extrem uniformly distributed. $\endgroup$
    – Liviu Nicolaescu
    Commented Mar 15, 2021 at 18:56

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Note that the random vector $G/\|G\|$ is uniformly distributed on the unit sphere in $\mathbb R^n$, where $G$ is a standard Gaussian random vector in $\mathbb R^n$.

So, given an $n$-tuple $(x_1,\dots,x_n)$ of unit vectors in $\mathbb R^n$, the probability that the random hyperplane separates $x_1$ from $x_2,\dots,x_n$ is $$p_n:=2P(X_1>0,X_2<0,\dots,X_n<0),$$ where $X_j:=x_j\cdot G$ for $j=1,\dots,n$, so that the covariance matrix of $(X_1,\dots,X_n)$ is the Gram matrix of $(x_1,\dots,x_n)$.

Schläfli (1858) obtained a differential recurrence relation for probabilities of the form $p_n/2$; see e.g. Plackett and references there. Using this recurrence for $n\le4$, $p_n$ can be expressed as an integral of elementary functions. As pointed out by Ruben. p. 213, "For dimensionality greater than three (spherical tetrahedra, spherical pentahedra, etc.) the areas can no longer be expressed in terms of elementary functions".

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