Five random points on a sphere

Solution. We apply Wendel’s general formula, already used with $n = 4$ in Chapter 27. For $n$ iid uniform points on the unit sphere $S^{d-1} \subset \mathbb{R}^d$, the probability that their convex hull contains the origin is $$\mathbb{P}_{n, d} \;=\; 1 - \frac{1}{2^{n-1}} \sum_{k = 0}^{d - 1} \binom{n - 1}{k}.$$ For $(n, d) = (5, 3)$, $$\mathbb{P}_{5, 3} = 1 - \frac{1}{2^4} \Big( \binom{4}{0} + \binom{4}{1} + \binom{4}{2} \Big) = 1 - \frac{1 + 4 + 6}{16} = 1 - \frac{11}{16}.$$ Therefore $$\boxed{\mathbb{P}(\text{5 points' hull contains centre of sphere}) = \frac{5}{16}.} \qedhere$$

Sketch of Wendel’s argument. By antipodal symmetry, the configuration $(\varepsilon_1 P_1, \ldots, \varepsilon_n P_n)$ for $\varepsilon \in \{\pm 1\}^n$ has the same joint distribution. Among the $2^n$ such sign patterns, the hull contains the origin iff no closed hemisphere contains all chosen points. A counting lemma from linear algebra (Wendel 1962) shows that for $n$ points in general position on $S^{d-1}$, exactly $\sum_{k = 0}^{d - 1} \binom{n-1}{k}$ of the $2^{n-1}$ (unordered) sign patterns place all points in some closed hemisphere. The ratio gives the “not contained” probability.

The pattern over small $(n, d)$ is:

For fixed dimension $d$, $\mathbb{P}_{n, d} \to 1$ as $n \to \infty$: many points are (almost) always in general position so the hull fills the ball. ■

import numpy as np
from scipy.optimize import linprog
# Origin is in conv{P_1,...,P_5} iff there exist weights w_i >= 0 summing
# to 1 with sum w_i * P_i = 0. Use LP feasibility directly.
N, count = 10**4, 0
for _ in range(N):
    pts = np.random.randn(3, 5); pts /= np.linalg.norm(pts, axis=0)
    A_eq = np.vstack([pts, np.ones((1, 5))])
    b_eq = np.array([0., 0., 0., 1.])
    res = linprog(np.zeros(5), A_eq=A_eq, b_eq=b_eq,
                  bounds=[(0, None)] * 5, method='highs')
    if res.success:
        count += 1
print(f"sim: {count/N:.5f}   exact: {5/16:.5f}")
# sim: 0.31210   exact: 0.31250