Stevens’s covering theorem

Solution. The complement of the union of arcs consists of the gaps between consecutive arc-endpoints that are not covered. The whole circle is covered iff every such gap is $\leq 0$, equivalently iff the distances between consecutive arc centres (when sorted cyclically) are all $\leq a$.

For $n=1$, the covering probability is $0$ when $a<1$ and $1$ when $a=1$. This case is handled separately to avoid the ambiguous endpoint power $0^0$.

Suppose now that $n\geq2$. Anchor the cyclic order at the centre of a labelled arc and let $G_1,\ldots,G_n$ be the clockwise gaps between successive centres. Rotating that centre to $0$ leaves $n-1$ independent uniform positions, so the gap vector is uniform on the simplex $G_i\geq0$, $\sum_iG_i=1$.

The circle is covered iff every $G_i\leq a$. For a fixed set of $k$ gaps, the event that each exceeds $a$ translates those coordinates by $a$. The remaining simplex has total length $1-ka$ and dimension $n-1$, so its probability is $(1-ka)_+^{n-1}$, where $(y)_+=\max(y,0)$. Inclusion–exclusion over the events $\{G_i>a\}$ gives Stevens’s formula: $$\mathbb{P}(\text{covered}) = \sum_{k=0}^{n}(-1)^k\binom{n}{k}(1-ka)_+^{n-1}.$$ Terms with $ka\geq1$ vanish, so this is equivalent to the usual sum truncated at $\lfloor1/a\rfloor$. For $(n, a) = (2, \tfrac12)$: $\sum = 1 - 2 \cdot \tfrac12 + 0 = 0$ (two arcs of length $\tfrac12$ exactly fill the circle only on a measure-zero configuration). For $(n, a) = (3, \tfrac12)$: $\sum = 1 - 3 \cdot (\tfrac12)^2 + 0 = \tfrac14$. ■

import numpy as np
# For n arcs of length a, covered iff all gaps between sorted centres are <= a
# (where "wrap-around" gap is c_1 + 1 - c_n).
n, a, N = 3, 0.5, 10**6
centres = np.sort(np.random.rand(n, N), axis=0)
gaps = np.diff(centres, axis=0)
wrap = centres[0] + 1 - centres[-1]
all_gaps = np.vstack([gaps, wrap[None, :]])
covered = (all_gaps <= a).all(axis=0)
print(f"sim: {covered.mean():.5f}   exact (Stevens): 0.25")
# sim: 0.24963   exact: 0.25000