n random points on a circle, all in a semicircle

Solution. Fix a labelling of the points. Call the event $S$ “there exists some semicircle containing $P_1, \ldots, P_n$.” For each $i$, let $E_i$ denote the event “the semicircle whose starting endpoint is at $P_i$ (measured clockwise) contains all the other $P_j$.”

Claim. The events $E_1, \ldots, E_n$ are mutually exclusive and $S = E_1 \cup \cdots \cup E_n$.

Exclusive: if $S$ occurs, there is an arc of length $\pi$ containing all points. The unique minimal such arc has both endpoints equal to points of the configuration; the starting-endpoint $P_i$ (in cyclic order) is determined, and only that $E_i$ occurs.

Covering: if $S$ occurs, some $E_i$ occurs by the same construction.

Probability of each $E_i$. By rotational symmetry, fix $P_i$ at angle $0$. The other $n - 1$ points must all lie in the semicircle from $0$ to $\pi$ (clockwise). Each is independently uniform on the full circle, so the probability it lies in a chosen semicircle is $\tfrac12$. Hence $$\mathbb{P}(E_i) = \left( \tfrac12 \right)^{n-1}.$$

Assemble. By mutual exclusivity, $$\mathbb{P}(S) = \sum_{i=1}^{n} \mathbb{P}(E_i) = n \cdot \left( \tfrac12 \right)^{n-1}.$$ So $$\boxed{\mathbb{P}(n \text{ random points all in a semicircle}) = \frac{n}{2^{n-1}}.} \qedhere$$

import numpy as np
for n in [3, 4, 5]:
    theta = np.sort(np.random.rand(n, 10**6) * 2*np.pi, axis=0)
    gaps = np.vstack([np.diff(theta, axis=0),
                      2*np.pi - theta[-1] + theta[0]])
    p = (gaps.max(axis=0) > np.pi).mean()
    print(f"n={n}: sim={p:.5f}   exact={n / 2**(n-1):.5f}")
# n=3: sim=0.74926   exact=0.75000
# n=4: sim=0.49987   exact=0.50000
# n=5: sim=0.31298   exact=0.31250