All Points Within an Arc
$N$ points are chosen uniformly at random on the circumference of the unit circle. What is the probability that all $N$ points lie within some arc of angular measure $x$ radians, where $0 \leq x \leq \pi$?
Derive the general formula, then evaluate it for the specific case $N = 4$ and $x = \frac{\pi}{2}$.
Hints
- Think about which point acts as the 'anchor' for the containing arc -- by symmetry, each of the $N$ points is equally likely to be leftmost.
- Once you fix the anchor, the remaining $N-1$ points are independent; each lands in an arc of length $x$ with probability $\frac{x}{2\pi}$.
- Multiply the per-anchor probability by $N$ (the number of anchor choices). The mutual exclusivity of anchor events holds only when $x \leq \pi$, which is why the problem enforces that constraint.
Worked Solution
How to Think About It: The tricky part of this problem is that the arc does not have a fixed position -- you are asking whether there exists *some* arc of length $x$ that contains all $N$ points. That sounds hard, but there is a slick trick: the worst-case arc is always the one anchored at one of the $N$ points. So you can condition on the leftmost (or any) point and ask whether the remaining $N-1$ points fall within $x$ radians clockwise of it. This turns a geometric existential problem into a simple independence calculation.
Quick gut check: with $N = 4$ and $x = \pi/2$ (a quarter-circle), each additional point has a $\frac{1}{4}$ chance of landing in that quarter. Three independent points each at $\frac{1}{4}$, scaled by 4 anchoring choices, gives $4 \times (\frac{1}{4})^3 = \frac{1}{16}$. That is small but not tiny -- four points all cramming into a quarter of the circle is a genuinely rare event.
Quick Estimate: For $N = 4$, $x = \pi/2$: each of the 3 non-anchor points must fall within a $\frac{\pi/2}{2\pi} = \frac{1}{4}$ fraction of the circle. Probability $\approx (\frac{1}{4})^3 = \frac{1}{64}$, times 4 (for which point acts as anchor) $= \frac{4}{64} = \frac{1}{16} \approx 6.25\%$. Sanity check: for $x = 2\pi$ (whole circle) this formula gives $N \cdot 1^{N-1} = N$, which is wrong -- meaning the formula only holds for $x \leq \pi$ (for larger arcs, overlapping cases need inclusion-exclusion). Good, consistent with the constraint.
Approach: Anchor the configuration at one of the $N$ points via a symmetry argument, then use independence of the uniform distribution.
Formal Solution:
Label the $N$ points $\theta_1, \ldots, \theta_N$ drawn i.i.d. from $\text{Uniform}[0, 2\pi)$. We want:
$P(\text{all points lie in some arc of length } x)$
Step 1: Reduction via the minimum. Define $\theta_{(1)} \leq \theta_{(2)} \leq \cdots \leq \theta_{(N)}$ as the order statistics. All $N$ points fit in an arc of length $x$ if and only if the smallest arc containing all points has angular span $\leq x$. The smallest such arc, when we "anchor" at each point in turn, has span equal to:
$\Delta = \min_{k} \left[ (\theta_{(k+1)} - \theta_{(k)}) \mod 2\pi + \cdots \right]$
More directly: the points all fit in an arc of length $x$ if and only if there exists some rotation such that all $N$ points fall in $[0, x]$. Equivalently, at least one of the $N$ "gaps" (arcs between consecutive points going all the way around) has length $\geq 2\pi - x$.
Step 2: Symmetry and anchoring. By symmetry, each of the $N$ points is equally likely to be the "leftmost" anchor of the tightest containing arc. Anchor at point $\theta_{(1)} = 0$ (without loss of generality via rotation invariance). Then all points lie in $[0, x]$ if and only if the remaining $N - 1$ points each independently fall in $[0, x]$.
Since each point is $\text{Uniform}[0, 2\pi)$, the probability that one point falls in $[0, x]$ is $\frac{x}{2\pi}$.
Conditioning on point $i$ being the minimum (probability $\frac{1}{N}$ by symmetry across all $N$ points), the remaining $N - 1$ points must all fall in the arc $[\theta_i, \theta_i + x]$, each with probability $\frac{x}{2\pi}$ independently.
Union over all $N$ choices of anchor (these events are mutually exclusive when $x \leq \pi$, since two distinct points cannot both be leftmost anchors of non-overlapping arcs of length $x$):
$P = N \cdot \left(\frac{x}{2\pi}\right)^{N-1}$
This holds for $0 \leq x \leq \pi$ -- when $x > \pi$, different anchor choices produce overlapping containment regions, so inclusion-exclusion is needed.
Step 3: Evaluate for $N = 4$, $x = \frac{\pi}{2}$.
$P = 4 \cdot \left(\frac{\pi/2}{2\pi}\right)^{3} = 4 \cdot \left(\frac{1}{4}\right)^{3} = 4 \cdot \frac{1}{64} = \frac{4}{64}$
Answer:
$\boxed{P = N \left(\frac{x}{2\pi}\right)^{N-1}}$
For $N = 4$, $x = \frac{\pi}{2}$: $P = \dfrac{1}{16}$.
Intuition
The core trick here is turning an existential statement ('there exists some arc containing all points') into a computable probability using symmetry. The key observation is that the optimal containing arc is always anchored at one of the $N$ points -- so you can enumerate exactly $N$ disjoint cases, compute each probability by independence, and add them up. This anchor-and-condition technique appears constantly in geometric probability: whenever you need to handle a 'best-case rotation or shift,' look for the extremal point that naturally pins the configuration down.
The constraint $x \leq \pi$ is subtle but important. When $x > \pi$, two different anchor points can produce overlapping containment arcs, so the $N$ cases are no longer mutually exclusive and the simple $N \cdot (\cdot)^{N-1}$ formula overcounts. In practice, this kind of circular-geometry problem -- where boundary effects and overlapping regions create hidden case splits -- shows up in clustering analysis on directional data, coverage problems in communications (how many receivers cover a transmitter?), and round-robin scheduling. Whenever you see 'uniform on a circle + some coverage condition,' check whether the arc length is above or below $\pi$ before applying your formula.