The Chance That Random Points Fit in a Semicircle

Scatter $n$ points independently and uniformly at random on the circumference of a circle. What's the probability that all $n$ points lie within some semicircle (a half-circle arc doesn't have to be fixed in advance — any half-circle that happens to contain everyone counts)? For $n=3$ this is a well-known interview question, and the clean way to solve it generalizes instantly to any $n$.

The pivot trick

Instead of trying to describe "some semicircle contains all the points" directly — which is awkward because the semicircle's position is itself unknown — flip the question around: for each point, ask whether the semicircle starting at that point and going clockwise contains all the other points. If all $n$ points fit in some semicircle, exactly one of them must be the "leading edge" — the point such that going clockwise a half-turn from it sweeps up every other point. This reframes the event in terms of each point taking a turn as the pivot.

For a fixed point acting as the pivot, the other $n-1$ points each independently need to land within the clockwise half-circle starting at the pivot — each has probability $\tfrac12$ of doing so, independently, since the points are uniform and a semicircle is exactly half the circumference. So the chance that a specific point is a valid pivot (all others clockwise-following it within a half turn) is $\left(\tfrac12\right)^{n-1}$.

Why you can just add the probabilities

Here's the key structural fact: for any given random configuration of $n$ points on a circle, at most one point can be a valid pivot — if the whole set already fits in a semicircle, there's a unique "first" point at the counterclockwise edge of that arc, and no other point can also be a valid clockwise-starting pivot for the same fit. Because these $n$ events (point 1 is the pivot, point 2 is the pivot, ..., point $n$ is the pivot) are mutually exclusive, the probability that some point is a valid pivot is just their sum:

$$P(\text{all } n \text{ points fit in a semicircle}) = n \cdot \left(\frac12\right)^{n-1} = \frac{n}{2^{n-1}} .$$

In plain English: there are $n$ candidates for "the point that starts the arc," each with an independent $\left(\tfrac12\right)^{n-1}$ chance of actually working, and since only one can ever work at a time, you just add up the $n$ chances.

Worked example: $n = 3$ points

$P = \dfrac{3}{2^{3-1}} = \dfrac{3}{4} = 0.75$. So three random points on a circle land within some semicircle 75% of the time — higher than most people's gut guess, because "some semicircle, wherever it happens to be" is a much easier target than "a fixed semicircle chosen in advance."

Worked example: $n = 5$ points

$P = \dfrac{5}{2^{4}} = \dfrac{5}{16} = 0.3125$. As $n$ grows, cramming every point into a shrinking effective target gets harder fast: the probability decays exponentially in $n$ (the $2^{n-1}$ in the denominator dominates the linear $n$ in the numerator), so by $n=10$ the chance is already down to $10/512 \approx 2\%$.

What this means in practice

The pivot trick — reframe a "does some arrangement exist" probability as "exactly one of n candidates succeeds," then sum mutually exclusive probabilities — is a reusable move for circular or cyclic interview questions (seating arrangements, coverage problems, "is there a majority direction" questions), and it avoids the much messier direct approach of integrating over the semicircle's unknown starting angle.