Probability a Random Triangle Contains the Center of a Circle
Pick three points independently and uniformly at random on the unit circle. They form a triangle $T$.
- What is the probability that $T$ contains the center of the circle?
- Give a clean argument using circular orderings and complementary events.
- Now suppose the three points are drawn i.i.d. uniformly on the circumference of an ellipse instead. Does the answer change? Why or why not?
Hints
- Think about when the triangle does NOT contain the center. What simple geometric condition characterizes that?
- The triangle fails to contain the center exactly when all three points fit inside some semicircle. Use the complementary probability $1 - P(\text{all in a semicircle})$.
- Anchor a semicircle at each of the three points. Show that $P(\text{both other points in that semicircle}) = 1/4$, and that these three events are (almost surely) mutually exclusive.
Worked Solution
How to Think About It: The triangle formed by three points on a circle contains the center if and only if the center is "inside" the triangle -- meaning the three arcs between consecutive points each subtend less than $\pi$ (i.e., each arc is less than a semicircle). Equivalently, the triangle contains the center if and only if no semicircle contains all three points. This complementary framing is the key insight: it is much easier to compute the probability that all three points lie in some semicircle, then subtract from 1.
Before doing any math, here is a quick sanity check: by symmetry, the triangle is "small" (not containing the center) more often than not, so we expect the probability to be less than $1/2$. For three random points the answer turns out to be exactly $1/4$.
Quick Estimate: Fix one point at angle $0$ (by rotational symmetry this is free). The other two points have angles $\theta_1, \theta_2$ uniform on $[0, 2\pi)$. The triangle contains the center roughly when the points are "spread out" around the circle. If you imagine throwing two more darts at random, about $3/4$ of the time they cluster enough that some semicircle catches all three. So $P(\text{contains center}) \approx 1/4$. That is the exact answer.
Approach: We use the complementary event: $P(\text{contains center}) = 1 - P(\text{all three points lie in some semicircle})$.
Formal Solution:
Label the three points $P_1, P_2, P_3$ with angles $\theta_1, \theta_2, \theta_3$ drawn i.i.d. uniformly on $[0, 2\pi)$.
*Step 1: Complementary event.* The triangle contains the center if and only if no semicircle (arc of length $\pi$) contains all three points. So:
$$P(\text{contains center}) = 1 - P(\text{some semicircle contains all three points}).$$
*Step 2: Compute $P(\text{all in some semicircle})$.* For each point $P_i$, define the event $A_i$ = "all three points lie in the semicircle starting at $P_i$ and going clockwise for $\pi$." By symmetry, $P(A_i) = (1/2)^2 = 1/4$ for each $i$, because the other two points each independently must land in a specific half of the circle.
The events $A_1, A_2, A_3$ are not disjoint, but the event "all three in some semicircle" equals $A_1 \cup A_2 \cup A_3$. However, notice that if all three points lie in a semicircle, then the point that is the "most clockwise" boundary of the smallest arc containing all three defines a unique $A_i$ (the semicircle anchored at that point covers the other two). More precisely, at most one $A_i$ can hold unless two points coincide (probability 0). So the events are almost surely mutually exclusive:
$$P(A_1 \cup A_2 \cup A_3) = P(A_1) + P(A_2) + P(A_3) = 3 \times \frac{1}{4} = \frac{3}{4}.$$
*Step 3: Final answer.*
$$P(\text{contains center}) = 1 - \frac{3}{4} = \frac{1}{4}.$$
*Part 3: Extension to an ellipse.* The answer does not change: $P = 1/4$. Resist the quick affine argument. An ellipse is an affine image of a circle, and affine maps do preserve "this point is inside this triangle," but they distort arc length, so they do NOT carry the uniform law on the circle to the uniform law on the ellipse. What the circle proof actually used is central symmetry, and every ellipse has it: for a boundary point $P$ the antipode $-P$ is also on the ellipse, so the line through the center cuts the perimeter into two arcs of equal length. Each of the other two points therefore lands in an anchored half with probability $1/2$, each anchored event again has probability $1/4$, and the same disjointness gives $1 - 3/4 = 1/4$. Central symmetry is what matters, not roundness: on the boundary of an equilateral triangle, where central symmetry fails, simulation gives about $0.245$ instead.
Answer: The probability that the triangle contains the center is $\boxed{1/4}$. The same holds for an ellipse -- not by affine invariance, but because an ellipse is centrally symmetric, which is all the anchoring argument needs.
Intuition
The core trick is flipping the question: instead of asking when the triangle "traps" the center, ask when it fails to -- which happens exactly when all three points huddle into a semicircle. This complementary viewpoint turns a tricky geometric containment problem into a clean calculation about uniform order statistics on a circle. The anchoring argument (fix one point and ask where the others land) is a workhorse technique in geometric probability that shows up constantly in stochastic geometry, coverage problems, and even in quant settings like analyzing whether a set of hedging instruments "spans" a risk space.
The ellipse extension is a lesson in spotting which hypothesis is load-bearing. The tempting move is "an ellipse is an affine image of a circle, so we are done" -- but affine maps, while they do preserve containment, stretch arc length, so uniform-on-the-circle does not map to uniform-on-the-ellipse. What the proof really needs is central symmetry: every boundary point's antipode splits the perimeter in half. Any centrally symmetric curve gives $1/4$; drop that symmetry and the answer moves. Knowing which assumption is doing the work is what lets you reuse a result instead of re-deriving it.