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.* Any ellipse is the image of a circle under an affine transformation (a linear map plus a translation). Affine maps preserve the property "a point is inside a triangle" because they map lines to lines and preserve "betweenness." A uniform distribution on the circle maps to a uniform distribution on the ellipse under this transformation. Therefore the probability is unchanged: $P = 1/4$ for the ellipse as well.
Answer: The probability that the triangle contains the center is $\boxed{1/4}$. The same holds for an ellipse, because the problem is invariant under affine transformations.
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 beautiful illustration of affine invariance. Many geometric probability results on circles carry over to ellipses for free because affine maps preserve convexity, incidence, and uniform distributions on boundaries. Recognizing when a result is affine-invariant saves you from re-deriving things from scratch -- a useful instinct in modeling where you often change coordinates or transform distributions.