Why Random Triangles Catch the Circle's Center ¼ of the Time

Drop three points on a circle, uniformly and independently, and join them into a triangle. Sometimes the triangle contains the circle’s center. Usually it doesn’t. Run the experiment a million times with a fixed seed, as the video does, and the tally lands on 25.06 %. The true answer is exactly one quarter, and the way you get there is one of the best arguments in elementary probability.

Freeze two points, watch the third

Three moving points is two too many. So pin $P_1$ and $P_2$ wherever they happen to be and slide $P_3$ around the rim. The positions where the triangle catches the center are not scattered. They form a single arc: the arc directly opposite $P_1$ and $P_2$, the one you get by drawing both diameters and looking across the circle.

The reason is a half-circle argument. If $P_3$ lands anywhere else, all three points fit inside one half of the circle, and the diameter bounding that half separates the center from the triangle. If $P_3$ lands on the opposite arc, no half-circle can hold all three points, so the triangle has to straddle the middle. Every miss hides in a half-circle. That is the whole geometric content of the problem.

Averaging the arc

Since $P_3$is uniform, the probability of a catch, given the first two points, is just the opposite arc’s share of the rim. That share moves: it matches the arc between $P_1$ and $P_2$, which after spinning $P_1$ to the top is uniform between 0 and half the circle. The average of a quantity spread evenly between 0 and $\tfrac12$ is $\tfrac14$. Done, honestly, with one small integral hidden in the word “average”.

Two coins in a geometry costume

A quarter is a suspiciously round answer for an average over infinitely many configurations. When that happens, you usually computed a simple thing the hard way. Here is the re-description that exposes it. To pick a uniform point, you can first draw a random diameter, then flip a fair coin for which end you keep. Do that for $P_1$ and $P_2$, and place $P_3$ before touching the coins.

Freeze the frame. Two diameters cut the rim into four arcs, and $P_3$sits in one of them. The two coins produce four equally likely triangles, and the winning condition from Fig. 1 says a triangle catches the center exactly when the kept endpoints sit opposite $P_3$’s arc. Flip either coin and the winning arc jumps to a different one of the four. So the four outcomes claim the four arcs, one each, and exactly one outcome wins. In every configuration:

No averaging, because there is nothing left to average: the conditional answer is the same constant everywhere. The 4 was never geometric. It is two fair coins.

The same quarter counts acute triangles

There is a bonus hiding in the half-circle picture. An inscribed triangle contains the center exactly when every arc between neighboring corners stays under half a turn, and by the inscribed angle theorem each corner sees the opposite arc at half its width. Arcs under a half turn mean angles under 90°. So “contains the center” and “acute” are the same property in different clothes: a random inscribed triangle is acute with probability $\tfrac14$, right-angled with probability zero (that boundary case is Thales’ theorem), and obtuse the remaining three quarters of the time. Most random triangles are lazy, sprawled things.

One dimension up

The coin argument does not know what dimension it lives in. Four random points on a sphere, one tetrahedron, same question: park the fourth point, draw three random diameters, flip three coins. Eight outcomes, one winner, so the answer is $\tfrac18$ — a result that once sat at the hard end of a famous competition paper. On the circle the pattern continues for $n$ points:

Feed it $n=3$ and the quarter comes back. Feed it $n=4$ and the whole thing collapses to a coin flip. Finding the four coins inside that one-half is a good exercise, and the video ends by leaving it to you. If you enjoy chance colliding with geometry, the widest-street view of support vector machines scratches a related itch: there too, almost all of the configuration turns out not to matter.