Bertrand's Chord Paradox

"Draw a random chord on a circle. What's the probability it's longer than the side of an inscribed equilateral triangle?" This sounds like a well-defined question with one answer. It isn't. Depending on how you generate the "random" chord, you can rigorously derive three different answers — $\tfrac13$, $\tfrac12$, and $\tfrac14$ — and each derivation is completely correct given its own procedure. The paradox isn't a trick; it's a lesson that "choose a random chord" doesn't specify a unique probability distribution until you say how you're choosing.

Method 1: random endpoints — answer $\tfrac13$

Fix one endpoint of the chord anywhere on the circle, then pick the second endpoint uniformly at random around the circumference. An inscribed equilateral triangle with one vertex at the fixed point divides the rest of the circle into an arc of $240°$ that's "too close" and gives a short chord, and the remaining structure works out so that the chord is longer than the triangle's side exactly when the second point lands in the middle third of the circle's arc opposite the first point. That arc is $\tfrac13$ of the full circumference, so $P = \tfrac13$.

Method 2: random midpoint — answer $\tfrac14$

Pick a point uniformly inside the disk and let it be the chord's midpoint (the chord is then the unique one perpendicular to the radius through that point). The chord is longer than the triangle's side exactly when the midpoint falls inside the circle of radius $\tfrac{R}{2}$ concentric with the original — a smaller disk with a quarter of the area. Since the midpoint is uniform over the full disk's area, $P = \dfrac{\pi (R/2)^2}{\pi R^2} = \dfrac14$.

Method 3: random radial distance — answer $\tfrac12$

Pick a random direction (a radius of the circle), then pick a point on that radius uniformly between the center and the edge, and draw the chord through that point perpendicular to the radius. The chord is longer than the triangle's side exactly when the point falls in the inner half of the radius (distance less than $R/2$ from center), which by construction has probability $P = \tfrac12$, since the point was chosen uniformly along the radius's length rather than uniformly by area.

Worked example: why the numbers really do differ

Take $R=1$. A chord is longer than the inscribed triangle's side ($\sqrt3$) exactly when its perpendicular distance from the center is less than $\tfrac12$. Method 2 places points uniformly by area, so the region "distance $<\tfrac12$" is a disk of area $\pi(0.5)^2=0.785$ against total area $\pi(1)^2=3.14$, ratio $0.25$. Method 3 places points uniformly by length along the radius, so "distance $<\tfrac12$" is just half of the unit-length radius, ratio $0.5$ directly — same geometric threshold, different sampling measure, different answer. The chord's geometric condition never changed; only the probability measure over which chords are "equally likely" did.

What this means in practice

The practical lesson shows up constantly in quant work: "assume X is random" is not a complete model until you specify the distribution and the process that generates it — a "random" spread, a "random" fill price, or a "random" default time can be modeled several defensible ways that give materially different probabilities, and interviewers use Bertrand's paradox precisely to test whether you reflexively ask "random according to what procedure?" before computing anything.