Published 13 September 2026

Bertrand paradox

Choose a random chord of a circle. What is the probability that it is longer than the side of an inscribed equilateral triangle? The phrase “random chord” has not specified a distribution. Three natural experiments give one third, one half, and one quarter.

The circle here has radius one. A chord whose midpoint is distance r from the center has length 2√(1 − r²). The equilateral triangle’s side is √3, so the chord is longer precisely when r < 1/2. The geometry is fixed; the distribution of r changes between experiments.

Choose a sampling rule

Use the equilateral-side button to restore the exact √3 threshold after editing it. The circle shows up to forty background chords and one inspected chord. The table counts all sampled chords for each rule. A small displayed collection is not silently used as the denominator for a larger sample.

The thick chord’s midpoint and perpendicular center distance let you check the length formula. Solid and dashed background lines distinguish threshold successes without relying only on color. All comparisons concern a strict inequality; the continuous models assign equality probability zero.

Independent uniform endpoints: one third

Choose two points independently and uniformly around the circumference, then join them. By rotational symmetry, fix the first endpoint and describe the second by an angle uniform from zero to 2π. Chord length is 2 sin(θ/2). It exceeds √3 when 2π/3 < θ < 4π/3, an interval occupying one third of the full angular range.

The implementation draws the equivalent midpoint-distance distribution r = |cos(πU)|, with U uniform between zero and one, and an independent uniform orientation. This preserves the endpoint experiment’s chord distribution. It does not claim that midpoint distance is uniform.

Uniform distance from the center: one half

Choose an orientation uniformly and a perpendicular distance r uniformly from zero to one. Draw the chord perpendicular to that radius at that distance. Exactly half of the allowed distances lie below one half, so the probability is 1/2.

This experiment weights equal radial intervals equally. It therefore puts more midpoint probability near the center than a sampler that weights equal disk areas equally. Rotation symmetry alone does not distinguish those two choices: both are rotation invariant.

Uniform midpoint in the disk: one quarter

Choose a midpoint uniformly by disk area, then draw the perpendicular chord. The favorable midpoints lie inside the radius-one-half disk. Its area is one quarter of the full disk area, giving probability 1/4.

To sample this rule, use r = √U, not r = U. A disk of radius r contains fraction r² of the total area; inverting that CDF gives the square root. The center itself has probability zero; assigning it a random diameter does not affect the continuous distribution.

Change the threshold

For a length threshold L between zero and two, set d = √(1 − L²/4). The three success probabilities are 2 arcsin(d)/π, d, and d². At threshold zero they are all one, and at two they are all zero. Agreement at these endpoints does not imply agreement elsewhere.

Sample proportions fluctuate around their own model probabilities. Raising sample size preserves old draws and reduces typical Monte Carlo variation; it cannot make three distinct population probabilities converge to one common answer. The answer to the original question requires an operational sampling rule, not a vote among plausible fractions.

Reference and connection

Probability lecture notes from the Hebrew University, in the uniform-distribution section, discuss the endpoint-angle and radial-distance constructions. The disk-area calculation follows the same geometric principle. Continue to Borel’s paradox for a related distinction involving the neighborhoods used in continuous conditioning.