Bertrand's Paradox Calculator

See the three classic answers — 1/3, 1/2 and 1/4 — for the probability that a random chord of a circle is longer than the inscribed equilateral triangle's side, and compute a specific chord's length from its midpoint distance.

Quick Facts

Method
Chord > triangle side ⟺ midpoint distance d < R/2
Triangle side = R√3; chord length = 2√(R² − d²). Probability answers: 1/3, 1/2, 1/4.

Your Results

Calculated
P — random endpoints
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Method 1: two uniform endpoints
P — random radius
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Method 2: uniform point on a radius
P — random midpoint
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Method 3: uniform midpoint in disk
Your chord length
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2√(R² − d²)

Ready

Enter a radius and a midpoint distance, then calculate.

What Bertrand's paradox is

Bertrand's paradox, posed by Joseph Bertrand in 1889, is a classic problem in geometric probability. Draw a circle and inscribe an equilateral triangle in it. Now pick a "random" chord of the circle. What is the probability that the chord is longer than a side of that triangle? The paradox is that three perfectly reasonable ways of choosing a "random" chord give three different answers — 1/3, 1/2, and 1/4 — because the phrase "at random" has not been pinned down. The problem is a standard illustration that a probability is undefined until the sampling mechanism is specified.

The geometry that drives every answer

For a circle of radius R, the inscribed equilateral triangle has side length R√3 (about 1.732 R). Every chord is fully described by the distance d of its midpoint from the center: a chord with midpoint distance d has length 2√(R² − d²). Setting 2√(R² − d²) > R√3 and solving gives d < R/2. So the single geometric fact behind the whole paradox is: a chord is longer than the triangle's side exactly when its midpoint lies within half a radius of the center. The three answers differ only because the three methods spread chord midpoints across the disk differently.

The three classic methods

  • Random endpoints (answer 1/3). Fix one endpoint on the circle and choose the other endpoint uniformly around the circumference. The chord beats the triangle's side when the second point lands on the far 120° arc. 120°/360° = 1/3.
  • Random radius (answer 1/2). Pick a radius at random, then pick the chord's midpoint uniformly along that radius (the chord is perpendicular to it). The chord is long enough when the midpoint is in the inner half of the radius, i.e. d < R/2. That is 1/2 of the radius, so the probability is 1/2.
  • Random midpoint (answer 1/4). Choose the chord's midpoint uniformly over the whole disk area. The favourable region is the disk of radius R/2, whose area is π(R/2)² = πR²/4 — one quarter of the full disk. So the probability is 1/4.

Why the answers are all "correct"

None of the three methods is wrong; they simply sample different probability distributions over chords. Jaynes (1973) argued that if the answer must be invariant under scaling, rotation, and translation of the circle, the random-radius method (1/2) is singled out, and physical experiments that drop straws onto a circle tend to reproduce it. But the pedagogical point stands: "pick a random chord" is not one experiment, it is a family of experiments, and the probability depends on which one you run.

Frequently Asked Questions

Why does Bertrand's paradox have three different answers?
Because "a random chord" is ambiguous. Random endpoints on the circle gives 1/3, a random radius with a uniform midpoint along it gives 1/2, and a uniform midpoint over the disk gives 1/4. Each method samples a different distribution of chords, so each yields a different, internally consistent probability. The lesson is that a probability question is not well-posed until the sampling procedure is fixed.
When is a chord longer than the inscribed equilateral triangle's side?
The triangle's side is R√3. A chord with midpoint a distance d from the center has length 2√(R² − d²), which exceeds R√3 exactly when d < R/2. So a chord beats the triangle's side whenever its midpoint lies within half a radius of the center — this single condition is what all three probability calculations measure.
Does the answer depend on the circle's radius?
No. All three probabilities (1/3, 1/2, 1/4) are pure numbers that do not depend on R, because the favourable condition d < R/2 scales with the circle. The radius only matters if you want an actual length — the triangle side R√3 or a specific chord length 2√(R² − d²).

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