Joseph Bertrand posed this question in 1889: draw an equilateral triangle inside a circle and pick a chord of the circle “at random”. What is the probability that the chord is longer than a side of the triangle? He gave three reasonable-looking arguments that reach three different answers, which is why the problem is called a paradox.
The resolution is that “at random” is not a single instruction for infinitely many chords. Choosing two random points on the circumference, choosing a random distance along a random radius, and choosing a random midpoint inside the disc are three different probability distributions over chords. Each one is internally consistent, and each gives its own answer: one third, one half and one quarter.
The calculator runs all three methods side by side, reports the exact and simulated probabilities, the length of the triangle’s side for your radius and the average chord length under the chosen method, and draws a sample of chords with the long ones highlighted, so you can see why the methods disagree.
How to use the Bertrand Paradox Calculator
- How the chord is chosen
- Circle radius — Only changes the side length and average chord length – the probabilities do not depend on the radius.
- Simulated chords per method
Everything is calculated in your browser, so results appear instantly and nothing you type is sent to a server.
How it works
For each method the tool draws thousands of chords on a unit circle, measures each one and counts those longer than √3 (the triangle side when r = 1). Lengths scale with the radius, so the probabilities do not depend on r; the radius only changes the side length and the average chord length shown.
Worked example
Take a circle of radius 1, so the triangle side is √3 ≈ 1.7321. With random endpoints the long chords are those whose second endpoint falls on a 120° arc: 120/360 = 33.33%. With a random point on a radius, the chord is long when the point lies within 0.5 of the centre: 0.5/1 = 50%. With a random midpoint, the long-chord region is a disc of radius 0.5, area π·0.25 out of π·1 = 25%. The average chord lengths are 4/π ≈ 1.2732, π/2 ≈ 1.5708 and 4/3 ≈ 1.3333.
Why the three methods disagree
Each method spreads chords differently. Random endpoints produce many short chords near the edge; random midpoints put most midpoints far from the centre, where chords are short; the random-radius method spreads the midpoints evenly along each radius. Different spreads give different shares of long chords, so the “paradox” is a lesson that a probability needs a precisely stated experiment.
Formula
Random endpoints: P = 1/3 · Random point on a radius: P = 1/2 · Random midpoint in the disc: P = (r/2)² ÷ r² = 1/4 · triangle side = r√3
Worked example
Inputs — How the chord is chosen: Compare all three methods; Circle radius: 1; Simulated chords per method: 5,000.
Results — Exact probability (chosen method) = 33.33 %; Simulated probability = 32.98 %; Side of the inscribed triangle (r√3) = 1.7321; Average chord length with this method = 1.2732; All three methods = Method Theory Simulated
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Random endpoints 33.33% 32.98%
Random point on a radius 50% 50.76%
Random midpoint in the disc 25% 24.68%.
Key terms
- Chord
- A straight segment joining two points of a circle.
- Inscribed equilateral triangle
- A triangle with all three corners on the circle; its side is r√3.
- Principle of indifference
- The rule of giving equal probability to outcomes when there is no reason to prefer one – ambiguous for infinite sets of outcomes.
Frequently asked questions
What is Bertrand's paradox?
A question about random chords of a circle that has three different correct answers – 1/3, 1/2 and 1/4 – depending on how a "random chord" is defined.
Which answer is right?
Each is right for its own sampling method. The question is incomplete until you say how the chord is chosen.
Why is the random-endpoint answer 1/3?
Fix one endpoint; the chord is longer than the triangle side when the other endpoint lands on the opposite third of the circle, an arc of 120° out of 360°.
Why is the random-radius answer 1/2?
The chord is long when its midpoint lies closer to the centre than half the radius, and a uniform point on the radius does that half of the time.
Why is the random-midpoint answer 1/4?
Long chords have midpoints inside a circle of radius r/2, whose area is a quarter of the whole disc.
Is there a "natural" choice?
E. T. Jaynes argued that the random-radius method (1/2) is the only one unchanged when the circle is moved or rescaled, but most textbooks present the problem as a warning to define the sampling method.