How this instrument works
Bertrand's Paradox, posed by Joseph Bertrand in 1889, asks a deceptively simple question: if you draw a random chord on a circle, what's the probability it's longer than a side of the equilateral triangle inscribed in that circle? The paradox is that 'random chord' doesn't pin down a single procedure for generating one, and three equally reasonable-sounding methods for choosing a chord at random give three different, all mathematically valid, answers — 1/3, 1/2, and 1/4.
The random-endpoints method picks two points independently and uniformly on the circle's edge and connects them; this gives probability 1/3, because the second point must land within the 120-degree arc directly opposite the first to produce a chord longer than the triangle's side. The random-radial-point method instead fixes a direction, picks a random point along that radius, and draws the chord through it perpendicular to the radius; this gives probability 1/2, since the triangle's side crosses the radius exactly at its midpoint. The random-midpoint method picks a point uniformly at random anywhere inside the circle and uses it as the chord's midpoint; this gives probability 1/4, the area ratio between a circle of half the radius and the full circle.
The paradox isn't a flaw in probability theory — it's a lesson about the phrase 'chosen at random' being meaningless until you specify the actual sampling procedure. Each of the three methods above is a perfectly consistent, well-defined way to generate a uniformly random chord under its own assumptions; they simply encode different notions of 'uniform,' and there's no method-independent fact of the matter about which one is the 'true' random chord.
- Choose one of the three random-chord methods from Random-chord method: random endpoints, random radial point, or random midpoint.
- Read P(chord longer than the inscribed equilateral triangle's side) for that method's probability.
- Switch between the three methods to see the paradox directly: the same phrase, 'a random chord,' produces three different, equally defensible probabilities.
- There are no other inputs — each method's probability is a fixed constant, independent of the circle's actual radius.
Worked example — comparing all three methods on the same circle
Selecting the random-endpoints method — picking two independent uniformly random points on the circle's edge and connecting them — P(chord longer than the inscribed equilateral triangle's side) reads 0.333333, exactly 1/3. The reasoning: fix the first endpoint anywhere, then the triangle's vertex nearest it splits the remaining circle into a 120-degree 'far' arc and a 240-degree 'near' arc; the chord is longer than the triangle's side exactly when the second endpoint lands in that 120-degree arc, which is 1/3 of the full circle.
Switching to the random-radial-point method — fixing a radius, picking a uniformly random point along it, and drawing the perpendicular chord through that point — the same question now reads 0.5 exactly. Here the triangle's side crosses the fixed radius exactly at its midpoint, so a chord is longer than the triangle's side exactly when the random point falls in the inner half of the radius, which happens with probability 1/2. Switching again to the random-midpoint method gives 0.25 exactly: the chord is longer than the triangle's side exactly when its midpoint falls inside a concentric circle of half the radius, whose area is 1/4 of the full circle's area. Same circle, same triangle, three different well-defined answers.
Questions
Which of the three methods is the 'correct' one?
None of them — that's the entire point of the paradox. All three are legitimate, well-defined procedures for generating a chord at random; they just correspond to different notions of what 'uniformly random' means for a chord (uniform over pairs of endpoints, uniform over a fixed radius, or uniform over the circle's interior for the midpoint). The phrase 'a random chord' is genuinely ambiguous until one of these procedures (or another) is specified, so no single number is the universally correct probability.
Why do the three methods give different answers to the same-sounding question?
Because each method induces a different probability distribution over the set of all possible chords, even though every method individually treats its own inputs (endpoint pairs, radial position, or midpoint location) as uniformly distributed. A distribution that's uniform in one of those parameterizations is generally not uniform in the others — picking chord midpoints uniformly across the disk, for instance, does not correspond to picking chord endpoints uniformly around the circle's edge, since the change of variables between the two isn't measure-preserving.
Does the circle's actual radius change any of these probabilities?
No — all three probabilities (1/3, 1/2, and 1/4) are pure ratios that depend only on the geometry of the inscribed equilateral triangle relative to the circle, not on the circle's absolute size. Scaling the circle up or down scales the triangle and every chord length by the same factor, leaving every probability in this calculator unchanged.
Is Bertrand's Paradox the same as Bertrand's Box Paradox?
No — despite both being named after Joseph Bertrand and both being famous probability paradoxes, they're unrelated puzzles. This page covers the random-chord paradox, about the ambiguity of 'a random chord' on a circle. This site's separate Bertrand's Box Paradox calculator instead covers a conditional-probability puzzle about drawing coins from boxes, with a single well-defined answer (2/3) once the setup is fixed — no ambiguity of method involved there at all.
What's the practical lesson from this paradox?
Whenever a probability problem describes an object as being 'chosen at random,' the answer can depend entirely on the specific sampling procedure implied, and different procedures that all sound equally natural in words can produce genuinely different distributions. In applied work, this means the exact random process — not just the word 'random' — needs to be nailed down before a probability question about it has one right answer.