The Same Chord, Three Probabilities
Draw a chord of a circle at random. What's the chance it comes out longer than the side of the triangle that fits snugly inside? The honest answer is 1/3, or 1/2, or 1/4 — and which one you get is decided entirely by what you meant by at random. Run all three and watch them refuse to agree.
Inside every circle sits one largest equilateral triangle, and its side has a fixed length: for a circle of radius 1 it is exactly √3 ≈ 1.732. A chord beats that length for a clean geometric reason — a chord is longer than the triangle's side exactly when its midpoint lies closer to the centre than half the radius. So the whole question is really: where do your random chords put their midpoints? Three reasonable people answer that three different ways, and each is internally flawless. This is Bertrand's paradox (Joseph Bertrand, 1889): not a trick, but a demonstration that the phrase “at random” is not yet a question.
Why the three clouds look different
Flip to Midpoints above. The same rule governs every method — a chord is “long” iff its midpoint lands in the inner half-radius disk (drawn faintly) — but the three ways of randomising scatter the midpoints with completely different densities. What settles each answer is how the midpoint's distance from the centre is distributed. With random endpoints that distance leans toward the rim, and only a third of chords reach inside the inner disk (1/3). With a random radius the distance is spread evenly from centre to rim — so exactly half fall inside (1/2), and by area that packs the dots toward the middle. Uniform-over-the-disk gives flat areal density, so the inner disk claims only its fair area share, a quarter (1/4), and the rim, with more room, gets the rest. The paradox was never about chords. It was about which distribution the word “random” had silently named.
Is one answer “the” answer? — throw a straw
Bertrand left it there. In 1973 E. T. Jaynes asked a sharper question: the problem never said how big the circle is or where it sits, so a well-posed answer ought not to depend on those either — it should survive moving and rescaling the circle. Only one of the three distributions has that invariance, and it is the 1/2 one. You can see it physically: drop long straight straws across the plane and keep the ones that happen to cross the circle. Their chords land on 1/2 — and they keep landing there no matter how you slide or shrink the circle underneath them.
→ 1/2 (invariant)
Lines that miss the circle are dropped; crossing chords are coloured long / short. The circle is shown off-centre on purpose — the fraction doesn't care.
This does not crown 1/2 as cosmically correct. It says: if you demand an answer blind to the circle's size and place, one exists and it is 1/2. Demand something else — points on the rim, a hand-dropped midpoint — and the other answers are exactly as valid. The lesson that outlived the puzzle: a probability question isn't finished until the mechanism of the randomness is named.
The check — every number here is recomputed, not asserted
- The geometry. Inscribed equilateral triangle side = R√3; a chord whose midpoint is distance d from the centre has length 2√(R²−d²), which exceeds R√3 iff d < R/2. Both identities are checked to machine precision.
- A) random endpoints → 1/3. Fix one endpoint; the chord is long iff the other lands on the opposite third of the rim. Exact 1/3; the live counter and an offline 4,000,000-trial Monte Carlo both converge to it.
- B) random radius → 1/2. Midpoint at uniform distance along a random radius is long iff d < R/2 — half the time. Exact 1/2.
- C) midpoint uniform in the disk → 1/4. Long iff the midpoint sits in the inner half-radius disk, whose area is (1/2)² = 1/4 of the whole.
- D) thrown straw → 1/2, and invariant. The translation- and rotation-invariant measure on lines, conditioned on crossing the circle, makes the perpendicular distance uniform, identical to method B. The conditional law of d/R is scale invariant. The verifier compares growing throwing windows with the 1/2 limit and checks a circle and throwing window shifted together, with the radius fixed. (A small finite-window bias is named there, not hidden.)
All eleven checks pass in research/bertrand-paradox/verify.mjs — run it yourself; no network, no dependencies. The live estimates above are the same simulation, drawn in your browser.