Bertrand Paradox Calculator — Random Chord Probabilities
The Bertrand paradox (1889) asks: what is the probability that a randomly chosen chord of a circle is longer than the side of the inscribed equilateral triangle? Three equally defensible definitions of "random chord" give answers of 1/3, 1/2, and 1/4 — all correct under their own assumptions. Enter a radius and optional threshold to compare all three.
Probability by the random-radius method: pick a random radius, then a uniform point on it
- 1
Ratio r = L ÷ 2R
1.7321 ÷ (2 × 1) = 0.866025 - 2
1 − r²
1 − 0.866025² = 0.25The chord exceeds L when its midpoint falls outside a smaller concentric circle of radius r. - 3
P(chord > L) — Method 2
√(0.25) = 0.5000
How does this calculator work?
The Bertrand paradox: P(random chord > R√3 in a circle of radius R) = 1/3 by Method 1 (random endpoints), 1/2 by Method 2 (random radius point), or 1/4 by Method 3 (random disk midpoint). All three are valid. For general L: M1 = 1−(2/π)arcsin(L/2R), M2 = √(1−(L/2R)²), M3 = 1−(L/2R)².
Formula
How this is calculated
The paradox lies in the ambiguity of "uniformly random" in continuous geometry. Each method defines a different probability measure, and each is mathematically self-consistent.
Method 1 (random endpoints): Two points are chosen uniformly and independently on the circumference. The chord length is 2R sin(θ/2) where θ is the central angle. P(chord > L) = 1 − (2/π) arcsin(L/2R). For the inscribed triangle side L = R√3, this gives 1/3.
Method 2 (random radius): A radius direction is chosen at random; a point is then chosen uniformly along that radius. A chord perpendicular to the radius at distance d from the centre has length 2√(R²−d²). P(chord > L) = √(1−(L/2R)²). This gives 1/2 at L = R√3.
Method 3 (random midpoint): A point is chosen uniformly inside the disk as the chord midpoint. The chord is longer than L when the midpoint falls within a smaller concentric circle of radius √(R²−L²/4). P = 1−(L/2R)². This gives 1/4 at L = R√3. There is no single "correct" answer — the paradox demonstrates that a probability is only well-defined once the sample space is fully specified.
Frequently asked questions
All three are mathematically correct under their own definition of randomness. There is no single right answer — the paradox illustrates that "random" in geometry requires an explicit probability measure. Different physical processes that generate random chords correspond to different methods.
A chord exceeds the inscribed triangle side (R√3) if and only if the central angle it subtends exceeds 120°. At this special threshold the three methods give the memorably clean answers 1/3, 1/2, and 1/4, making the paradox particularly vivid.
No. All three probabilities depend only on the ratio L/2R, not on R or L individually. Scaling the circle scales both equally, leaving the probability unchanged. The paradox is scale-invariant.
TG we-Calculate Editorial Team. (2026). Bertrand Paradox Calculator — Random Chord Probabilities [Online calculator]. TG we-Calculate. https://we-calculate.com/calculator/betrand-paradox-calculator
TG we-Calculate Editorial Team. "Bertrand Paradox Calculator — Random Chord Probabilities." TG we-Calculate. 2026. https://we-calculate.com/calculator/betrand-paradox-calculator.
TG we-Calculate Editorial Team, "Bertrand Paradox Calculator — Random Chord Probabilities," TG we-Calculate, 2026. [Online]. Available: https://we-calculate.com/calculator/betrand-paradox-calculator
@misc{wecalculate_betrand_paradox_calculator, title = {Bertrand Paradox Calculator — Random Chord Probabilities}, author = {{TG we-Calculate Editorial Team}}, howpublished = {\url{https://we-calculate.com/calculator/betrand-paradox-calculator}}, year = {2026}, note = {TG we-Calculate} }
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