The Bertrand's paradox calculator above solves the random chord problem: given a circle, how likely is a randomly chosen chord to be longer than a given length? It reports the answer under each of the three classic methods of choosing a chord at random, because the problem as usually stated has three legitimate answers and no way to prefer one of them. It is a different problem from Bertrand's box paradox, which is a conditional-probability puzzle about coins and does have a single answer; the two share only the name of their author.
Arb Digital publishes it because the paradox is not a trick and not a mistake. Joseph Bertrand raised it in 1889 to show that the phrase "at random" is not self-explanatory, and that a probability question is not fully specified until the mechanism producing the outcome is described. The three answers are all correct answers to three different, equally reasonable questions.
What This Bertrand's Paradox Calculator Does
It computes, for a circle of your chosen radius and a threshold length you set, the probability that a random chord exceeds that length under each of three sampling methods. The classic statement takes the threshold to be the side of the inscribed equilateral triangle, which equals the radius times the square root of three, and produces the famous answers of one third, one half and one quarter.
Alongside the exact formulas it runs an optional simulation, drawing chords by each method and counting how many clear the bar. The simulation uses a fixed seed so the figures are reproducible, and it exists to make the point that these are not competing approximations of one true value: each method genuinely produces its own answer, repeatably.
The threshold is adjustable, because the three-way disagreement is not an artefact of the triangle. Set the threshold to the radius and the answers become two thirds, about 0.866 and 0.75 — still three different numbers. What the triangle gives you is a set of memorable fractions, not a special case.
How to Use It
- Set the circle radius, or leave it at one, since the answers do not depend on scale.
- Choose the threshold length — the triangle side, the radius, or a value you type.
- Read the three probabilities in the grid. They will disagree, and that is the result rather than an error.
- Turn on the simulation to see each method converge on its own answer independently of the formula.
- Read the working panel for the geometric reason each method gives the number it gives.
The Three Methods and How They Are Calculated
Method one, random endpoints. Fix one end of the chord anywhere on the circle and choose the other end uniformly around the circumference. The chord length is 2r sin(θ/2), where θ is the angle between the two points. It exceeds L when θ lies between 2α and 2π − 2α, with α = arcsin(L / 2r), so the probability is 1 − 2α/π. For the triangle side, α is sixty degrees and the answer is one third.
Method two, random radial point. Choose a radius of the circle, then choose a point uniformly along it, and take the chord perpendicular to the radius at that point. A chord at distance d from the centre has length 2√(r² − d²), which exceeds L when d is less than √(r² − L²/4). Since d is uniform on [0, r], the probability is √(r² − L²/4) / r. For the triangle side that distance is half the radius, and the answer is one half.
Method three, random midpoint. Choose a point uniformly at random inside the disc and take the chord for which it is the midpoint. The chord beats L when the point lies within √(r² − L²/4) of the centre, and because the point is uniform over area the probability is the ratio of areas, (r² − L²/4) / r². For the triangle side that is a quarter.
Cut the Knot's page on Bertrand's Paradox works through all three constructions with diagrams and arrives at the same one third, one half and one quarter.
Why There Is No Single Correct Answer
Each method is a perfectly reasonable reading of "choose a chord at random", and each imposes a different uniform distribution. Method one spreads probability uniformly over pairs of endpoints. Method two spreads it uniformly over distances from the centre. Method three spreads it uniformly over the area of the disc. These are genuinely different measures on the same set of chords, and nothing in the phrase "at random" tells you which to use.
The underlying issue is that the principle of indifference — treat equally possible cases as equally probable — does not survive a change of parametrisation. If two quantities are related non-linearly, spreading probability uniformly over one does not spread it uniformly over the other. The Stanford Encyclopedia of Philosophy's article on Interpretations of Probability sets out the general form of this difficulty, using a factory example in which side length, face area and volume give three different answers for the same event.
So the paradox is not a puzzle awaiting a clever resolution. It is a demonstration that "pick one at random" is an incomplete specification, and that a probability model requires the generating mechanism to be described. Once the mechanism is stated, the ambiguity disappears entirely and the answer is unique.
The Jaynes Argument, and What It Does and Does Not Settle
E. T. Jaynes proposed in 1973 that one answer can be singled out by requiring the distribution to be invariant under transformations the problem does not specify. His reasoning: if you toss straws onto a circle drawn on the ground, the process knows nothing about where the circle is, how big it is, or how it is oriented. So the distribution over chords should be unchanged by translation, rotation and scaling. Only method two satisfies all three, giving one half.
Cut the Knot's follow-up discussion of Bertrand's Paradox sets out the invariance argument and notes that the answer was checked experimentally by physically throwing straws at a circle.
This is worth taking seriously, and it is worth being precise about its scope. Jaynes answered a different, better-specified question: not "what is the probability for a random chord" but "what is the probability for a chord produced by this particular physical process". Given that process, the answer is unique — which is exactly the point Bertrand was making. The invariance argument does not show that method two was always the right reading of the original ambiguous question; it shows that specifying a mechanism removes the ambiguity. Presenting one half as the answer to Bertrand's original question overstates what the argument establishes.
Where the Same Trap Appears Outside Geometry
The structure recurs whenever a probability is assigned by appealing to symmetry without specifying a mechanism. Ask for a random point on the surface of a sphere and sampling latitude and longitude uniformly clusters points at the poles, because equal ranges of latitude do not cover equal areas. Ask for a random rotation, a random triangle, or a random subset and the same question arises: uniform with respect to what?
In applied work the trap usually shows up in simulation. A model that draws parameters "uniformly" is making a choice of parametrisation that carries real assumptions, and a uniform prior on a rate is not a uniform prior on its reciprocal. The results can shift substantially, and the shift is invisible unless the parametrisation is written down. Our random number generator and probability calculator are the practical starting points, and the birthday paradox calculator covers a different way intuition about randomness fails.
For the geometry itself, our circle calculator handles radius, area and circumference, and the arc length calculator covers the arc and sector measurements that method one depends on.
Arb Digital builds analysis that states its sampling assumptions rather than leaving them implicit.
Browse All Free Tools Talk To Our TeamCommon Mistakes to Avoid
- Looking for the one right answer — the question as usually stated does not have one, and all three methods are valid readings of it.
- Confusing it with Bertrand's box paradox — that is a coins-in-boxes puzzle with a single answer of two thirds and no ambiguity at all.
- Treating Jaynes as having resolved it — his argument answers a better-specified question, which is a different achievement from picking a winner among the three.
- Assuming the triangle threshold is special — the three methods disagree for every threshold; the triangle just gives memorable fractions.
- Sampling uniformly without saying in what — uniform over angles, over radius and over area are three different distributions on the same objects.
Related Free Tools From Arb Digital
Solve the coin puzzle with the Bertrand's box paradox calculator, work through the sibling version with the boy or girl paradox calculator, compute event probabilities with the probability calculator, measure the circle with the circle calculator, or draw samples with the random number generator. The full free online tools hub lists every statistics tool we publish.
Frequently Asked Questions
It asks how likely a randomly chosen chord of a circle is to be longer than the side of the inscribed equilateral triangle. Three reasonable ways of choosing a chord at random give one third, one half and one quarter.
All of them, for the question each method actually answers. The original wording does not specify how the chord is chosen, and until it does the problem has no unique solution.
No. The box paradox is a conditional-probability puzzle about coins in boxes with a single correct answer of two thirds. The two problems share only the name of the mathematician who posed them.
He showed that requiring invariance under translation, rotation and scaling singles out one half. That answers a more precisely specified physical question rather than showing the original ambiguous question had a unique answer.
Each spreads probability uniformly over a different thing: pairs of endpoints, distance from the centre, or area of the disc. Those are genuinely different distributions on the same set of chords.
Not when the threshold is expressed relative to the radius. All three probabilities depend only on the ratio of the threshold length to the radius, so the results are scale-free.
The three methods still disagree. The triangle side is used because it produces the memorable fractions one third, one half and one quarter, not because anything special happens there.
That a probability question is incomplete until the mechanism generating the outcome is described. Appealing to symmetry alone does not fix a distribution, because symmetry depends on how the problem is parametrised.
This page explains a well-known problem in probability for educational purposes. The three answers it reports are all correct for their respective sampling methods; none is presented as the single solution.
