The Bertrand Paradox: When Randomness Needs a Definition

In 1889, Joseph Bertrand posed a deceptively simple question:

A chord is drawn at random inside a circle. What is the probability that it is longer than the side of the inscribed equilateral triangle?

Three equally natural ways to answer give three different probabilities: $\frac{1}{3}$, $\frac{1}{2}$, and $\frac{1}{4}$. Each is mathematically correct. The paradox lives in the word random.

The Setup

Fix a circle of radius $R$. The inscribed equilateral triangle has side length $R\sqrt{3}$. A chord of the circle is longer than this when it passes close enough to the centre.

Precisely: a chord is "long" (longer than $R\sqrt{3}$) if and only if its midpoint lies within distance $R/2$ of the centre.

Now: how do you pick a chord at random?

Method 1 — Random Endpoints  P = 1/3

Pick two points independently and uniformly on the circumference. Connect them.

By rotational symmetry, fix one endpoint and let the other vary. The chord is longer than the triangle's side when the arc between the two points subtends an angle between $60°$ and $300°$ — a fraction $\frac{2}{3}$ of the circle. Wait, that gives $\frac{2}{3}$, not $\frac{1}{3}$.

Let me be more careful. The inscribed equilateral triangle has vertices $120°$ apart. The chord from a fixed point is longer than the triangle's side when the second point falls in the arc opposite to it — specifically the middle third of the circle, spanning $120°$ out of $360°$. That gives:

$P_1 = \frac{120°}{360°} = \frac{1}{3}$

Method 2 — Random Radius Point  P = 1/2

Pick a random radius (direction) and then a random point uniformly along that radius. Draw the chord through that point, perpendicular to the radius.

By symmetry we may fix the radius direction. The point is at distance $t$ from the centre, where $t \sim \text{Uniform}(0, R)$. The perpendicular chord at distance $t$ has half-length $\sqrt{R^2 - t^2}$, so full length $2\sqrt{R^2 - t^2}$.

The chord is longer than $R\sqrt{3}$ when $2\sqrt{R^2 - t^2} > R\sqrt{3}$, i.e. when $t < R/2$. That event has probability:

$P_2 = \frac{R/2}{R} = \frac{1}{2}$

Method 3 — Random Midpoint  P = 1/4

Every chord has a unique midpoint inside the circle (the diameter has midpoint at the centre). Pick a point uniformly at random inside the circle and use it as the midpoint of the chord.

The chord is long when its midpoint lies within $R/2$ of the centre — inside an inner circle of radius $R/2$. The probability is the ratio of areas:

$P_3 = \frac{\pi (R/2)^2}{\pi R^2} = \frac{1}{4}$

The Playground

Three valid methods. Three different probabilities. See them diverge in real time — each panel accumulates chords drawn by its own method.

The yellow triangle is the inscribed equilateral triangle. The dashed inner circle in Method 3 shows the region where midpoints of long chords fall.

The visual difference is immediate. Method 2 (Random Radius) produces many long chords — nearly half the screen is green. Method 3 (Random Midpoint) produces the fewest. Same circle. Same triangle. Same question. Different answers.

Toggle the ⬤ Midpoints view to see where the chord midpoints fall:

• Method 1 — midpoints cluster near the boundary. Short chords (endpoints close together) have midpoints far from the centre.
• Method 2 — midpoints cluster near the centre. Uniform in radius packs more points into small annuli.
• Method 3 — midpoints are uniformly spread across the disk.

Why Density Matters: A Polar Coordinates Warm-Up

Before resolving the paradox, here is a concrete reason to care about where points fall when you sample "uniformly."

Imagine placing $N$ points evenly spaced in angle on two concentric circles — one of radius $r$, one of radius $2r$. Both carry exactly $N$ points. Yet the spatial gap between adjacent points is $\frac{2\pi r}{N}$, which grows linearly with $r$. Double the radius, double every gap. The angular distribution is uniform, but the spatial distribution is not.

This is exactly the issue with Method 2. Sampling a point uniformly along a radius ($r \sim \text{Uniform}(0,R)$) gives every small interval $[r, r+dr]$ the same number of midpoints. But a thin annulus at radius $r$ has area $2\pi r\,dr$ — growing with $r$ — so the density per unit area is $\propto 1/r$: highest at the centre, thinning toward the boundary. That clustering in the midpoints view is not an artifact; it is the direct consequence of the sampling method.

Density Functions: The Precise Language

Each method secretly defines a probability density function (PDF) for the chord's midpoint distance $r$ from the centre. That PDF determines the probability.

Method 1 (Random Endpoints):

$f_1(r) = \frac{2}{\pi \sqrt{R^2 - r^2}}, \quad 0 \le r < R$

Diverges as $r \to R$ — density is highest near the boundary.

Method 2 (Random Radius Point):

$f_2(r) = \frac{1}{R}, \quad 0 \le r \le R$

Uniform in $r$, but not uniform in area ($\propto 1/r$ per unit area toward the centre).

Method 3 (Random Midpoint):

$f_3(r) = \frac{2r}{R^2}, \quad 0 \le r \le R$

The only density that is uniform in area. The cleanest scatter in the midpoints view.

Each probability follows directly from integrating its density from $0$ to $R/2$ (the long-chord condition):

$P_k = \int_0^{R/2} f_k(r)\,dr \quad \Longrightarrow \quad \tfrac{1}{3},\; \tfrac{1}{2},\; \tfrac{1}{4}$

The three answers are not in conflict — they are consequences of three different densities on the same geometry.

Try It: Override the Density

If the three methods all produced the same density, they would give the same probability. You can verify this in the playground below. Use the Density override panel to force all three panels to sample from $f_1$, $f_2$, or $f_3$ regardless of their geometric method. All three converge to the same value. Or type any expression in r and R into the Custom f(r) input to define your own density and see what probability it implies.

Conclusion

The Bertrand paradox shows that "pick a random chord" is not a well-posed question. Each method is a choice of parametrization of the sample space — by endpoint angles, by radius, by midpoint coordinates — and a uniform distribution over one parametrization is non-uniform over another. The geometry is the same. The triangle is the same. What differs is the density.

This is not a pathology of one contrived example. Whenever you write "pick a random $X$," you are implicitly choosing a probability measure on the space of all $X$'s. Different choices are valid; they just give different answers. The Bertrand paradox makes that invisible choice visible.

Further Reading

• Joseph Bertrand, Calcul des probabilités (1889) — the original source
• E. T. Jaynes, "The Well-Posed Problem" (1973) — argues Method 2 is the unique scale-and-translation-invariant solution
• Wikipedia: Bertrand paradox (probability)