Bertrand's paradox simulator

The key fact

In a circle of radius r, the inscribed equilateral triangle has side √3 r. A chord whose midpoint is a distance d from the centre has length 2√(r² − d²), so

chord > √3 r ⇔ d < r/2

Every method for picking a chord is a method for picking its midpoint, and the three classic methods give that midpoint three different distributions.

The three methods, worked out

Random endpoints (1/3). Pick two points on the circle independently. Fix the first one and draw the triangle from it. The chord is longer than a side when the second point lands on the arc opposite the first vertex, which is one third of the circle. So P = (2π/3) ÷ 2π = 1/3.

Random radius (1/2). Pick a radius, then a point on it uniformly, and draw the chord through that point at right angles to the radius. The chord is long when the point is in the half of the radius nearer the centre: P = (r/2) ÷ r = 1/2.

Random midpoint (1/4). Pick a point uniformly anywhere in the disc and use it as the chord's midpoint. The chord is long when the point is inside the circle of radius r/2, which has a quarter of the area: P = π(r/2)² ÷ πr² = 1/4.

What the simulation shows

With the default 10,000 chords per method, the three estimates land close to 0.333, 0.5 and 0.25, each within a percentage point or so. The exact values do not change; the simulated ones differ a little on every run. The standard error column tells you how much variation to expect: about 0.0047 for the first method at n = 10,000.

The pictures make the difference visible. Random endpoints give many short chords near the edge. Random radius spreads the midpoints evenly along each radius. Random midpoint pushes midpoints towards the edge, because there is more area there, so it gives the fewest long chords.

How to interpret it

Bertrand used the problem to argue that probabilities on infinite sets of outcomes are meaningless unless the sampling method is stated. The modern reading is the same: "uniformly at random" is only well defined once you say what is uniform. For a finite list of outcomes there is an obvious choice; for chords of a circle there are several.

The same issue appears in real work whenever a model assumes a "uniform" prior. Uniform in a parameter is not uniform in a function of it. A prior flat in a standard deviation is not flat in the variance.

Common mistakes

• Picking a midpoint by choosing a random distance from the centre uniformly and calling it uniform over the disc. That is the random-radius method, not the random-midpoint one; uniform over the disc needs distance = r√U.
• Thinking one method is "wrong". Each is a valid probability model; they answer different questions.
• Reading a single simulation as exact. Compare the difference with the standard error before drawing conclusions.