Problem — Bertrand's paradox

Problem

In a circle of radius $R$ a chord is drawn “at random”. What is the probability that the chord is longer than $R\sqrt{3}$ (the side of the inscribed equilateral triangle)? Show that three natural definitions of “random chord” give different answers, and discuss what this teaches the experimental method.

Solution

The chord exceeds $R\sqrt{3}$ when its midpoint is closer than $R/2$ to the centre. The three “natural” constructions give three different values.

Method 1 — random endpoints on the circumference. Fix one endpoint and choose the second uniformly on the circumference. The chord exceeds the side of the inscribed equilateral triangle when the second endpoint falls in the opposite arc, which spans one third of the whole circumference: $P_1 = \frac{1}{3}$

Method 2 — midpoint uniform in the disc. Choose the midpoint of the chord uniformly over the area of the disc. The condition (distance from the centre $< R/2$) picks out a concentric disc of radius $R/2$: $P_2 = \frac{\pi (R/2)^2}{\pi R^2} = \frac{1}{4}$

Method 3 — distance from the centre uniform along the radius. Choose a direction and the distance $d$ of the midpoint from the centre uniformly in $[0,R]$. The condition $d < R/2$ gives: $P_3 = \frac{R/2}{R} = \frac{1}{2}$

Why there is no contradiction. The phrase “random chord” is not defined until the operational procedure by which the chord is drawn is specified. Each method samples a different space with a different distribution, and so answers a different question.

Lesson for the experimental method. A random quantity (or a measurement) makes sense only if it is defined operationally by how it is drawn or observed. Without a sampling protocol the question is ill-posed: the operational definition precedes the numerical answer.

$\ev{P_1 = \tfrac{1}{3},\quad P_2 = \tfrac{1}{4},\quad P_3 = \tfrac{1}{2}}$

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Topics: Metodo sperimentale