The Random Chord

Question

A chord of a circle is "chosen at random." What is the probability that it is longer than a side of the equilateral triangle inscribed in the circle? Defend your answer, including whether the question is even well-posed.

Accepted Answer

The question is not well-posed: "random chord" has no canonical meaning, and three equally natural procedures give three different answers. This is Bertrand's paradox. Throughout, a chord is longer than the triangle side iff its midpoint lies within distance frac12 (half the radius) of the center. Method 1, random endpoints (P = frac13). Fix one endpoint; the inscribed triangle with a vertex there splits the circle into three equal arcs. The chord beats the side iff the second endpoint lands in the far arc: frac13. Method 2, random radius and offset (P = frac12). Choose a direction, then slide the chord to a distance d sim U[0,1] from the center along it. Long iff d < frac12: probability frac12. Method 3, random midpoint in the disk (P = frac14). A midpoint determines the chord. Long iff the midpoint falls in the concentric disk of radius frac12, whose area is frac14 of the whole: frac14. Each method is uniform, over arc length, over offset, over area, but uniformity over one parametrization is non-uniformity over another, because the maps between them are nonlinear. (Jaynes argued that requiring invariance under translation and scaling of the experimental setup singles out Method

The Random Chord

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