Problem — Bertrand's three answers

Text

Prove that the three answers of Bertrand’s paradox are $1/3$, $1/4$ and $1/2$, using an explicit parametrisation for each.

Solution

Set-up. The paradox asks for the probability that a «random» chord of a circle is longer than the side of the inscribed equilateral triangle. The answer depends on how «random» is defined: each parametrisation gives a different geometric probability measure.

• Random endpoints on the circumference ($1/3$): fix one endpoint and parametrise the other with the uniform angle; the chord exceeds the side when the second endpoint falls in the opposite arc, one third of the circumference wide.
• Random midpoint in the disc ($1/4$): parametrise the chord by its midpoint uniform over the area of the circle; the condition is that the midpoint falls in the concentric circle of half the radius, whose area is $1/4$ of the total.
• Random radius and distance from the centre ($1/2$): choose a direction and the distance of the midpoint from the centre uniform along the radius; the chord exceeds the side when this distance is less than half the radius, i.e. with probability $1/2$. For each case write the favourable event in the chosen parametrisation and compute the ratio of the measures to obtain $1/3$, $1/4$, $1/2$. The moral: without specifying the drawing mechanism, «at random» is not well defined.

Links

Topics: Probabilita
Concepts: Paradosso di bertrand · Probabilita geometrica
Skills: Calcolo probabilita · Dimostrare · Ragionare per casi
Exercise type: Problema probabilita