Qual Quant (2012) 46:1709–1714
Bertrand’s paradox: is there anything else?
Vesna Jevremovic · Marko Obradovic
Published online: 6 September 2011
© Springer Science+Business Media B.V. 2011
Abstract In this paper we give some other solutions of the well-known Bertrand’s para-
dox. The random choice of a chord is done by choosing two independent points under some
uniform distributions. We obtained results that differ from the classic ones.
Keywords Random choice of a point · Geometric probability
The problem was originally posed by Joseph Bertrand in Calcul des probabilités, in 1888:
Consider an equilateral triangle P Q R inscribed in a circle and choose at random a chord
AB of the circle. What is the probability that the chord is longer than a side of the triangle?
This problem is called paradox since it seems that it has three different solutions. These solu-
tions are well known nowadays and one could ﬁnd them in almost every book on probability
theory. We give here a brief survey of these solutions.
In the ﬁrst solution, the point A of the chord is chosen at random on the circle, and we
rotate the triangle so the point A becomes one of its vertex, say P. The point B of a chord is
then chosen at random on the circle. If the point B belongs to the arc QR, then the chord AB
is longer than the side of the triangle (Fig. 1). Therefore the probability is one third, since it
equals l/L,wherel is the length of the arc QR, while L is the perimeter of the circle.
The second solution consists of choosing at random a point T in the circle and construct-
ing a chord AB so that T is its midpoint. The chord AB will be longer than the side of the
triangle PQR if the point T is within the circle inscribed in the triangle (Fig. 2). This circle
is concentric with the given one and its radius is
of the radius of the given circle. The
V. Jevremovic (
) · M. Obradovic
Faculty of Mathematics,
University of Belgrade, Belgrade, Serbia