Critical Studies/Book Reviews

Picture a circle with an equilateral triangle inscribed within it. Now, choose a chord within the circle at random. What is the probability that this randomly chosen chord is longer than the sides of the equilateral triangle? That is the question Bertrand [1889] asked, and he proposed three different methods for deriving the answer. According to the first method, two points on the circumference of the circle are first chosen at random and a chord is drawn between them. The triangle can be rotated such that one of its vertices coincides with an endpoint of the chord. Notice that the chord is longer than its sides iff (if and only if) the other endpoint of the chord lies on the arc of the circle between the other two vertices of the triangle. And since the length of this arc is |$ 1/3 $| of the entire circle, according to this first method, the answer to the question above is |$ 1/3 $|⁠. But there is a second way of deriving the answer. According to the second method, we choose a point on the circumference at random first and then consider the radius of the circle extending from that chosen point to the centre of the circle. Now, we further choose a point on this radius at random and construct the chord through this second point such that the chord is perpendicular to the radius. The triangle can then be rotated again so that one of its sides is perpendicular to the radius and notice that the chord is longer than its sides iff the second chosen point on the radius is anywhere between the centre of the circle up to but excluding the point, where the radius bisects the side of the triangle. Since this interval is |$ 1/2 $| of the radius, the answer to the question above, according to this second method, is instead |$ 1/2 $|⁠. But the event, for which we are trying to calculate the probability, is essentially the same across these two methods; yet they yield different answers. And to make matters worse, there is a third method. According to this last method, we imagine a smaller circle inscribed within the bigger one such that the radius of the smaller circle is |$ 1/2 $| of the radius of the bigger one. Then, choose a chord at random and notice that the chord is longer than the sides of the triangle iff the midpoint of the chord is in the smaller circle. Since the area of the smaller circle is |$ 1/4 $| the area of the bigger one, the answer to the question above, according to this method, is instead |$ 1/4 $|⁠. There we have it: three different methods giving three different answers to what seems to be one and the same event: a randomly chosen chord being longer than the sides of an equilateral triangle inscribed in a circle. Worse still, all three methods seem to be perfectly legitimate in deriving an answer, but a single event cannot have multiple probabilities. This vexing puzzle is now known as Bertrand’s paradox, and ever since its publication about a hundred and fifty years ago, it has spurred numerous discussions and inspired philosophers and mathematicians alike in coming up with novel solutions and insights.

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Publication Details

Journal
Philosophia Mathematica
Published
2026-07-17
DOI
https://doi.org/10.1093/philmat/nkag013
Primary Topic
Renaissance and Early Modern Studies
Type
article
Field-Weighted Citation Impact
0.00
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article

Critical Studies/Book Reviews

Nicholas Shackel
Philosophia Mathematica
Renaissance and Early Modern Studies
article

Critical Studies/Book Reviews

Nicholas Shackel
article en

Abstract

Picture a circle with an equilateral triangle inscribed within it. Now, choose a chord within the circle at random. What is the probability that this randomly chosen chord is longer than the sides of the equilateral triangle? That is the question Bertrand [1889] asked, and he proposed three different methods for deriving the answer. According to the first method, two points on the circumference of the circle are first chosen at random and a chord is drawn between them. The triangle can be rotated such that one of its vertices coincides with an endpoint of the chord. Notice that the chord is longer than its sides iff (if and only if) the other endpoint of the chord lies on the arc of the circle between the other two vertices of the triangle. And since the length of this arc is |$ 1/3 $| of the entire circle, according to this first method, the answer to the question above is |$ 1/3 $|⁠. But there is a second way of deriving the answer. According to the second method, we choose a point on the circumference at random first and then consider the radius of the circle extending from that chosen point to the centre of the circle. Now, we further choose a point on this radius at random and construct the chord through this second point such that the chord is perpendicular to the radius. The triangle can then be rotated again so that one of its sides is perpendicular to the radius and notice that the chord is longer than its sides iff the second chosen point on the radius is anywhere between the centre of the circle up to but excluding the point, where the radius bisects the side of the triangle. Since this interval is |$ 1/2 $| of the radius, the answer to the question above, according to this second method, is instead |$ 1/2 $|⁠. But the event, for which we are trying to calculate the probability, is essentially the same across these two methods; yet they yield different answers. And to make matters worse, there is a third method. According to this last method, we imagine a smaller circle inscribed within the bigger one such that the radius of the smaller circle is |$ 1/2 $| of the radius of the bigger one. Then, choose a chord at random and notice that the chord is longer than the sides of the triangle iff the midpoint of the chord is in the smaller circle. Since the area of the smaller circle is |$ 1/4 $| the area of the bigger one, the answer to the question above, according to this method, is instead |$ 1/4 $|⁠. There we have it: three different methods giving three different answers to what seems to be one and the same event: a randomly chosen chord being longer than the sides of an equilateral triangle inscribed in a circle. Worse still, all three methods seem to be perfectly legitimate in deriving an answer, but a single event cannot have multiple probabilities. This vexing puzzle is now known as Bertrand’s paradox, and ever since its publication about a hundred and fifty years ago, it has spurred numerous discussions and inspired philosophers and mathematicians alike in coming up with novel solutions and insights.

Philosophia Mathematica
Innlandet Hospital Trust (NO)
Openalex Percentile: Top 1%
Renaissance and Early Modern Studies
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