### Bertrand’s Chord Problem

The history of probability theory has been influenced strongly by paradoxes, results that seem to defy intuition. Many of these have been reviewed in a recent book by Prakash Gorroochurn . We will have a look at Bertrand’s Paradox (1889), a simple result in geometric probability. Let’s start with an equilateral triangle and add an inscribed circle and a circumscribed circle. It is a simple geometric result that the radius of the outer circle is twice that of the inner one. Bertrand’s problem may be stated thus:

Problem: Given a circle, a chord is drawn at random. What is the probability that the chord length is greater than the side of an equilateral triangle inscribed in the circle?

We will consider three ways of drawing a chord in the outer circle:

1. Fix the end-points of the chord.

2. Choose the chord centre on a fixed diameter.

3. Fix the mid-point of the chord.

We will find that the probability differs for the three methods of choice.

(1) Fix the end-points of the chord. On the basis of symmetry, we may choose one endpoint of the chord to coincide with a vertex of the equilateral triangle. Since the other endpoint (e.g., A or B) must be on one of three arcs, all of equal length, we may argue that each arc is equally probable. The chord length is less than the side of the triangle for the arcs adjacent to the first endpoint (e.g. point A) and exceeds it for points on the centre arc (e.g., point B). Thus, the probability of this event is .

(2) Choose the chord centre-point on a fixed diameter. We first choose a diameter. Then we select a point at random on it, and draw the chord through the point and normal to the diameter. Since the inner circle has diameter half that of the outer circle, there is a 50% chance that the point falls within the inner circle (e.g., point B). Thus, the probability of the chord length exceeding the triangle side is ½.

(3) Fix the mid-point of the chord. We fix the chord by choosing its midpoint. Excepting the centre point of the circle, only one chord has a given midpoint. If the point falls outside the inner circle (e.g., point A), the chord length is less than the side of the triangle. Otherwise (e.g., point B) it exceeds it.

But the area of the inner circle is one quarter that of the outer circle, since the radius is half. Therefore, the chance of the point being within the inner circle is 25% and the probability of the event is ¼.

Discussion

We have found three answers for the probability P of the event:

1. Fixed end-points of the chord:       P = .

2. Chord centre on a fixed diameter:  P = ½.

3. Fixed mid-point of the chord:         P = ¼.

Which is the correct result? We cannot say: the question is not well-posed and the answer depends on the method used to choose the chord.

Difficulties with results such as Bertrand’s chords problem gave rise to dissatisfaction with classical probability theory and acted as an impetus for the later development of an axiomatic foundation of the subject. David Hilbert recognized this need and identified an axiomatic basis for probability theory as one of his famous 23 problems for the twentieth century.

Poincaré, and later Edwin T. Jaynes, argued that the solution should be insensitive to changes in the position, orientation and size of the circle and triangle. Jaynes used this invariance to obtain a probability of ½. He also found that this agreed with actual experiments dropping sticks onto circles. However, this was not universally accepted and the debate continues today.

Final Remark

Imagine the plane covered with all possible lines. Drop a circular form, such as a hula-hoop, on the plane. A set of lines intersect the circle. What proportion of them exceed the length of an inscribed equilateral triangle?  We hope to use this approach to develop another solution to Bertrand’s Problem. If successful, details will follow.

Sources

• Bertrand, Joseph, 1889: Calculs de Probabilités. Gaultier-Villars, Paris.

• Gorroochurn, Prakash, 2012: Classic Problems of Probability. Wiley, ISBN: 978-1-118-06325-5