Striking the right chord

Paula Rowińska

14 March 2019

• Click to share on X (Opens in new window)
• Click to share on Facebook (Opens in new window)
• Click to share on Reddit (Opens in new window)
• Click to share on LinkedIn (Opens in new window)
• Click to share on Threads (Opens in new window)
• Click to share on Mastodon (Opens in new window)

Go ahead and try to solve this little problem! Or, even better, grab a friend (or enemy) and think about the problem together.

The first thing you might try to do is compute the length of the sides of the inscribed equilateral triangle: you should find it is $\sqrt{3}$. Let’s denote the length of the random chord by $L$. To find the probability that $L$ is bigger than $\sqrt{3}$, which we shall denote $\mathbb{P}(L>\sqrt{3})$, we need to divide the number of ‘desired’ outcomes (chords such that $L>\sqrt{3}$) by the number of all possible outcomes (chords of any length). I would not recommend drawing all possible chords of the circle; for problems with infinitely many possible outcomes we need a slightly different approach than simply counting infinitely many objects.

Geometric probability

Luckily, we can use a very handy tool called geometric probability. It allows us to calculate probabilities in terms of size: length, area, volume etc.

For example, imagine that we want to calculate the probability that a real number $M$ chosen at random between $0$ and $1$ is bigger than $0.6$. Remember that we can represent real numbers on a line, so we identify picking the number $M$ with picking a point on a line segment of length $1$. In this case we need to divide the length of `desired’ outcomes by the length of all outcomes:

$$
\mathbb{P}(M>0.6)=\frac{0.4}{1.0}=0.4
$$

Too easy?

Unit cube intersecting the unit sphere

Your chosen numbers $A$, $B$ and $C$ represent the coordinates of the point $(A,B,C)$ in the region $[0,1]^3$, ie in the solid unit cube with one vertex at $(0,0,0)$. Hence the volume of all possible choices of three numbers equals the volume of the cube: exactly $1$. What is the volume of desired outcomes? We want $A^2+B^{\hspace{1pt}2}+C^{\hspace{1pt}2} \leq 1$, which describes the solid unit sphere centred at $(0,0,0)$ with volume $4\mathrm{\pi}/{3}$. Since $A$, $B$ and $C$ are bigger than $0$, we actually take only $(1/2)^3=1/8$ of this sphere, the part which intersects the cube, so the `desired’ volume equals $1/8 \cdot 4\mathrm{\pi}/{3}= \mathrm{\pi}/{6}$. Now we can compute our probability:

$$
\mathbb{P}(A^2+B^{\hspace{1pt}2}+C^{\hspace{1pt}2} \leq 1)=\frac{\pi/6}{1}=\frac{\mathrm{\pi}}{6}\approx 0.524
$$

Striking a chord

We can pick endpoints of the chord at random

Now we are ready to tackle the original problem. Because we only care about the length of the chord, the problem is rotationally symmetric, which means that we can rotate the circle and the triangle however we fancy without changing the answer. To make things easy, let us assume that the chord starts at one of the triangle’s vertices. We pick randomly its other end $X\hspace{1pt}$ somewhere on the circumference of the circle. The vertices of our triangle divide the circle into three arches of equal length. The chord is longer than the side of the triangle if and only if $X\hspace{1pt}$ lies somewhere in the blue arc, so

\[\mathbb{P}(L>\sqrt{3})=\frac{1}{3}.\]

Maybe you chose your random chord by a different method. To be safe we should check other methods to ensure we get the same answer.

Discord to our harmony

We can pick a chord at random perpendicular to a radius

Every chord is perpendicular to a unique radius (except a diameter which is perpendicular to two radii), so using the symmetry of the problem let us draw the top vertical radius and inscribe an equilateral triangle with one side perpendicular to this radius. We pick a random point $X$ on the radius and draw the chord parallel to this triangle’s side, ie perpendicular to the chord. The chord is longer than the side of the inscribed triangle if and only if it intersects the radius somewhere in the blue segment between the triangle’s side and the centre of the circle. This segment is exactly half of the radius (check!), so
\[\mathbb{P}(L>\sqrt{3})=\frac{1}{2}.\]

Wait… it seems that we made a mistake, because we got a different answer! Well, luckily we can think of more methods of choosing a random chord, surely if we try another one we will find out which answer is correct.

We can pick the midpoint of the chord at random

What if we pick at random a point $X\hspace{1pt}$ inside the circle, which will become the midpoint of the chord? We can draw a radius going through this point and construct the unique chord perpendicular to this radius. OK, you got me! Uniqueness has one exception: if we happen to choose the centre of the circle, this is the midpoint of infinitely many chords (diameters, in fact), but they all have the same length, so this does not cause a problem.

Our chord is longer than the side of the triangle if and only if its midpoint is closer to the centre of the circle than to its circumference, ie it lies in the blue disk. Look again at the previous method to understand why this is the case. Now we just need to divide the area of the blue disk by the total area of the circle, which gives us
\[\mathbb{P}(L>\sqrt{3})=\frac{\mathrm{\pi}\cdot(1/2)^2}{\mathrm{\pi}\cdot 1^2}=\frac{1}{4}\]

This has not fixed our problem at all; instead of confirming whether $1/3$ or $1/2$ is the right answer, we have got yet another different solution.

What is going on?

Joseph Bertrand. Image: public domain.

Do not waste your time trying to find mistakes in these three solutions. Since 1889, when the French mathematician Joseph Bertrand described this paradox in Calcul des probabilités, generations of mathematicians have come to the same conclusion: the probability that a random chord is longer than a side of an inscribed equilateral triangle equals $1/2$, $1/3$, $1/4$, or maybe something else, depending on the method of generating the random chord.

This example warns us that analysing probabilities without a clear statement of the problem can lead to disaster. To work with random events, we need to build models, which can be simple or very complex. Different models can give us different answers, as we just observed.

‘Choose a chord of the circle at random’ is not a strict, mathematical formulation. For mathematicians the phrase ‘at random’ must be followed by specification of some probability distribution, a function assigning a number between $0$ and $1$ to all possible outcomes. The formulation of our question allows for many possible probability distributions, we have just mentioned three of them. However, we can state the problem differently: `Pick independently at random two points on the unit circle, uniformly distributed on the circumference, and draw a chord between them. What is the probability that the chord is longer than $\sqrt{3}$?’ Now we can obtain the unique answer $1/3$. Different formulations would give us different answers, as we saw before.

Which solution is correct then? All of them. This is the beauty of mathematics: we can model the same phenomenon in many different ways, depending on our needs, as long as we clearly state our assumptions. We get different answers to different questions, so we do not have any contradictions here. If $1/2$ were equal to $1/3$ and $1/4$, we would face the beginning of the end of mathematics. Thankfully, long ago, mathematicians realised the importance of making sure their problems were strictly defined and well posed.

I believe that everyone should see this example at the beginning of their mathematical journey. I used to think that maths teachers developed a weird obsession with unnecessary formalities and I could not understand why I had to waste so much time writing down all the `obvious’ facts. After getting a few different answers to Bertrand’s problem I never complained about mathematical rigour again.

Paula Rowińska

As a PhD student at Imperial College, London, Paula uses maths and stats to study the impact of wind energy on electricity prices. The title of her TEDx talk `Let’s have a maths party!’ seems to summarise her two favourite activities.

@PaulaRowinska   paularowinska.wordpress.com/    All articles by Paula