Bertrand's book
Joseph Bertrand described the chord problem in his textbook "Calcul des probabilités", published in 1889. He gave three solutions to show that the question is badly posed, because it does not say how to pick the chord.
Draw a chord of a circle at random. What is the chance it is longer than the side of the inscribed equilateral triangle? Three careful solutions give 1/3, 1/2 and 1/4. All three are right, because "a random chord" means nothing until you say how you pick it.
The French mathematician Joseph Bertrand posed this problem in his 1889 book "Calcul des probabilités". He wanted to show that the words "at random" do not, by themselves, define a probability.
Take a circle of radius R with an equilateral triangle inscribed in it. The triangle's side is √3 · R ≈ 1.73 R.
A chord is a segment joining two points of the circle. The longest chord is a diameter, 2R long. Short chords run close to the edge.
Call a chord long if it is longer than the triangle's side. That is the same as its midpoint lying closer to the centre than R/2.
A chord is drawn at random. What is the probability that it is long?
"There is one answer, you just have to compute it correctly"
In fact three correct calculations give three different numbers. None of them contains a mistake. Each one describes its own way of picking the chord.
Each method looks natural. Rounds 2 and 3 show the answer each one gives.
Lock in your answer before reading on. Then compare it with the three ways of picking a chord.
Your answer will be compared with the three solutions here.
Pick a method and draw chords. Long chords are shown in magenta. A single chord shows how it was built. A hundred chords show that each method has its own share of long ones.
The bar shows your share. The tick shows the exact answer for that method: 1/3, 1/2 or 1/4. The more chords, the closer the bar gets to the tick.
Each method keeps its own chords. Switch between methods and compare the pictures: method 2 fills the centre more densely, method 1 crowds near the edge.
A chord is long exactly when its midpoint lies inside the smaller circle of radius R/2. So everything depends on where the midpoints land. Below are the midpoints for each method and the distribution of their distance from the centre.
Lower charts: the line is the exact density of the midpoint's distance from the centre, the dots are the simulation. Horizontal axis: distance from 0 to R; vertical axis: density.
Each answer has a short geometric argument. The calculations are all fine. They describe three different experiments.
Turn the circle so the first endpoint is at the top. The chord is long if the second endpoint lands on the lower arc between the triangle's other two vertices. That arc is 120° out of 360°, a third of the circle.
The chord is perpendicular to the radius and crosses it at the chosen point. The triangle's side crosses the radius exactly halfway, at distance R/2 from the centre. The chord is long if the point is closer to the centre, on the inner half of the radius. The chance is 1/2.
The chord is long if its midpoint lies inside the circle of radius R/2. That circle is inscribed in the triangle. Its radius is half as big, so its area is a quarter: π(R/2)² = πR²/4. A random point of the disc lands there with chance 1/4.
A probability is always computed for a specific experiment. "A random chord" names the outcome but not how it was produced. The three methods are three different experiments, so the answers differ.
The same thing shows up elsewhere. In the two-children problem the answer is 1/3 or 1/2 depending on how you learned that one child is a boy. In the Monty Hall problem switching pays because the host knows where the car is and always opens a door with a goat. If he opened a door at random and it happened to show a goat, the odds would be 1/2 and 1/2.
The three densities from round 3 differ, so the area under them from 0 to R/2 differs too. Method 1 often produces chords near the edge, which are short. Method 3 rarely puts the midpoint near the centre, because there is little area there.
Is one of the three answers "the most natural"? The physicist Edwin Jaynes argued that it is. You can also invent as many new methods as you like, and the same problem appears even with a random point in a square.
Jaynes reasoned like this. The problem says nothing about where the circle is or how big it is. So the answer should not change if the circle is rotated, shifted or scaled. He called this a principle of maximum ignorance: if we do not know something, the solution should not depend on it.
Only one distribution of lines passes this test, and it gives the same answer as method 2: 1/2. Methods 1 and 3 fail when the circle is shifted. Physically, this distribution arises when long straws are thrown onto the circle from far away: a straw "does not know" where the circle is. The simulation here does exactly that. Change the circle's size: the share of long chords stays near 50%.
Straw centres are spread uniformly over the square, with random directions. Because the square is finite, the exact share is slightly above 1/2, by no more than half a percentage point.
This is an argument for 1/2. It does not prove the others wrong. If the experiment really works like method 1 or 3, the right answer is 1/3 or 1/4.
Each row is one more "natural" way to draw a chord at random. The exact answer comes from a formula, and a simulation of 100,000 chords checks it. Press "Check" to run the simulation again.
| How we pick the chord | Exact |
|---|
Take a random radius and set the distance from the centre to the chord to d = R · uk, with u uniform from 0 to 1. With k = 1 this is method 2, with k = 1/2 it is method 3. The chord is long when uk < 1/2, which happens with chance (1/2)1/k.
The square contains a smaller square with the same centre and half the side. What is the chance that a random point lands in it? If you pick the coordinates x and y uniformly, the chance is the share of the area: 1/4. If instead you pick a random direction from the centre and go a random fraction of the way to the edge, the point lands in the small square when that fraction is under one half. The chance is 1/2.
The same 1/4 and 1/2 as in methods 3 and 2. The second procedure crowds points near the centre, just like the chord midpoints in method 2.
An even simpler example. A factory makes squares with a side from 1 to 2 cm. What is the chance the side is under 1.5 cm? If the side is random, 1/2. If the area is random between 1 and 4 cm², the area must be under 2.25 cm², and the chance is 1.25/3 ≈ 41.7%.
Joseph Bertrand described the chord problem in his textbook "Calcul des probabilités", published in 1889. He gave three solutions to show that the question is badly posed, because it does not say how to pick the chord.
The same book contains the "box paradox". There are three boxes: one with two gold coins, one with two silver coins and one with a gold and a silver coin. You draw a gold coin at random. The chance that the other coin in that box is also gold is 2/3. Intuition usually says 1/2. It is a close relative of the Monty Hall problem.
That was the title of a 1973 paper by the physicist Edwin T. Jaynes. He argued that if the answer must not change under rotations, shifts and changes of scale, Bertrand's problem has a single solution, 1/2. Not everyone agrees that these requirements are compulsory.
In 1933 Andrey Kolmogorov set out probability theory on axioms. In it a probability is always defined on a specific space of outcomes. Bertrand's paradox is often cited as an example of why such a precise definition is needed.
There are endless ways to pick a chord, and the answer can be any number from 0 to 1. You only need to choose the right distribution of the distance from the centre to the chord, as in round 5. Bertrand's three methods are just the best known.
Jaynes's straw throwing uses the same distribution of lines as Buffon's needle. It obeys Crofton's formula: a random line hits a convex shape with a probability proportional to its perimeter. This branch of mathematics is called integral geometry.
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