Bertrand's Paradox

What’s going on

A chord is a straight segment that joins two points of a circle. Draw an equilateral triangle inside the circle, with its corners on the circle. Now choose a chord “at random”. What is the probability that it is longer than a side of the triangle?

The question sounds complete, but it is not. “At random” does not say how the chord is chosen, and different reasonable ways give different answers. The scene above lets you try three of them. Press Draw one to watch a single chord being built, then Draw 1,000 to see the fraction settle.

Way 1: random ends — one third

Pick two random points on the circle and join them. Turn the picture so that a corner of the triangle sits at the first point. The chord is longer than the side exactly when the second point lands on the arc opposite that corner, between the other two corners. That arc is one third of the circle, so the answer is 1/3.

Way 2: random radius — one half

Pick a random radius, pick a random point on it, and draw the chord through that point at right angles to the radius. Turn the triangle so that one side is also at right angles to this radius. That side crosses the radius exactly half-way from the centre. The chord is longer than the side when the point lies on the inner half of the radius, so the answer is 1/2.

Way 3: random midpoint — one quarter

Pick a random point anywhere inside the circle and draw the chord that has this point as its midpoint. The chord is long when its midpoint is closer to the centre than half the radius, that is, inside a small circle of half the radius. That small circle has a quarter of the area, so the answer is 1/4.

So which is right?

All three. Each calculation is correct for its own method. Notice what they agree on: a chord is longer than the side exactly when its midpoint lies inside the small dashed circle. What differs is how the methods scatter the midpoints. Press Show midpoints. The random midpoint spreads them evenly, so a quarter land in the small circle, which covers a quarter of the disc. Random ends crowd more of them towards the centre (a third), and the random radius even more (a half).

A probability question only has an answer once the random experiment is described completely. That sounds like nit-picking, but it is exactly what goes wrong in real surveys and experiments, where the way you sample decides what you find.

You can also tap anywhere inside the circle to place a midpoint yourself and see the chord it makes.

Try it yourself

1. Longer than the radius, way 1

With random ends, what is the probability that the chord is longer than the radius?

Answer

Two out of three. A chord exactly as long as the radius cuts off an arc of 60°, because together with the centre it makes an equilateral triangle. So the chord is shorter than the radius only when the second end lands within 60° of the first, on either side: that is 120° out of 360°, a third of the circle. It is longer the other two thirds of the time.

2. Longer than the radius, way 3

Now use random midpoints. What is the probability that the chord is longer than the radius?

Answer

Three quarters. Take the radius as 1. A chord of length 1 is at distance √(1 − ¼) = √3/2 from the centre. A chord is longer than 1 when its midpoint is closer than √3/2, so inside a circle of radius √3/2. Its area is (√3/2)² = ¾ of the whole disc.

3. Straws on the floor

A circle is painted in the middle of a large floor, and you drop long straight straws at random all over the floor, with random directions. Some straws cross the circle and cut a chord. Which of the three methods does this match?

Answer

Way 2, random radius, so about half of the chords are long. For each direction, the straws are spread evenly across the floor, so the distance of a straw's line from the centre is spread evenly too. That is exactly what picking a point evenly on a radius does.

Go deeper

• Joseph Bertrand put this puzzle in his book on probability in 1889, as a warning that “at random” must always be made precise.
• In 1973 the physicist Edwin Jaynes argued that if nothing is known about the circle’s size or position, only one method treats all circles fairly, and that method gives 1/2.