What a random chord is

Monte Carlo methods, as the first essay on them explained, turn a probability into a frequency by doing the same thing many times. There is a precondition that is usually so obvious that it goes unsaid: “the same thing” must have been described. It is a precondition that can quietly fail, and a question put by Joseph Bertrand in his Calcul des probabilités of 1889 is the standard demonstration.

Take a circle. Inscribe an equilateral triangle in it, whose side has length $\sqrt 3$ when the radius is 1. Choose a chord of the circle at random. What is the chance that the chord is longer than the side of the triangle?

Here are three answers, each from a reasoning that is hard to fault. By symmetry, fix one end of the chord at a corner of the triangle, and choose the other end at random on the rim. The chord is longer than the side exactly when the other end falls on the arc opposite that corner, a third of the rim: the chance is 1/3. Or, by symmetry, fix the direction of the chord, so that its midpoint lies on a fixed radius. It is longer than the side exactly when its midpoint lies in the inner half of the radius: the chance is 1/2. Or, a chord is determined by its midpoint, so choose the midpoint at random in the disc. The chord is longer than the side exactly when its midpoint lies inside a concentric circle of half the radius, which has a quarter of the area: the chance is 1/4.

The three pictures look different. With two random ends, the chords crowd towards the rim, and there is a thin bright ring. With a random point on a radius, the disc is shaded nearly evenly. With a random midpoint, the chords are thinnest across the centre, which is left almost clear, since a small central disc holds few midpoints. None of these is a mistake or an approximation, and if a computer is told to carry out each one, each gives its own answer every time.

One condition, three readings #

The three calculations agree about which chords are long. A chord is long exactly when its distance $d$ from the centre satisfies $d < 1/2$. That part involves no probability at all. A chord of a circle is determined by a direction and a distance from the centre, and each procedure spreads direction evenly. All of the disagreement is about the law of $d$.

These laws can be read directly from the arguments. Two random ends subtend a uniformly random angle, and $d = \cos(\theta/2)$, which gives the density $2/(\pi\sqrt{1 - d^2})$, heaped up towards $d = 1$, near the rim. A random point on a radius has uniform $d$, with density 1. A random midpoint in the disc falls within distance $d$ with probability $d^2$, with density $2d$, thin near the centre, because a small disc has little area. The areas to the left of $d = \tfrac12$ are $1/3$, $1/2$ and $1/4$. Sampling 200,000 chords from each lands on the curves, so the arithmetic is secure, and the paradox has nothing to do with arithmetic.

Bertrand’s conclusion was that the question is not well posed. “None of the three is false, none is correct, the question is badly set.” That is exactly right as far as it goes. A probability is a measure on a set of outcomes. For finite sets, there is often one natural choice, spreading mass evenly among possibilities. For a continuum there is no such thing as evenly spreading mass, until a coordinate has been named. A uniform distribution in one variable is not uniform in a function of that variable. A distribution even in $d$ is not even in $d^2$, and $d^2$ is the natural variable if what is being spread is area.

The disagreement extends over everything. At any length other than the side of the triangle, the three procedures give another three numbers. Their means are $4/\pi = 1.273$, $\pi/2 = 1.571$ and $4/3 = 1.333$. They are all confirmed by sampling. So “random chord” is not a slightly ambiguous description of a single object that needs tidying up. It names three unrelated objects, and there are infinitely many others: any choice of law for $d$ produces another, which can answer any value at all between 0 and 1. It is the same warning that an ambiguous sentence about two children raises in a discrete setting, and it is similar to what Monty Hall’s problem depends on, in the door that was not opened. In each case the answer depends on a procedure and the sentence did not specify it.

An argument for one answer #

For about eighty years, that was the accepted conclusion. Then in 1973 Edwin Jaynes published a paper called The Well-Posed Problem. He pointed out that one could take the question seriously as a question about a physical situation. Imagine someone tossing straws at a circle from far away. The result does not depend in any serious way on the size of the circle or on its exact position, since the person throwing the straws does not know those things very precisely. So a sensible rule should not change if the circle is moved or scaled. Jaynes argued that this requirement picks out one law.

His test is a good one to run. Suppose chords are drawn on a large circle by some procedure. Now take a small circle placed anywhere inside it. The chords of the large circle that hit the small circle give chords of the small circle. If the rule really describes “a random chord” without reference to any particular circle, then those chords should follow the same rule on the small circle.

That test is exact, with 400,000 samples each. Under the random-radius rule, the share of long chords, as judged by the small circle, stays at about 0.50 for all four small circles tried, centred or not, and of radius 0.5, 0.3 or 0.25. The midpoint rule is badly affected. It gives 0.248 on the centred small circle and 0.486 on the off-centre one. Two random ends is more subtle: it stays close to one half on these small circles, between 0.475 and 0.491, but it is not at one half, and its share depends on the small circle’s place and size. Only one rule passes.

This has a clean mathematical form. The measure $dp\,d\theta$ on lines, uniform in direction and in signed distance from a point, is the unique measure on lines that does not change under rotations and translations of the plane, up to a constant. The same measure is what turns length into a count of crossings in Crofton’s formula, and it is what makes Buffon’s needle give $\pi$. It is also a measure that behaves properly under scaling. So the random-radius rule is not one option among others in this view. It is the restriction, to lines through the circle, of the only measure on all lines that has these symmetries.

The simulated throw of the figure is just a sample from that measure: 2,000 straws at random direction and offset, of which 643 cross the circle, with a share of long chords of 0.493. Jaynes says he had the experiment performed, with broom straws and a circle on the floor, and he reported something near one half. The answer is $1/2$.

Why a small enough circle always says one half #

The numbers in that table have a pattern that is more interesting than the verdict. The midpoint rule gives 0.248 on the centred circle, and then 0.363, 0.473 and 0.486 as the small circle moves away from the centre and shrinks. The rule of two ends sits close to one half everywhere. Why should a rule that is not invariant get so close?

Every one of these rules can be written as a density on lines, relative to the invariant measure $dp\,d\theta$. For two random ends the density is proportional to $1/\sqrt{1 - p^2}$, and for the midpoint rule it is proportional to $|p|$, where $p$ is the distance from the large circle’s centre. Now look through a small circle. Across its width, a smooth density hardly changes. So lines through a small circle are close to being evenly distributed in offset, whichever smooth density they came from, and the share of long chords is then close to one half. The midpoint rule fails on the centred circle only because its density has a zero and a kink there, at $p = 0$, exactly where the small circle sits. Anywhere else, it is smooth, and as the small circle shrinks the share goes to one half.

So there is a stronger result behind Jaynes’s argument. A thrower who cannot aim to within the size of the circle produces a smooth density at that scale, and then the invariant measure emerges by itself. The result is not much affected by exactly how the straws are thrown. The idea is older than Jaynes. Henri Poincaré used it in his 1896 treatment of the roulette wheel, where a smooth distribution of initial speeds gives nearly equal chances to red and black, whatever that distribution is, because a small change of speed moves the ball through many slots. That is now called the method of arbitrary functions, and it accounts for why coins, dice and straws have stable frequencies. It is a physical answer to a question that looks purely conceptual: one half is what comes out because the circle is small compared with the thrower’s lack of control.

Is that the end of it? #

It would be wrong to claim that invariance settles all such questions, and Jaynes did not claim it. His argument answers a question with a physical setting and with no privileged circle. If instead a chord is produced in some other way, the answer depends on that. A rotating spinner with one fixed end genuinely gives the first procedure. A dart thrown at a board, with the chord being the one with the dart’s hit as midpoint, gives the third. The paradox is not “resolved” by Jaynes in the sense of showing the other two were mistakes. What he showed is that one of the three answers has a privileged status when nothing else is specified, and that the status comes from symmetry.

Critics have pointed out that invariance can be chosen in more than one way, and that a different group of transformations picks out a different measure. Bas van Fraassen discussed this in a general form with his cube factory: a machine produces cubes whose side lies between 1 and 2, so the volume lies between 1 and 8, and a uniform distribution in side is not uniform in volume, with no symmetry to tell which is right. Unlike the chord problem, there is no group acting there, and Jaynes’s argument has nothing to get hold of. So the method works exactly when the problem carries a natural group of symmetries, as this one does. Even then, the choice of group has to be justified from the setting.

This is related to the difficulties of the Bayesian approach, which needs a prior. A prior chosen for its invariance is what Harold Jeffreys proposed in 1939 for that reason, and it gives the same kind of answer as Jaynes’s. A method of assigning probabilities that does not depend on the choice of coordinate is a very attractive thing, and in many practical settings it does not exist or it has awkward consequences. So it isn’t surprising that this example still has a lively literature. There is also the observation that the paradox only occurs in continuous spaces. On a finite set with no structure, the symmetric group already gives one uniform distribution.

What it means for sampling #

The practical lesson is direct. In any Monte Carlo calculation, the first sentence of the method says what is uniformly distributed, and that sentence is a real modelling choice. Importance sampling makes that choice on purpose, and elsewhere it is made by accident. Sampling a surface or a set of rotations shows this at once: choosing latitude and longitude evenly does not spread points evenly over a sphere, since it crowds them at the poles. Sampling a direction by choosing two independent random angles does not give a random direction. Choosing a rotation by three random Euler angles does not give a random rotation. The right way to sample is the way that respects the group acting on the object being sampled, and it is often not the most obvious parameterisation.

The same issue arises for random lines through a region. A simulation of tomography, of radiation crossing a medium, or of lines that a dust must intercept, has to use the invariant measure, or it will answer a different question. In stereology, an area estimated from random line probes will be biased if the probes are not drawn from that measure, and the correction factor can be read from the three curves of this essay. The physical setting is usually the thing that decides what is correct. A beam fired from one fixed source gives chords with a common end, and the first of the three laws applies, not the second.

What the numbers rest on #

The exact values $1/3$, $1/2$ and $1/4$, the laws for the distance from the centre and the means were all derived above, and they were tested by sampling up to 200,000 chords. The invariance test uses 400,000 chords per procedure, with sampling error of about 0.003 in the shares. These are finite samples and four particular small circles, so they establish no invariance in general. They show failure of it for two of the rules, and they are consistent with it for the third, which is a theorem about the invariant measure.

The throw of straws is a simulation of a uniform choice of direction and offset, and not an experiment; it would not be evidence about straws even if it agreed. Jaynes’s account of a physical test is from his paper. It was informal and is not a controlled experiment. The discussion of van Fraassen and Jeffreys is a summary of positions and has been kept brief. The claim that there are infinitely many other laws is easy: any density on $[0, 1]$ for $d$ defines a chord rule.

Still open: when does a problem choose its own measure? #

The mathematical part is settled, and what remains is philosophical and methodological, though with real content. When a problem is stated with a symmetry group, under what conditions is there a unique invariant probability? For a compact group acting transitively, Haar measure answers this. For the group of motions of the plane acting on lines, there is a unique invariant measure up to scale, but it is infinite, and turning it into a probability requires restricting it to a set of lines, such as those hitting a particular convex shape. Then the answer depends on the shape in a known way. Beyond these cases, it is not clear what to do, and whether a principled general rule exists for choosing a measure, in a form that statisticians would accept for a prior, is still debated.

There is also a question of geometric probability with no philosophy in it at all. Given a convex body, what law of chord length results from the invariant measure? For a disc and for some polygons, there are exact formulas. For most shapes only moments are known, and the problem of recovering a shape from its chord-length law, a problem that arises in the analysis of materials, is known to be not always unique. Examples are known of different convex shapes with the same chord-length distribution, and it is not known how far that non-uniqueness extends.

A sentence that needs another sentence #

“A random chord” names no distribution, and three natural readings give 1/3, 1/2 and 1/4, with all three calculations correct and all three simulations convergent. Each spreads evenly a different quantity: the rim, the radius or the area. Their laws for the distance from the centre are different at every point. Jaynes observed that only the second is unchanged when the circle is moved or rescaled, because it is the unique invariant measure on lines, and so when nothing else is known about how the chord arises, one half has a reason behind it.

Bertrand wanted to show that a phrase could be empty, and Jaynes showed that it could be given content by saying what must not matter. Both are lessons for anyone about to write “choose at random” in the first line of a simulation. What should be uniform is never free of charge, and there is often no way to ask about it except by looking at the symmetry of the question.