Bertrand's Chord Paradox, math that surprises you lesson

Draw a circle and an equilateral triangle with its three corners on the circle. Now choose a chord of the circle at random. What is the chance that the chord is longer than a side of the triangle? Three natural ways of choosing the chord give three different answers, one third, one half and one quarter. This page works out each answer exactly and explains why the question has no single answer until it says how the chord is chosen.

Key terms

Bertrand’s chord paradox

Definition Bertrand’s chord paradox starts from a circle with center $O$ and radius $R$ and an equilateral triangle whose three corners lie on the circle. Each side of the triangle has length $\sqrt{3}\,R$ . The question is the probability that a chord chosen at random is longer than a side of the triangle. Joseph Bertrand posed it in 1889 and compared three procedures for choosing the chord.

Random endpoints. Choose two points of the circle independently, each uniformly, so that each point is as likely to land in one arc as in any other arc of the same length, and join them. The probability is $1/3$ .
Random distance along a radius. Choose a direction from $O$ uniformly, choose a distance $d$ uniformly from $0$ to $R$ , and draw the chord through the point at distance $d$ from $O$ in that direction, perpendicular to that direction. The probability is $1/2$ .
Random midpoint. Choose a point $M$ uniformly from the disk, which means that regions of equal area are equally likely to contain it, and draw the chord whose midpoint is $M$ . The probability is $1/4$ .

The resolution is that the words “chosen at random” do not say how likely each set of chords is. The three procedures can produce the same chords but give them different probabilities, so they are three different probability models, and each answer is correct for its own procedure. Probability theory is not contradicted, and none of the three is the one correct meaning of a random chord. The question has a definite answer only once a procedure is stated. Three special cases have probability $0$ and change none of the answers. These are a chord exactly as long as a side, a chosen distance of exactly $R$ , which gives a single touching point instead of a chord, and a midpoint exactly at $O$ , where any diameter may be drawn.

In plain words “At random” sounds like a complete instruction, but it is not. Each procedure is a different machine for making chords, and the machines make long chords at different rates. The circle, the triangle and the length test stay the same, and only the chances change.

Example Take $R = 2$ . The sides of the triangle then have length $2\sqrt{3} \approx 3.46$ , and each side lies at distance $1$ from the center. A chord is longer than a side exactly when it passes within distance $1$ of the center, as the next card shows.

For random endpoints, turn the picture so that the first endpoint is a corner of the triangle. Turning changes no lengths, and the second endpoint is still uniform on the circle. The chord is longer than a side exactly when the second endpoint lies on the arc between the other two corners that does not contain the first endpoint, and that arc is one third of the circle, so the probability is $1/3$ .

For a random distance, the chord is longer exactly when $d < 1$ , which is half of the interval from $0$ to $2$ , so the probability is $1/2$ .

For a random midpoint, the chord is longer exactly when $M$ lies within distance $1$ of the center. That smaller disk has area $\pi$ and the whole disk has area $4\pi$ , so the probability is

\frac{\pi \cdot 1^2}{\pi \cdot 2^2} = \frac{1}{4}.

Random chord

Definition A chord of a circle is a line segment whose two endpoints are different points of the circle, and a diameter is a chord through the center. Let the circle have center $O$ and radius $R$ , and let a chord lie at distance $d$ from $O$ , where $0 \le d < R$ . The perpendicular from $O$ to the chord meets it at its midpoint. When $d > 0$ , the right triangle formed by $O$ , the midpoint and one endpoint has a hypotenuse of length $R$ and legs $d$ and half the chord, and when $d = 0$ the chord is a diameter of length $2R$ . In both cases the chord has length

L = 2\sqrt{R^2 - d^2}.

Every point $M$ inside the circle other than $O$ is the midpoint of exactly one chord, the chord through $M$ perpendicular to the segment $OM$ , and $O$ is the midpoint of every diameter. A random chord is a chord produced by a stated random procedure. The words “a chord chosen at random” name no procedure, so on their own they do not say how likely any set of chords is.

In plain words Long chords pass close to the center, and short chords stay near the edge. How often a random chord is long depends on how often the procedure puts chords near the center, and that is exactly where procedures differ.

Example Take $R = 5$ . A chord at distance $d = 3$ from the center has half length $\sqrt{25 - 9} = 4$ , so $L = 8$ . For instance, the chord with midpoint $(3, 0)$ runs from $(3, -4)$ to $(3, 4)$ . At $d = 0$ the chord is a diameter of length $10$ , and at $d = 4$ it has length $6$ . At $d = 5/2$ , half the radius,

\begin{aligned} L &= 2\sqrt{25 - \tfrac{25}{4}} \\[0.4em] &= 5\sqrt{3} \approx 8.66, \end{aligned}

which is the side of the equilateral triangle with corners on this circle. Since $L$ shrinks as $d$ grows, a chord is longer than a side exactly when $d < R/2$ , the fact used in all three procedures.

Geometric probability

Definition Geometric probability finds probabilities by measuring. Suppose a point is chosen uniformly at random from a region $G$ of finite positive size, where size means length for a segment or an arc and area for a plane region. Uniformly means that the probability of landing in a part of $G$ is proportional to the size of that part, so for every part $A$ that has a size,

P(A) = \frac{\text{size of } A}{\text{size of } G}.

The rule needs two things. The point must be uniform, and size must be measured in the way the uniformity is meant, by length for a point on a segment or arc and by area for a point in a disk. A part of size zero, such as a single point of a segment or a curve inside a disk, has probability $0$ .

In plain words When every spot is equally likely, the chance of landing in a part is the fraction of the whole that the part takes up. The fraction depends on what is being measured, so “uniform by length” and “uniform by area” are different instructions.

Example For random midpoints in a disk of radius $2$ , the disk of radius $1$ has area $\pi$ out of $4\pi$ , so it gets probability $1/4$ . The ring between radius $1$ and radius $2$ has area $3\pi$ and gets $3/4$ , although it covers the same length of every radius as the inner disk. Under the random distance procedure the same two halves of the radius get $1/2$ each, because there the distance is uniform by length. For random endpoints the favorable arc has length $\tfrac13 \cdot 4\pi$ out of $4\pi$ , which gives $1/3$ . The circle of radius $1$ , where a chord is exactly as long as a side, has area $0$ , so “longer than a side” and “at least as long as a side” have the same probability in all three procedures.

Probability model

Definition A probability model of a random experiment has two parts. The sample space is the set of all possible outcomes. The probability rule gives each event, a set of outcomes, a number from $0$ to $1$ , so that the whole sample space gets $1$ and the probability of a union of disjoint events, finitely many or an infinite sequence, is the sum of their probabilities. When there are infinitely many outcomes, such as all chords of a circle, every single outcome may have probability $0$ , and the rule is given for events such as arcs, intervals and regions, often through lengths or areas. Two models can share a sample space and still give the same event different probabilities, so a probability question has a definite answer only after its model is fixed.

In plain words A model is the full description of the experiment, both what can happen and how likely each kind of result is. Listing the possible outcomes is only half of it.

Example Bertrand’s three procedures share one sample space, the set of chords of the circle, and every chord can come out of each of them. Their rules differ. The table gives, for each procedure, the probability that the chord’s distance from the center falls in each quarter of the radius.

Distance from the center	Random endpoints	Random distance	Random midpoint
$0$ to $R/4$	$0.161$	$0.25$	$0.0625$
$R/4$ to $R/2$	$0.172$	$0.25$	$0.1875$
$R/2$ to $3R/4$	$0.207$	$0.25$	$0.3125$
$3R/4$ to $R$	$0.460$	$0.25$	$0.4375$

The random distance procedure gives each quarter $1/4$ by design. For random midpoints the quarters are rings whose areas are in the ratio $1 : 3 : 5 : 7$ , because the disk of radius $kR/4$ has area $\pi k^2 R^2/16$ , so they get $1/16$ , $3/16$ , $5/16$ and $7/16$ . For random endpoints, let $\varphi$ be the angle at the center between the radii to the two endpoints, taken between $0$ and $180$ degrees. With the first endpoint fixed and the second uniform, $\varphi$ is uniform from $0$ to $180$ degrees, and the chord’s distance from the center is $R\cos(\varphi/2)$ . That distance is below $R/4$ when $\varphi/2$ is more than the angle whose cosine is $1/4$ , about $75.52$ degrees, so when $\varphi$ is more than about $151.04$ degrees, which has probability about $(180 - 151.04)/180 \approx 0.161$ . The other rows come the same way. The first two rows of each column add up to that procedure’s answer to the paradox, $1/3$ , $1/2$ and $1/4$ .

Check yourself

Keep the three procedures, but ask for the probability that the chord is longer than the radius $R$ , which is the side of a regular hexagon with its corners on the circle. What does each procedure give?

Show the answer

A chord is longer than $R$ when $2\sqrt{R^2 - d^2} > R$ , that is, when $R^2 - d^2 > R^2/4$ , which means

d < \frac{\sqrt{3}}{2}\,R.

For a random distance the probability is $\sqrt{3}/2 \approx 0.866$ . For a random midpoint it is the area fraction $(\sqrt{3}/2)^2 = 3/4$ . For random endpoints, turn the picture so that the first endpoint is a corner of the hexagon. The chords to its two neighboring corners are sides of length exactly $R$ , and the chord is longer than $R$ when the second endpoint lies on the arc between those two neighbors that passes the opposite corner. That arc is $4$ of the $6$ equal arcs between corners, so the probability is $2/3$ . The three answers again differ, $2/3$ , about $0.866$ and $3/4$ .

A program meant to simulate random midpoints picks a direction uniformly and then a distance $d = RU$ from the center, where $U$ is a random number chosen uniformly between $0$ and $1$ . About what fraction of its chords are longer than a side of the triangle, and how should the distance be chosen instead?

Show the answer

The distance $d = RU$ is uniform along the radius, so the program runs the random distance procedure, and about half of its chords are longer than a side, not a quarter. Its midpoints crowd toward the center. Points within $R/2$ of the center get probability $1/2$ , although they fill only $1/4$ of the area.

For points that are uniform by area, the probability that the distance is below $x$ must be the area fraction $x^2/R^2$ for every $x$ from $0$ to $R$ . The choice $d = R\sqrt{U}$ does this, because $R\sqrt{U} < x$ exactly when $U < x^2/R^2$ , which has probability $x^2/R^2$ . Then

\begin{aligned} P\left(d < \tfrac{R}{2}\right) &= P\left(U < \tfrac14\right) \\[0.4em] &= \tfrac14. \end{aligned}

Another correct method chooses the two coordinates of a point independently and uniformly from $-R$ to $R$ , which is uniform by area on the square around the disk, and discards the points that fall outside the disk.

Replace the triangle by a square with its corners on the circle, whose sides have length $\sqrt{2}\,R$ . Find the three probabilities. Two of them agree. Do those two procedures describe the same random chord?

Show the answer

A chord is longer than $\sqrt{2}\,R$ when $2\sqrt{R^2 - d^2} > \sqrt{2}\,R$ , that is, when $R^2 - d^2 > R^2/2$ , which means $d < R/\sqrt{2}$ . For a random distance the probability is $1/\sqrt{2} \approx 0.707$ . For a random midpoint it is $(1/\sqrt{2})^2 = 1/2$ . For random endpoints, put the first endpoint at a corner of the square. The chord is longer than a side when the second endpoint lies on the arc between the two neighboring corners that passes the opposite corner, which is half the circle, so the probability is $1/2$ .

Random endpoints and random midpoints both give $1/2$ here, but they are still different models. For the triangle they gave $1/3$ and $1/4$ , and in the table of the last card they differ in every quarter of the radius. Two models can agree on one event and disagree on others, so a single matching answer does not make them the same.

Bertrand's Chord Paradox

Video 1, chapter 13

Key terms