Bertrand's Paradox: Three Probabilities from One Question
Bertrand's Paradox asks: what is the probability that a randomly chosen chord of a circle is longer than the side of the inscribed equilateral triangle? The answer depends entirely on what you mean by "randomly chosen." Enter a circle radius below and explore how three equally valid sampling procedures - random endpoints, random radial point, and random midpoint - give three completely different probabilities: 1/3, 1/2, and 1/4.
Formula
Worked example
For r = 10: side = 10*sqrt(3) = 17.3205 units; critical distance = 5 units. Method 1: favorable arc = 120 of 360 deg, P = 1/3. Method 2: favorable length = 5 of 10, P = 1/2. Method 3: inner area = pi*25 = 78.54; full area = pi*100 = 314.16; P = 78.54/314.16 = 1/4. Check: chord at d=5 is 2*sqrt(100-25) = 2*sqrt(75) = 17.3205 = side length. Confirmed.
What is Bertrand's Paradox?
Bertrand's Paradox is a foundational puzzle in probability theory first posed by French mathematician Joseph Bertrand in 1889. The problem asks: given a circle with an inscribed equilateral triangle, what is the probability that a randomly chosen chord of the circle is longer than a side of the triangle? The surprising answer is that there is no single answer. Bertrand demonstrated three different but equally reasonable procedures for choosing a chord at random, each yielding a different probability - 1/3, 1/2, or 1/4. The paradox is not a logical contradiction, but rather a demonstration that "random" is not self-defining: you must specify the probability space (the procedure that generates the chord) before the question has a unique answer.
The Geometry: Why r/2 Is the Critical Distance
An equilateral triangle inscribed in a circle of radius r has its three vertices on the circumference, and its side length equals r*sqrt(3) (approximately 1.732r). A chord is longer than this side if and only if its midpoint lies within r/2 of the center. This follows from the chord-midpoint formula: chord length = 2*sqrt(r^2 - d^2), where d is the distance from the center to the midpoint. Setting this greater than r*sqrt(3) and solving gives d < r/2. This single geometric fact - the critical distance is exactly half the radius - drives all three probability computations, and explains why the inner circle in Method 3 has area exactly (1/2)^2 = 1/4 of the full disk.
The Three Sampling Methods and Their Probabilities
Method 1 (random endpoints, P = 1/3): Pick two independent uniform random points on the circumference. Fix one as an endpoint; orient the triangle so one vertex coincides with it. The chord exceeds the side only if the second endpoint falls on the 120-degree arc between the other two vertices, which is 1/3 of the full circumference. Method 2 (random radial point, P = 1/2): Choose a random point along a radius (uniform on [0, r]). Draw the perpendicular chord through it. The chord exceeds the side when the point is within r/2 of the center - covering exactly half the radius - so P = 1/2. Method 3 (random midpoint, P = 1/4): Pick any point uniformly inside the disk as the chord midpoint. The chord exceeds the side when the midpoint falls inside a concentric circle of radius r/2. That inner circle has area (r/2)^2 * pi = r^2*pi/4, which is 1/4 of the full disk area pi*r^2, so P = 1/4.
Jaynes' Invariance Argument and Why 1/4 May Be Canonical
The paradox exposes a limitation of the classical "principle of indifference," which holds that without a reason to prefer one outcome, all outcomes should have equal probability. The problem is that "equally likely" depends on how you parameterize the sample space: uniform over angles, distances, or areas are all defensible and incompatible. Statistician Edwin Jaynes (1973) proposed a resolution: the true answer should be invariant under all rigid motions of the circle (translations and rotations). Only Method 3 - uniform over area - satisfies this invariance criterion. Under Jaynes' argument, P = 1/4 is the canonical answer. Modern probability theory frames the full resolution differently - Bertrand's question is simply under-specified without a stated sampling mechanism - but the paradox remains a cornerstone example in foundations of probability, Bayesian prior selection, and the philosophy of statistical inference.
Bertrand's Paradox: Three Methods at a Glance
| Method | Physical procedure | Sample space | Probability |
|---|---|---|---|
| Method 1: Random endpoints | Two darts at a circular rim | Uniform over arc angle (0 to 360 deg) | 1/3 = 33.33% |
| Method 2: Random radial point | Spin a diameter; pick a point on it | Uniform over distance (0 to r) | 1/2 = 50.00% |
| Method 3: Random midpoint | Drop a pin anywhere in the disk | Uniform over area (disk of radius r) | 1/4 = 25.00% |
| Jaynes' canonical answer | Invariant under all rotations and translations | Method 3 satisfies this uniquely | 1/4 = 25.00% |
All three methods apply the principle of indifference, yet yield different probabilities because they define "random chord" differently.
Frequently asked questions
What exactly is Bertrand's Paradox?
Bertrand's Paradox is a classic probability puzzle showing that "choose a chord at random" is ambiguous. Depending on the procedure used to generate the chord, the probability that it exceeds the side of the inscribed equilateral triangle can be 1/3, 1/2, or 1/4 - three equally valid mathematical answers to the same question.
Which of the three probabilities is the "correct" answer?
All three are correct given their respective sampling procedures. Edwin Jaynes argued that Method 3 (random midpoint, P = 1/4) is canonical because it is the only method whose distribution is invariant under all translations and rotations of the circle. In practice, the right answer depends on the physical or experimental process that actually generates the chord.
Why does the triangle side length equal r * sqrt(3)?
An equilateral triangle inscribed in a circle of radius r has each side subtending a 120-degree central angle. The chord-length formula gives: 2r * sin(60 deg) = 2r * (sqrt(3)/2) = r * sqrt(3). For r = 10, the side is about 17.32 units.
Why is the critical distance exactly r/2?
A chord is longer than the triangle side (r*sqrt(3)) when its midpoint is at distance d from the center satisfying 2*sqrt(r^2 - d^2) > r*sqrt(3). Squaring both sides: 4(r^2 - d^2) > 3r^2, so r^2 > 4d^2, giving d < r/2. The threshold is exactly the midpoint of the radius.
Does the probability depend on the size of the circle?
No. All three probabilities (1/3, 1/2, 1/4) are pure geometric ratios, scale-invariant regardless of whether r is 1 mm or 1 km. The actual chord lengths and the critical distance r/2 scale linearly with r, but the probability ratios are dimensionless constants.
What is the principle of indifference and how does it relate to the paradox?
The principle of indifference assigns equal probability to outcomes when there is no reason to prefer one over another. Bertrand's Paradox shows that for continuous sample spaces this principle is ill-defined: uniform over angles, over distances, and over areas are all "equal" in their own sense, yet they are mathematically incompatible and give different probabilities.
Is there a real-world experiment corresponding to each method?
Yes. Method 1 corresponds to throwing two darts at the circular rim. Method 2 corresponds to dropping a pin randomly along a diameter. Method 3 corresponds to dropping a pin anywhere inside the disk and using that landing point as the chord midpoint. Each physical procedure genuinely produces a different probability.