Bertrand Paradox Calculator

About the Bertrand Paradox Calculator

Bertrand's paradox, posed by Joseph Bertrand in 1889, asks a deceptively simple question: if you draw a random chord of a circle, what is the probability that it is longer than the side of an inscribed equilateral triangle? The paradox is that the answer depends on how you define "random chord" — three natural methods yield three different probabilities: 1/3, 1/2, and 1/4.

Method 1 (Random Endpoints) picks two points uniformly on the circumference and connects them, giving P = 1/3. Method 2 (Random Radial Point) picks a random radius and a random distance along it, then draws the perpendicular chord, giving P = 1/2. Method 3 (Random Midpoint) picks a random point inside the disk as the chord's midpoint, giving P = 1/4.

This calculator lets you compare all three methods analytically and through Monte Carlo simulation. You can also generalize from equilateral triangles to any regular inscribed polygon. The paradox demonstrates that "random" is not self-defining — the choice of probability measure changes the answer, a foundational issue in geometric probability and measure theory.

When This Page Helps

Bertrand's paradox is a cornerstone of probability education that highlights the importance of precise problem formulation. This calculator makes all three methods tangible through side-by-side comparison, visual bars, and Monte Carlo simulation that students can run repeatedly.

It's invaluable for courses on probability, measure theory, and the philosophy of randomness. The generalization to arbitrary inscribed polygons extends the learning beyond the classic triangle case, helping build deeper geometric intuition.

How to Use the Inputs

1. Select the chord generation method: Random Endpoints, Random Radial Point, or Random Midpoint.
2. Set the circle radius and the number of sides for the inscribed regular polygon (3 = triangle, the classic case).
3. Use the polygon presets to quickly switch between common shapes.
4. Review the probability outputs for the selected method and the comparison table for all three methods.
5. Enter simulation trial count and click Simulate to run a Monte Carlo verification.
6. Compare the theoretical and simulated results in the output cards.

Example Calculation

Tips & Best Practices

• Start with the triangle (3 sides) to see the classic paradox, then try other polygons to see how probabilities change.
• Run the simulation multiple times — each run gives slightly different results, illustrating sampling variability.
• Method 3 (random midpoint) is often considered the most physically realistic when throwing objects at a circle.
• As polygon sides increase toward infinity, all three methods' probabilities converge toward 100%.
• Use this paradox to understand why probability questions must specify the sampling mechanism.
• Jaynes' maximum entropy approach favors Method 2 — discuss this in the context of Bayesian vs. frequentist perspectives.

Historical Background

Joseph Bertrand introduced this paradox in his 1889 textbook "Calcul des probabilités" to argue that classical geometric probability was ill-defined without specifying a probability measure. His challenge undermined the principle of indifference — the assumption that without specific information, all outcomes should be treated as equally likely — because "equally likely" depends on how you parameterize the space of chords.

The paradox remained philosophically unresolved for decades. In 1973, physicist E.T. Jaynes proposed a resolution based on the principle of maximum ignorance (maximum entropy), arguing that the correct answer should be invariant under scale and positional transformations. His analysis favored Method 2 (P = 1/2), but this interpretation is not universally accepted.

The Three Methods in Detail

**Method 1 (Random Endpoints):** Choose two points uniformly on the circle's circumference. Fix one point; the other must fall in the arc subtended by the triangle's opposite side (1/3 of the circumference) for the chord to exceed the side length. P = 1 - 2/3 = 1/3.

**Method 2 (Random Radial Point):** Choose a random radius and a random point uniformly along it. The chord perpendicular to the radius at that point is longer than the triangle side when the point is closer to the center than r/2. P = 1/2.

**Method 3 (Random Midpoint):** Choose a random point uniformly in the disk. The chord with this midpoint is longer than the triangle side when the midpoint falls inside a concentric circle of radius r/2, which has area πr²/4. P = (πr²/4)/(πr²) = 1/4.

Implications for Modern Probability

Bertrand's paradox fundamentally shaped the development of measure-theoretic probability in the 20th century. Kolmogorov's axiomatization (1933) resolved the issue by requiring every probability model to explicitly specify a σ-algebra and probability measure. The paradox remains a powerful teaching tool for illustrating why mathematical rigor matters in probability and why intuitive appeals to symmetry or fairness can be misleading.

Frequently Asked Questions

• Why do different methods give different answers?

  Each method defines "random chord" using a different probability distribution. Random endpoints use uniform distribution on the circumference, random radial uses uniform distance along a radius, and random midpoint uses uniform distribution over the disk area. These are distinct probability spaces.

• Which method is "correct"?

  None is uniquely correct — that's the paradox. The answer depends on the physical process generating random chords. If you physically throw straws onto a circle, the midpoint method (P = 1/4) most closely matches experiments, but the "right" method depends on context.

• What does this paradox teach us?

  It demonstrates that specifying a probability problem requires defining the sample space and probability measure precisely. "Choose at random" is ambiguous when applied to continuous geometric objects — you must state the method of selection.

• How does the polygon affect the probability?

  As the number of sides increases, the inscribed polygon's side length decreases, making it easier for random chords to exceed it. In the limit of infinite sides, the polygon approaches the circle, and almost all chords are "longer" than the vanishingly small side.

• Is there a maximum entropy solution?

  Jaynes argued in 1973 that the Method 2 answer (1/2) is uniquely correct because it's the only solution invariant under translations and rotations — it satisfies the maximum ignorance principle. However, this is debated.

• How accurate is the simulation?

  With 10,000 trials, the simulation typically agrees with the theoretical value to within 1-2 percentage points. Increasing to 100,000 trials reduces the standard error by about √10 ≈ 3.16×.