Bertrand's Paradox Chord Simulator
Print PageThe Paradox Demystified
Joseph Bertrand introduced this paradox in 1889. It demonstrates that probabilities cannot be calculated without defining the random variables model exactly.
The three methods correspond to three different distributions of chord midpoints inside the circle:
- **Method 1** gives a higher concentration of midpoints towards the center.
- **Method 2** yields midpoints uniformly distributed along a radius.
- **Method 3** gives midpoints uniformly distributed across the 2D circle area.
Because the midpoint distributions differ, the resulting probabilities (1/3, 1/2, 1/4) are all mathematically valid for their respective models.
Bertrand’s Paradox is a famous foundational problem in continuous geometric probability theory first posed by French mathematician Joseph Bertrand in his 1889 treatise Calcul des probabilités. The paradox considers a circle of radius R containing an inscribed equilateral triangle (with side length L = R · √3 ≈ 1.73205 R) and asks a deceivingly simple question: What is the probability (P) that a randomly chosen chord of the circle is longer than the side of the inscribed equilateral triangle?
The paradox reveals that in continuous sample spaces, the phrase “chosen at random” is ill-defined without specifying a precise geometric selection mechanism. Depending on which mathematically valid definition of “random” is chosen, Bertrand proved there are three distinct, correct answers:
- Method 1: Random Endpoints (The Angle Method):
P1 = 1 ÷ 3 = 0.333333 (33.33%). - Method 2: Random Radius (The Radial Midpoint Method):
P2 = 1 ÷ 2 = 0.500000 (50.00%). - Method 3: Random Midpoint (The Concentric Area Method):
P3 = 1 ÷ 4 = 0.250000 (25.00%).
Our free online Bertrand’s Paradox Calculator allows you to simulate thousands of random chord draws across all three geometric methods, visualize probability density functions, and explore E.T. Jaynes’ 1973 Maximum Ignorance Invariance Principle.
Master Comparison of Bertrand’s 3 Geometric Solutions
The table below displays the mathematical definitions, geometric construction rules, chord length conditions, and exact probabilities for each of the three classic selection methods:
| Selection Method Name | Geometric Random Selection Mechanism | Condition for Chord Length > R√3 | Exact Calculated Probability | E.T. Jaynes Invariance Classification (1973) |
|---|---|---|---|---|
| Method 1: Random Endpoints | Fix one endpoint at triangle vertex A; pick second endpoint B uniformly along circumference (0 to 360°) | Endpoint B falls within the opposite 120° arc subtended by triangle sides | 1 ÷ 3 ≈ 33.33% | Rotational invariant, but NOT translationally invariant |
| Method 2: Random Radius | Choose a radius uniformly at random; pick a midpoint uniformly along the radius (0 to R) | Midpoint distance from circle center is less than R ÷ 2 (inner half of radius) | 1 ÷ 2 = 50.00% | UNIQUE Solution meeting Scale & Translational Invariance |
| Method 3: Random Midpoint | Choose a point anywhere inside the circle area (πR2) uniformly at random to serve as chord midpoint | Midpoint falls inside smaller concentric circle of radius r = R ÷ 2 (Area = πR2 ÷ 4) | 1 ÷ 4 = 25.00% | Area-uniform, but NOT scale invariant when tossing long straws |
Step-by-Step Proofs for Methods 1, 2, and 3
Method 1: Random Endpoints (The Circumference Angle Method)
Fix one endpoint of the chord at a vertex A of the inscribed equilateral triangle. To complete the chord, select a second endpoint B uniformly at random along the circle’s circumference (total angle 360°). The inscribed equilateral triangle divides the circumference into three equal arcs of 120° each. The chord AB will be longer than side L = R · √3 if and only if endpoint B lies on the arc opposite to vertex A.
P1 = Arc Length ÷ Total Circumference = 120° ÷ 360° = 1 ÷ 3 ≈ 33.33%.
Method 2: Random Radius (The Radial Distance Method)
Select a radius of the circle uniformly at random, and choose a point along this radius uniformly at random to serve as the midpoint of the chord. The chord is constructed perpendicular to the radius at this midpoint. The side of the inscribed equilateral triangle bisects the radius at a distance of exactly R ÷ 2 from the center. For the chord to be longer than L = R · √3, its midpoint must lie closer to the center than R ÷ 2.
P2 = Target Radius Segment ÷ Total Radius = (R ÷ 2) ÷ R = 1 ÷ 2 = 50.00%.
Method 3: Random Midpoint (The Concentric Circle Area Method)
Choose a point anywhere inside the circle of radius R uniformly at random to serve as the midpoint of a unique chord. For the chord to be longer than L = R · √3, its midpoint must lie strictly inside a smaller concentric circle of radius r = R ÷ 2 that is tangent to the sides of the inscribed triangle.
Area of outer circle: Aouter = π · R2.
Area of inner concentric circle: Ainner = π · (R ÷ 2)2 = π · R2 ÷ 4.
P3 = Ainner ÷ Aouter = (π · R2 ÷ 4) ÷ (π · R2) = 1 ÷ 4 = 25.00%.
1973 E.T. Jaynes Invariance Principle & Physical Straw Experiment
In 1973, American physicist Edwin Thompson Jaynes proposed a celebrated physical resolution to Bertrand’s Paradox based on the Principle of Maximum Ignorance. Jaynes argued that if a physical experiment (such as throwing long broom straws at a circle drawn on the floor from a distance) is truly random, the calculated probability must remain invariant under translation (shifting) and scaling (resizing) of the circle.
| Invariance Property | Method 1 (Endpoints) | Method 2 (Radius – Jaynes Choice) | Method 3 (Area Midpoint) |
|---|---|---|---|
| Rotational Invariance (Rotating Circle) | YES | YES | YES |
| Scale Invariance (Resizing Radius R) | NO | YES | NO |
| Translational Invariance (Shifting Circle) | NO | YES | NO |
Jaynes proved mathematically that Method 2 (P = 1/2 = 50.0%) is the unique probability distribution that satisfies both scale and translational invariance. When long straws are tossed randomly onto a target circle, the observed physical frequency converges to 50.0%.
History & Philosophy: 1889 Joseph Bertrand’s Calcul des probabilités
1889 Joseph Bertrand & Continuous Geometric Probability
In 1889, French mathematician Joseph Louis François Bertrand published Calcul des probabilités to caution mathematicians against blindly applying the Principle of Indifference to infinite or continuous sample spaces. While discrete probabilities (such as rolling dice or drawing cards) have clear finite outcomes, continuous spaces require a fully specified density function or physical mechanism before “uniform randomness” can be evaluated.
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Frequently Asked Questions (FAQ)
What is the correct answer to Bertrand’s Paradox?
There is no single “correct” answer in pure mathematics without specifying the random selection mechanism. The answers are 1/3 (33.33%) for random endpoints, 1/2 (50.0%) for random radius, and 1/4 (25.0%) for random midpoints. However, in physical real-world straw-tossing experiments, 1/2 (50.0%) is the unique solution that satisfies Jaynes’ translational invariance.
Why does the definition of “random” change the probability?
In continuous geometry, an infinite number of points or lines exist. “Randomly selecting a chord” can mean selecting uniform random angles along the perimeter, uniform radial distances from the center, or uniform points in two-dimensional space. Each mechanism imposes a different probability density function (PDF).
How long is the side of an inscribed equilateral triangle?
For a circle of radius R, the side length of an inscribed equilateral triangle is L = R · √3 ≈ 1.73205 R.