Free Bertrand's Paradox Calculator

The Bertrand paradox is a famous problem in geometric probability that demonstrates how an apparently simple question can generate multiple contradictory yet mathematically valid answers. At the heart of the paradox is a circle with an inscribed equilateral triangle. The question is: if you draw a random chord inside the circle, what is the probability that this chord is longer than a side of the triangle? Depending on how the random chord is generated, three different probabilities emerge. This Bertrand Paradox Simulation tool lets you interactively explore each method and see the principle of indifference in action.

The Setup of the Paradox

Imagine a circle of any size. Inside it, draw an equilateral triangle whose vertices touch the circle’s circumference. Each side of the triangle is a chord of the circle. Now consider all possible chords that can be drawn across the circle. The paradox asks for the chance that a randomly selected chord exceeds the length of one triangle side. The catch is that “random chord” is not a uniquely defined concept—different random processes give different probabilities.

The Three Classic Solutions

The original paradox, posed by Joseph Bertrand, introduces three distinct sampling methods. Each method follows a natural randomization procedure, and each produces a different probability. The tool simulates all three, allowing you to visualize the results with any number of samples.

1. Random Endpoints

Fix one endpoint of the chord at a vertex of the inscribed triangle. Choose the second endpoint uniformly at random along the circumference. The chord will be longer than the triangle’s side only when the second endpoint falls on the arc of the circle that lies opposite the chosen vertex. That arc spans exactly one‑third of the total circumference because the equilateral triangle divides the circle into three equal arcs. Therefore, the probability is:

$$P_{\text{endpoints}} = \frac{1}{3} \approx 33.33\%$$

This is the first valid answer.

2. Random Radial Point

For this method, consider all chords that are parallel to a given side of the triangle. Their midpoints lie on a radius that is perpendicular to that side. A chord is longer than the triangle’s side if its midpoint falls between the circle’s center and the point where the triangle side intersects that radius. Because the side cuts the radius exactly at its midpoint, exactly half of these midpoints yield a chord longer than the side. The probability becomes:

$$P_{\text{radial}} = \frac{1}{2} = 50\%$$

Again, the reasoning is internally consistent.

3. Random Midpoint

Every chord can be identified uniquely by its midpoint (provided the midpoint is not the center of the circle). By selecting a point uniformly at random inside the circle and treating it as the midpoint of a chord, we get a third sampling scheme. For the chord to be longer than the triangle’s side, the midpoint must lie inside a smaller concentric circle whose radius is half the original radius. The area of this smaller circle is:

$$A_{\text{small}} = \pi \left(\frac{r}{2}\right)^{2} = \frac{\pi r^{2}}{4}$$

The original circle has area $A_{\text{large}} = \pi r^{2}$. Hence the probability is the ratio of these areas:

$$P_{\text{midpoint}} = \frac{A_{\text{small}}}{A_{\text{large}}} = \frac{1}{4} = 25\%$$

Thus the third answer.

Why the Principle of Indifference Is Not Enough

In each of the three methods, the chords are drawn “at random” and every possible chord has an equal chance of being selected in its own sampling framework—this satisfies the statistical principle of indifference. The principle says that when we have no reason to favor one outcome over another, we assign equal probabilities. Yet the three approaches give three different answers. The explanation lies in the infinite set of possible chords. When the sample space is infinite, specifying “random” does not uniquely define a probability measure; the method of selection matters. Bertrand’s paradox shows that without a precise description of the generating process, the question is ambiguous. All three answers are correct under their respective sampling schemes, and other schemes could produce still other values.

Simulating the Paradox with This Calculator

This Probability Paradox Calculator enables you to run each of the three random‑chord experiments interactively. You can choose the number of trials, observe the frequencies, and compare them to the theoretical probabilities of 33.33 %, 50 % and 25 %. The simulation also visualizes the geometric constraints—the arcs, the radial segments, and the concentric midpoints region—making the reasoning concrete. By experimenting with different sample sizes, you can see how the empirical results converge to the predicted values for each method.

The tool serves as a hands‑on way to understand that the same question can have multiple valid probabilities when the sampling procedure is not fully specified. It illustrates why the principle of indifference alone cannot resolve ambiguities in continuous probability spaces and highlights the need for an unambiguous definition of randomness in any real‑world application.