Created by Davide Borchia
Reviewed by Anna Szczepanek, PhD and Adena Benn
Last updated: Jan 18, 2024

Bertrand's paradox begins with a simple set-up and question and leads you to make you doubt your confidence in mathematics! Are you ready to embark on this journey of exploration of statistics? No wrong answers: and we mean it; there will be no wrong answers!

• The solutions to Bertrand's paradox chord problem; and
• Why the probability of Bertrand's paradox solutions may have multiple correct answers (Bertrand's paradox and the principle of indifference).

Bertrand's paradox is a problem of probability where a single question results in three different, yet correct, answers, which depend on the approach used to answer.

We define Bertrand's paradox starting with a circle. Inside the circle, we inscribe an equilateral triangle, and we notice that the sides of the triangles are chords: Bertrand's paradox asks us what is the probability of a random chord drawn in the circle being longer than the side of the triangle.

🙋 The circle theorems calculator can help you introduce the concept of the chord of a circle if needed!

## The solution of Bertrand's paradox?

It's time to draw some chords: Bertrand's paradox asks us to find the probability of a chord being longer than the side of the triangle. And how do we obtain our chords? We can use three different approaches:

• Chords drawn starting from random endpoints;
• Chords drawn starting from a random radial point; and
• Chords drawn starting from a random midpoint.

Let's find the probability of Bertrand's paradox solution in each of these situations!

#### Bertrand's paradox solution with random endpoints

In this case, we notice that the chords in a circle are longer than the side of the triangle if, after we fix one of the endpoints in one of the triangles' vertexes, the other endpoint falls between the other two vertexes. You can see this situation in the image below.

PLACEHOLDER IMAGE

In this case, as the circumference is divided into three equally long arches, we can safely assume that a random point on the circle will fall in the desired region of the circumference, on average, one time out of three. The solution to Bertrand's paradox is a probability of $33.33\%$. Looks reasonable, right?

#### Solution of Bertrand's paradox: chords defined by a random radial point

We can identify chords that satisfy Bertran's paradox request by looking at the triangle's side. Look at all the chords parallel to one of the sides; their midpoints belong to the radius perpendicular to the side. The chord is longer than the side of the triangle if its midpoint falls between the center of the circle and the triangle's side, and vice-versa.

PLACEHOLDER IMAGE

As the side cuts the radius at mid-length, Bertrand's paradox's solution is a probability of $50\%$. Wait... is it different from the previous result? But the math checks out. What's going out?

#### Bertrand's paradox's solution with chords defined by a random midpoint

We can find a solution to Bertrand's paradox using a neat property of chords: they are univocally defined by their midpoint (if you need to refresh the concept, visit our midpoint calculator). Hence, we can sample chords by choosing points at random in the circle and consider each point the midpoint of a chord. When do these chords satisfy Bertrand's paradox request? It turns out that the midpoint must fall in a circle with a radius equal to half the radius of the original circle.

PLACEHOLDER IMAGE

We need a bit of math: the area of the original circle is $A = \pi\cdot r^2$, while the area of the circle where the midpoints define valid chords is $A = \pi\cdot (r/2)^2 = \pi \cdot r^2 /4$: each point has a $25\%$ probability of being in the smaller circle. This is another valid answer. Three solutions, three correct answers.

## Explaining Bertrand's paradox and the principle of indifference

Yes. Let's sum up the previous results: we found three solutions to Bertrand's paradox:

• $33.33\%$ for chords drawn from random endpoints;
• $50\%$ for chords drawn from a random radial point; and
• $25\%$ for chords drawn from a random midpoint.

What changed in these approaches? How did this arrive at three different results?

In all these cases, we sample over an infinite number of chords, each of them with an equal probability of being "drawn" (we say that the chords have a uniform distribution). This satisfies the so-called principle of indifference: each outcome is as likely as the others due to a lack of additional information.

Think of a six-faced dice: each face has a probability $1/6$ of being the outcome of a toss since we can't model the complex variables that affect the outcome of the toss. This is the principle of indifference in action.

How is it that even when we applied the principle of indifference in all the methods, we got different answers to the same question? Bertrand argued that when sampling over an infinite number of choices, asking for "randomness" is not enough to restrict the answer to a specific value. The different ways we sampled the chords (the way we generated them), albeit generating equally probable chords (random chords), change the outcome, and the original question posed by Bertrand remains unanswered without ambiguity.

We conclude by saying that Bertrand is right: all the answers are acceptable unless we are more specific in our question!

## How to simulate Bertrand's paradox

With our tool, you can play with Bertrand's paradox: its solutions, the explanations of the results, and simulations of the three possible approaches with any number of samples.

In the first part, you can challenge yourself, your friends, or your students to find the result for each possible approach.

🙋 Bertrand didn't stop here: he devised a plethora of paradoxes, in particular in the field of probability. We created a tool to allow you to simulate the intriguing Bertrand's box paradox; try it!

## FAQ

Bertrand's paradox is a probability problem highlighting the possible ambiguities that stem from sampling over an infinite domain of possibilities. Bertrand's paradox looks at the probability of a random chord in a circle being longer than the side of the inscribed triangle. It turns out that, depending on how we sample that infinite domain, the result can be either 25%, 33.33%, or 50% (in the original paradox), and in theory, any other value.

### What are the solutions to Bertrand's paradox?

In its original formulation, Bertrand's paradox has three solutions. The problem is finding the probability of a random chord of a circle being longer than the side of the inscribed triangle.

• If we sample the chords using the endpoints, we find that the probability is 33.33%.
• If we use the property that half the chords parallel to a triangle's side are longer than the side, the probability is 50%.
• If we sample the midpoint of the chords, we find that the probability is 25% (all the points are in the inscribed circle).

### How can I find the solution to Bertrand’s paradox using the endpoint?

To find the solution to Bertrand's paradox using the endpoint:

1. Draw the inscribe triangle to a circle.
2. Select one of the triangle's vertexes as starting point of the chord.
3. Count how many chords ending in a random point on the circumference would be longer than the triangle's side.
4. Only the 1/3 long section of the circumference opposite to the starting point contains points that create chords longer than the triangles's side: Bertrand's paradox has solution 1/3 (33.33% probability of the chord being longer than the side).

### How can Bertrand's paradox have three solutions?

Bertrands's paradox has three solutions because we don't specify the way we sample the infinite number of chords of a circle, thus leaving space for ambiguity. In the original solution of the paradox, Bertrand proposed three ways of sampling chords:

• Endpoints;
• Midpoints.

Each approach gives different results, as we found three valid but different distributions of the chords in our circle. When sampling over infinite spaces, the principle of indifference is not enough to give us a definitive answer.

### What is the probability in Bertrand's paradox?

The probability in Bertrand's paradox is not well-defined if we consider the original formulation of the problem: the probability of a random chord in the circle being longer than the side of the inscribed triangle. If this is the only information we have, we can't say precisely what the answer is: we need to specify how we sample these chords.

Davide Borchia
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Every chord is uniquely identified by its midpoint. Given a random point in the circle, what is the chance that the chord with such point as midpoint is longer than the inscribed triangle's side? Make a guess after taking a look at the picture below!
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