Bertrand’s paradox challenges our understanding of probability by posing a seemingly straightforward question with multiple correct answers. The scenario begins with a circle inscribed with an equilateral triangle, where the sides of the triangle act as chords within the circle. The pivotal question of Bertrand’s paradox is: What is the probability of a randomly drawn chord being longer than the side of the inscribed triangle?

## Solutions to Bertrand’s Paradox Chord Problem

### 1. Chords drawn from Random Endpoints

When considering chords drawn from random endpoints, the probability of a chord being longer than the triangle’s side is found to be 33.33%. This result is based on fixing one endpoint at a triangle’s vertex and observing that the other endpoint falls within a specific section of the circumference.

### 2. Chords defined by a Random Radial Point

By examining chords parallel to one side of the triangle and selecting midpoints, a different solution emerges. In this case, the probability becomes 50%, as the midpoints fall within a specific region defined by the circle’s center and the triangle’s side.

### 3. Chords defined by a Random Midpoint

Taking advantage of the property that chords are uniquely defined by their midpoints, another solution is derived. The probability, in this case, is 25%, as each random midpoint has a 25% chance of falling within a circle with a radius equal to half the original circle’s radius.

## Explaining Bertrand’s Paradox and the Principle of Indifference

Bertrand’s paradox reveals the complexities that arise when sampling over an infinite set of possibilities. Despite applying the principle of indifference, which assumes equal probabilities for each outcome, three different solutions are obtained. Bertrand argued that the way chords are sampled, even when they have an equal chance of being selected, influences the final probability. The principle of indifference alone is insufficient to yield a definitive answer without additional specificity in the question.

You can explore Bertrand’s paradox further with our tool, simulating its solutions, explanations, and conducting simulations with any number of samples. Challenge yourself and others to find the result for each approach.

## FAQ:

A: Bertrand’s paradox is a probability problem exploring the ambiguities arising from sampling over an infinite domain. It focuses on the probability of a random chord in a circle being longer than the side of an inscribed triangle, with possible solutions of 25%, 33.33%, or 50%.

Q: What are the solutions to Bertrand’s paradox?

A: In its original form, Bertrand’s paradox has three solutions:

1. 33.33% for chords drawn from random endpoints.
2. 50% for chords drawn from a random radial point.
3. 25% for chords drawn from a random midpoint.

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

A: To find the solution using the endpoint, choose one vertex of the inscribed triangle as the starting point for the chord. The probability of the chord being longer than the side is 33.33%, as observed in a specific section opposite the starting point.

Q: Why does Bertrand’s paradox have three solutions?

A: Bertrand’s paradox has three solutions because it doesn’t specify the method of sampling chords from an infinite set. The three proposed approaches (endpoints, radial points, midpoints) lead to different valid distributions, emphasizing the importance of specificity when dealing with infinite spaces.

Q: What is the probability in Bertrand’s paradox?

A: The probability in Bertrand’s paradox is not precisely defined without specifying the chord sampling method. The original question, regarding the probability of a random chord being longer than the triangle’s side, requires additional details for a definitive answer.