Bertrand's Box Paradox Calculator - Solve Conditional Probability Problems

What is Bertrand's Box Paradox?

The Origins of the Paradox
Mathematical Foundation
Why It's Counterintuitive

Bertrand's Box Paradox, first described by Joseph Bertrand in 1889, is a classic problem in probability theory that demonstrates how our intuition about conditional probability can be misleading. The paradox involves three boxes with different configurations of gold and silver balls, and asks us to calculate the probability that a second ball is gold, given that the first ball drawn was gold.

The Classic Setup

In the original formulation, there are three boxes: Box 1 contains two gold balls, Box 2 contains one gold ball and one silver ball, and Box 3 contains two silver balls. A box is chosen at random, and then a ball is drawn from that box. If the first ball is gold, what is the probability that the second ball from the same box is also gold?

Many people initially think the answer is 1/2 (50%), reasoning that since we know the ball came from either Box 1 or Box 2 (Box 3 is eliminated), and these seem equally likely, there's a 50% chance it came from Box 1 (which would make the second ball definitely gold) and a 50% chance it came from Box 2 (which would make the second ball definitely silver).

Classic Box Configurations

Box 1: GG (2 gold balls)
Box 2: GS (1 gold, 1 silver)
Box 3: SS (2 silver balls)

Step-by-Step Guide to Solving the Paradox

Applying Bayes' Theorem
Conditional Probability Calculation
Common Misconceptions

The Correct Mathematical Approach

The key insight is that not all gold balls are equally likely to be drawn. Box 1 has two gold balls, so there are two ways to draw a gold ball from it, while Box 2 has only one gold ball, so there's only one way to draw a gold ball from it. This means that given we drew a gold ball, it's twice as likely to have come from Box 1.

Using Bayes' theorem, we can calculate: P(Box 1 | Gold ball drawn) = P(Gold ball | Box 1) × P(Box 1) / P(Gold ball). Since each box is equally likely to be chosen initially (probability 1/3), and the probability of drawing gold from Box 1 is 1, from Box 2 is 1/2, and from Box 3 is 0, we get:

P(Gold ball) = (1/3 × 1) + (1/3 × 1/2) + (1/3 × 0) = 1/3 + 1/6 = 1/2

Therefore: P(Box 1 | Gold ball) = (1 × 1/3) / (1/2) = 2/3, and P(Box 2 | Gold ball) = (1/2 × 1/3) / (1/2) = 1/3

Since the second ball is definitely gold if we're in Box 1 and definitely silver if we're in Box 2, the probability that the second ball is gold is 2/3 × 1 + 1/3 × 0 = 2/3 or approximately 66.67%.

Key Probability Results

P(Second ball is gold) = 2/3 ≈ 66.67%
P(Box 1 | Gold drawn) = 2/3
P(Box 2 | Gold drawn) = 1/3

Real-World Applications of Bertrand's Paradox

Medical Diagnosis and Testing
Quality Control in Manufacturing
Financial Risk Assessment

Medical Applications

The principles behind Bertrand's Paradox are crucial in medical diagnosis. When a patient tests positive for a disease, the probability that they actually have the disease depends not just on the test accuracy, but also on the disease prevalence in the population. This is exactly analogous to how the probability of drawing from Box 1 depends on the relative numbers of gold balls in each box.

Quality Control

In manufacturing, when a defective product is found, determining which production line it came from requires similar conditional probability reasoning. If different production lines have different defect rates, the line with higher defect rates is more likely to be the source of any particular defective item.

This paradox also appears in machine learning and data science, particularly in classification problems where we need to understand the posterior probability of class membership given observed features.

Practical Applications

Disease diagnosis with positive test results
Defect source identification in manufacturing
Classification algorithms in machine learning

Common Misconceptions and Correct Methods

The Equiprobability Fallacy
Importance of Prior Information
Monte Carlo Verification

Why Intuition Fails

The most common error is assuming that Box 1 and Box 2 are equally likely sources of the gold ball. This equiprobability assumption ignores the fact that Box 1 has twice as many gold balls as Box 2, making it a more likely source of any randomly drawn gold ball.

Another misconception is thinking about this as a simple 50-50 choice between two remaining boxes. The key insight is that we're not choosing between boxes; we're updating our beliefs about which box we drew from based on the evidence (the gold ball).

Simulation Verification

Monte Carlo simulations provide an excellent way to verify the theoretical result. By running thousands of trials where we randomly select boxes, draw balls, and track outcomes, we can empirically demonstrate that the probability converges to 2/3 rather than 1/2.

Theory vs. Intuition

Theoretical probability: 66.67%
Simulation results typically: 66.5% - 66.8%
Intuitive (wrong) answer: 50%

Mathematical Derivation and Advanced Examples

Formal Bayes' Theorem Application
Generalized Box Configurations
Connection to Other Paradoxes

Formal Mathematical Treatment

Let's define the events formally: Let B₁, B₂, B₃ represent the events that Box 1, 2, or 3 was chosen, and let G represent the event that a gold ball was drawn. We want to find P(second ball is gold | G).

Using the law of total probability: P(G) = P(G|B₁)P(B₁) + P(G|B₂)P(B₂) + P(G|B₃)P(B₃) = 1·(1/3) + (1/2)·(1/3) + 0·(1/3) = 1/2

By Bayes' theorem: P(B₁|G) = P(G|B₁)P(B₁)/P(G) = (1·1/3)/(1/2) = 2/3, and P(B₂|G) = P(G|B₂)P(B₂)/P(G) = (1/2·1/3)/(1/2) = 1/3

Generalization to n Boxes

The paradox can be extended to any number of boxes with arbitrary configurations. The key principle remains the same: boxes with more gold balls are more likely sources of any observed gold ball, proportional to their gold ball count.

For n boxes with gᵢ gold balls and sᵢ silver balls in box i, if we observe a gold ball, the probability it came from box j is: P(Bⱼ|G) = gⱼ / Σᵢ₌₁ⁿ gᵢ

Mathematical Formulas

P(B₁|G) = 2/3 using Bayes' theorem
P(B₂|G) = 1/3 using Bayes' theorem
General formula: P(Bⱼ|G) = gⱼ / Σgᵢ

Original Bertrand Paradox

Symmetric Configuration

Extended Box Problem

Minimal Paradox