Introduction

Bertrand’s Box Paradox is a captivating probability problem that challenges our common sense understanding of odds. In this comprehensive article, we’ll delve into the formulation, rules of the game, and the counterintuitive solution to this intriguing puzzle. Let’s explore the paradox step by step and understand why the intuitive answer might lead us astray.

The Setup

Imagine three boxes, each containing a mix of gold and silver coins:

1. Box 1: Two gold coins
2. Box 2: Two silver coins
3. Box 3: One gold coin and one silver coin

The boxes are shuffled, setting the stage for Bertrand’s Box Paradox.

Now, if you randomly select a box and draw a coin from it, discovering it to be gold, what is the probability that the other coin in the box is also gold? At first glance, it might seem logical to assume a 50% chance, but as we’ll unravel, this common sense intuition is incorrect.

Bertrand’s Insightful Solution

Bertrand addressed the paradox by challenging the common-sense solution. His key insight involves imagining the selection of a box without drawing a coin. The problem is symmetric, allowing us to switch gold for silver without altering probabilities.

Consider the three possible outcomes leading to a gold coin being drawn. If you chose the box with a gold and silver coin, there’s no chance of the other coin being gold after the first extraction. This eliminates one of the three gold coins. The remaining two coins, if drawn, result in the other coin being gold. Each of these outcomes has a probability of 1/3.

Therefore, the correct solution to Bertrand’s Box Paradox is that the probability of finding a gold coin after the first extracted coin is gold is 2/3, not 1/2 as one might intuitively expect.

Bayes’ Rule and Bertrand’s Box Paradox

Mathematicians extended the understanding of Bertrand’s Box Paradox using Bayes’ rule. This statistical tool allows us to calculate conditional probability without explicitly considering possible outcomes. The formula for Bayes’ rule is as follows:

P(GGG)=P(G)P(GGG)⋅P(GG)​

Here:

• P(GGG) is the posterior probability, representing the chance of finding a gold coin given that the first coin drawn is gold.
• P(GGG) is the likelihood, denoting the probability of drawing gold from a box containing two gold coins (which is 1).
• P(GG) is the prior probability, indicating the likelihood of choosing the box with two gold coins (1/3).
• P(G) is the normalization factor, the probability of drawing a gold coin (1/2).

Plugging in these values, we find that P(GGG)=2/3, confirming the paradox’s solution.