Bertrand Box Paradox Calculator

About the Bertrand Box Paradox Calculator

The Bertrand box paradox, proposed by Joseph Bertrand in 1889, is one of the most famous thought experiments in probability theory. Three boxes each contain two coins: one box has two gold coins (GG), one has one gold and one silver (GS), and one has two silver coins (SS). You pick a box at random, draw one coin, and it's gold. What is the probability that the other coin in the same box is also gold?

Most people instinctively answer 1/2, reasoning that the gold coin must have come from either the GG or GS box (50-50 chance). But the correct answer is 2/3. The key insight is that you're equally likely to have drawn any of the three gold coins in the problem — two of which are in the GG box — so two out of three gold-draw scenarios lead to the other coin being gold.

This calculator lets you explore the paradox analytically and through Monte Carlo simulation. Vary the box configuration, see the coin-level enumeration that reveals the correct answer, and run thousands of simulated trials to verify the mathematics empirically. It's a powerful demonstration of why conditioning on evidence must account for the probability of observing that evidence from each hypothesis.

When This Page Helps

The Bertrand box paradox is a staple of probability education, yet many students and even professionals initially get it wrong. This interactive calculator makes the correct reasoning transparent by enumerating every possible coin draw and showing exactly why 2/3 beats the intuitive 1/2.

The Monte Carlo simulation feature is especially valuable — seeing the simulated proportion converge to the theoretical answer provides visceral confirmation of the mathematics. This calculator is ideal for probability courses, self-study, and anyone exploring Bayesian reasoning.

How to Use the Inputs

1. Select a box configuration: Classic (GG, GS, SS), 4-box variant, or the all-gold-heavy variant.
2. Examine the visual box display showing the coin colors in each box.
3. Review the analytical probability outputs — compare the correct answer with the common naïve answer.
4. Study the enumeration table to see every possible gold-coin draw and what the other coin would be.
5. Enter a number of simulation trials and click Run Simulation to verify the theoretical result empirically.
6. Compare the probability bars to see how the correct, naïve, and simulated probabilities differ.

Example Calculation

Tips & Best Practices

• Think at the coin level, not the box level — there are 3 gold coins, not 2 gold options.
• The same reasoning applies to the Monty Hall problem: you're conditioning on revealed information.
• Use the simulation with increasing trial counts (100, 1000, 10000) to watch convergence.
• Try the alternative box configurations to build intuition about how structure affects conditional probability.
• This is a direct application of Bayes' theorem: posterior ∝ prior × likelihood.
• Teaching tip: have students predict the answer before seeing the calculation — the surprise enhances learning.

The Classic Setup

Joseph Bertrand presented this problem to illustrate subtleties in conditional probability. Three closed boxes sit before you. Box 1 contains two gold coins, Box 2 contains one gold and one silver, and Box 3 contains two silver coins. You pick a box uniformly at random, reach in without looking, and draw one coin. It's gold. What's the probability the remaining coin in the same box is also gold?

The crucial insight is that "I drew gold" is more informative than it first appears. It tells you something about which box you picked. The GG box was certain to produce a gold coin, while the GS box had only a 50% chance. By Bayes' theorem, after observing gold, the GG box becomes twice as likely as the GS box.

Formal Bayesian Analysis

Let H_GG, H_GS, H_SS be the events that you picked each box. Let E be the event of drawing gold.

P(H_GG | E) = P(E | H_GG) × P(H_GG) / P(E) = 1 × (1/3) / (1/2) = 2/3 P(H_GS | E) = P(E | H_GS) × P(H_GS) / P(E) = (1/2) × (1/3) / (1/2) = 1/3

So given that you drew gold, there's a 2/3 chance you're in the GG box and a 1/3 chance you're in the GS box. The SS box is eliminated entirely.

Connection to Other Paradoxes

The Bertrand box paradox belongs to a family of conditional probability puzzles that share the same mathematical structure. The Monty Hall problem, the three prisoners problem, and various medical testing scenarios all require careful Bayesian updating. In each case, the common error is treating unequally likely hypotheses as equally likely after conditioning on evidence.

Understanding this family of problems builds strong probabilistic intuition that transfers to real-world decision making — from medical diagnosis (base rate neglect) to legal reasoning (prosecutor's fallacy) to everyday risk assessment.

Frequently Asked Questions

• Why isn't the answer simply 1/2?

  The 1/2 answer incorrectly treats the GG and GS boxes as equally likely given a gold draw. But you're twice as likely to draw gold from the GG box (certainty) than from the GS box (50% chance), so the GG box is weighted 2:1 over the GS box after observing gold.

• How is this related to the Monty Hall problem?

  Both paradoxes demonstrate that updating probabilities after observing evidence often yields counterintuitive results. In both cases, the naïve 50-50 split ignores how the evidence was generated. The mathematical structure is closely related through Bayesian updating.

• Does the Monte Carlo simulation always converge to 2/3?

  Yes, by the law of large numbers, the simulated proportion approaches 2/3 as the number of trials increases. With 10,000 trials, results typically fall within 1-2 percentage points of 66.67%.

• What changes with different box configurations?

  Changing the number or composition of boxes changes the conditional probability. With 4 boxes (GG, GS, SG, SS), the probability becomes 2/4 = 1/2 if GS and SG are treated as separate boxes with identical content — but the structure of the problem determines the answer.

• What is the key lesson of this paradox?

  The paradox teaches that conditional probability must account for the likelihood of the evidence under each hypothesis. You can't just count hypotheses — you must weight them by how likely each one is to produce the observed outcome.

• Who was Joseph Bertrand?

  Joseph Bertrand (1822-1900) was a French mathematician and economist. He contributed to probability theory, number theory, and thermodynamics. His 1889 book "Calcul des probabilités" introduced several famous paradoxes including this one.