Bertrand's Box Paradox Simulator
Print PageWhy is the Probability 2/3 and not 1/2?
It is easy to think that once you see a Gold coin, you are in either Box 1 (GG) or Box 3 (GS), making it a 50/50 chance. However, you must count the individual coins rather than the boxes. There are 3 Gold coins in total across all boxes:
1. Box 1 Gold Coin A (other is Gold)
2. Box 1 Gold Coin B (other is Gold)
3. Box 3 Gold Coin C (other is Silver)
Since the drawn Gold coin is equally likely to be any of these 3, in 2 out of 3 cases (Coin A or Coin B), the remaining coin is also Gold. Therefore, the probability is 2/3 (66.67%).
Free Interactive Bertrand’s Box Paradox Simulator & Probability Solver
The Bertrand’s Box Paradox Simulator by CalculatorRank is an interactive probability modeling utility designed for data scientists, statistics educators, mathematics students, and puzzle enthusiasts. First formulated by French mathematician Joseph Bertrand in 1889, this classic conditional probability paradox demonstrates how human intuition frequently stumbles when calculating probabilities with partial information.
The problem presents three identical boxes with two coin drawers each:
- Box 1 (GG): Contains 2 Gold coins.
- Box 2 (SS): Contains 2 Silver coins.
- Box 3 (GS): Contains 1 Gold coin and 1 Silver coin.
If you choose a random box and open a drawer revealing a Gold coin, what is the probability that the other drawer in that same box also contains a Gold coin? While intuition commonly falls into the Equally Likely Fallacy and assumes a 50/50 chance (1/2), rigorous mathematical proofs and Monte Carlo simulations prove that the true probability is exactly 2/3 (≈ 66.67%).
Interactive Monte Carlo Simulation Capabilities
Our simulation engine allows you to test empirical frequencies against mathematical theory:
- Single Step Trial (1 Draw): Visually step through individual random box selections and coin drawer revelations.
- Batch Simulations (1,000 to 10,000 Trials): Run high-speed Monte Carlo simulations to observe empirical convergence toward the theoretical 66.67% (2/3) success rate.
- Complete Outcome Tracking: Real-time tallies of Total Boxes Selected, Valid Gold-First Cases, Discarded Silver-First Cases, and Second-Coin Successes.
How to Use the Paradox Simulator
- Select Trial Volume: Click Draw Coin (1 Trial) for manual demonstration, or click Run 1,000 Trials / Run 10,000 Trials for rapid automated execution.
- Track Valid Gold Cases: When the first drawer revealed is silver, the trial is automatically discarded (since the paradox conditions are not met).
- Review Success Rate: Observe the calculated empirical success percentage (
Second Coin was Gold / Valid Cases) converging on 66.67%. - Reset Statistics: Click Reset Statistics to clear previous run tallies and initiate a fresh test sequence.
Master Coin Drawer Sample Space Breakdown
The fundamental key to solving Bertrand’s Box Paradox is counting individual coins rather than entire boxes. Across the 3 boxes, there are 6 total drawers containing 3 Gold coins and 3 Silver coins:
| Box & Contents | Drawer A Coin | Drawer B Coin | Prior Selection Chance | First Coin Drawn is Gold? | Other Drawer Coin Outcome |
|---|---|---|---|---|---|
| Box 1 (GG – Double Gold) | Gold 1 (G1) | Gold 2 (G2) | 1/3 (33.33%) | YES (G1 or G2) | GOLD (2 of 3 valid gold draws) |
| Box 2 (SS – Double Silver) | Silver 1 (S1) | Silver 2 (S2) | 1/3 (33.33%) | NO (Discarded) | N/A (Invalid state) |
| Box 3 (GS – Mixed) | Gold 3 (G3) | Silver 3 (S3) | 1/3 (33.33%) | YES (Only if G3 drawn) | SILVER (1 of 3 valid gold draws) |
Mathematical Solutions: Counting vs. Bayes’ Theorem
Method 1: Direct Sample Space Counting
When you open a drawer and reveal a Gold coin, you are holding one of 3 equally likely physical Gold coins:
- Scenario A: You drew Gold Coin 1 (Box 1). The opposite drawer holds Gold Coin 2. → Success (Gold)
- Scenario B: You drew Gold Coin 2 (Box 1). The opposite drawer holds Gold Coin 1. → Success (Gold)
- Scenario C: You drew Gold Coin 3 (Box 3). The opposite drawer holds Silver Coin 3. → Failure (Silver)
Because 2 out of the 3 possible starting Gold coins belong to Box 1, P(Other is Gold) = 2 ÷ 3 ≈ 66.67%.
Method 2: Rigorous Bayesian Updating Formula
Let B1 = Box 1 (GG), B2 = Box 2 (SS), and B3 = Box 3 (GS).
- Prior Probabilities:
P(B1) = P(B2) = P(B3) = 1/3 - Likelihood of drawing Gold:
P(Gold | B1) = 1.0,P(Gold | B2) = 0.0,P(Gold | B3) = 0.5 - Total Probability of Gold:
P(Gold) = (1.0 × 1/3) + (0.0 × 1/3) + (0.5 × 1/3) = 1/3 + 1/6 = 1/2
Applying Bayes’ Theorem to determine P(B1 | Gold):
P(B1 | Gold) = [P(Gold | B1) × P(B1)] ÷ P(Gold) = (1.0 × 1/3) ÷ (1/2) = (1/3) ÷ (1/2) = 2/3 (≈ 66.67%).
Observing a Gold coin provides decisive evidence that makes Box 1 twice as probable as Box 3.
Bertrand’s Box vs. Monty Hall & Three Prisoners Paradoxes
Bertrand’s Box Paradox is mathematically isomorphic to several classic probability puzzles:
| Probability Paradox | Intuitive Incorrect Fallacy | Correct Probability | Isomorphic Equivalence |
|---|---|---|---|
| Bertrand’s Box Paradox (1889) | 1/2 = 50% (Box 1 vs Box 3) | 2/3 ≈ 66.67% | Drawing Gold identifies 1 of 3 possible gold coins. |
| Monty Hall Problem (1975) | 1/2 = 50% (Stay vs Switch) | 2/3 ≈ 66.67% (Switching) | Host revealing a goat provides identical conditional filtering. |
| Three Prisoners Problem (1959) | 1/2 = 50% (Pardon chance) | 2/3 ≈ 66.67% (Other prisoner) | Warden naming an executed inmate updates conditional priors identically. |
Historical Context: Joseph Bertrand (1889)
French mathematician and educator Joseph Louis François Bertrand introduced this problem in his 1889 work Calcul des probabilités. Bertrand constructed the paradox to demonstrate how informal intuition fails when individuals confuse the probability of picking a box (which is 1/3) with the conditional probability of selecting a specific coin once an outcome is observed.
Today, Bertrand’s Box Paradox remains the standard benchmark for exposing the Equally Likely Fallacy—the cognitive bias of assuming two remaining unknown outcomes must be equally split (50/50).
Frequently Asked Questions (FAQ)
Why is the answer to Bertrand’s Box Paradox 2/3 and not 1/2?
Because there are 3 total Gold coins across the three boxes. Two of those Gold coins are located inside Box 1 (GG), while only one is in Box 3 (GS). When you draw a Gold coin, there is a 2 in 3 chance that you drew from Box 1, meaning the other drawer is also Gold.
Is Bertrand’s Box Paradox identical to the Monty Hall Problem?
Yes. Mathematically, Bertrand’s Box Paradox is structurally identical (isomorphic) to the Monty Hall Problem. Revealing a Gold coin provides the exact same conditional probability update as the game show host revealing a goat behind an unchosen door.
What is the Generalized Formula for K Gold Coins?
If Box 1 contains K Gold coins, Box 2 contains K Silver coins, and Box 3 contains 1 Gold coin and K − 1 Silver coins, the probability that the chosen box is Box 1 given that a Gold coin was drawn is: P = K ÷ (K + 1).
What is the Equally Likely Fallacy?
The Equally Likely Fallacy is the cognitive bias where people assume that having two remaining possibilities (Box 1 vs. Box 3) means they must be 50/50. It ignores the fact that Box 1 had twice as many opportunities to produce the observed Gold coin as Box 3.