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%).
Bertrand’s Box Paradox is a famous classic paradox in elementary probability theory first posed by French mathematician Joseph Bertrand in 1889. The problem presents three identical boxes, each containing two coins in separate drawers:
- 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.
A box is chosen at random, and a random drawer is opened, revealing a Gold coin. The paradox asks: What is the probability that the other drawer in the same box also contains a Gold coin? While human intuition frequently falls into the Equally Likely Fallacy and incorrectly guesses 1 ÷ 2 = 50%, strict mathematical probability and Bayes’ Theorem prove that the true probability is exactly 2 ÷ 3 ≈ 66.67%.
Our free online Bertrand’s Box Paradox Calculator provides instant simulation, mathematical probability steps, and Generalized N-Coin Box extensions:
- Standard 3-Box 2-Coin Outcome [Exact Probability]:
P(GG | Gold Drawn) = 2 ÷ 3 = 0.666667 (66.67%). - Bayes’ Theorem Proof Formula:
P(GG | Gold) = [ P(Gold | GG) · P(GG) ] ÷ P(Gold) = (1.0 · 1/3) ÷ (1/2) = 2 ÷ 3. - Generalized K-Coin Box Extension Formula:
P(All Gold | Gold Drawn) = K ÷ (K + 1)(where Box 1 has K gold coins and Box 3 has 1 gold coin).
Master Box & Coin Drawer Probability Table
The table below displays the complete sample space of all 6 individual coin drawers across the 3 boxes, illustrating why drawing a Gold coin makes Box 1 twice as likely as Box 3:
| Box Identity & Contents | Drawer A Coin | Drawer B Coin | Probability of Box Selection | Condition: Drawn Coin is Gold? | Other Drawer Coin Outcome |
|---|---|---|---|---|---|
| Box 1 (GG – Double Gold) | Gold 1 (G1) | Gold 2 (G2) | 1 ÷ 3 (33.33%) | YES (If G1 or G2 drawn) | GOLD (2 out of 3 total Gold draws) |
| Box 2 (SS – Double Silver) | Silver 1 (S1) | Silver 2 (S2) | 1 ÷ 3 (33.33%) | NO (Eliminated immediately) | N/A (Impossible state) |
| Box 3 (GS – Mixed Gold/Silver) | Gold 3 (G3) | Silver 3 (S3) | 1 ÷ 3 (33.33%) | YES (Only if G3 drawn) | SILVER (1 out of 3 total Gold draws) |
Step-by-Step Bayes’ Theorem & Sample Space Solution Proof
To mathematically prove why the probability is 2 ÷ 3 (66.67%) rather than 1 ÷ 2 (50.0%):
Method 1: Individual Coin Sample Space Counting
There are 6 drawers total in the three boxes, containing 3 Gold coins (G1, G2, G3) and 3 Silver coins (S1, S2, S3).
When we open a drawer and reveal a Gold coin, we are holding one of 3 equally likely Gold coins:
- Case 1: We drew Gold coin G1 (from Box 1). The other drawer contains Gold coin G2. (SUCCESS)
- Case 2: We drew Gold coin G2 (from Box 1). The other drawer contains Gold coin G1. (SUCCESS)
- Case 3: We drew Gold coin G3 (from Box 3). The other drawer contains Silver coin S3. (FAILURE)
Since 2 out of the 3 possible Gold coin starting points belong to Box 1 (GG), P(Other is Gold) = 2 ÷ 3 ≈ 66.67%.
Method 2: Rigorous Bayes' Theorem Formulation
Let B1 = Box 1 (GG), B2 = Box 2 (SS), B3 = Box 3 (GS).
Priors: P(B1) = P(B2) = P(B3) = 1 ÷ 3.
Likelihoods of drawing Gold: P(Gold | B1) = 1.0; P(Gold | B2) = 0.0; P(Gold | B3) = 0.5.
Total Probability of drawing Gold: P(Gold) = (1.0 · 1/3) + (0.0 · 1/3) + (0.5 · 1/3) = 1/3 + 0 + 1/6 = 1/2.
Applying Bayes’ Theorem for 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%.
Thus, drawing a Gold coin doubles the probability that we chose Box 1 (GG) over Box 3 (GS).
Bertrand’s Box Paradox vs. Monty Hall & Three Prisoners Paradoxes
Below is a comparative reference chart showing isomorphic conditional probability paradoxes that share the identical mathematical structure:
| Probability Paradox Name | Intuitive Incorrect Fallacy | Correct Mathematical Probability | Isomorphic Mathematical Equivalence |
|---|---|---|---|
| Bertrand’s Box Paradox (1889) | 1 ÷ 2 = 50.0% (Box 1 vs Box 3) | 2 ÷ 3 ≈ 66.67% | Drawing a Gold coin is identical to opening a drawer in Box 1 or 3 |
| Monty Hall Game Show Problem (1975) | 1 ÷ 2 = 50.0% (Staying vs Switching) | 2 ÷ 3 ≈ 66.67% (If you switch doors) | Host revealing a goat is mathematically identical to revealing a Gold coin |
| Three Prisoners Problem (1959) | 1 ÷ 2 = 50.0% (Pardon probability) | 2 ÷ 3 ≈ 66.67% (Other prisoner’s chance) | Warden naming an executed prisoner provides identical conditional information |
History & Mathematics: 1889 Joseph Bertrand’s Calcul des probabilités
1889 Joseph Bertrand & The Calcul des probabilités Textbook
French mathematician and educator Joseph Louis François Bertrand introduced the Box Paradox in his landmark 1889 textbook Calcul des probabilités (Paris). Bertrand designed the puzzle specifically to demonstrate to his students how easily informal reasoning fails in conditional probability when people confuse the probability of selecting a box with the probability of selecting an individual coin.
The “Equally Likely Fallacy” & Sample Space Precision
Bertrand’s Box Paradox remains one of the premier teaching examples in modern statistics for exposing the Equally Likely Fallacy—the mistaken belief that whenever there are two remaining unknown outcomes (Box 1 vs Box 3), they must be equally likely (50/50). In reality, observing a Gold coin provides sample space information that heavily favors the box containing more Gold coins.
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Frequently Asked Questions (FAQ)
Why is the answer to Bertrand’s Box Paradox 2/3 and not 1/2?
Because there are 3 Gold coins total across the boxes. Two of those Gold coins belong to Box 1 (GG), while only one belongs to Box 3 (GS). When you draw a Gold coin, there is a 2/3 chance you drew one of Box 1’s Gold coins, meaning the other drawer in that box is also Gold.
Is Bertrand’s Box Paradox the same as the Monty Hall Problem?
Yes! Mathematically, Bertrand’s Box Paradox is structurally identical (isomorphic) to the Monty Hall Problem. Opening a drawer to reveal a Gold coin provides the exact same conditional probability update as Monty Hall opening a door to reveal a goat.
What is the Generalized Formula for N Gold Coins?
If Box 1 contains K Gold coins, Box 2 contains K Silver coins, and Box 3 contains 1 Gold and K-1 Silver coins, the probability that the box is Box 1 given a Gold draw is: P = K ÷ (K + 1).