Bertrand's Box Paradox Calculator
Three boxes each hold two coins. Draw one coin at random from a randomly chosen box. If the drawn coin is Gold, what is the probability the remaining coin in that box is also Gold?
Bayes' Theorem Results
Sample Space Method
Probability tree: each path shows the route from box selection to coin observation
Define any three-box scenario. Enter the number of gold coins in each box (each box holds exactly 2 coins), choose the observed coin, and calculate the posterior probability.
Run a simulation of random box and coin selections. Only trials where the first drawn coin is Gold are counted. Watch the observed frequency converge toward 2/3 as the number of trials increases.
Run the Classic Setup or Custom Boxes tab first, then return here for the full derivation.
No calculation yet — use the Classic Setup or Custom Boxes tab first.
What is Bertrand's Box Paradox?
Bertrand's Box Paradox is a conditional probability puzzle in which observing a gold coin changes the probability of the selected box being the gold-gold box from 1/3 to 2/3. The French mathematician Joseph Bertrand posed the problem in his 1889 book Calcul des probabilités. It remains one of the most frequently taught examples in probability theory because its answer contradicts the intuition of most people encountering it for the first time.
The setup: three identical boxes each hold two coins. Box 1 holds two gold coins (GG). Box 2 holds one gold and one silver coin (GS). Box 3 holds two silver coins (SS). A box is chosen at random and a single coin is drawn from it at random. The coin is observed to be gold. The question is: what is the probability that the other coin in the same box is also gold?
The intuitive but incorrect answer is 1/2. The reasoning goes: since you know you drew from either Box 1 or Box 2 (Box 3 has no gold coins), and these two boxes each have one more coin, either gold or silver, there should be a 50/50 chance. This reasoning is wrong because it counts boxes rather than individual coins in the sample space, as the correct solution below shows.
The Mathematics: Formulas and Methods
Three methods produce identical answers. Each approaches the same problem from a different angle, making the paradox a useful teaching tool for connecting sample spaces, conditional probability, and Bayesian inference.
Conditional Probability Formula
P(A|B) = P(A ∩ B) / P(B)
P(GG box | Gold drawn) =
P(Gold drawn ∩ GG box) / P(Gold drawn)
= (1/3 × 1) / (1/2)
= 2/3
Bayes' Theorem
P(A|B) = P(B|A) × P(A) / P(B)
P(GG | Gold) =
P(Gold|GG) × P(GG) / P(Gold)
= 1 × (1/3) / (1/2)
= 2/3
Sample Space Method
Gold coins: G1, G2 (Box 1), G3 (Box 2)
Each equally likely to be drawn.
P(other coin is Gold | drew gold) =
|{G1, G2}| / |{G1, G2, G3}|
= 2 / 3
Total Probability (denominator)
P(Gold drawn) =
P(Gold|Box1)×P(Box1)
+ P(Gold|Box2)×P(Box2)
+ P(Gold|Box3)×P(Box3)
= 1×(1/3) + (1/2)×(1/3) + 0×(1/3)
= 1/2
How to Solve Bertrand's Box Paradox — Step by Step
To solve Bertrand's Box Paradox correctly, you must analyze the sample space of individual coins rather than the sample space of boxes. Here is the complete solution.
Box 1 = {G1, G2} (two gold coins). Box 2 = {G3, S1} (one gold, one silver). Box 3 = {S2, S3} (two silver coins). Each box is equally likely to be chosen, so P(Box 1) = P(Box 2) = P(Box 3) = 1/3.
The full sample space has 6 equally probable outcomes: {G1, G2, G3, S1, S2, S3}. Each coin has probability 1/6 of being the drawn coin. This is the critical step that most incorrect solutions skip.
The drawn coin is Gold. This rules out S1, S2, and S3. The reduced sample space is {G1, G2, G3}, each still equally probable (1/3 conditional probability).
P(Gold) = P(Gold|Box1)×P(Box1) + P(Gold|Box2)×P(Box2) + P(Gold|Box3)×P(Box3) = 1×(1/3) + (1/2)×(1/3) + 0×(1/3) = 1/2.
P(Box 1 | Gold) = P(Gold|Box1)×P(Box1) / P(Gold) = (1×1/3) / (1/2) = 2/3. P(Box 2 | Gold) = (1/2×1/3) / (1/2) = 1/3. P(Box 3 | Gold) = 0 / (1/2) = 0.
If you are holding a box known to be either Box 1 or Box 2, the probability that it is Box 1 (GG) is 2/3. Since Box 1 has two gold coins, the other coin in the box is definitely gold. So P(other coin = Gold | drew Gold) = P(Box 1 | Gold) = 2/3.
Result: P(other coin = Gold | first coin = Gold) = 2/3 ≈ 0.667 — not 1/2. This result is confirmed by all three methods (sample space, conditional probability formula, and Bayes' theorem), and by the Monte Carlo simulation in the calculator above.
Worked Examples and Variations
Example 1 — Classic Gold-Silver Setup (the paradox)
{G1 from Box1, G2 from Box1, G3 from Box2} — three equally likely outcomes.
G1 → other is G2 (gold). G2 → other is G1 (gold). G3 → other is S1 (silver). Two out of three lead to gold.
P = 2/3 ≈ 0.667. The intuitive answer of 1/2 is wrong because it treats coins G1 and G2 as a single outcome rather than two distinct ones.
Common error: People reason “I know it’s Box 1 or Box 2, so it’s 50/50.” This counts boxes, not coins. Box 1 contributes two paths to the gold-coin event; Box 2 contributes only one. Bayes' theorem correctly weights these contributions.
Example 2 — Symmetric Silver Case
By symmetry, the answer is identical: P(other = Silver | drew Silver) = 2/3. The silver coins S1 (Box 2), S2, and S3 (Box 3) form the relevant sample space. S2 and S3 are both in Box 3 (SS), giving 2 out of 3 paths to the all-silver box.
Result: P = 2/3. The paradox is symmetric — whichever color coin you draw, the probability that the other coin in the same box matches it is 2/3.
Example 3 — Modified Four-Box Variation
Gold coins: G1, G2 (Box1), G3, G4 (Box2), G5 (Box3) — 5 gold coins total. Drawing G1, G2, G3, or G4 leads to the other coin being gold (4 paths). Drawing G5 leads to silver (1 path). P(other = Gold | drew Gold) = 4/5 = 0.80.
Result: P = 4/5. Adding a second GG box increases the posterior probability. The custom calculator tab handles this class of variation for any box configuration.
Example 4 — Bayesian Interpretation for Teaching
The prior is uniform: P(Box1) = P(Box2) = P(Box3) = 1/3. The likelihoods are P(G|B1) = 1, P(G|B2) = 1/2, P(G|B3) = 0. After applying Bayes, the posterior is P(B1|G) = 2/3, P(B2|G) = 1/3, P(B3|G) = 0. This example demonstrates how Bayesian updating converts prior beliefs into posterior beliefs when evidence arrives, which is the foundation of Bayesian inference.
Visualizing Bertrand's Box Paradox: Probability Tree
A probability tree maps every path from box selection to coin observation. Each leaf represents a coin draw. The tree makes it clear why there are 3 paths to a gold first coin, and why 2 of those paths lead to a gold second coin.
Bertrand's Box Paradox vs. the Monty Hall Problem
Both problems produce a counterintuitive 2/3 answer and both rely on the same mechanism: conditioning on new information that is not independent of the unknown variable. Yet the two problems have a structural difference worth noting.
| Property | Bertrand's Box | Monty Hall Problem |
|---|---|---|
| Origin | Joseph Bertrand, 1889 | Marilyn vos Savant, 1990 (Parade) |
| Setup | 3 boxes, 2 coins each; draw one coin | 3 doors, 1 prize; host reveals empty door |
| New information | Coin color observed at random | Host deliberately reveals a losing door |
| Key insight | Count coins, not boxes, in the sample space | Host action carries asymmetric information |
| Answer | 2/3 the other coin matches | 2/3 gain by switching doors |
| Mathematical tool | Conditional probability / Bayes' theorem | Conditional probability / Bayes' theorem |
| Common wrong answer | 1/2 | 1/2 |
The Monty Hall problem is often described as the more famous of the two, but Bertrand's Box is arguably the cleaner mathematical construction because the information is revealed passively (you simply observe the coin color) rather than through a deliberate agent action. For a thorough treatment of both problems, the Introduction to Probability by Blitzstein and Hwang (free online) devotes a section to each.
Common Mistakes in Solving Bertrand's Box Paradox
The most common error is using the wrong sample space. Instead of treating each of the 6 individual coins as equally likely outcomes, people treat each of the 3 boxes as equally likely outcomes after observing the coin color. This collapses the two gold coins in Box 1 into a single event, which understates Box 1's contribution.
| Mistake | What Goes Wrong | Correction |
|---|---|---|
| Counting boxes not coins | Treats GG box as one outcome; it should be two (G1, G2) | List all 6 individual coins; condition on gold coins only |
| Wrong prior after observation | Assumes P(Box1) = P(Box2) = 1/2 after seeing gold | Apply Bayes: posterior P(B1|G) = 2/3, P(B2|G) = 1/3 |
| Conflating "could be" with equal probability | "It could be Box 1 or Box 2, so 50/50" | Prior probabilities are not equal after conditioning |
| Ignoring how the gold coin was drawn | Forgets that each gold coin is equally likely to be drawn | Box 1 has twice as many gold coins, so double the probability of being the source |
Complete Formula and Entity Reference
The table below covers every key concept and formula in Bertrand's Box Paradox, structured for direct reference during study or exam preparation.
| Term | Symbol / Formula | Definition | Role in the Paradox |
|---|---|---|---|
| Conditional Probability | P(A|B) = P(A∩B)/P(B) | Probability of A given B has occurred | Core formula: P(GG box | drew gold) |
| Bayes' Theorem | P(A|B) = P(B|A)P(A)/P(B) | Updates probability based on new evidence | Computes posterior probability for each box |
| Prior Probability | P(Box i) = 1/3 | Probability before any observation | Each box equally likely before drawing |
| Likelihood | P(Gold | Box i) | How probable the evidence is given each hypothesis | P(G|B1)=1, P(G|B2)=1/2, P(G|B3)=0 |
| Posterior Probability | P(Box i | Gold) | Updated probability after observing gold coin | P(B1|G)=2/3, P(B2|G)=1/3, P(B3|G)=0 |
| Marginal Probability | P(Gold) = 1/2 | Total probability of observing a gold coin | Denominator in Bayes' theorem calculation |
| Sample Space | Ω = {G1,G2,G3,S1,S2,S3} | All possible individual coin draws | Must count coins, not boxes |
| Bayesian Updating | prior → posterior | Process of revising beliefs when evidence arrives | Observing gold coin shifts belief toward GG box |
💻 Simulation Code: Python and R
The following code lets you verify the 2/3 result computationally. Both examples run 100,000 simulated draws and count the fraction of gold-first trials where the second coin is also gold.
import random
# Bertrand's Box simulation in Python
# Boxes: GG=2 gold, GS=1 gold 1 silver, SS=2 silver
boxes = [
['gold', 'gold'],
['gold', 'silver'],
['silver', 'silver']
]
gold_first = 0
both_gold = 0
trials = 100_000
for _ in range(trials):
box = random.choice(boxes) # pick a box
coins = box.copy()
random.shuffle(coins) # randomize coin order
first, second = coins[0], coins[1]
if first == 'gold':
gold_first += 1
if second == 'gold':
both_gold += 1
# Expected result: ~0.6667
print(f"P(other=Gold | first=Gold) = {both_gold/gold_first:.4f}")
# Bertrand's Box simulation in R
set.seed(42)
trials <- 100000
boxes <- list(c("gold","gold"), c("gold","silver"), c("silver","silver"))
results <- replicate(trials, {
box <- sample(boxes, 1)[[1]]
coins <- sample(box) # shuffle within box
c(first = coins[1], second = coins[2])
})
gold_first_idx <- results["first",] == "gold"
gold_first <- sum(gold_first_idx)
both_gold <- sum(results["second", gold_first_idx] == "gold")
cat("P(other=Gold | first=Gold) =", round(both_gold / gold_first, 4), "\n")
# Expected: ~0.6667
Related Tools and Guides on Statistics Fundamentals
Bertrand's Box sits at the intersection of probability theory and Bayesian reasoning. The following resources from Statistics Fundamentals cover the prerequisite and adjacent topics in depth.
Frequently Asked Questions
Bertrand's Box Paradox is a conditional probability puzzle proposed by French mathematician Joseph Bertrand in 1889. Three boxes each contain two coins: Box 1 holds two gold coins (GG), Box 2 holds one gold and one silver coin (GS), and Box 3 holds two silver coins (SS). A box is chosen at random and a coin drawn from it at random turns out to be gold. The question is: what is the probability the other coin in the same box is also gold? The answer is 2/3, not 1/2 as most people initially guess, making it a classic example of how conditional probability can be counterintuitive.
The answer is 2/3 because you must analyze the sample space of individual coins, not just boxes. There are three gold coins in total: G1 and G2 in Box 1, and G3 in Box 2. Each has probability 1/3 of being the drawn coin. Two of those three scenarios (drew G1 or drew G2) place you in Box 1, meaning the other coin is definitely gold. Only one scenario (drew G3) places you in Box 2, where the other coin is silver. The 1/2 answer is wrong because it treats Box 1 as a single outcome, ignoring that it contributes two equally probable paths to the gold-first event.
Bayes' theorem updates the probability assigned to each box after observing the coin. Starting with equal priors P(B1) = P(B2) = P(B3) = 1/3, the likelihoods are P(Gold|B1) = 1, P(Gold|B2) = 1/2, and P(Gold|B3) = 0. The marginal probability P(Gold) = 1/3 + 1/6 + 0 = 1/2. Applying Bayes: P(B1|Gold) = (1 × 1/3) / (1/2) = 2/3. Since Box 1 contains two gold coins, the probability the other coin is gold equals P(B1|Gold) = 2/3.
The full sample space consists of 6 equally probable individual coin draws: G1 and G2 from Box 1, G3 and S1 from Box 2, and S2 and S3 from Box 3. Each has probability 1/6 of being drawn. After conditioning on the observed event "drew a gold coin," the sample space reduces to {G1, G2, G3}, each with conditional probability 1/3. The key insight is that G1 and G2 both reside in Box 1, so two out of three elements in the reduced sample space lead to the conclusion that the other coin is also gold.
Both problems use conditional probability and produce a counterintuitive 2/3 answer, but they differ structurally. In Bertrand's Box, information is revealed by a random process (you simply observe which coin you drew). In the Monty Hall Problem, a knowledgeable host deliberately chooses which door to open, and this deliberate choice carries asymmetric information. Bertrand's Box is the simpler construction because it requires no assumption about how the information provider behaves. Both problems are corrected by applying Bayes' theorem to the correct sample space.
The conditional probability reasoning in Bertrand's Box appears in medical testing (a positive test result updates the probability of disease, weighted by base rates), forensic evidence interpretation (DNA evidence is weighed against prior probabilities), machine learning (naive Bayes classifiers use exactly this update logic), and quality control (observing a defective item updates the probability that a particular production line is the source). The common thread is using Bayes' theorem to update beliefs when new information arrives, rather than treating all remaining possibilities as equally likely.
Yes. Take three opaque bags. Fill one with two gold coins, one with a gold and a silver coin, and one with two silver coins. Shuffle the bags so you cannot tell them apart. Randomly pick a bag and draw one coin without looking inside. Record trials where the first coin is gold. In those trials, record whether the second coin drawn from the same bag is also gold. Over many repetitions, you will observe the ratio approaching 2/3. The Monte Carlo simulation in the calculator above models this exact experiment. Physical experiments with playing cards or colored blocks produce the same result.
Yes. The classic 1/3 prior for each box is what gives the 2/3 answer. If the GG box were chosen with probability p rather than 1/3, the posterior probability would be P(GG|Gold) = p / (p + q/2), where q is the prior probability of the GS box and the SS box contributes nothing. For non-uniform priors, the Custom Boxes tab in the calculator does not yet support non-uniform priors, but applying Bayes' theorem directly with your chosen priors gives the correct answer for any prior distribution over boxes.