Bertrand's Box Paradox Calculator
Bertrand's box paradox asks a deceptively simple question: you pick a box at random from three boxes (one holding two gold coins, one holding two silver coins, one holding one gold and one silver coin), draw a coin and see it is gold - what is the probability the other coin in that box is also gold? Most people answer 1/2, but the correct answer is 2/3. This calculator shows you exactly why, using Bayes' theorem, and lets you generalise the result to any number of all-gold, all-silver, or mixed boxes.
What is Bertrand's box paradox?
Bertrand's box paradox was introduced by the French mathematician Joseph Bertrand in his 1889 book Calcul des probabilites. The classic version presents three identical-looking boxes: the first holds two gold coins (GG), the second holds two silver coins (SS), and the third holds one gold coin and one silver coin (GS). You choose a box at random, reach in without looking, and pull out one coin. It is gold. What is the probability that the other coin in the same box is also gold? Most people answer 1/2, reasoning that the box must be either GG or GS, and those are equally likely. The correct answer is 2/3. The reasoning error is subtle and is shared with the Monty Hall problem: once you condition on observing a gold coin, the two candidate boxes are no longer equally probable sources, because the GG box had twice as many gold coins to offer.
Why the answer is 2/3, not 1/2 - the coin-counting argument
The key is to think about individual coins rather than whole boxes. Originally, all six coins are equally likely to be the one you draw. Label the coins GG1 and GG2 (in the all-gold box), SS1 and SS2 (all-silver), G (gold side) and S (silver side) in the mixed box. Given you drew gold, only three coins could be in your hand: GG1, GG2, or G. Each of these three is equally likely to be the one you are holding. If it is GG1, the other coin in the box is GG2 (gold). If it is GG2, the other is GG1 (gold). If it is G, the other is S (silver). Two out of three cases give a second gold coin, so P = 2/3. Bayes' theorem reaches the same answer formally. Letting B1 = GG box, B2 = SS box, B3 = GS box, and G = gold coin drawn: P(B1 | G) = P(G | B1) x P(B1) / P(G) = 1 x (1/3) / (1/2) = 2/3. P(G) = 1 x (1/3) + 0 x (1/3) + (1/2) x (1/3) = 1/2.
Generalising to any number of boxes
The two-coin-per-box structure allows a clean generalisation. Suppose you have n1 all-gold boxes, n2 all-silver boxes, and n3 mixed boxes. You observe a gold coin. The probability the other coin is gold is: P(match | gold) = (2 x n1) / (2 x n1 + n3). The all-silver boxes drop out entirely: once you have drawn gold, they could not have been your source. Adding more all-gold boxes pushes the probability toward 1; adding more mixed boxes pulls it toward 0. The custom mode of this calculator lets you explore any combination. For the classic setup (n1 = n2 = n3 = 1) the formula gives 2 / (2 + 1) = 2/3, confirming the classic result.
Real-world applications of this reasoning
The Bayesian reasoning behind Bertrand's paradox appears across many practical fields: Medical diagnosis: when a patient tests positive for a rare disease, the prior probability of having the disease (like the GG box's frequency) matters as much as the test's accuracy. A positive test does not mean 50-50; rare diseases combined with imperfect tests often produce more false positives than true ones. Forensic evidence: the probability that a DNA match points to the suspect rather than a chance coincidence depends on the population frequency of the profile - the same Bayesian weighting used here. Quality control: inspecting a component drawn from a mixed batch and finding it defective updates your estimate of which supplier batch it came from, proportionally to each supplier's defect rate. In all these cases, the intuitive error is identical: treating the remaining possibilities as equally likely after observing new evidence, when they should be weighted by how likely each was to produce that evidence.
Connection to the Monty Hall problem
Bertrand's box paradox and the Monty Hall problem are members of the same family of conditional-probability puzzles. In the Monty Hall problem a car is behind one of three doors; a host who knows where the car is opens a losing door; should you switch? The correct answer is to switch (probability 2/3 of winning), and the reasoning is identical to Bertrand's: new information does not simply split the remaining options equally - it updates probabilities in proportion to how likely each option was to produce that information. Both problems share the same intuition-defeating structure, and both are resolved by applying Bayes' theorem rather than naively counting remaining possibilities.
Common configurations at a glance
| Setup | Observed coin | P(other matches) | Intuitive guess | Paradox present? |
|---|---|---|---|---|
| 1 GG, 1 SS, 1 GS (classic) | Gold | 2/3 = 66.7% | 1/2 = 50% | Yes |
| 1 GG, 1 SS, 1 GS (classic) | Silver | 2/3 = 66.7% | 1/2 = 50% | Yes |
| 2 GG, 1 SS, 1 GS | Gold | 4/5 = 80.0% | 2/3 = 66.7% | Yes |
| 1 GG, 1 SS, 2 GS | Gold | 2/4 = 50.0% | 1/3 = 33.3% | Yes |
| 3 GG, 1 SS, 1 GS | Gold | 6/7 = 85.7% | 3/4 = 75% | Yes |
| 2 GG, 0 SS, 0 GS | Gold | 1 = 100% | 1/2 = 50% | Yes |
| 0 GG, 1 SS, 2 GS | Gold | 0 = 0% | 1/2 = 50% | Yes |
| 1 GG, 1 SS, 0 GS | Gold | 1 = 100% | 1/2 = 50% | Yes |
All results follow from Bayes' theorem. The classic setup with one box of each type gives the famous 2/3 answer. Use these as quick presets to verify your intuition.
Frequently asked questions
Why do most people say the answer is 1/2?
Because they frame the problem at the box level: after seeing gold, the SS box is eliminated, leaving only GG and GS. With two boxes remaining, people split the probability 50-50. The mistake is treating the two remaining boxes as equally probable sources of the coin. The GG box had two gold coins to offer, while GS had only one, so GG is twice as likely to be where you reached. Counting at the coin level rather than the box level gives the correct 2/3.
Does it matter which coin I pick from the box?
No. Because you cannot distinguish the two coins by feel and you draw at random, each coin in the chosen box is equally likely to come out. The calculation assumes uniform random selection within the chosen box, and that assumption is exactly what makes the GG box produce a gold coin with probability 1 while the GS box produces gold with probability 1/2.
What if I draw a silver coin instead?
By symmetry the answer is also 2/3. If you draw silver, the GG box is eliminated. The SS box had two silver coins to offer and the GS box had only one, so SS is twice as likely to be your source - giving a 2/3 chance the other coin is also silver. This calculator lets you switch the observed coin to verify the symmetric result.
How does the generalised formula work?
With n1 all-gold boxes, n2 all-silver boxes, and n3 mixed boxes, the probability the remaining coin matches a drawn gold coin is (2 x n1) / (2 x n1 + n3). The denominator counts all gold coins that could have been drawn: two from each GG box plus one from each mixed box. The numerator counts only those gold coins that come from a GG box - where the other coin is also gold. All-silver boxes are irrelevant once gold is observed because they contain no gold coins.
Is this the same reasoning as in a medical diagnostic test?
Yes, structurally. When a doctor knows a disease has a certain prevalence (like the frequency of GG boxes) and a test has a known sensitivity, a positive result does not mean 50% probability of having the disease. The probability must be weighted by how likely each hypothesis was to produce the observed result - exactly Bayes' theorem as used here. The Bertrand paradox is a clean, easy-to-visualise model for this kind of reasoning.
Is this the same as the Monty Hall problem?
Structurally yes. In the Monty Hall problem a car is behind one of three doors; a host who knows the answer opens a losing door; should you switch? The correct strategy is to switch (2/3 probability of winning), and the mathematics is identical to Bertrand's box: new information (the opened door, or the drawn coin colour) does not simply split remaining options equally - it updates them in proportion to how likely each was to generate that information.
What is the probability the other coin matches if there are no mixed boxes?
If there are no mixed (GS) boxes and you have at least one GG box, drawing a gold coin guarantees you are in a GG box, so the probability the other coin is also gold is 1 (100%). The formula (2 x n1) / (2 x n1 + 0) = 1 confirms this. Similarly, with no mixed boxes, drawing silver guarantees you are in an SS box.