About

Bertrand's Box Paradox, formulated by Joseph Bertrand in 1889, is a veridical paradox in elementary probability theory. Three boxes each contain two coins: one holds two gold coins, one holds one gold and one silver, and one holds two silver. You choose a box at random, draw one coin, and it is gold. The probability the remaining coin in that box is also gold is 23, not 12 as most people guess. The error stems from conflating "choosing a box" with "choosing a coin." There are three equally likely gold coins you could have drawn. Two of those three reside in the Gold-Gold box. This tool lets you run the experiment manually or simulate up to 100,000 trials to watch the empirical frequency converge to the Bayesian answer.

The paradox is structurally identical to the Monty Hall problem and demonstrates why naive enumeration of outcomes without weighting by likelihood leads to systematic error. Misapplying this reasoning in clinical diagnostics or quality control produces materially wrong risk estimates. This simulator applies no approximations. It computes exact conditional probabilities and validates them empirically.

Formulas

The solution applies Bayes' theorem directly. Given that the drawn coin is gold (G), compute the posterior probability that the chosen box is Box A (containing two gold coins):

P(A | G) = P(G | A) ⋅ P(A)P(G)

Where the total probability of drawing gold is computed via the law of total probability:

P(G) = 3∑i=1 P(G | B_i) ⋅ P(B_i) = 1 ⋅ 13 + 12 ⋅ 13 + 0 ⋅ 13 = 12

Substituting:

P(A | G) = 1 ⋅ 1312 = 23

Where A = event that Box A (Gold-Gold) was chosen, G = event that the drawn coin is gold, B_i = event of choosing box i, P(G | A) = 1 because both coins in Box A are gold, P(A) = 13 is the prior probability of selecting Box A.

Reference Data

Box	Coin 1	Coin 2	P(pick this box)	P(draw gold \| this box)	P(this box & draw gold)	P(this box \| drew gold)
Box A (GG)	Gold	Gold	13	1	13	23
Box B (GS)	Gold	Silver	13	12	16	13
Box C (SS)	Silver	Silver	13	0	0	0
Total P(draw gold)						12

Frequently Asked Questions

The error arises from conditioning on boxes instead of coins. After drawing a gold coin, you know it is one of three specific gold coins in the system: two in Box A (GG) and one in Box B (GS). Two of those three gold coins have a gold companion. Therefore P(other coin is gold) = 23. Thinking "I picked a box with at least one gold, so it is either GG or GS - 50/50" ignores that the GG box is twice as likely to produce the gold coin you observed.

Both paradoxes share identical Bayesian structure. In Monty Hall, the host revealing a goat is analogous to drawing a gold coin - it provides evidence that updates the prior. The key mechanism is the same: the "favorable" option (car door / GG box) is more likely to produce the observed evidence than the "unfavorable" option (goat door / GS box). Both yield a 23 posterior probability for the favorable outcome.

By the Central Limit Theorem, the standard error of a proportion estimate is √p(1 − p)n. For p = 23 and n = 1000, the standard error is approximately 0.015, meaning the observed proportion will typically fall within ±0.03 of 0.667. At n = 10,000, the error drops to about 0.005. This tool lets you run up to 100,000 trials.

Yes. The specific 23 result depends on the symmetric setup of exactly two coins per box with the given distribution. Changing the composition (e.g., adding a third coin to one box) alters the likelihoods in Bayes' formula. The paradox's pedagogical value lies in its minimal, symmetric structure that maximally exposes the intuition failure.

If the draw within a box is not uniformly random, the posterior changes. For example, if you always draw the "top" coin from Box B and it is always the gold one, then P(G | B) = 1 instead of 12, and the posterior becomes 12. The standard formulation assumes uniform random draw, which is critical to the result.