Bertrand's box paradox

Bertrand's box paradox is a veridical paradox in probability theory, first proposed by French mathematician Joseph Bertrand in his 1889 treatise Calcul des probabilités. The setup involves three identical boxes: one containing two gold coins (GG), one with two silver coins (SS), and one with one gold coin and one silver coin (GS). A box is chosen uniformly at random, and then a coin is drawn at random from it, revealing a gold coin; the question is the conditional probability that the remaining coin in the selected box is also gold.^[1]^[2] Intuitively, many reason that since the SS box is eliminated and a gold coin was drawn, the box must be either GG or GS with equal probability (1/2 each), implying a 1/2 chance the other coin is gold.^[3] However, this overlooks the differing likelihoods of drawing a gold coin from each box: the prior probability of selecting GG or GS is 1/3 each, but the probability of drawing gold from GG is 1, while from GS it is 1/2. Thus, the total probability of drawing a gold coin is (1/3)(1) + (1/3)(1/2) + (1/3)(0) = 1/2, and by Bayes' theorem, the posterior probability that the box is GG given a gold coin is [(1/3)(1)] / (1/2) = 2/3, meaning the probability the other coin is gold is 2/3.^[2]^[4] This paradox is "veridical" because the counterintuitive result (2/3) is correct, highlighting common errors in conditional probability, such as failing to account for the number of ways an event can occur.^[3] It predates and shares structural similarities with the Monty Hall problem, both illustrating how additional information updates probabilities non-uniformly.^[4] Experimental simulations consistently confirm the 2/3 probability, reinforcing its pedagogical value in teaching Bayesian reasoning.^[2]

Problem Statement

The setup

Bertrand's box paradox was posed by the French mathematician Joseph Bertrand in his 1889 book Calcul des Probabilités, serving as a veridical paradox to demonstrate counterintuitive aspects of probability. The setup involves three identical boxes, each containing two coins. One box holds two gold coins (denoted GG), another contains two silver coins (SS), and the third has one gold coin and one silver coin (GS). A box is selected at random—meaning each has an equal chance of being chosen—and then a coin is drawn at random from the selected box.^[5]

The question and intuitive answer

The core question in Bertrand's box paradox, following the random selection of one of the three boxes and the drawing of a gold coin from it, is the probability that the remaining coin in that box is also gold. This conditional probability query arises because a gold coin could have been drawn from either the box containing two gold coins or the box containing one gold and one silver coin, while the all-silver box is ruled out by the observation.^[6] Intuitively, many individuals conclude that this probability is 1/2, reasoning that the gold coin is equally likely to have originated from the two-gold-coin box or the mixed box, making the unseen coin equally likely to be gold or silver. This line of thought treats the two possible boxes as symmetric given the gold draw, overlooking nuances in how the observation conditions the probabilities across the setups.^[6] Joseph Bertrand introduced this paradox in his 1889 treatise to illustrate how everyday intuitive reasoning about chances can diverge sharply from rigorous probabilistic analysis, thereby underscoring the need for formal methods in evaluating uncertainties.

Solution

Enumerative approach

The enumerative approach to solving Bertrand's box paradox involves directly counting the possible outcomes that lead to drawing a gold coin, treating each individual gold coin as an equally likely draw given the random selection of a box and then a coin within it.^[7] There are three boxes—one containing two gold coins (GG), one containing two silver coins (SS), and one containing one gold and one silver coin (GS)—and a box is chosen at random before drawing a coin at random from it.^[8] Across these boxes, there are exactly three gold coins in total: two in the GG box and one in the GS box. Assuming each gold coin is equally likely to be the one drawn (since boxes are chosen with equal probability 1/3, and coins within a box with equal probability), the drawn gold coin originates from the GG box in two out of these three cases.^[2] In the two instances where it comes from the GG box, the remaining coin in that box is gold; in the single instance where it comes from the GS box, the remaining coin is silver. Thus, the probability that the other coin is gold is 2/3.^[9] This counting can be visualized through a listing of the equiprobable paths leading to a gold draw, labeling the gold coins distinctly for clarity (e.g., G1 and G2 in GG, G3 in GS):

Path 1: Choose GG box, draw G1 (other coin: G2, gold).
Path 2: Choose GG box, draw G2 (other coin: G1, gold).
Path 3: Choose GS box, draw G3 (other coin: silver).

Each path has probability (1/3) × (1/2) = 1/6, but since only these three paths result in drawing gold (the SS box yields no gold), they are the relevant outcomes, each equally likely at 1/3 conditional on drawing gold. Two of the three paths lead to the other coin being gold, confirming the 2/3 probability.^[7] This direct enumeration avoids conditional probability formulas and highlights the GG box's greater contribution to gold draws due to having two such coins.^[8]

Bayesian formulation

In the Bayesian formulation of Bertrand's box paradox, the problem is framed in terms of updating beliefs about the selected box based on the observed draw of a gold coin, using prior probabilities and conditional likelihoods. The three boxes—GG (two gold coins), SS (two silver coins), and GS (one gold and one silver coin)—are assumed to be equally likely a priori, reflecting the random selection process: P(GG) = P(SS) = P(GS) = \frac{1}{3}. This prior distribution captures the initial uncertainty before any coin is drawn.^[10] The likelihoods represent the probability of observing a gold coin given each box type: P(G \mid GG) = 1, since both coins are gold; P(G \mid SS) = 0, as neither is gold; and P(G \mid GS) = \frac{1}{2}, due to the single gold coin. To find the marginal probability of drawing a gold coin, the law of total probability is applied:

P(G) = \sum P(G \mid B_i) P(B_i) = P(G \mid GG) P(GG) + P(G \mid SS) P(SS) + P(G \mid GS) P(GS) = 1 \cdot \frac{1}{3} + 0 \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{2},

where B_i denotes the box types. This computation normalizes the evidence across all possible boxes.^[10] Bayes' theorem then updates the prior to the posterior probability that the GG box was selected given the gold coin observation:

P(GG \mid G) = \frac{P(G \mid GG) P(GG)}{P(G)} = \frac{1 \cdot \frac{1}{3}}{\frac{1}{2}} = \frac{2}{3}.

This posterior emphasizes how the GG box becomes twice as likely as the GS box after observing gold, due to its higher likelihood of producing that evidence. The formulation highlights the role of priors in resolving conditional probabilities, confirming the counterintuitive result of \frac{2}{3} for the probability that the remaining coin is also gold.^[10]

Why It's Paradoxical

Common errors in reasoning

One common error in reasoning about Bertrand's box paradox arises when individuals treat the remaining possible boxes—the all-gold (GG) box and the mixed (GS) box—as equally likely sources of the drawn gold coin, leading to an intuitive probability of 1/2 that the second coin is also gold.^[11] This overlooks the fact that the GG box has twice the likelihood of producing a gold coin compared to the GS box, since it contains two gold coins rather than one, making the GG box the more probable origin given the observation.^[12] As a result, the posterior probability favors the GG box at 2/3, but this base rate difference is often ignored in favor of assuming equiprobability among the surviving boxes. A related mistake involves failing to properly account for the coin-drawing process itself, where reasoners assume the observation of a gold coin equally "selects" between the GG and GS boxes without considering the varying probabilities of drawing gold from each.^[11] This error stems from a resistance to updating probabilities based on the new information from the draw, instead viewing the boxes as fixed equiprobable cases even after conditioning on the gold outcome.^[13] Such miscounting of the sample space—treating the three boxes as the primary units rather than the six equally likely coins—reinforces the flawed 1/2 intuition.^[14] These errors are underpinned by the representativeness heuristic, a cognitive bias where people assess probabilities based on how closely an event resembles a prototype or stereotype, often neglecting base rates or prior probabilities.^[15] In the paradox, the gold coin observation is seen as representatively indicating a "gold-containing box" without weighting the differing likelihoods from each box's composition, leading to judgments that prioritize descriptive similarity over statistical structure.^[16]

Resolution of the apparent contradiction

The apparent contradiction in Bertrand's box paradox arises from an intuitive but incorrect assumption that, after drawing a gold coin, the remaining possibilities—selecting the GG box or the GS box—are equally likely, leading to a 1/2 probability for the other coin being gold.^[17] However, the correct resolution involves reframing the probability space through proper conditioning on the event of drawing a gold coin, which alters the sample space and reveals that the GG box is twice as likely. The key insight is that conditioning on the gold coin draw effectively considers the individual coins rather than the boxes alone. There are three possible gold coins in the setup: two from the GG box and one from the GS box. Since the box is initially chosen with equal probability (1/3 each) and then a coin is drawn randomly from it, the two gold coins in the GG box each have a prior probability of (1/3) × (1/2) = 1/6 of being selected, while the single gold coin in the GS box also has probability 1/6. Normalizing over the conditioned sample space of these three equally likely gold coins yields a 2/3 probability that the drawn gold coin came from the GG box (and thus the other coin is gold). This approach ties back to the earlier enumerative or Bayesian calculations without altering them, emphasizing that the sample space post-conditioning consists of these three distinct draws rather than two boxes.^[18] To align intuition with this result, consider labeling the gold coins to distinguish them—call the two in the GG box G1 and G2, and the one in the GS box G3. The conditioned event of drawing a gold coin now corresponds to the equally probable outcomes {G1, G2, G3}, where G1 or G2 implies the other coin is gold, while G3 implies silver. Thus, the probability is 2/3, as the GG box contributes two of the three possible outcomes. This labeling analogy highlights the unequal representation of the GG box in the conditioned space, resolving the paradox by showing it is veridical: the mathematical result is correct, and intuition reconciles once the probabilities of the draws are properly accounted for rather than assuming equal box likelihoods after the observation.

Experimental Evidence

Key studies

One of the primary empirical investigations into intuitive responses to Bertrand's box paradox was conducted by psychologists Maya Bar-Hillel and Ruma Falk in 1982. They presented a variant of the paradox to 53 psychology freshmen enrolled in an introductory probability course, using three cards instead of boxes and coins: one with red on both sides (RR), one with white on both sides (WW), and one with red on one side and white on the other (RW, where red is treated analogously to gold and white to silver). Participants were asked the probability that the other side of a drawn card showing red (or gold) is also red (or gold). Of the respondents, 35 (66%) incorrectly estimated the probability as 1/2, only 3 (6%) correctly identified it as 2/3, and the remaining 15 (28%) expressed uncertainty.^[19] This study, published in the journal Cognition, underscored the prevalence of the intuitive 1/2 error among non-experts, even in an educational setting, with correct responses remaining notably low.^[19] The research has notable limitations, including its small sample size of 53 participants, which restricts generalizability. Conducted in 1982, it is also dated, and no major large-scale replications or updated empirical studies on intuitive responses to the paradox have been identified in the literature since 2000.^[19]

Implications for probability intuition

Experimental studies on Bertrand's box paradox reveal significant challenges in human probabilistic reasoning, with participants frequently arriving at incorrect intuitive answers despite the problem's apparent simplicity. This low success rate underscores a pervasive difficulty in applying conditional probability correctly.^[19] The errors observed in such studies stem from well-documented cognitive biases, including base-rate neglect, where individuals disregard the prior probabilities of selecting each box (each equally likely at \frac{1}{3}), and an overreliance on the representativeness heuristic, leading them to focus narrowly on the drawn gold coin as representative of the mixed box rather than considering the full sample space of possible outcomes. These biases cause reasoners to effectively assume two equally likely scenarios—drawing from the gold-gold box or the mixed box—ignoring that the gold-gold box contributes two ways to draw a gold coin compared to one from the mixed box. As a result, the paradox serves as a stark illustration of how intuitive judgments can systematically deviate from normative Bayesian principles.^[19] Beyond individual errors, the paradox has broader implications for understanding limitations in probability intuition across diverse populations. It is commonly integrated into educational curricula to demonstrate Bayes' theorem and the mechanics of conditional probability, helping learners recognize how evidence updates priors in non-intuitive ways. For example, the setup encourages exploration of tree diagrams or enumeration to resolve the apparent 50-50 split, fostering deeper conceptual grasp.^[7] However, the empirical foundation remains sparse, relying heavily on the 1982 study with its small, homogeneous sample. This highlights a gap in contemporary research, pointing to the value of modern replications using larger, more diverse participant pools—potentially incorporating cross-cultural or online methods—to assess the persistence of these biases and their interaction with factors like education level or numeracy. Such updated investigations could refine our knowledge of cognitive vulnerabilities in probabilistic thinking and inform improved teaching strategies.^[19]

Similar probability puzzles

The Boy or Girl paradox, also known as the two children problem, involves a family with two children where at least one is known to be a boy; the probability that both are boys is 1/3.^[20] This counterintuitive result arises from conditioning on partial information about the children's genders, similar to how Bertrand's box paradox updates probabilities based on drawing a single gold coin from an unknown box. The paradox was first posed by Martin Gardner in his "Mathematical Games" column in Scientific American.^[20] The Monty Hall problem presents a contestant with three doors, behind one of which is a car and behind the others goats; after selecting a door and the host revealing a goat behind another, switching doors yields a 2/3 probability of winning the car.^[21] This unequal probability post-revelation mirrors the skewed odds in Bertrand's paradox after observing a gold coin, both illustrating how initial equal chances become asymmetric upon new evidence. The problem gained prominence through Marilyn vos Savant's "Ask Marilyn" column in Parade magazine in 1990.^[21] The Three Prisoners problem features three prisoners A, B, and C, one of whom will be pardoned at random; the warden informs A that B is not pardoned, leading A to believe his survival chance is 1/2, but it remains 1/3 while C's rises to 2/3.^[20] This setup parallels Bertrand's box paradox in identifying the "other" outcome among symmetric possibilities, with the counterintuition stemming from misjudging the implications of the revealed information. It was introduced by Martin Gardner in the same 1959 Scientific American column as the Boy or Girl paradox.^[20] These puzzles share a mathematical structure with Bertrand's box paradox, where the 2/3 probability serves as a template for updating beliefs about hidden states given observed evidence, often leading to errors in intuitive reasoning about conditional probabilities.^[20]

Generalizations of the paradox

The paradox can be generalized to an arbitrary number of boxes with varying compositions of gold and silver coins, where probability updates are performed using Bayes' theorem, often resulting in counterintuitive conditional probabilities that emphasize the role of base rates in hierarchical setups.^[7] In the original formulation, Bertrand extended the setup to 300 boxes: 100 containing two gold coins (GG), 100 with two silver coins (SS), and 100 with one gold and one silver coin (GS). A box is selected uniformly at random, and one coin is drawn at random from it. Given that the drawn coin is gold, the probability that the other coin in the same box is also gold is \frac{2}{3}. This arises because, among the expected 150 gold coins drawn, 100 originate from GG boxes and 50 from GS boxes, so the conditional probability of being in a GG box is \frac{100}{150} = \frac{2}{3}.^[1] More generally, consider g GG boxes, s SS boxes, and m GS boxes, for a total of n = g + s + m boxes, each equally likely to be chosen. A coin is then drawn uniformly from the selected box. The unconditional probability of drawing gold is \frac{g}{n} \cdot 1 + \frac{m}{n} \cdot \frac{1}{2} = \frac{g + \frac{m}{2}}{n}. By Bayes' theorem, the posterior probability of having selected a GG box given a gold coin is P(\text{GG} \mid \text{gold}) = \frac{g}{g + \frac{m}{2}}. The SS boxes do not affect this conditional probability, as they contribute no gold draws. This formula illustrates how imbalances in box types lead to fractions like \frac{2}{3} in the classic case (g=1, m=1) or other non-obvious values, underscoring the paradox's scalability and the need to account for multiple sources of evidence in Bayesian updates.^[7] For instance, with two GG boxes, one GS box, and one SS box (g=2, m=1, s=1, n=4), the probability of drawing gold is \frac{2 + 0.5}{4} = \frac{2.5}{4} = \frac{5}{8}. Thus, P(\text{[GG](/page/GG)} \mid \text{gold}) = \frac{2}{2 + 0.5} = \frac{2}{2.5} = \frac{4}{5}. This result, while higher than \frac{2}{3}, remains counterintuitive for many, as the presence of more GG boxes intuitively suggests a near-certain match, yet the mixed box still dilutes the probability through its half-contribution to gold draws. Such extensions add depth for advanced readers by revealing why base rates—here, the relative abundances of box types—dominate intuitive equiprobability assumptions in conditional inference.^[7]