Bertrand's box paradox is averidical paradox in elementaryprobability theory. It was first posed byJoseph Bertrand in his 1889 workCalcul des Probabilités.

There are three boxes:

1. a box containing two gold coins,
2. a box containing two silver coins,
3. a box containing one gold coin and one silver coin.

A coin withdrawn at random from one of the three boxes happens to be a gold. What is the probability the other coin from the same box will also be a gold coin?

A veridical paradox is a paradox whose correct solution seems to be counterintuitive. It may seem intuitive that the probability that the remaining coin is gold should be⁠1/2⁠, but the probability is actually⁠2/3⁠.^[1] Bertrand showed that if⁠1/2⁠ were correct, it would result in a contradiction, so⁠1/2⁠ cannot be correct.

This simple but counterintuitive puzzle is used as a standard example in teaching probability theory. The solution illustrates some basic principles, including theKolmogorov axioms.

The problem can be reframed by describing the boxes as each having one drawer on each of two sides. Each drawer contains a coin. One box has a gold coin on each side (GG), one a silver coin on each side (SS), and the other a gold coin on one side and a silver coin on the other (GS). A box is chosen at random, a random drawer is opened, and a gold coin is found inside it. What is the chance of the coin on the other side being gold?

The following reasoning appears to give a probability of ⁠1/2⁠:

• Originally, all three boxes were equally likely to be chosen.
• The chosen box cannot be boxSS.
• So it must be boxGG orGS.
• The two remaining possibilities are equally likely. So the probability that the box isGG, and the other coin is also gold, is ⁠1/2⁠.

The reasoning for the 2/3 is as follows:

• Originally, all six coins were equally likely to be chosen.
• The chosen coin cannot be from drawerS of boxGS, or from either drawer of boxSS.
• So it must come from theG drawer of boxGS, or either drawer of boxGG.
• The three remaining possibilities are equally likely, so the probability that the drawer is from boxGG is ⁠2/3⁠.

Bertrand's purpose for constructing this example was to show that merely counting cases is not always proper. Instead, one should sum the probabilities that the cases would produce the observed result.

A survey of psychology freshmen taking an introductory probability course was conducted to assess their solutions to the similar three-card problem. In the three-card problem, three cards are placed into a hat. One card is red on both sides, one is white on both sides, and one is white on one side and red on the other. If a card pulled from the hat is red on one side, the probability of the other side also being red is⁠2/3⁠.

53 students participated and were asked what the probability of the other side being red were. 35 incorrectly responded with⁠1/2⁠; only 3 students correctly responded with⁠2/3⁠.^[2]

Other veridical paradoxes of probability include:

• Boy or Girl paradox
• Monty Hall problem
• Three Prisoners problem
• Two envelopes problem
• Sleeping Beauty problem

The Monty Hall and Three Prisoners problems are identical mathematically to Bertrand's Box paradox. The construction of the Boy or Girl paradox is similar, essentially adding a fourth box with a gold coin and a silver coin. Its answer is controversial, based on how one assumes the "drawer" was chosen.

• Nickerson, Raymond (2004).Cognition and Chance: The psychology of probabilistic reasoning, Lawrence Erlbaum. Ch. 5, "Some instructive problems: Three cards", pp. 157–160.ISBN 0-8058-4898-3
• Michael Clark,Paradoxes from A to Z, p. 16;
• Howard Margolis,Wason, Monty Hall, and Adverse Defaults.

• Estimating the Probability with Random Boxes and Names, a simulation

• Probability theory paradoxes
• Probability problems

• Articles with short description
• Short description matches Wikidata