Bertrand's box paradox

For other paradoxes by Joseph Bertrand, see Bertrand's paradox (disambiguation).

The paradox starts with three boxes, the contents of which are initially unknown

Bertrand's box paradox is a veridical paradox in elementary probability theory. It was first posed by Joseph Bertrand in his 1889 work Calcul 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.

Choose a box at random. From this box, withdraw one coin at random. If that happens to be a gold coin, then what is the probability that the next coin drawn from the same box is also 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 be ⁠1/2⁠, but the probability is actually ⁠2/3⁠. Bertrand showed that if ⁠1/2⁠ were correct, it would result in a contradiction, so ⁠1/2⁠ 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 the Kolmogorov axioms.

Solution

Bertrand's box paradox: the three equally probable outcomes after the first gold coin draw. The probability of drawing another gold coin from the same box is 0 in (a), and 1 in (b) and (c). Thus, the overall probability of drawing a gold coin in the second draw is ⁠0/3⁠ + ⁠1/3⁠ + ⁠1/3⁠ = ⁠2/3⁠.

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 faulty reasoning appears to give a probability of ⁠1/2⁠:

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

The flaw is in the last step. While those two cases were originally equally likely, the fact that you are certain to find a gold coin if you had chosen the GG box, but are only 50% sure of finding a gold coin if you had chosen the GS box, means they are no longer equally likely given that you have found a gold coin. Specifically:

• The probability that GG would produce a gold coin is 1.
• The probability that SS would produce a gold coin is 0.
• The probability that GS would produce a gold coin is ⁠1/2⁠.

Initially GG, SS and GS are equally likely $\left(\mathrm {i.e.,P(GG)=P(SS)=P(GS)} ={\frac {1}{3}}\right)$. Therefore, by Bayes' rule the conditional probability that the chosen box is GG, given we have observed a gold coin, is:

$\mathrm {P(GG\mid see\ gold)={\frac {P(see\ gold\mid GG)\times {\frac {1}{3}}}{P(see\ gold\mid GG)\times {\frac {1}{3}}+P(see\ gold\mid SS)\times {\frac {1}{3}}+P(see\ gold\mid GS)\times {\frac {1}{3}}}}} ={\frac {\frac {1}{3}}{\frac {1}{3}}}\times {\frac {1}{1+0+{\frac {1}{2}}}}={\frac {2}{3}}$

The correct answer of ⁠2/3⁠ can also be obtained as follows:

• Originally, all six coins were equally likely to be chosen.
• The chosen coin cannot be from drawer S of box GS, or from either drawer of box SS.
• So it must come from the G drawer of box GS, or either drawer of box GG.
• The three remaining possibilities are equally likely, so the probability that the drawer is from box GG 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; and the two methods are equivalent only if this probability is either 1 or 0 in every case. This condition is applied correctly by the second solution method, but not by the first.

Experimental data

In a survey of 53 Psychology freshmen taking an introductory probability course, 35 incorrectly responded ⁠1/2⁠; only 3 students correctly responded ⁠2/3⁠.

Related problems

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.

References

• 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.

External links

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