Bertrand's box paradox

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

Bertrand's box paradox is a paradox of elementary probability theory, first posed by Joseph Bertrand in his 1889 work Calcul des probabilités.

There are three boxes:

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

The question is to calculate the probability, after choosing a box at random and withdrawing one coin at random, if that happens to be a gold coin, of the next coin drawn from the same box also being a gold coin. It may seem that the probability that the remaining coin is gold is 1/2, but the probability is actually 2/3.^[1]

It is not a "true" paradox, because it has a well-defined single answer.

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 03 + 13 + 13 = 23.

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.

Alternatively, one can simply note that the chosen box has two coins of the same type 2/3 of the time. So, regardless of what kind of coin is in the chosen drawer, the box has two coins of that type 2/3 of the time. In other words, the problem is equivalent to asking the question "What is the probability that I will pick a box with two coins of the same color?".

Bertrand's point in 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 correctly applied in the second solution method, but not in the first.^{[citation needed]}

The paradox as stated by Bertrand[]

It can be easier to understand the correct answer if you consider the paradox as Bertrand originally described it. After a box has been chosen, but before a box is opened to let you observe a coin, the probability is 2/3 that the box has two of the same kind of coin. If the probability of "observing a gold coin" in combination with "the box has two of the same kind of coin" is 1/2, then the probability of "observing a silver coin" in combination with "the box has two of the same kind of coin" must also be 1/2. And if the probability that the box has two like coins changes to 1/2 no matter what kind of coin is shown, the probability would have to be 1/2 even if you hadn't observed a coin this way. Since we know his probability is 2/3, not 1/2, we have an apparent paradox. It can be resolved only by recognizing how the combination of "observing a gold coin" with each possible box can only affect the probability that the box was GS or SS, but not GG.^{[citation needed]}

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.^[2]

References[]

^ "Bertrand's box paradox". Oxford Reference. doi:10.1093/oi/authority.20110803095501915.
^ Bar-Hillel, Maya; Falk, Ruma (1982). "Some teasers concerning conditional probabilities". Cognition. 11 (2): 109–22. doi:10.1016/0010-0277(82)90021-X. PMID 7198956. S2CID 44509163.

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

[Oxford-1] "Bertrand's box paradox". Oxford Reference. doi:10.1093/oi/authority.20110803095501915.

[Bar-Hillel-2] Bar-Hillel, Maya; Falk, Ruma (1982). "Some teasers concerning conditional probabilities". Cognition. 11 (2): 109–22. doi:10.1016/0010-0277(82)90021-X. PMID 7198956. S2CID 44509163.

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