Bertrand's box paradox
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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:
- a box containing two gold coins,
- a box containing two silver coins,
- 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.[1] 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
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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 . Therefore, by Bayes' rule the conditional probability that the chosen box is GG, given we have observed a gold coin, is:
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.[citation needed]
Experimental data
[edit]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]
Related problems
[edit]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
[edit]- ^ "Bertrand's box paradox". Oxford Reference.
- ^ 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
[edit]- Estimating the Probability with Random Boxes and Names, a simulation