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Bertrand's box paradox

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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.

A coin withdrawn at random from the three boxes happens to be a gold coin. If you now examine the *other* coin in that box, what is the probability it 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 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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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 . 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

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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]

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Other veridical paradoxes of probability include:

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

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  1. ^ "Bertrand's box paradox". Oxford Reference.
  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.
  • 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.
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