Bertrand's box paradox
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.
A coin withdrawn at random from the three boxes happens to be a gold coin.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
[edit]
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 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 reasoning for the 2/3 P(GG∣see gold)=P(see gold∣GG)×13P(see gold∣GG)×13+P(see gold∣SS)×13+P(see gold∣GS)×13=1313×11+0+12=23is 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.
However, both reasonings have the same flaw. The selection of one box from the three boxes and the selection of one drawer and its coin of the selected box have already occurred. The question asks what is the probability of an outcome after it has occurred. But the probabilities 1/2 and 2/3 are answers to the question of what is the probability of the outcome before it has occurred.
Before the box is selected, there is one sample space consisting of three elements: {box GG, box GS, and box SS}. After a box is selected, there are three sample spaces, each with only one element: {box GG}, {box GS} and {box SS}. The probability for each is simple: divide the number of elements that meet the criterion by the total number of elements in the sample space.
Therefore, for sample space {Box GG} the probability is 1/1 = 1 while for the other two sample spaces {box GS} and {box SS} the probability is 0/1 = 0.
A simple proof of the correct answer of 0 or 1 is as follows based on two facts: a) once a box has been selected, the contents of its drawers do not change; b) from the axiomatic probability theorem P(A | A) = 1 and its corollary P(A | not-A) = 0, the probability that a gold coin is gold is 1. So, open the other drawer of the box selected to reveal whether the coin in it is gold with probability 1 or silver (not gold) with probability 0. Now you know that before it was revealed that was the same probability. It’s just that we don’t know which of 1 or 0 it is—until we reveal the coin.
There are other puzzles, e.g., those listed below, that claim to be veridical, counter-intuitive or paradoxical. They reach that conclusion by using the same faulty reasoning: ask what the probability is of an outcome after it has occurred and answer with the probability of that outcome before it has occurred. These types of puzzles, problems, or paradoxes have other errors in their reasoning and there are other proofs for the correct probability of 0 or 1.[1]
Experimental data
[edit]In a survey of 53 Psychology freshmen taking an introductory probability course, 35 incorrectly responded 1/2; only 3 students incorrectly responded 2/3.[2] None correctly responded 0 or 1.
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.
- Kicab Castaneda-Mendez (2024), "Taking a Chance in the Classroom: Puzzling Probabilities of Probability Puzzles," CHANCE, 37:3, 60-65, DOI: 10.1080/09332480.2024.2415844 https://doi.org/10.1080/09332480.2024.2415844
External links
[edit]- Estimating the Probability with Random Boxes and Names, a simulation