Bertrand's box paradox

Bertrand's box paradox is a classic paradox of elementary probability theory. It was first posed by Joseph Bertrand in his "Calcul des probabilités", published in 1889. There are three boxes: a box containing two gold coins, a box with two silver coins, and a box with one of each. After choosing a box at random and withdrawing one coin at random that happens to be a gold coin, it may seem that the probability that the remaining coin is gold has a probability of 1/2; in fact, the probability is actually 2/3.

In a 1950 article, Warren Weaver introduced a simple way to conduct the experiment on people: the boxes are replaced by cards, and gold-and-silver coins are replaced by red-and-black markings on the sides of the cards. In what Martin Gardner has called the three-card swindle, a card is drawn from a hat, and if a red mark is shown, the dealer bets the victim even money that the other side is also red. The victim is convinced that the bet is fair, but the dealer makes money in the long run by winning 2/3 of the time.

These simple but slightly counterintuitive puzzles are used as a standard example in teaching probability theory. Their solution illustrates some basic principles, including the Kolmogorov axioms.

Box version

You have three boxes, each with one drawer on each of two sides, A&B. Each drawer contains a coin. One box has a gold coin on both sides (GG), one a silver coin on both sides (SS), and the third gold on one side and silver on the other (GS). You choose a box at random, open one drawer, and find a gold coin. What is the chance of the coin on the other side being silver?

The correct answer is one-third; the coin you see is equally likely to be any of the three gold coins, only one of which is opposite a silver coin. However, there is a tendency to fall into the following fallacious reasoning, which has been compared to the Monty Hall problem:

*You cannot be looking at the box SS; so you must be looking at GG or GS
*You were equally likely to pick either one.
*So there must be a 50/50 chance of GG or GS now.

In reality, though, you were not choosing boxes, but drawers. Now that you have a gold coin, all the possibilities are as follows:
*You chose G of GS, and the other drawer contains a silver coin (⅓)
*You chose G₁ of GG, and the other drawer contains a gold coin (⅓)
*You chose G₂ of GG, and the other drawer contains a gold coin (⅓)This provides for a combined ⅔ chance of the other coin being a gold coin, and thus, the chance of the coin in the other drawer being silver is ⅓.

In other words, in order to get gold coin in the first chosen drawer, you only have three possibilities:
*GG, A side, gold coin on the other side.
*GG, B side, gold coin on the other side
*GS, A side (gold coin side), silver coin on the other side.
*SS won't be taken into account since we got a gold coin.

*Therefore only 1 out of 3 shows silver coin.

Card version

Suppose you have three cards:
*a "black card" that is black on both sides,
*a "white card" that is white on both sides, and
*a "mixed card" that is black on one side and white on the other.You put all of the cards in a hat, pull one out at random, and place it on a table. The side facing up is black. What are the odds that the other side is also black?

The answer is that the other side is black with probability 2/3. However, common intuition suggests a probability of 1/2.

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

Preliminaries

To solve the problem, either formally or informally, we must assign probabilities to the events of drawing each of the six faces of the three cards. These probabilities could conceivably be very different; perhaps the white card is larger than the black card, or the black side of the mixed card is heavier than the white side. The statement of the question does not explicitly address these concerns. The only constraints implied by the Kolmogorov axioms are that the probabilities are all non-negative, and they sum to 1.

The custom in problems when one literally pulls objects from a hat is to assume that all the drawing probabilities are equal. This forces the probability of drawing each side to be 1/6, and so the probability of drawing a given card is 1/3. In particular, the probability of drawing the double-white card is 1/3, and the probability of drawing a different card is 2/3.

In our question, however, you have already selected a card from the hat and it shows a black face. At first glance it appears that there is a 50/50 chance (ie. probability 1/2) that the other side of the card is black, since there are two cards it might be: the black and the mixed. However, this reasoning fails to exploit all of your information; you know not only that the card on the table "has" a black face, but also that one of its black faces is facing you.

olutions

Intuition

Intuition tells you that you are choosing a card at random. However, you are actually choosing a face at random. There are 6 faces, of which 3 faces are white and 3 faces are black. Two of the 3 black faces belong to the same card. The chance of choosing one of those 2 faces is 2/3. Therefore, the chance of flipping the card over and finding another black face is also 2/3. Another way of thinking about it is that the problem is not about the chance that the other side is black, it's about the chance that you drew the all black card. If you drew a black face, then it's twice as likely that that face belongs to the black card than the mixed card.

Labels

One solution method is to label the card faces, for example numbers 1 through 6.ref|Label16 Label the faces of the black card 1 and 2; label the faces of the mixed card 3 (black) and 4 (white); and label the faces of the white card 5 and 6. The observed black face could be 1, 2, or 3, all equally likely; if it is 1 or 2, the other side is black, and if it is 3, the other side is white. The probability that the other side is black is 2/3.

Bayes' theorem

Given that the shown face is black, the other face is black if and only if the card is the black card. If the black card is drawn, a black face is shown with probability 1. The total probability of seeing a black face is 1/2; the total probability of drawing the black card is 1/3. By Bayes' theorem,ref|Bayes the conditional probability of having drawn the black card, given that a black face is showing, is:$frac\{1cdot1/3\}\{1/2\}=2/3.$

Eliminating the white card

Although the incorrect solution reasons that the white card is eliminated, one can also use that information in a correct solution. Modifying the previous method, given that the white card is not drawn, the probability of seeing a black face is 3/4, and the probability of drawing the black card is 1/2. The conditional probability of having drawn the black card, given that a black face is showing, is:$frac\{1/2\}\{3/4\}=2/3.$

ymmetry

The probability (without considering the individual colors) that the hidden color is the same as the displayed color is clearly 2/3, as this holds if and only if the chosen card is black or white, which chooses 2 of the 3 cards. Symmetry suggests that the probability is independent of the color chosen. (This "can" be formalized, but requires more advanced mathematics than yet discussed.)

Experiment

Using specially constructed cards, the choice can be tested a number of times. By constructing a fraction with the denominator being the number of times "B" is on top, and the numerator being the number of times both sides are "B", the experimenter will "probably" find the ratio to be near 2/3.

Note the logical fact that the B/B card contributes significantly more (in fact twice) to the number of times "B" is on top. With the card B/W there is always a 50% chance W being on top, thus in 50% of the cases card B/W is drawn, card B/W virtually does not count. Conclusively, the cards B/B and B/W are not of equal chances, because in the 50% of the cases B/W is drawn, this card is simply "disqualified".

Related problems

*Boy or Girl
*Monty Hall problem

Notes and references

# Bar-Hillel and Falk (page 119)
# Nickerson (page 158) advocates this solution as "less confusing" than other methods.
# Bar-Hillel and Falk (page 120) advocate using Bayes' Rule.

*Bar-Hillel, M. A., and R. Falk (1982). " [http://dx.doi.org/10.1016/0010-0277(82)90021-X Some teasers concerning conditional probabilities] ". "Cognition", "11", 109–122.
*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 [http://harrisschool.uchicago.edu/About/publications/working-papers/pdf/wp_05_14.pdf Wason, Monty Hall, and Adverse Defaults] .