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Bertrand's box paradox

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+#redirect [[Bertrand's box paradox]]
-'''''Three Cards''''' is a simple but slightly counterintuitive puzzle used as a standard example in [[probability theory]]. Its solution illustrates some basic principles, including the [[Kolmogorov axioms]].
+{{rcat shell|
-Suppose you have three cards:
+{{R to related}}
-*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 face down 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 [[probability|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 of the table ''has'' a black face, but also that at least one of its black faces is facing you.
-== Solutions ==
-===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
-:<math>\frac{1\cdot1/3}{1/2}=2/3.</math>
-===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
-:<math>\frac{1/2}{3/4}=2/3.</math>
-=== Symmetry ===
-The probability (without considering the individual colors) that the hidden color is the same is 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 this probability is [[Statistically independent|independent]] of the color chosen.  (This ''can'' be formalized, but requires more advanced mathematics than yet discussed.)
-=== Experiment ===
-Try it. Construct the cards (using pieces of paper with "B" and "W" written on the sides, unless you really want to color in the sides), and try it a number of times. Construct 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". You'll ''probably'' find the ratio is near 2/3.
-==See also==
-*[[Boy or Girl]] is a very similar problem.  The probabilities are different but could be simulated by having two mixed cards instead of one.
-*[[Monty Hall problem]]
-==Notes and references==
-#{{note|53}} Bar-Hillel and Falk p.119
-#{{note|Label16}} Nickerson (p.158) advocates this solution as "less confusing" than other methods.
-#{{note|Bayes}} Bar-Hillel and Falk (p.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&ndash;122.
-*Nickerson, Raymond (2004). ''Cognition and Chance: The psychology of probabilistic reasoning'', Lawrence Erlbaum. Ch. 5, "Some instructive problems: Three cards", pp.157&ndash;160. ISBN 0-8058-4898-3
-==External links==
-*[http://www.flaguide.org/tools/math/fault/fault2B.php Example from an assessment test] Example 4 (includes answer).
-[[Category:Probability theory paradoxes]]