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67. Silver and Gold:

Bertrand’s Probability Paradox

The Question:

In front of you there are three boxes, each with two drawers. The two drawers of one box each contain a gold coin. The two drawers of another box each contain a silver coin. And the third box has one drawer in which there is a silver coint, the other drawer contains a gold coin.

You choose one box at random, and then one drawer at random. It contains a gold coin. What’s the chance that the other drawer of that box also contains a gold coin?

50 percent?


The Paradox:

If you said yes, you are in large, though not necessarily good company. Like you, those who answered in the affirmative, usually reason as follows: since they did not draw a silver coin, the box they chose could not have been the one with the two silver coins. Hence, what they chose was either the gold-gold box, or the gold-silver box. Since there is a fifty-fifty chance that one chose either of the two boxes, there must be a fifty-fifty chance that the second drawer contains either a silver or a gold coin. Sounds reasonable…but it is wrong!



This is one more of the apparent conundrums that the French mathematician Joseph Bertrand (1822-1900) utilized in his book Calcul des probabilités (Probability calculus), to demonstrate how naïve people – or at least people not yet introduced to the mysteries of statistics -- can be misled.

Bertrand was also active in the field of economics where he developed a theory of competition and interactions among firms. His model of oligopoly contrasted with the model of his fellow Frenchman, Antoine Augustin Cournot. (See the chapter on Bertrand’s Economic Paradox.)



Once one has eliminated, as one must, the silver-silver box, there are three possibilities of drawing a gold coin: one possibility is to draw the gold coin from the gold-silver box, and there are two possibilities to draw either of the gold coins of the gold-gold box. So, there is one chance out of two that one chose the gold-silver box, but there are two chances out of three that the box one chose is the gold-gold box. Hence, as soon as one has drawn a gold coin, the chances of drawing another gold coin are two out of three.  


Technical supplement:

In the 1980s, two psychology professors from Hebrew University of Jerusalem , Maya Bar-Hillel and Ruma Falk, conducted an experiment among their first year students. They presented them with a hat that contained three palying cards: one was red on both sides, one was white on both sides, and one was red on one side, and white on the other. A card was drawn out of the hat and laid on the table without revealing the other side.

The question was: if it is red, what are the chances that the other side is also red? (The experiment also works if the face of the drawn card is white.) And now it is obvious why the correct answer is 2/3. “Clearly, an all-red card is twice as likely to show a red face up as a card that only has one red side.” [i]

Nevertheless, 35 of the the 53 students answered incorrectly and only 3 out of the 53 gave the correct answer of two thirds. (Strangely, the paper did not report what the remaining 15 students answered.) The authors called the tendency of most people to give the incorrect answer the “fallaciousness of attributing posterior equiprobability to the remainign events”.


Bar-Hillel and Falk present another illustration of Bertrand’s Probability Paraodox.  If one sees Mr. Smith – known to be a father of two children – walk in the street with a boy whom he introduces as his son, what is the probability that his other child is also a boy?

Now, if Mr. Smith presents the boy as his eldest child, what is the probability that Mr. Smith’s other child is also a boy?

In the first case, the possibilities are boy & boy, girl & girl, or boy & girl. Since girl & girl is ruled out, there are two chances out of the three that if the child accompanying Mr. Smith is a boy, the other child would also be a boy.

In the second case, the probabilty changes! The reason is that when listing the eldest child first, the possibilities now are boy & boy, girl & girl, boy & girl, and girl & boy. Girl & girl is ruled out. But so is girl & boy. What remains are the possibilities, boy & boy and boy & girl. So, the probability that the younger child is also a boy, is …one half!


© George Szpiro, 2019


[i] Bar Hillel, Maya and Ruma Falk, “Some teasers concerning conditional probabilities”, Cogniton, 11 (1982) 109-122.

Corrections, comments, observations:    

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