Bertrand's Box Paradox - Box Version

Box Version

There are three boxes, each with 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 flaw is in the last step. While those two cases were originally equally likely, the fact that you could not have found a silver coin if you had chosen the GG box, but could if you had chosen the GS box, means they do not remain equally likely. 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.

So the probability that the chosen box is GG becomes:

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.

Alternatively, one can simply note that the chosen box has two coins of the same type 2⁄₃ of the time. So, regardless of what kind of coin is in the chosen drawer, the box has two coins of that type 2⁄₃ of the time. In other words, the problem is equivalent to asking the question "What is the probability that I will pick a box with two coins of the same color?".

Bertrand's point in 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 correctly applied in the second solution method, but not in the first.