Solving the Bertrand’s Box Problem

A Coin Paradox

The Paradox

Someone presents us with three identical boxes. Inside one of the boxes there are two gold coins; inside another box there are two silver coins. The third box holds inside one gold coin and one silver coin. We do not know which coins are in which box.

Now, let us choose a box at random. We reach without looking under the cloth covering the coins and take out one of the coins. Now we can look. It is gold.

So, we can be sure that the box we chose cannot be the box containing the two silver coins. It must be either the box containing two gold coins or the box containing one gold coin and one silver coin. So, the other coin must be either a gold coin or a silver coin.

Given what we now know, what is the probability that the other coin in the box is also gold, and what odds should we take to bet on it?

This is essentially the so-called “Bertrand’s Box” Paradox, first proposed by Joseph Bertrand in 1889.

After withdrawing the gold coin, there are only two options left. One is the box containing the two gold coins, and the other is the box containing one gold and one silver coin. It seems intuitively clear that each of these boxes is equally likely to be the one we chose at random, and that therefore the chance it is the box with two gold coins is 1/2, and the chance that it is the box containing one gold and one silver coin is also 1/2. The probability that the other coin is a gold coin must, therefore, be 1/2.

Is this right? In fact, the answer is not 1/2. The probability that the other coin is also gold is 2/3. But this seems counter-intuitive. How can looking at one of the coins make the other coin likely to be of the same type of metal?

To understand why, let us look a little closer. Three equally likely scenarios might have led to us choosing that shiny gold coin. Let us separately label all the coins in the boxes to make this clear.

In the box containing two gold coins, there will be Gold Coin 1 and Gold Coin 2. These are both gold coins, but they are distinct, different coins.

In the box containing the gold and silver coins, we have Gold Coin 3, which is a different coin to Gold Coin 1 and Gold Coin 2. There is also what we might label Silver Coin 3 in the box with Gold Coin 3. This silver coin is distinct and different to what we might label Silver Coin 1 and Silver Coin 2, which are in the box containing two silver coins, which was not selected.

So, here are the equally likely scenarios when we withdrew a gold coin from the box.

We chose Gold Coin 1.
We chose Gold Coin 2.
We chose Gold Coin 3.
We do not know which of these gold coins we withdrew from the box.

If it was Gold Coin 1, the other coin in the box is also gold.
If it was Gold Coin 2, the other coin in the box is also gold.
If it was Gold Coin 3, the other coin in the box is silver.
Each of these possible scenarios is equally likely, so the probability that the other coin is gold is 2/3 and the probability that the other coin is silver is 1/3.

Before withdrawing the gold coin, the chance that the box we had selected was that containing two gold coins was 1/3. By revealing the gold coin, however, we not only excluded the box containing two silver coins but also introduced the new information that we could potentially have chosen a silver coin (if the selected box was that containing one gold and one silver coin) but did not.

That made it more likely (twice as likely) that the box you withdrew the gold coin from was that containing the two gold coins than the box containing one gold and one silver coin.

This is the solution to the Bertrand’s Box Paradox.