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What’s in the box? Betting on the Bertrand’s Box paradox.

November 26, 2017

You are presented with three identical boxes. You are made aware that one of the boxes contains two gold coins, another contains two silver coins, and the third contains one gold coin and one silver coin. You do not know which box contains which.

Now, choose a box at random. Reach without looking under the cloth covering the coins and take out one of the coins. Now you can look. It is gold.

So you can be sure that the box you 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.

Withdrawing the gold coin from the box doesn’t provide you with the information to identify which of these two boxes it is. So the other coin must either be a gold coin or a silver coin.

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

This is essentially the so-called ‘Bertrand’s Box’ paradox, first proposed by Joseph Bertrand in 1889 in his opus, ‘Calcul des probabilités’.

 

Spoiler alert (Solution)

After withdrawing the gold coin, there are only two boxes 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 you 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. Therefore, the probability that the other coin is gold must be 1/2.

This sounds right, but it is in fact the wrong answer.

In fact, there are three equally likely scenarios that might have led to you 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 you withdrew a gold coin from the box.

a. You chose Gold Coin 1.

b. You chose Gold Coin 2.

c. You chose Gold Coin 3.

You do not know which of these gold coins you 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 (i.e. each has a probability of being the true state of the world of 1/3), so the  probability that the other coin is gold is 2/3 and the probability that the other coin is silver is 1/3. So, if you are offered even money about the other coin being gold, the edge is very much with you.

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

And that is the solution to the Bertrand’s Box paradox.

 

 

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