I’m a sucker for probability puzzles. Here’s an interesting and unusual one I found on the site of Amos Storkey:

You are taking part in a game show. The host introduces you to two envelopes. He explains carefully that you will get to choose one of the envelopes, and keep the money that it contains. He makes sure you understand that each envelope contains a cheque for a different sum of money, and that in fact, one contains twice as much as the other. The only problem is that you don’t know which is which.

The host offers both envelopes to you, and you may choose which one you want. There is no way of knowing which has the larger sum in, and so you pick an envelope at random (equiprobably). The host asks you to open the envelope. Nervously you reveal the contents to contain a cheque for 40,000 pounds.

The host then says you have a chance to change your mind. You may choose the other envelope if you would rather. You are an astute person, and so do a quick sum. There are two envelopes, and either could contain the larger amount. As you chose the envelope entirely at random, there is a probability of 0.5 that the larger check is the one you opened. Hence there is a probability 0.5 that the other is larger. Aha, you say. You need to calculate the expected gain due to swapping. Well the other envelope contains either 20,000 pounds or 80,000 pounds equiprobably. Hence the expected gain is 0.5×20000+0.5×80000-40000, ie the expected amount in the other envelope minus what you already have. The expected gain is therefore 10,000 pounds. So you swap.

Does that seem reasonable? Well maybe it does. If so consider this. It doesn’t matter what the money is, the outcome is the same if you follow the same line of reasoning. Suppose you opened the envelope and found N pounds in the envelope, then you would calculate your expected gain from swapping to be 0.5(N/2)+0.5(2N)-N = N/4, and as this is greater than zero, you would swap.

But if it doesn’t matter what N actually is, then you don’t actually need to open the envelope at all. Whatever is in the envelope you would choose to swap. But if you don’t open the envelope then it is no different from choosing the other envelope in the first place. Having swapped envelopes you can do the same calculation again and again, swapping envelopes back and forward ad-infinitum. And that is absurd.

That is the paradox. A simple mathematical puzzle. The question is: What is wrong? Where does the fallacy lie, and what is the problem?

Storkey proposes a somewhat involved Bayesian explanation of what is wrong. I think a much simpler explanation is possible.

Let’s go back and examine the player’s reasoning. He opened the first envelope and found 40,000 pounds. Then as Storkey puts it “You need to calculate the expected gain due to swapping. Well the other envelope contains either 20,000 pounds or 80,000 pounds equiprobably.”

No. The player is in no position to calculate expected values because he does not know what the sample space is. He does not know that 20,000 and 80,000 pounds are both possible, much less equiprobable.

To make this concrete, imagine we are the hosts, and we prepared the envelopes. We put 20,000 in one, 40,000 in the other. The player observes 40,000 in the first envelope. If from that he infers that there is a fifty percent chance of 80,000 in the other envelope he is just mistaken. There is no chance of it.

The player cannot compute expectations without knowing the sample space. We, the hosts, know the sample space and can easily compute expected values. The player has two pure strategies available, call them Hold and Swap. A player using the Hold strategy opens the envelope and then does not swap. The expected value of this strategy is 30,000. A player using the Swap strategy opens the first envelope and then swaps. The expected value of this strategy is 30,000. Any mixed strategy will be a weighted sum of Hold and Swap, and hence will have expected value of 30,000. Observing the first value does not provide the player any useful information. He cannot perform conditional probability computations of the form P(Y|X=40,000) because he does not know what values of Y are admissible.