Amos Storkey

Money Trouble

This is one of my favourite brain teasers. It was first introduced to me some years ago in a Cambridge pub garden during (if my memory serves me well) the Bayesian Methods workshop at the Neural Networks and Machine Learning session at the Newton Institute. There are many interpretations, but needless to say, I prefer the Bayesian ones.

The scene

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 0.5x20000+0.5x80000-40000, 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?

My answer

There have been comments made by many people on this problem, most of which provide good solutions to the problem, but some of which are just plain wrong! Those which are right generally amount to much the same idea. My attempt at an answer can be found on the Two Envelope Paradox Solution page.

Amos Storkey 2000-2005.