Mathematical paradoxes -- The Two Envelopes Problem

The problem

Suppose you are given two envelopes with money in them. You are told that you may pick and keep only one of them. You are also told that one envelope contains twice as much money as the other, but not which envelope has more money. You choose one of the envelopes, and you find that it contains $10. Now, you are given the option of taking the other envelope instead. Is there any advantage to switching envelopes? How much money can you expect to gain/lose?

You might like to spend some time trying to figure it out. If I offered you the chance to buy the other envelope for $12, would you take it?

Math background

You need the ability to add and multiply. Dividing by 2 might help also. That's it! (It's great that some very complex problems don't need much knowledge of what most people think of as "math.")

The paradox

There are two common answers.

Answer 1: There are two options for what's in the other envelope. It either contains $5 or $20, with a 50% chance of each. On average, then, if you switch envelopes you will get ($5 + $20)/2 = $12.50. Keeping the $10 envelope only gets you $10. Therefore, you should switch, and you will get on average $12.50, for an average gain of $2.50.


Answer 2: Since you picked the first envelope randomly, it's impossible to gain by switching. You might have taken the higher number, and you might have taken the lower number. If you took the higher one, then you stand to lose the difference between the two envelopes. If you took the lower one, then you stand to gain that difference. Since there was a 50/50 chance of taking each, the differences balance out, and the average gain from switching envelopes is $0.

Which do you think is correct?

A more precise statement of the problem

1. There are two envelopes with dollar amounts x and 2x. We'll say that envelope L has x and envelope H has 2x


2. There is a probability of 50% of choosing each envelope.


3. In one of the envelopes, the amount is $10.

The question: what is the average of the amount you would gain if you chose envelope L and the amount you would lose if you chose envelope H?

You might like to take a second to think about it again. Does that change your idea of what the answer is?

The solution

As for the Monty Hall problem, stating the problem more precisely makes it easier to figure out what the answer is. If you took envelope L, then you will gain x dollars if you switch. If you took envelope H, then you will gain x - 2x = -x dollars if you switch. (That is, you lose x dollars.) The average is (x + -x)/2 = 0.

But we can't accept this answer until we know why the other possible answer is wrong. After all, it's pretty intuitive. There is either double the amount in the other envelope, or there's half, and there is a 50% chance of each, right?

Well, no. To be precise (and you really have to be!), there is either a 100% chance of the other envelope having double the money as the one you chose, or there is a 100% chance of the other envelope having half. But that's not the same as saying there is a 50% chance of each. If you did an experiment where you had a constant number for the first envelope chosen (call it $10), and then either doubled it or halved it for the other envelope, then you would indeed benefit from switching envelopes. But that's not what is happening in the Two Envelopes Problem. That is a different problem entirely.


There's another way of looking at it. If you say,

1. I have a chance of gaining $10, but I only have a chance of losing $5,

then you are assigning undue weight to the gains. It's more accurate to say

2. If the other envelope has $5 in it, then I stand to lose $5 by switching. But if I'd chosen the other envelope first, I'd stand to gain $5.


3. If the other envelope has $20 in it, then I stand to gain $10 by switching. But if I'd chosen the other envelope first, I'd stand to lose $10 by switching.

Do you see the difference? If you make the first statement, then you are assigning double the weight to the third statement compared to the second. You are saying something like "there is a 2/3 chance of statement 1 being correct, and a 1/3 chance of statement 2 being correct." (The 50% chance pertains only to whether or not you took H or L, so you can't even guess that statements 2 and 3 each have a 50% probability.) However, you simply don't know which of 2 and 3 is correct, so you can not assign odds to them. Because the first statement is assigning odds where it is impossible to do so, it isn't true.

Was your first idea right? Did you change your mind to the right answer after the problem was reworded?

Conclusion

In the Two Envelopes Problem, there is on average no gain to be had by switching envelopes. You will either gain x dollars or lose x dollars from the best-case scenario, and there's a 50% chance of each.

This is probably a great way to scam people. Put $10 and $20 in two envelopes, then let them look in one. Ask them to pay a bit under $12.50 if they chose the $10 envelope and want to switch, or a bit under $30 if they chose the $20 envelope. They'll think it's a great deal--on average, how can they lose?--but your gains from the people who choose the $20 envelope first will outweigh your losses from the people who chose the $10 envelope first. (Of course, you have to stipulate that they can't keep the first envelope they chose, or they're just going to take your money and run. Maybe. Some people don't like gambling.)

In practical terms, the amount of money you can offer is limited. If Alice offers two envelopes to Bob, and Bob opens an envelope to find $10 000, and he knows Alice's net worth is only $25 000, then he knows he must have opened the envelope with more money. For a similar reason, the amounts in both envelopes should be even numbers. I don't know if there's a way around this. Maybe selecting the amounts according so a distribution that puts emphasis on low values would work... but knowledge of the distribution might allow a savvy investor to beat the system. I think this is possibly a good research opportunity for someone (maybe me, after I'm done the current batch of papers I'm working on). The problem is because money isn't infinitely divisible. You can't offer someone 1/9 dollars. It's not obvious how finely divisible money has to be to make this scheme work.