Monday, October 24, 2011

Why you should not bother to switch the envelope

You are presented with two envelopes, and are told that one contains twice as much money as the other. You are to choose one envelope and earn the money in it. You pick one envelope, and upon opening it, find that it contains y dollars. You have the option to keep that envelope, or to switch to the other envelope. Should you switch?

Here is the paradox: there is a 1/2 chance that the other envelope contains 2y dollars, and a 1/2 chance that it contains y/2. If you switch the envelope, your expected earning is 1/2 × 2y + 1/2 × y/2 = 1.25y, which is greater than y—your expected earning should you choose to stay with the current envelope. To earn more money, you should, therefore, switch.

It does not make sense, does it? Intuitively it should not matter whether you switch or not, because the envelope that you end up with is equally likely to be the one with more money or the one with less. So what is the logic flaw in the paradoxical argument above?

Let's say that one envelope contains x dollars and the other 2x dollars. When you pick one envelope, you have an equal chance of picking either one, i.e., P(y is x) = P(y is 2x) = 1/2. Should you choose not to switch, your expected earning is P(y is x) y(y is x) + P(y is 2x) y(y is 2x) = 1/2 × x + 1/2 × 2x = 1.5x. Here is an important subtlety: I've written y(y is ...) to denote the revealed amount y upon opening the envelope, in the two different events of y being either 2x or x; that amount is not the same for the two events! In other words, you cannot write the expected earning as P(y is x) y + P(y is 2x) y = y, because y is different in the two different events of "y is x" and "y is 2x"!

Should you switch the envelope, your expected earning is P(y is x) 2 × y(y is x) + P(y is 2x) 1/2 × y(y is 2x) = 1/2 × 2x + 1/2 × x = 1.5x, which is exactly the same as that should you keep your original choice.

The paradox is thus resolved by defining the right set of events ("y is x" and "y is 2x") and realizing that the revealed amount y is not a constant for these different events, but a function whose value varies between the events: y(y is x) = x, while y(y is 2x) = 2x.

Wait a minute, some may say. If you are told that the other envelope has equal probability of containing 2y or y/2 dollars, after you open the first envelope and find within it y dollars, you should obviously switch. So what is the difference between this setup and the original one?

Well, the two setups are quite different. To clarify their difference, it is helpful to consider how the second setup can be correctly implemented. A simple way is to begin with an empty envelope, and have another person put in it either 2y or y/2 with equal probability after you open the first envelope. No matter how it is implemented, the second setup requires making changes to the envelopes after the first envelope is opened, and the change depends on the amount in the opened envelope. So you will be playing a different game than the original one: in the original setup, the two envelopes are prepared beforehand and are not adjusted after your first pick.

The difference between the second and the original setup thus reveals the true fallacy of the paradox: it confuses the expectation of a two-step procedure with that of its second step alone. The two steps are: 1) pick one envelope out of the two, find out the amount within, 2) decide whether to keep it or switch to the other. Your expected earning depends on your action in the first step as well as on the second.

After opening the first envelope and finding y dollars in it, there are two possibilities for the initial set-up of envelopes: {y, y/2} and {2y, y}, and they are equally probable: P({y, y/2}) = P({2y,y}) = p. The expected earning, should you decide to switch on the second step, is

   P({y,y/2})[P(1st is y|{y,y/2}) y/2 + P(1st is y/2)|{y,y/2}) y
+ P({2y,y})[P(1st is y|{2y,y}) 2y + P(1st is 2y)|{2y,y}) y] = 2.25 p y

The expected earning of not switching on the second step is

   P({y,y/2})[P(1st is y|{y,y/2}) y + P(1st is y/2)|{y,y/2}) y/2] 
P({2y,y})[P(1st is y|{2y,y}) y + P(1st is 2y)|{2y,y}) 2y] = 2.25 p y

Again, there is no difference either you keep your pick or switch.

There is actually another subtlety. In the above analysis, I should have written P(y is x|y) and P(y is 2x|y) to denote the conditional probabilities that the picked envelope contains x or 2x, given that the revealed amount is y. I have taken P(y is ...|y) = P(y is ...), with the assumption that seeing what is in the envelope does not tell you any information on whether it is x or 2x. This is not necessarily true when you have additional information available. For example, if you know that neither envelope contains more than 1 million dollars, and your opened envelope contains over half a million dollars, you should definitely not switch. On the other hand, if your opened envelope contains 5 cents, and realistically you know that there is no way for the other envelope to have 2.5 cents (physically impossible), you should definitely switch since you now know for sure that the other envelope has 10 cents in it.


  1. You'll like the Monty Hall problem for Stat 101.

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