Two Envelopes

The Concept of “Expected Outcome”

To understand the following problem better, let us introduce the concept of expected outcome of a game. Suppose that in a game you have two possible outcomes, $A_1$ and $A_2$. These outcomes correspond to amounts of money $x_1$ and $x_2$ that you can win (positive value) or lose (negative value) when the corresponding event occurs. Then the expected outcome is defined as

$$E=x_1\cdot P[A_1]+x_2\cdot P[A_2].$$For example, if you play a game with a friend where you toss a fair coin and your friend will either give you 5 RMB if the coin comes up heads or you will give your friend 10 RMB if the coin turns up tails, then the expected outcome is $$E=5\cdot\frac{1}{2}+(-10)\cdot\frac{1}{2}=-\frac{5}{2},$$i.e., on average you will lose 5 RMB. A game where the expected outcome is zero can be sad to be fair.



Note 1: Of course, this can be generalized to games with more than two possible outcomes.

Note 2: We will introduce a more formal notion of expectation later, but for now this is sufficient for our purposes. It is assumed that the outcomes of the game are events in a suitable probability space $(S,\mathcal{F},P)$.


The Classical Problem of the Two Envelopes

Consider the following situation: a friend has filled two envelopes with money. One of the envelopes contains twice as much money as the other. You may pick one envelope and keep the money that it contains. After you have chosen the envelope, however, and before you open it, your friend stopes you and gives you the chance to switch the envelopes, i.e., choose the other one.

At first the question appears non-sensical; after all you have no information about the two envelopes and you just chose one; why then should you switch?


The Argument for Switching

Suppose that you open the envelope that you first selected and find the amount $x$ within. Then there are two options: you could take this amount, or you could try the other envelope. With probability $1/2$ the other envelope contains $x/2$, while with the same probability it contains $2x$ RMB. The expected amount is therefore $$\frac{1}{2}\cdot \frac{x}{2}+\frac{1}{2}\cdot 2x=\frac{5}{4}x>x.$$Therefore, on average, it seems that the correct choice is to switch. This is true for any value of $x$, so you don’t really need to open the envelope. In fact, you can just accept the offer to switch.

This seems strange, since once you have switched, you are faced with the same situation as before. Again, the above analysis should apply and you would be better off switching again, setting off a never-ending loop.


A Little Knowledge is a Dangerous Thing...

At first, intuition tells us that it shouldn't matter if we switch or not, since we have no knowledge about the two identical, closed envelopes. But then the above argument, which appears to be based on an evidently correct mathematical discussion causes us to doubt our intuition. But contrary to the Monty Hall Paradox or the Two Children Problem, here the mathematical tratement causes problems, rather than resolve them.

It turns out that the problems are caused by the vague phrasing of the problem. One key piece of information that is missing is how the amounts in the envelopes are determined in the first place. So in order to understand the problem properly, we need to clearly formalize it instead of relying on the ad hoc approach used above. 

Some Exercises

Try to answer the following questions:

  1. Suppose that your friend has put $3x=30$ RMB into the envelopes and tells you this at the beginning of the game. One of the envelopes contains 20 RMB, the other contains 10 RMB. You choose an enevelope and, without opening it, are given the opportunity to switch. If you open the envelope anyway and take the money, what is the expected amount you win? What is the expected amount you would gain by switching?
  2. Suppose that you do not know what the amounts in the envelopes are. You select the first envelope and open it. It contains a 20 RMB. Then either you are playing the game with (10,20) RMB in the envelopes or the game with (20,40) RMB in the envelopes. If the probability of playing the (10,20) game is $p_1$ and the probability of the (20,40) game is $p_2$, then what is the expected gain when you switch envelopes?
  3. Answer the above question in the case where you find $x$ RMB in the envelope that you first select and $p_1=p_2=1/2$.
  4. The assumption that $p_1=p_2=1/2$ may seem the natural choice if there is no information on how the amounts were selected (see the Principle of Insufficient Reason). However, explain why it would not be possible to construct a probability space where, for example, the pairs (1,2), (2,4), (4,8), (8,16),… are all equally likely.

It follows from this discussion that when you open the first envelope and find $x$ RMB, you need to know how likely it is that you are playing a game with $(x,2x)$ or a game with $(x/2,x)$ RMB. It is not possible to define a probability space where all amounts are possible and all pairs of values are equally likely.

So does this mean that the Two Envelope Paradox is meaningless? Actually, while the classical way pf phrasing it is too vague, there exists a way of expressing the paradox in a rigorous manner. Watch the video below to see how it is done.