Question: Two envelopes, each containing a check, are placed in front of you. You are to choose one of the envelopes, open it, and see the amount of the check. At this point you can either accept that amount or you can exchange it for the check in the unopened envelope. What should you do? Is it possible to devise a strategy that does better than just accepting the first envelope?

Let A and B, A<B, denote the (unknown) amounts of the checks, and note that the strategy that randomly selects an envelope and always accepts its check has an expected return of (A+B)/2. Consider the following strategy: Let F(x) be any strictly increasing distribution function. Randomly choose an envelope and open it. If the discovered check has value x then accept it with probability F(x), and with probability 1-F(x) exchange it.

a) Show that if you employ the latter strategy, then your expected return is greater than (A+B)/2. Hint: Condition on whether the first envelope has value A or B.

b) Show that for any x, the expected return under the x-strategy is always at least (A+B)/2, and that it is strictly larger than (A+B)/2 if x lies between A and B.

c) Let X be a continuous random variable on the whole real line with cdf F(x), and consider the following strategy: Generate the value of X, and if X=x then employ the x-strategy of part (b). Show that the expected return under this strategy is greater than (A+B)/2. Compare it with the expected return from part (c).

d) Consider the following paradoxical/cyclical reasoning, for the same situation, but when you know in advance that one of the envelopes has twice as much money as the other: “Suppose I choose first envelope. Suppose it has $100. Then the other has either $200 or $50, with equal probabilities (since I had a 50-50 chance to select this particular envelope out of the two). Then the expected amount of money in the second envelope is ($200+$50)/2=$125, so it looks like I have to take the second envelope. Now, without opening it, I apply the same reasoning: the second envelope has either $50 or $200. Then the first envelope has either ($25, $100) with equal chances, or ($100, $400) with equal chances, which yield $156.25 in expectation. First, it looks like I have to take the first envelope then. Second, it seems that if I apply this reasoning n times, the expected amount in an envelope, which I consider right now, becomes $100*(1.25)ⁿ. That’s weird!” In other words, the paradox is in: 1) The other envelope always looks more attractive (since the expected sum of money there is 1.25 times greater); 2) By switching back and forth, the expected sum of money in any envelope goes to infinity. Can you find a flaw in these arguments (resolve the paradox)? Do solutions (d)-(c) help?