Consider this: there are two envelopes on the table containing some cash. You don’t know how much it is, but you know that one contains twice the amount as the other. You were given one of them at random. You have two choices: open the one you got and take the cash, or switch to the other. What will you do?
Expected Value
Naturally, you will go for the expected value of a switch, and if that is greater than the money at hand, you will make the exchange. Let the amount of money inside your envelope be X. You doubt the other envelope has 2X or X/2 amount. And they are equally likely (probability of half). The Expected value, in that case, is (1/2)x2X + (1/2)x(X/2) = X + (1/4)X > X. So, you switch, right?
The problem with the problem
No, you will not. Think about the issue with a swap. After making the switch, you can make the same expected value calculations for the original envelope, draw the same conclusion, and switch back—and you do it forever!
The problem, in reality, involves two conditional probabilities and needs to be solved separately. First, you assume that the second envelope contains less than the first. In the second case, you consider the opposite, i.e., the second envelope holds more than the first. You multiply the respective probabilities to get the answer.
E(2nd) = V(2nd | 1st < 2nd)xP(1st < 2nd) + V(2nd | 1st > 2nd) x P(1st > 2nd).
For X and 2X, E(2nd) = 2X(1/2) + X(1/2) = 1.5X.
Naturally, if X and 2X are the two options, the average value with the first envelope is not X, but (X+2X)/2 = 1.5X
They match, and no need to switch!
