| The flaw in the argument for switching comes in its second line: "1. I denote by A the amount in my selected envelope. 2. The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2." If we're gonna be taking expected values we need to assume that the monetary amounts in the two envelopes are generated from a probability distribution on the nonnegative reals. Then the probability distribution for the amount of money in the smaller envelope is going to be some kind of curve (call it S), and obviously the probability distribution for the amount of money in the bigger envelope is going to be the same curve only "stretched out" and "squashed" by a factor of two (we'll call it B). Now if you say "my envelope contains exactly A dollars," you can tell what the probability is that your envelope is the smaller envelope by comparing the relative heights of the two curves at the value A; let's denote these heights by S(A) and B(A). It is certainly possible that the two heights are the same, in which case it's fifty-fifty that you have the bigger or smaller envelope, the rest of the logic holds, and you should indeed switch, getting an expected return of 5/4 * A. But it is obviously impossible that in general the two heights are the same for any given A, because then the two probability distributions are the same -- and that can't be true, because B is a squashed, stretched-out version of S, and for a variety of fairly obvious reasons you can't squash and stretch out a finite curve on the positive reals and get the same curve (unless that curve is 0 everywhere). And so we can't conclude that in general you should switch, which is good because if you're not allowed to look at the money before you switch it obviously doesn't matter whether you do or not (unless the people running the game are messing with you). |