Two Envelopes Paradox

The paradox claims that it is always better to switch because the expected utility of switching is greater than for not switching. This expected utility is derived using a decision theory table. Say x is the expected utility of not switching. The calculation yields 0.5(2x) + 0.5(x/2) = 1.25x which comes from the thinking that there is a 50-50 probability of having either of the lesser or greater valued envelopes, and the benefit of switching with the lesser is 2x, while the benefit of switching with the greater is x/2.

The problem is that this decision theory table does not account for the fact that we are working in a bounded range of values (since there are only two). Basically, if you have the higher one, there is no chance that you will double your money, and if you have the lower one, there is no chance that you will halve your money. This can also be seen in the fact that the expected benefit of taking both should be either 3x or 1.5x, not 2.5x.

We can reevaluate the situation by using definite values of expected utility. Say y is the utility of getting the lesser value. Then the expected utility of swapping is: 0.5(y) + 0.5(2y) = 1.5y And this is the same as the expected utility of not swapping.

The paradox assumes an unbounded range of values. We can create a situation where this assumption is correct. Say a rich person allows you to switch envelopes as many times as you like, and every time you say you want to switch, he or she opens the envelope you gave them, and randomly doubles or halves the value, then gives you back an envelope with a check of this new value. In this case, the paradox's conclusion is true, and it is best to continue switching for as long as possible. While on average, you will very likely end up with about the same amount as you started with, increasingly larger quantities will become accesssible. Say you start with $1 and choose to do n switches. Then you can get anywhere from 2^n to 2^(-n) dollars. To find the expected utility, we calculate the average of the dollar amounts over all possible amounts, weighted by the probability of getting that amount. To do this, we create a variable y that runs from -n/2 to n/2 (we assume an even number of switches for convenience) and then Utility(y)=2^(2y) and Probability(y) = [n_C_(n/2 - y)]/2^n, so that the weighted average becomes:

Judging by the huge value of expected utility, we assume that this average diverges in the limit of large n, so it is best to continue swapping as long as possible.