# the envelope problem

this problem is deeply upsetting to me because i don’t see what the fuss is.

i’m in the camp that thinks it’s a trivial problem and that the paradox stems from not understanding how to set up an expected value calculation correctly. clearly it doesn’t matter if you switch.

the “paradox” describes an entirely different problem: one with 3 envelopes of cash. your first envelope is the one with the positive amount X, and you are given the choice between two indistinguishable envelopes which you know contain either 2X or X/2, respectively.

the correct calculation for the two envelope problem is as follows: knowing the amounts in the envelope are X and 2X, you have two possibilities for the amount in “the other envelope” after choosing your first. here they are:

1. half the time, you have chosen X first, and so the other envelope has 2X.
2. the other half of the time, you have chosen 2X first, so the other envelope has X.

there’s that nice commutative property of addition, so it’s clear that both of these have the same expected value of (3/2)X.