This question is regarding the two envelope problem.
http://en.wikipedia.org/wiki/Two_envelopes_problem
It seems to me the very simple solution to this problem is to make the sum of all the money in the envelopes (SUM) = X. No matter how you distribute X between the two envelopes, when you calculate the probabilistic value of money in the envelopes, it's always the same.
For example:
E1 = (A)X
E2 = (1-A)X
Initial choice probabilistic value calculation: (1/2)(A)X + (1/2)(1-A)X = (1/2)X
The probabilistic value of the second envelope would be: (1/2)(1-A)X + (1/2)(A)X = (1/2)X
Doesn't this resolve the problem? It seems too simple, can someone tell me what I'm missing?