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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?

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    You are offering a non-paradoxical solution. That's fine. But the paradox is not just in the formulation of the problem but also in a particular proposed solution that seems sound but leads to a paradoxical answer. – Ittay Weiss Feb 25 '15 at 00:42
  • If you're bored with this problem, try your hand at the Sleeping Beauty problem which offers a true paradox of different answers: http://en.wikipedia.org/wiki/Sleeping_Beauty_problem – Alex R. Feb 25 '15 at 00:47

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