The two envelopes puzzle

I show you two sealed envelopes and say that one envelope has x dollars and the other has 2x but don’t say what x is. It is not possible to determine which envelope has the larger amount without opening them. I hand you one envelope (at random – whatever that means) and say that you have two options:

1. Open the envelope that you have and keep whatever you find.
2. Exchange envelopes with me, open the envelope that you receive from the exchange, and keep the amount you find. End of game.

This is clearly wrong. What is wrong? It is not sufficient to answer that the right analysis is something else. The question is what is wrong with the above.

Let us suppose, per the problem, that we draw a sample y using a given probability density function (pdf). We place in one envelope the amount y and in the other 2y. One of the envelopes is picked at random (with selection probabilities of .5) and is opened to reveal that it contains x. What is the expected value of the other envelope?

Now, given the above information, the other envelope contains either x/2 or 2x. In the first case y was x/2; in the second it was x. Let p(x) be the pdf in question. Then the probabilities are:

        P(y=x/2) = p(x/2)/(p(x) + p(x/2)) 
        P(y=x)   = p(x)  /(p(x) + p(x/2))

        E(other) = P(y=x/2)*(x/2) + P(y=x)*(2x)
                          2*p(x) + p(x/2)/2
                  = x  *  ---------------
                          p(x) + p(x/2)

        E(switch) = P(y=x)*x - P(y=x/2)*(x/2)
                          p(x) - p(x/2)/2
                  = x  *  ---------------
                          p(x) + p(x/2)

        E(switch) = .5*x - .5*(x/2) =.25*x.

E(switch) = Mean (P(y=x)*x)       - Mean(P(y=x/2)*(x/2))
          = Mean (P(y=x))*Mean(x) - Mean(P(y=x/2)*Mean(x/2)
          = .5*Mean(x) -.5*Mean(x/2)
          = .25*Mean(x)