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July 2005
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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.
You analyze: I have an unknown amount x in my hand. If I exchange I get x/2 or 2x with equal probability. The expected value of the exchange is (1/2)((x/2) + (2x)) or 1.25y. Expected value from not exchanging is x. Therefore it is better to exchange.

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))
According the expectation of the contents of the other envelope is
        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)
And the expected gain from switching is
        E(switch) = P(y=x)*x - P(y=x/2)*(x/2)
        
                          p(x) - p(x/2)/2
                  = x  *  ---------------
                          p(x) + p(x/2)
Now comes the switcheroo. We do not know what these probabilities are because we do not know what the pdf is. However we do know that the mean of these probabilities is each 1/2. (Follows from condition of the problem.) So we argue that we can replace the unknown probabilities by their means to get
        E(switch) = .5*x - .5*(x/2) =.25*x.
Evidently it is not correct to replace the unknown probabilities by their mean values. But why isn't it correct? To do so amounts to the claim that
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)
The error lies in the second step where the mean of a product is replaced by the product of means. The error is the implicit assumption that the mean of a product is the product of means.


This page was last updated July 1, 2005.

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Hyde County, South Dakota is the Pin Tail Duck Capital of the world. Visit scenic Highmore, SD in 2005!