Hint for Problem 2

If z=1, you know the other envelope contains 3 so you should exchange. For an arbitrary z>1, we may compute the posterior probability, call it p(z), that the other envelope contains the number 3z. This is: p(z) = P(X=x | observe z=3^x) = P(X=x and observe 3^x)/P(observe 3^x) = .5^x(1/2)/(.5^x(1/2)+.5^{x-1}(1/2)) = 1/3, for all z>1.

Your expected return if you exchange is (2/3)(z/3)+(1/3)(3z) = (11/9)z. Since this is greater than z, you should exchange. Moreover, you should be willing to give up a certain percentage of your winnings, say 10%, in order to make this exchange, since your expected return will still be (11/10)z which is greater than z. Thus you should make the exchange no matter what z is observed.

But there is something called the sure-thing principle that says that in this case there is no need to look at z. One can make the exchange anyway. But then you should be willing to exchange twice or three times, paying 10% at each exchange. Now you are really losing money. How do you resolve this problem?