### 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?