Suppose there are two boxes with money in them. The amount of money is a positive real number of dollars...

>And yet E(x) = 1.25*E(y).

No, it's 1.25*y, not 1.25 times expectation of y.
And E(y) = 1.25*x as well, so the choice is trivial.
This also proves that expectation fails with this premise.

Well, exactly. The expectation only works if y ISN'T variable. So it doesn't work.

>the strange man was using a gamma distribution to determine the amount of money
>it wasn't impossible
Guess you just missed out on $10000000 user.

For any possible value of y, you can easily compute that E(x) = 5*y/4. So why does it matter that you don't know exactly what y is?

>For any possible value of y, you can easily compute that E(x) = 5*y/4. So why does it matter that you don't know exactly what y is?
Because if you don't know what y is, the problem becomes symmetric and E(x) is not 5*y/4. It's (x+y)/2. Of course knowing what one of the envelopes contains changes the dependency of E(x), since you break symmetry.

depends on money.

imagine instead of 2 the koefficient is 1000
you open box and there is $10, it is obvious that you should check another box for $10000 and if other box is $0.01 - no big deal

but if you open box and there is $1000000000, I would keep it despite chance if 1000 billion, because $1 billion is enough for me

Logic vs math

Pic related