Two envelopes problem

Quote: DorothyGale

A man is presented with two envelopes full of money. One of the envelopes
contains twice the amount as the other envelope. One the man has chosen
his envelope, opened and counted it, he is given the option of changing
it for the other envelope. The question is, is there any gain to the man
in changing the envelope?

It would appear that by switching the man would have a 50% chance of doubling
his money should the initial envelope be the lesser amount and a 50% chance
of halving it if the initial envelope is the higher amount. Thus, let x be
the amount contained in the initial envelope and y be the value of changing it:

y = 0.5*(x/2) + 0.5*(2x)

Let’s say that the initial envelope contained $100 (so that x = $100). There should
be a 50% chance that the other envelope contains either 2 * $100 = $200 (2x)
or a 50% chance that the other envelope contains (1/2) * $100 = $50 (x/2).
In such a case, the value of the envelope is:

$125 = 0.5*($100/2) + 0.5*(2*$100)

This inequality can be shown by simplifying this equation:

y = (0.5)*(x/2) + (0.5)*(2x) = (5/4)x

This implies that the man would, on average, increase his wealth by 25% simply
by switching envelopes! How can this be?

It is not 50/50
If there is a limited amount in the envelop, there are chances only pair of (0.5X, X) exists, and pair of (X, 2X) never exist.

with limited amount(L) in the envelop,
no matter where the X lays between 1 and L, 0.5X is always be there....100% chance of having (0.5X, X)
if X is greater than 0.5L, there is no 2X...50% chance of having (X,2X)
so for an unknown X, EV of switching = 100%/150%*-0.5X + 50%/150%*+X = 0

without limited amount(L) in the envelop,
for any X(X still less than half infinity anyway), EV of switching = 50%*-0.5X + 50%*+X = +0.25X

X=$100,
for a limit of $200 or higher, the pair of ($100, $200) is possible, switching EV=+$25
with limited amount less than $200, the pair of ($100, $200) is impossible, switching EV=-$50

Quote: Wizard

We already have 19 pages about this problem.

Quote: 7up

Quote: Wizard

We already have 19 pages about this problem.

Yes, but most are base on 50/50. Not the same as mine.

Quote: MathExtremist

No, most are based on the fact that once you open up the envelope, the chances of the other amount being higher or lower are *not* 50/50 because they can't be. That's the answer to the paradox. See this and this.

Quote: MathExtremist

Here's looking at it in reverse. We know (intuitively and logically) that the EV of switching or keeping has to be the same. That means if I pick an envelope with value V, I can keep it and have V or switch and have EV(switch) = V = A*2V + B*V/2, where A+B = 1. That leads to A = 1/3, B = 2/3. In short, the chances of switching to a higher envelope is only 1/3. The question, dear reader, is why?