[scotty79]

Ad.1 If in the reasoning A is a fixed number then you can't assume that you have 1/2 probability that the other envelope contains more money. Probability depends on distribution and if you don't know the distribution it cannot be calculated. Fact that it cannot be calculated does not entitle you to assuming its 1/2.

If you have problem with that imagine stack of cards. You and your opponent pick one card. Whoever picks higher card wins. Until you see your card you have 1/2 probability of winning. But after you see you just picked 3 you see your probability of winning sharply changes (depending on what cards are left in the stack).

Ad.2 When you don't open your envelope you must treat amount in it (A) as random variable. Then all I wrote in post above applies.

It doesn't matter that you did not looked into an envelope. Because whether you got higher or lower amount is significant and you don't know which occurred you must split your reasoning and consider both cases. But upon splitting you must remember that condition you assume for each given branch of your reasoning must be taken into account. You can do this by pruning your random variable A with the appropriate condition and that influences its expected value. So by all means you can sum up expected values but not before appropriately adjusting them with the conditions you assumed for branches of your reasoning.

Consider solving equation x^2 + yx +1 = 0 While solving your may split you reasoning to three cases depending on whether y^2-4 is larger, smaller or equal to zero (even though you don't know what is the value of y). But when summing up solution you must remember that you assumed something about y in your branches so for the x you've found y can no longer be any real number.

[miloshh]

Ad 1 - Not sure what you're trying to say, all I said is that you don't know the distribution of A - if you knew it, you could make an informed decision on whether to switch.

Ad 2 - If you carry out the analysis you're trying to teach me, you'll find that the expected relative gain from switching is 1.25. Now explain that. (You'll also find that the expected absolute gain is 0 - that's not a paradox.)

[scotty79]

Let's denote expected value of money in envelope I picked with A.

If I want to switch I have to consider two cases. Let's label them e1 and e2. e1 means I picked envelope with smaller amount of money. e2 means I picked envelope with larger amount of money. Probability of each case is 1/2.

What is the expected value of money I'll have after switch?

Let's consider first case:

S(e1) = 2 * A(e1)

which mens two times the amount of money I had PROVIDED THAT I PICKED ENVELOPE WITH LESS MONEY.

A(e1) is not equal A If I denote exact amount of dollars in envelope with less money by X then A = 1/2 * X + 1/2 * 2 * X = 3/2 * X but A(e1) = X

Similarly considering case e2 I get expected value of money after switch equal

S(e2) = 1/2 * A(e2)

which means half of the amount of money I had PROVIDED THAT I PICKED ENVELOPE WITH MORE MONEY.

Again A(e2) is not equal A but 2 * X, if you prefer not to use X then A(e1) = 2/3 * A and A(e2) = 4/3 * A

To sum up probability of each case is 1/2 so S that I use to donate expected value of money I could have after switch is equal to:

S = 1/2 * 2 * X + 1/2 * 1/2 * 2 * X = 3/2 * X = A

or if you want without using X

S = 1/2 * 2 * 2/3 * A + 1/2 * 1/2 * 4/3 * A = A

So there is no gain or loss from switching. You can switch zero or more times without changing expected value of money you will get after opening your envelope.

Similar reasoning can be conducted for any amount of envelopes and any probability distribution and I guess it will to same conclusion. I'm not sure if observable symmetry of situation can be taken as proof of this conjecture but I sincerely hope so.

Paradox comes from using same symbol A to denote three different expected values of random variables which I denote above as A, A(e1) and A(e2)

If you have further doubts about above "relative" reasoning I'll gladly try to clear them up.

[miloshh]

You just painstakingly carried out the absolute analysis, which I know works.

Let X be the value in the envelope you have, and Y in the other one (X and Y are both random variables with well-known distributions). Then E[Y/X] = 1.25. That's what I wanted you to explain. You just keep saying that E[X] = E[Y], which I know.

Note that this paradox would not arise if X and Y were independent, since then E[Y/X] = E[Y] / E[X] = 1.

[scotty79]

What reasoning leads to claim that E[Y/X] = 1.25 ?

It can't be the reasoning from wikipedia about two envelopes problem because it has error in steps 4 and 5 so results in step 7 and further are nonsensical.

I painstakingly shown how this reasoning should look like to be right and that in fact E[Y/X] = E[Y] / E[X] = 1