Two envelopes problem

The two envelopes problem, also known as the exchange paradox, is a brain teaser, puzzle, or paradox in logic, probability, and recreational mathematics. It is of special interest in decision theory, and for the Bayesian interpretation of probability theory. Historically, it arose as a variant of the necktie paradox. The problem typically is introduced by formulating a hypothetical challenge of the following type:

It seems obvious that there is no point in switching envelopes as the situation is symmetric. However, because you stand to gain twice as much money if you switch while risking only a loss of half of what you currently have, it is possible to argue that it is more beneficial to switch. The problem is to show what is wrong with this argument.

Basic setup: You are given two indistinguishable envelopes, each of which contains a positive sum of money. One envelope contains twice as much as the other. You may pick one envelope and keep whatever amount it contains. You pick one envelope at random but before you open it you are given the chance to take the other envelope instead.

The switching argument: Now suppose you reason as follows:

What has gone wrong?

Many solutions have been proposed. Commonly one writer proposes a solution to the problem as stated, after which another writer shows that altering the problem slightly revives the paradox. Such sequences of discussions have produced a family of closely related formulations of the problem, resulting in a voluminous literature on the subject.

No proposed solution is widely accepted as definitive. Despite this it is common for authors to claim that the solution to the problem is easy, even elementary. However, when investigating these elementary solutions they often differ from one author to the next. Since 1987 new papers have been published every year.

The total amount in both envelopes is a constant ${\displaystyle c=3x}$, with ${\displaystyle x}$ in one envelope and ${\displaystyle 2x}$ in the other.
If you select the envelope with ${\displaystyle x}$ first you gain the amount ${\displaystyle x}$ by swapping. If you select the envelope with ${\displaystyle 2x}$ first you lose the amount ${\displaystyle x}$ by swapping. So you gain on average ${\displaystyle G={1 \over 2}(x)+{1 \over 2}(-x)={1 \over 2}(x-x)=0}$ by swapping.
Swapping is not better than keeping. The expected value ${\displaystyle \operatorname {E} ={\frac {1}{2}}2x+{\frac {1}{2}}x={\frac {3}{2}}x}$ is the same for both the envelopes. Thus there is no contradiction any more.

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