## Tuesday, April 10, 2012

### Two envelopes problem

The two envelopes problem, also known as the exchange paradox, is a brain teaser, puzzle or paradox in logic, philosophy, probability and recreational mathematics, of special interest in decision theory and for the Bayesian interpretation of probability theory. Historically, it arose as a variant of the necktie paradox.
A statement of the problem starts with:
Let us say 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 offered the possibility to take the other envelope instead.
It is possible to give arguments that show that it will be to your advantage to swap envelopes by showing that your expected return on swapping exceeds the sum in your envelope. This leads to the logical absurdity that it is beneficial to continue to swap envelopes indefinitely.
A large number of different solutions have been proposed. The usual scenario is that one writer proposes a solution that solves the problem as stated, but then some other writer discovers that by altering the problem a little the paradox is brought back to life again. In this way a family of closely related formulations of the problem is created which are then discussed in the literature.
It is quite common for authors to claim that the solution to the problem is easy, even elementary. However, when investigating these elementary solutions they are not always the same conclusions from one author to the next. Recently, several new papers are published every year regarding the problem and introducing new conclusions. Occasionally, new variants of the problem are introduced.

The Problem

The 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 offered the possibility to take the other envelope instead.
The switching argument: Now suppose you reason as follows:
1. I denote by A the amount in my selected envelope.
2. The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2.
3. The other envelope may contain either 2A or A/2.
4. If A is the smaller amount, then the other envelope contains 2A.
5. If A is the larger amount, then the other envelope contains A/2.
6. Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2.
7. So the expected value of the money in the other envelope is ${1 \over 2} (2A) + {1 \over 2} \left({A \over 2}\right) = {5 \over 4}A$
8. This is greater than A, so I gain on average by swapping.
9. After the switch, I can denote that content by B and reason in exactly the same manner as above.
10. I will conclude that the most rational thing to do is to swap back again.
11. To be rational, I will thus end up swapping envelopes indefinitely.
12. As it seems more rational to open just any envelope than to swap indefinitely, we have a contradiction.
The puzzle: The puzzle is to find the flaw in the very compelling line of reasoning above.

1. i am not sure but the equation should be 1/4(2A)+1/2(A/2)=3A/4<A

1. Each envelope shares the same probability of 1/2..But there is indeed something going wrong in step 7..In the first term A is the smaller amount while in the second term A is the larger amount. To mix different instances of a variable in the same formula like this is said to be illegitimate, so step 7 is incorrect, and this is the cause of the paradox. A correct alternative argument would have run on the following lines. Assume that there are only two possible sums that might be in the envelope. Denoting the lower of the two amounts by X, we can rewrite the expected value calculation as (1/2)X + (1/2)2X = (3/2)X

Here X stands for the same thing in every term of the equation. We learn that 1.5X is the average expected value in either of the envelopes. According to the correct calculation there is no need to swap envelopes, and certainly no need to swap indefinitely.

2. The flaw can easily be detected: The wording of steps 1 to 6 is ambiguous and offers two quite different ways to look at and to observe the situation. So you have to distinguish those two quite different and mutually exclusive scenarios: the "either/or-scenario" versus the "as-well-as-scenario". They both are addressed likewise by steps 1 to 6, so both are applicable, but both are mutually exclusive. The flaw lies in trying to apply the conclusion of the specific "as-well-as-scenario" of step 7 to the quite different "either/or-scenario" also. See http://en.wikipedia.org/wiki/User:Gerhardvalentin/TEP ;
Quite easy to understand and clear to see for anyone, but laborious and difficult to solve for mathematicians, if not even "unsolvable". Rather interesting, indeed :)

Regards, gev

3. Nice comment!!