# User:Gerhardvalentin/Two envelopes problem

## 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.

As this statement does not clearly distinguish between two quite different conceivable scenarios (of symmetry or asymmetry), it is possible to give arguments that show that in the asymmetric scenario it will be to your advantage to swap envelopes by showing that your expected return on swapping exceeds the sum in your envelope.  Not distinguishing these two quite different scenarios 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 from one author to the next. Currently, at least a couple of new papers are published every year. Also, from time to time new variants of the problem have been 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, meaning that one envelope contains some 2 units say ${\displaystyle \bullet \bullet }$, whereas the other envelope contains 1 such unit only, say ${\displaystyle \bullet }$. 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.

Now the two conceivable but quite different scenarios ("exclusive either/or"  versus "as well as") are:

I.) The switching argument in symmetry ("exclusive either/or-scenario"):

In absence of any "known predetermined amount A" and of any "known dependent amount B" – taking into account the given "either/or-scenario" of "indistinguishable envelopes" – you are supposed to reason as follows:

1. I denote by A the amount in my selected envelope. A could either be ${\displaystyle \bullet }$ or ${\displaystyle \bullet \bullet }$ likewise.
2. The probability that A is the smaller amount of ${\displaystyle \bullet }$ is 1/2, and that it is the larger amount of ${\displaystyle \bullet \bullet }$ is also 1/2.
3. The other envelope may contain either 2A in case that A=${\displaystyle \bullet }$ or A/2 in case that A=${\displaystyle \bullet \bullet }$.
Meaning if A should be ${\displaystyle \bullet }$, then B would be ${\displaystyle \bullet \bullet }$. And if A should be ${\displaystyle \bullet \bullet }$, then B would be ${\displaystyle \bullet }$.
4. If A is the smaller amount of ${\displaystyle \bullet }$, then the other envelope contains 2A (${\displaystyle \bullet \bullet }$).
5. If A is the larger amount of ${\displaystyle \bullet \bullet }$, then the other envelope contains A/2 (${\displaystyle \bullet }$).
6. Thus the other envelope contains 2A (=${\displaystyle \bullet \bullet }$) "IF" A=${\displaystyle \bullet }$, or A/2 (=${\displaystyle \bullet }$) "IF" A=${\displaystyle \bullet \bullet }$, both with probability 1/2 each.
(Please note that this "IF", in the symmetric scenario here, is an "essential constitutive pre-condition" and not just only a given logical consequence. So very limiting that "with probability 1/2 each" to being valid
not for "every A" or "whatever A is". But note also: This "IF" never is any "constitutive pre-condition" in the asymmetric scenario below, where it is just only a given logical consequence, but never a precondition.)
7. So the expected value of the money in each one of the two envelopes is (${\displaystyle \bullet \bullet }$)/2 + (${\displaystyle \bullet }$)/2 = (${\displaystyle \bullet \bullet \bullet }$)/2 or (3 units)/2 or (6 units)/4 in each one of the two envelopes.

So in this symmetric scenario of "indistinguishable envelopes", all you can say is that there obviously can be no argument whatsoever to swap envelopes.

II.) The switching argument in asymmetry   (predetermined A and dependent B,  where  B = A/2  "as well as"  B = 2A,  both with equal probability):

If (although completely inconsistent with the explicitly required premise of "indistinguishable envelopes"), it is known that after A already had been determined, the dependent amount B has been fixed afterwards to be either 2A or A/2 with equal probability ("as well as").  And only in this really asymmetric scenario  one can "suppose to know" which one of the envelopes contains some predetermined A and reason as follows:

1. I denote by A the (known to be predetermined) 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 the other envelope contains 2A.
5. If A is the larger amount the other envelope contains A/2.
6. Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2.
(Please note that no "IF", as in the symmetric scenario, is any "essential constitutive pre-condition" here in the asymmetic scenario. Here it is just only a given logical
consequence, but never limiting that "with probability 1/2 each" to being valid only "IF A is", but being valid for EACH and EVERY "A", whatever "A" is.)
7. So the dependent expected value of the money in the other envelope is

${\displaystyle {1 \over 2}(2predeterminedA)+{1 \over 2}\left({predeterminedA \over 2}\right)={5 \over 4}predeterminedA}$

8. This is greater than predetermined A, so I gain on average by swapping from predetermined A to dependent B.
9. But after the switch, swapping back from dependent B to predetermined A will give the original predetermined A again, i.e.   ${\displaystyle {4 \over 5}dependentB}$.

In this evidently asymmetric scenario of "predetermined A" and of "dependent B" it is wise to switch from A to B, but never to swap back from dependent B to predetermined A.

This argument of the asymmetric scenario (with a predetermined A and some dependent B) cannot be applied to the symmetric "either/or-scenario.
Ignoring this obvious fact, any preposterous attempt to though applying this incorrectly to the quite different variant "I.)" also, leads to the absurd conclusion that this understandably "works" for amounts of zero or infinite only.

### The puzzle

The puzzle is to find the flaw of not distinguishing those two quite different scenarios. It is incorrect not to distinguish those two quite different scenarios and to apply in step 7 the arguments of the asymmetric scenario to the quite different symmetric scenario, also.

In the asymmetric scenario it is known that the dependent amount forever must on average be larger than the determining amount.

But the really striking fact is that in both scenarios, whether a special predetermined amount did generate/evoke a dependent amount to be twice or half, and you know (as in the asymmetric scenario) which envelope contains the predetermined amount and which one contains the (on average larger) dependent amount, or this is unknown (as in the symmetric scenario), so for example by just placing two indistinguishable envelopes uniformly at random, then it is indeed given that in both scenarios, whatever is in the chosen envelope, the other looks to be equally likely to be half or twice. But in both scenarios of course half only if the first chosen envelope contains the larger amount, and twice only if the first chosen envelope contains the smaller amount. All of that is valid and true in both scenarios.

The crucial difference of those two scenarios is that this point ("IF") is valid and true obviously as an essential constitutive pre-condition in the symmetric scenario only, but,
although being "true as a logical consequence" in the asymmetric scenario also, not being there any essential constitutive pre-condition, but just only a given logical consequence.

That makes the difference between the symmetric and the asymmetric scenario: looking to be equally likely and being equally likely makes a great difference.

### The puzzle can also be solved under the microscope

Again, you are given two indistinguishable envelopes, each of which contains a positive sum of money. One envelope contains not twice as much as the other, but the twentyfold amount of the other, meaning that one envelope contains some 20 units say ${\displaystyle \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet }$, whereas the other envelope contains 1 such unit only, say ${\displaystyle \bullet }$. 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 other envelope contains 20A (=${\displaystyle \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet }$) "IF" A=${\displaystyle \bullet }$, or A/20 (=${\displaystyle \bullet }$) "IF" A=${\displaystyle \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet }$, both with probability 1/2 each.
(Please note that this "IF", in the symmetric scenario here, is an "essential constitutive pre-condition" and not just only a given logical consequence. So very limiting that "with probability 1/2 each" to being valid
not for "every A" or "whatever A is". But note also: This "IF" never is any "constitutive pre-condition" in the asymmetric scenario below, where it is just only a given logical consequence, but never a precondition.)

In symmetry (indistinguishable envelopes), the expected value of the money in each one of the two envelopes is

(${\displaystyle \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet }$)/2 + (${\displaystyle \bullet }$)/2 = (${\displaystyle \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet }$)/2 or (21 units)/2 or (42 units)/4 in each one of the two envelopes.

So in this symmetric scenario of "indistinguishable envelopes", all you can say is that there obviously can be no argument whatsoever to swap envelopes.

But in the asymmetric scenario with a predetermined amount A and some dependent amount B, the dependent expected value of the money in the other envelope is

${\displaystyle {1 \over 2}(20predeterminedA)+{1 \over 2}\left({predeterminedA \over 20}\right)={40,1 \over 4}predeterminedA}$, so slightly more than ${\displaystyle 10A}$

This is greater than predetermined A, so I gain on average by swapping from predetermined A to dependent B.
But after the switch, swapping back from dependent B to predetermined A will give the original predetermined A again, i.e. ${\displaystyle {4 \over 40,1}dependentB,}$
so slightly less than ${\displaystyle {1 \over 10}dependentB.}$.

In this evidently asymmetric scenario of "predetermined A" and of "dependent B" it is wise to switch from A to B, but never to swap back from dependent B to predetermined A.

Under this microscope you clearly see that in both scenarios, whatever is in the chosen envelope, the other looks to be equally likely to be 1/20 or the twenty-fold. But in both scenarios of course 1/20 only if the first chosen envelope contains the larger amount, and the twenty-fold only if the first chosen envelope contains the smaller amount. All of that is valid and true in both scenarios.

The crucial difference of those two scenarios is that this point ("IF") is valid and true obviously as an essential constitutive pre-condition in the symmetric scenario only, but,
although being "true as a logical consequence" in the asymmetric scenario also, not being there any essential constitutive pre-condition, but just only a given logical consequence.

That makes the difference between the symmetric and the asymmetric scenario: looking to be equally likely and being equally likely makes a great difference.