Revers probleming of the two envelopes paradox

Caution
I who am a Japanese wrote this page in English, but I am not so good at English.

I tried to create the most suitable problems for each solutions of the two envelope paradox. Variaty of the problems may reveal the non-credibility of each solutions.

The mathemaically standard solution and the most suitable problem

Solution

The paradoxical argument is as follows.

Let x be an amount of money in the chosen envelope.
Then the expected amount of money in the other envelope is as follows.
(1/2)(x/2) + (1/2)2x > x.
So, the other envelope is always more favorable.

But correct calculation of probability bring correct argument like below.

Let X and Y be random variables of the amounts of money in the chosen envelope and the other envelope respectivery.
Let x be the amount of money in the chosen envelope.
And Let p = P(X is lesser | X=x).
Then E(Y|X=x) = p(2x) + (1-p)(x/2) = (unknown rate r)x.
So there is no wonder even if the opposite envelope is more favorable for a value of the amount of money in the chosen envelope.
And if the opposite envelope is favorable for a value of the amount of money in the chosen envelope, the opposite envelope must be unfavorable for some value of the amount of money in the chosen envelope.
The equivalence of the two envelopes is surely kept.

Most suitable problem

There are two envelopes.
An unexpected amount of money is placed in one envelope, and twice that amount is placed in another envelope.
You chose one envelope at random.
Let x be the amount of money in your envelope.
Then the amount of money in the other envelope is equaly likely x/2 or 2x.
So the expected amount of money in the other envelope is (1/2)(x/2) + (1/2)(2x) = (5/4)x > x.
Therefore these envelopes are not equivalent.
<<Various description of the existence of a paradox follows.>>

The not three amounts theory and the most suitable problem

Solution

The paradoxical argument is as follows.

Let x be an amount of money in the chosen envelope.
Then E(Y|X=x)=(1/2)(x/2)+(1/2)2x > x.
So, the other envelope is always more favorable.

But the pairs of amount of money (x, x/2) and (x, 2x) belongs different games.
So, the correct argument shall be like below.

Let X, Y be random variables of amounts of money in the chosen envelope and the other envelope respectivly.
Let a be the lesser amount of money of one of pair of amounts of money.
Then E(Y|Min(X,Y)=a)=(1/2)a+(1/2)2a=(3/2)a.
And E(X|Min(X,Y)=a)=(1/2)a+(1/2)2a=(3/2)a.
So, the two envelopes are equally favorable.

Most suitable problem

There are two envelopes.
An unexpected amount of money is placed in the first envelope.
Depending on the result of hidden flipping of coin, twice the original amount of money or half the original amount of money is placed in the second envelope.
You chose one of these envelopes at random.
Let x be the amount of money in your envelope.
Then the amount of money in the other envelope is equaly likely x/2 or 2x.
So the expected amount of money in the other envelope is (1/2)(x/2) + (1/2)(2x) = (5/4)x > x.
Therefore these envelopes are not equivalent.
<<Various description of the existence of a paradox follows.>>

The inconsistent variable theory and the most suitable problem

Solution

The paradoxical argument is as follows.

Let x be an amount of money in the chosen envelope.
Then the expected amount of money in the other envelope is as follows.
(1/2)(x/2) + (1/2)2x > x.
So, the other envelope is always more favorable.

But the variable symbol x in the first term it denotes amount of the greater money, and it in the second term denotes amount of the lesser money.
So, the correct argument shall be like below.

Let A be amount of the lesser money in the two envelopes.
Then the expected amount of money in the other envelope is as follows.
(1/2)A + (1/2)2A = (3/2)A.
And the expected amount of money in the other envelope is same.
And the two envelopes are equaly favorable.

Most suitable problem

(Revised on July 21, 2017.)

There are two envelopes.
An amount of money, say $20 is placed in one envelope, and twice that amount, say $40 is placed in another envelope.
You chose one envelope at random.
Let x denote the amount of money in your envelope.
If your envelope contains the lesser amount the amount of money in the other envelope is 2x.
If your envelope contains the greater amount the amount of money in the other envelope is x/2.
Because your envelope contains equally likely the lesser amount or the greater amount, so the amount of money in the other envelope is equally likely x/2 or 2x.
So the expected amount of money in the other envelope is (1/2)(x/2) + (1/2)(2x) = (5/4)x > x.
Therefore these envelopes are not equivalent.
<<Various description of the existence of a paradox follows.>>

The theory of the mean values of the lesser amount of money and the greater amount of money. And the most suitable problem

Solution

The paradoxical argument is as follows.

Let x be an amount of money in the chosen envelope.
Then the expected amount of money in the other envelope is as follows.
(1/2)(x/2) + (1/2)2x > x.
So, the other envelope is always more favorable.

But the author of this argument forgot they were thinking of expected values.
And they also forgot that these expected values were under two different conditions.
So, the correct expectation formula shall be like below.

Let X and Y are random variables of the amounts of money in the chosen envelope and the other envelope respectivery.
E(Y)=(1/2)2E(X|X<Y) + (1/2)(1/2)E(X|X>Y).

The equivalence of the two envelopes can be verified like below.

Most suitable problem

There are two envelopes.
One has twice as much money as the other.
You chose one envelope at random.
You can change choice before opening the chosen envelope.
You think as follows.
If I change my choice then with a probability 1/2 I may lose a half of the amount of my money.
If I change my choice then with a probability 1/2 I may obtain the same amount as the amount of money which I have.
Therefore the potential gain is greater than zero and I should change my choice.

Conclusion

Among the above solutions, only the mathematically standard solution solves all of the above problems.
This suggests that the other solutions rather than the mathematically standard solution are dubious.

Return to the list of my pages written in English about the two envelopes problem