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2019/01/08 8:55:19
First edition 2014/10/11

Incantations used by equivalent-expectationian on the two envelopes problem

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

Composition of this page was changed on March 21, 2015, and February 26, 2017.

Variety of equivalent-expectationian on the two envelopes problem

Some people think the expectation of the amounts of money in the two envelopes must be same, even one of these envelopes has been opened. 
I call them the "equivalent-expectationian."
They spell various incantations to prove their hypothesis.

There is a variety of equivalent-expectationian.

Incantations spelled by ExpectedAmountEqualsChosenAmountian

Mathematical incantation spelled by ExpectedAmountEqualsChosenAmountian

Let x be the amount of money in the opened envelope.
The possible pairs of amounts are (x/2, x) and (x, 2x).
If we restrict the pairs to these two pairs from beginning,   then the prior expectation of the amount of the opened envelope is
p(3x/4) + (1-p)(3x/2).
If that expectation is equal to x, then
p = (2/3).
If that expectation is equal to x, then the expectation of the amount in the closed envelope is
E = p(x/2) + (1-p)2x = x.
Therefore, the two envelopes are equivalent !
This incantation has some mathematical flavor, so I think this has some dubiousness.

Naive incantation spelled by ExpectedAmountEqualsChosenAmountian

Let x be the amount in the opened envelope, and let Y be the random variable of the amount in the another envelope.
The expectation of the amount in the another envelope must be the same as the amount in the opened envelope.
Therefore Provability(Y= x/2) = 2/3, and Provability(Y=2x) = 1/3.
This incantation is very naive, but it contains the statement of why they think so. I think this incantation has no dubiousness.

Equivalence of these incantations

If the pairs of amounts are restricted to only two pairs (x/2, x) and (x, 2x) from beginning of the game, and the amount of the opened envelope is x, following propositions are mathematically equivalent. From this, we can deduce the following.

If one of the incantations mentioned above is an absolute incantation, then the rest also are absolute incantations.

Among these incantations, the naive incantation is most pure, because it contains no trick.

Incantations spelled by OpeningAnEnvelopeGivesNoInformationian

This paragraph was added on February 1, 2015. And its title was changed on March 21, 2015.
The title was re-changed on March 29, 2018.


Some people spell an incantation to apply the principle of "no information no change" to expectation.
Incantation 1
Before opening, it has already been known that the other envelope contains twice or half as much money as your envelope. 
So opening an envelope does not give new information about whether you should trade or not.

Incantation 2
Opening an envelope cannot change the amounts in the envelopes,
So it should not matter whether you keep or trade envelopes.
In a famous book, I found an incantation which is similar to incantation 1.
I found incantation 2 in a research paper of psychology.

Addition
A calculation which is not an incantation but may be a magic

This section was revised on February 26, 2017 with new title.

A calculation which appeared in English language Wikipedia

Recently, following expectation formula was added to the revision of 12:20, 28 May 2014 of the article "Two envelopes problem" in English language Wikipedia.
The amount in the opened envelope is 100.
If the pair of amounts is 50 and 100, the gain of trading is minus 50, and the average of the two amounts is 75.  · · · (1)
If the pair of amounts is 100 and 200, the gain of trading is plus 100, and the average of the two amounts is 150. · · · (2)
The expected return from trading is
E = (1/2)(100/150) + (1/2)(-50/75) = 0. · · · (3)
Therefore the two envelopes are equivalent !
The subject of this expectation formula
According to the paper which presented the above formula, the subject of this expectation is called "Success Factor", and as far as I understood it, the success factor is the ratio of the following two values.
  (1) The difference (positive or negative) of the amount which will be gotten after trading and the amount which has been gotten before trading
  (2) The mean value of the amount which will be gotten after trading and the amount which has been gotten before trading

When I express this in a mathematical form.
Let x be the chosen amount of money.
Let r be the ratio of the greater amount of money to the lesser amount of money in a pair of amount of money.
Then the success factor F is as follows.

For the case that the chosen amount is the lesser
 F = (rx - x)/((rx + x)/2) = 2(r - 1)/(r + 1). – – – constant value
For the case that the chosen amount is the greater
 F = (x/r - x)/((x + x/r)/2) = - 2(r - 1)/(r + 1). – – – constant value

If r = 2 then each of them is 2/3 or -2/3.

The case of the English language Wikipedia article "Two envelopes problem" at the revision of 12:20, 28 May 2014 is as follows. (← Revised on April 12, 2018.)
If the lesser has been chosen,  F = (200 - 100)/((200 + 100)/2) = 100/150 = 2/3.
If the greater has been chosen,   F = (50 - 100)/((100 + 50)/2) = -50/75 = -2/3

Indeed the above equation means nothing but the following theorem.
(Added on February 2017)
Let x, y denote two numbers.
Then if the ratio of x and y is constant, the ratio of (x - y) and (x + y) is constant.

This expectation formula is incomprehensible, because of the following reasons. In fact, is it a magic ?
(This paragraph was added on October 18, 2014, and was revised on February 28, 2016, September 4, 2016, and September 10, 2016)

I have notice the following tricks. These tricks make an illusion that the calculation formula "(1/2)(100/150) + (1/2)(-50/75)" is an expectation.

Similar calculation

This paragraph was added on March 19, 2015.

There was similar calculation formula in an article which was posted to a web page in 2009.
I will summarize it as follows.
You choose one envelope and find $20.
There are two situations { $10, $20 } and { $20 , $40 }.
Let Y be the lesser amount in each pair of amounts.
If you have 2Y, then you will lose Y after switching.
If you have Y, then you will earn more Y after switching.
You will either gain or lose Y.
Expected gain is (1/2)(-Y) + (1/2)(Y) = 0.
It is essentially the same as the above calculation formula except the following differences.

An article that presented exactory the same opinion 10 years ago

This paragraph was added on March 29, 2018. The title was revised on April 12, 2018.

An article represented exactly the same opinion in 2004.
I will summarize that opinion as follows.
Suppose that the amount of money in the chosen envelope is 10,000.
Then the expected gain by exchange is (1/2)(-2,500/7,500) + (1/2)(5,000/15,000) = 0.
And the expected loss by no-exchange is (1/2)(2,500/7,500) + (1/2)(-5,000/15,000) = 0.
From this, I realized that the people who advocate such an opinion are not as rare as I thought.


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