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The two envelopes problem
 Posted: 23 January 2012 09:14 AM [ Ignore ]   [ # 181 ]
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macgyver - 23 January 2012 06:18 AM

I haven’t read all 12 pages of this post so excuse me if someone has already said this but the problem with the explanation is that when you describe 2A and A/2 you are really describing two different A’s. A does not =A in this equation.  They are not the same. If one envelope has \$1 and the other has \$2 then the only way you could be trading for the A/2 envelope is if A=\$2, and the only way you could be trading for the 2A envelope is if A=\$1.

Am I missing something here. Doesn’t that pretty much clear up the paradox?

The amount in the selected envelope is denoted by A.

1 and 2 are two possible situations.

1. If A is the smaller amount, the other envelope has 2A
2. If A is the larger amount, the other envelope has A/2

A in 1 and 2 refer to the same amount, but in the context of two possible situations of which only one of the situations can happen.

Hence, there is no inconsistency of two different A’s.

[ Edited: 23 January 2012 09:20 AM by kkwan ]
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 Posted: 23 January 2012 09:22 AM [ Ignore ]   [ # 182 ]
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Axegrrl - 21 January 2012 12:50 AM
kkwan - 20 January 2012 09:07 PM

Hence, we are in a quandary. This is intriguing because…...

`
Yet again, you lose me at the ‘intrigue’ part

What’s intriguing about recognizing that we have know way of knowing what the ‘expected value’ is given the amount of (or lack of) information provided?

What’s intriguing about recognizing that 50/50 chances means that’s there’s no way of discerning whether or not we should switch?

`

It at least looks like there is a 50/50 chance of getting double or half of what is in the envelope you have.

On that information it is right to switch.

Stephen

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 Posted: 23 January 2012 10:09 AM [ Ignore ]   [ # 183 ]
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macgyver - 23 January 2012 06:18 AM

If one envelope has \$1 and the other has \$2 then the only way you could be trading for the A/2 envelope is if A=\$2, and the only way you could be trading for the 2A envelope is if A=\$1.

But if your envelope has \$1 in it, the other envelope might have 50 cents in it (as it happens it doesn’t) or it might have \$2 dollars in it ( as it happens it does).

The question is what is wrong with looking at it like that, apart from it gets us the wrong answer?

Stephen

[ Edited: 23 January 2012 10:27 AM by StephenLawrence ]
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 Posted: 23 January 2012 10:40 AM [ Ignore ]   [ # 184 ]
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StephenLawrence - 23 January 2012 10:09 AM

The question is what is wrong with looking at it like that, apart from it gets us the wrong answer?

Ask your self if by picking one envelope the total amount in the 2 envelopes together changes.

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.

In 4. the total amount is 3A, in 5. it is 1.5A. Isn’t that magic?

Defined as:

4a. If A is the smaller amount the other envelope contains 2A.
5b. If 2A is the larger amount the other envelope contains A.

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 Posted: 23 January 2012 11:12 AM [ Ignore ]   [ # 185 ]
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GdB - 23 January 2012 10:40 AM
StephenLawrence - 23 January 2012 10:09 AM

The question is what is wrong with looking at it like that, apart from it gets us the wrong answer?

Ask your self if by picking one envelope the total amount in the 2 envelopes together changes.

No.

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.
In 4. the total amount is 3A, in 5. it is 1.5A. Isn’t that magic?

No. It’s a question of comparing two possible scenarios, one which is the case and one which could be but isn’t.

Why should the totals match?

Defined as:

4a. If A is the smaller amount the other envelope contains 2A.
5b. If 2A is the larger amount the other envelope contains A.

What you are doing here is saying we should only look at the difference between the two envelopes and the fact that there is a 50/50 chance of it being gain or a loss.

I agree, but why we should is the question.

Stephen

[ Edited: 23 January 2012 11:20 AM by StephenLawrence ]
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 Posted: 23 January 2012 01:02 PM [ Ignore ]   [ # 186 ]
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kkwan - 20 January 2012 08:01 PM
domokato - 20 January 2012 10:51 AM

kkwan…what are you talking about? My point was that step 7 was calculating expected value using impossible values, not that expected value is not a reliable principle in decision theory. And since it is using impossible values, it is an invalid expected value calculation. Why was it posited using impossible values in the first place? Because of steps 4-6, and I’ve already explained what’s wrong with them.

The expected value was not calculated using impossible values. In fact, it was calculated using possible values as:

Thus, step 7 is impeccably valid and your objection is not justified.

Apparently you are having trouble following multiple threads of argument. I will stop responding for now

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 Posted: 23 January 2012 01:23 PM [ Ignore ]   [ # 187 ]
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GdB - 23 January 2012 10:40 AM

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.

In 4. the total amount is 3A, in 5. it is 1.5A. Isn’t that magic?

Take a Different problem.

You have £10 call that A.

You are offered double or half depending on the toss of a coin.

So if you get double, the total of the two amounts, what you have + what you will have is 3A

If you get half the total of the two amounts, what you have and what you will have is 1.5A .

Isn’t this magic?

Clearly not and in this example it is right to switch.

It’s not obvious that your explanation really does any explaining.

Stephen

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 Posted: 23 January 2012 01:24 PM [ Ignore ]   [ # 188 ]
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I’m gonna stuff those envelopes with IOUs, where you owe me a value. You either owe me double or half.  That’ll solve the problem…...

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 Posted: 23 January 2012 01:34 PM [ Ignore ]   [ # 189 ]
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It occurred to me that philosophically, both envelopes must be given equal weight regardless of the amount they contain.  It IS a Schrodinger problem !

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 Posted: 23 January 2012 05:55 PM [ Ignore ]   [ # 190 ]
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StephenLawrence - 23 January 2012 09:22 AM

It at least looks like there is a 50/50 chance of getting double or half of what is in the envelope you have.

On that information it is right to switch.

`
If there’s an equal chance that you’ll lose half of what you have or double it, why is “right” to switch?

What about 50/50 chances compels switching?

`

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 Posted: 23 January 2012 05:58 PM [ Ignore ]   [ # 191 ]
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StephenLawrence - 23 January 2012 01:23 PM

Take a Different problem.

You have £10 call that A.

You are offered double or half depending on the toss of a coin.

So if you get double, the total of the two amounts, what you have + what you will have is 3A

If you get half the total of the two amounts, what you have and what you will have is 1.5A .

Isn’t this magic?

Clearly not and in this example it is right to switch.

It’s not obvious that your explanation really does any explaining.

Stephen

`
Wait a minute, I thought the person in question is swapping the envelope they have (with A amount in it) for another one…..not adding to something they already have.

Isn’t that the situation?

Hell, if it’s guaranteed that you’re going to be ADDING to what you already have by picking either of the two envelopes (instead of swapping what you have and risking losing it), then well yeah, there’s no reasong NOT to switch!

But if that’s the scenario, this thread certainly wouldn’t be this long, would it? ;)

`

[ Edited: 23 January 2012 06:17 PM by Axegrrl ]
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 Posted: 23 January 2012 06:26 PM [ Ignore ]   [ # 192 ]
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kkwan - 21 January 2012 01:23 PM
GdB - 21 January 2012 03:34 AM

Assume one envelope has A dollars in it, the other 2A.

But if A is an unknown quantity, how can you assume one envelope has A dollars, the other 2A?

By doing so, you have considered A as a known quantity.

`
Ok, you’ve got to be kidding with that.

The entire REASON that labels like ‘A’ or ‘x’ or ‘y’ are used to begin with is to represent unknowns!  Otherwise, they’d use a specific number.

And how can we assume that one envelope has A dollars and the other 2A? (in other words, one has twice the amount as the other) because YOU stated that very thing!  GdB even bolded the part where you did.

*mind boggled*

kkwan, you seriously can’t be as thick as the above comment of yours makes you appear…....
`

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 Posted: 23 January 2012 06:43 PM [ Ignore ]   [ # 193 ]
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He is pretty damn thick. And he seems to be willfully so

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 Posted: 24 January 2012 12:29 AM [ Ignore ]   [ # 194 ]
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Axegrrl - 23 January 2012 05:55 PM
StephenLawrence - 23 January 2012 09:22 AM

It at least looks like there is a 50/50 chance of getting double or half of what is in the envelope you have.

On that information it is right to switch.

`
If there’s an equal chance that you’ll lose half of what you have or double it, why is “right” to switch?

Lets say you have £10 and you are offered £1,000 or £5 on the toss of a coin. What should you do?

Same principle applies to double or half, you can gain more than you can lose.

Stephen

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 Posted: 24 January 2012 12:33 AM [ Ignore ]   [ # 195 ]
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Axegrrl - 23 January 2012 05:58 PM
StephenLawrence - 23 January 2012 01:23 PM

Take a Different problem.

You have £10 call that A.

You are offered double or half depending on the toss of a coin.

So if you get double, the total of the two amounts, what you have + what you will have is 3A

If you get half the total of the two amounts, what you have and what you will have is 1.5A .

Isn’t this magic?

Clearly not and in this example it is right to switch.

It’s not obvious that your explanation really does any explaining.

Stephen

`
Wait a minute, I thought the person in question is swapping the envelope they have (with A amount in it) for another one…..not adding to something they already have.

Isn’t that the situation?

Hell, if it’s guaranteed that you’re going to be ADDING to what you already have by picking either of the two envelopes (instead of swapping what you have and risking losing it), then well yeah, there’s no reasong NOT to switch!

The post you are responding to is a criticism of Gdb’s solution. GdB’s idea was to add the first amount to the second amount in the two possible scenario’s being considered and to argue that they should match.

Stephen

[ Edited: 24 January 2012 01:06 AM by StephenLawrence ]
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