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The two envelopes problem
 Posted: 15 July 2012 08:29 AM [ Ignore ]   [ # 1396 ]
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GdB - 10 July 2012 10:41 PM

You must consider one TEP pair, and not just a ‘denotation of the amount you pick first’. By taking only one amount fixed, you consider two TEP pairs, but there is only one, the one in front of you. By considering only one amount you deprive yourself of the knowledge that, “of one TEP pair, if you pick the smallest amount, the other one is the biggest; if you pick the biggest one first, then the other one is the smallest”. This is what you deny. That if you pick one envelope there is only one other one, that is there lying in front of you. This is the knowledge you have, but you do not use in your ‘derivation’ of TEP.

Not quite so simple.

In the TEP, it is equally possible that either (X, 2X) or (X, 1/2X) are in the two envelopes but you don’t know which of the two equally possible events is actually in the envelopes. Also, if you pick one of the two indistinguishable envelopes, you don’t know the amount in it and neither do you know whether it contains the smaller or the larger amount.

As such, it is rational to start by denoting A as the amount in the selected envelope and proceed from steps 2 to 7.

I have two envelopes, one with a green piece of paper, the other with a red one.

1. If I have the envelope with the green one the other contains the red one.
2. If I have the envelope with the red one the other contains the green one.

I have two envelopes with different amounts, call them X and Y:

3. If I have the envelope with X, then the other one contains Y.
4. If I have the envelope with Y, then the other one contains X.

I have two envelopes with amounts X and 2X:

5. If I have the envelope with X, then the other one contains 2X.
6. If I have the envelope with 2X, then the other one contains X.

I have two envelopes according to TEP:
7. If I have the envelope with the smallest amount, then the other one contains the biggest.
8. If I have the envelope with the biggest amount, then the other one contains the smallest.

All your examples are equivalent to fixing the amounts as (X, 2X) which is not complete as per the TEP whereby either (X, 2X) or (X, 1/2X) are in the two envelopes.

As such, you proceed as explained above.

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 Posted: 15 July 2012 09:44 AM [ Ignore ]   [ # 1397 ]
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StephenLawrence - 10 July 2012 11:50 PM

It isn’t true that you must consider both these possibilities.

You know you have a pair, whatever that is, and you can just consider what would be the case if you had the larger or the smaller amount of a pair.

There is nothing wrong with doing it that way. The puzzle is why doesn’t your way work as well.

Both possibilities (X, 2X) and (X, 1/2X) must be considered as they both satisfy the requirement that one envelope contains twice the amount of the other as per the TEP.

By considering only one possible pair (X, 2X) or (X, 1/2X), it is tantamount to simplifying the TEP by omission.

The problem is, you don’t know which pair is actually in the two envelopes as both pairs are possible.

They are not equivalent as X is the smaller amount in (X, 2X) whereas X is the larger amount in (X, 1/2X) and the total amounts are 3X and 3/2X respectively.

Thus, if you select an envelope, you cannot proceed as though the amount in it is either the smaller or the larger amount as in one specific pair because there are two possible pairs to consider.

Hence, denoting the amount in the selected envelope as A and proceeding from steps 2 to 7.

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 Posted: 15 July 2012 10:26 AM [ Ignore ]   [ # 1398 ]
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kkwan - 15 July 2012 08:29 AM

In the TEP, it is equally possible that either (X, 2X) or (X, 1/2X) are in the two envelopes but you don’t know which of the two equally possible events is actually in the envelopes. Also, if you pick one of the two indistinguishable envelopes, you don’t know the amount in it and neither do you know whether it contains the smaller or the larger amount.

The are two amounts in the envelopes, not two different sets of variables. And as you write, it is the biggest or the smallest. Two amounts. Two possibilities. Not three: X/2, X and 2X.

As expected, you avoid the questions.

Where am I fixing amounts as (X,2X) when I am talking about green and red pieces of paper??? Only the last questions (7 and 8) concern TEP, so you should be able to answer all questions before. Show us in what way TEP differs from the (sometimes trivial) questions before. If you don’t, I know enough: you are not able to show how TEP differs from the previous cases.

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 Posted: 15 July 2012 02:02 PM [ Ignore ]   [ # 1399 ]
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kkwan - 15 July 2012 09:44 AM

Both possibilities (X, 2X) and (X, 1/2X) must be considered as they both satisfy the requirement that one envelope contains twice the amount of the other as per the TEP.

I’ve said what’s wrong with this. The player has the information that if he knew the two amounts, switching would make no difference whatever the two amounts are.

As in every case switching would make no difference, it follows that the player doesn’t need to know what the two amounts are to know that switching makes no difference.

The problem is, you don’t know which pair is actually in the two envelopes as both pairs are possible.

Millions of pairs are possible, we know we only have one of them, which is all we need to know.

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 Posted: 16 July 2012 05:09 PM [ Ignore ]   [ # 1400 ]
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Holy crapola, 94 pages. Is this a record?

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 Posted: 16 July 2012 10:53 PM [ Ignore ]   [ # 1401 ]
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domokato - 16 July 2012 05:09 PM

Holy crapola, 94 pages. Is this a record?

No. Free will still does better.

I just refuse to give up when I see that people make logical errors, especially when they seem too stubborn to acknowledge that. Call it a personality problem if you wish…

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 Posted: 17 July 2012 04:46 PM [ Ignore ]   [ # 1402 ]
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Nah, I’m right there with you. I just ran out of juice and expertise about a thousand pages ago

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 Posted: 17 July 2012 07:19 PM [ Ignore ]   [ # 1403 ]
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GdB - 15 July 2012 10:26 AM

The are two amounts in the envelopes, not two different sets of variables. And as you write, it is the biggest or the smallest. Two amounts. Two possibilities. Not three: X/2, X and 2X.

Not so. There are two equally possible events i.e. either (X, 2X) or (X, 1/2X) are in the two envelopes. In each mutually exclusive event, there are two amounts.

As expected, you avoid the questions.

Where am I fixing amounts as (X,2X) when I am talking about green and red pieces of paper??? Only the last questions (7 and 8) concern TEP, so you should be able to answer all questions before. Show us in what way TEP differs from the (sometimes trivial) questions before. If you don’t, I know enough: you are not able to show how TEP differs from the previous cases.

All your questions refer to two specific quantities which are assumed to be known, whether it is green/red paper or (X, 2X) but this is not so explicit in the TEP.

In the TEP, you have no definite knowledge wrt the actual amounts in the two envelopes. It could be either (X, 2X) or (X, 1/2X) as both satisfy the requirement that one envelope contains twice as much as the other.

For any finite value of X, they are not equivalent.

By omitting and not considering (X, 1/2X) it is tantamount to fixing (X, 2X) as the amounts in the two envelopes and consequently simplifying the TEP per se.

[ Edited: 17 July 2012 07:24 PM by kkwan ]
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 Posted: 17 July 2012 08:10 PM [ Ignore ]   [ # 1404 ]
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StephenLawrence - 15 July 2012 02:02 PM

I’ve said what’s wrong with this. The player has the information that if he knew the two amounts, switching would make no difference whatever the two amounts are.

As in every case switching would make no difference, it follows that the player doesn’t need to know what the two amounts are to know that switching makes no difference.

The problem is, the player does not have the specific information wrt what are actually the amounts in the two envelopes.

It could be either (X, 2X) or (X, 1/2X).

By considering both possibilities, it is shown in steps 1 to 7 that switching makes a difference over many independent repetitions of the TEP.

Millions of pairs are possible, we know we only have one of them, which is all we need to know.

Generally, it is either (X, 2X) or (X, 1/2X). However, we don’t know with certainty which pair is actually in the two envelopes and hence we cannot say “which is all we need to know” because we don’t know anything specifically at all. All we can say is, it is either this or that.

In other words, we are totally ignorant and as such, we must consider both pairs in the context of the TEP.

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 Posted: 17 July 2012 10:44 PM [ Ignore ]   [ # 1405 ]
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kkwan - 17 July 2012 07:19 PM
GdB - 15 July 2012 10:26 AM

The are two amounts in the envelopes, not two different sets of variables. And as you write, it is the biggest or the smallest. Two amounts. Two possibilities. Not three: X/2, X and 2X.

Not so. There are two equally possible events i.e. either (X, 2X) or (X, 1/2X) are in the two envelopes. In each mutually exclusive event, there are two amounts.

If you have two envelopes with two different amounts in front of you, you know there are only two amounts. If you describe them as (X, 2X) or (X, 1/2X) does not change that fact. ‘X’ is an unknown and could be any amount, so as Stephen already rightfully said, there is no reason to give (X,X/2) a special treatment, because it means nothing. Your juggling with (X, 2X) or (X, 1/2X) does not change the simple fact that there are two fixed amounts in two closed envelopes in front of you, one amount is the smallest and one is the biggest. You are not using that fact, i.e. you do not use all information available.

Compare:
1. I give you an envelope, and I have a second in my hand and say it contains twice or half the amount as in the envelope you have
with
2. I give you two envelopes, say that one contains twice the amount as the other and you may pick one envelope

You say these situations are equivalent: but in 1 the point is missing that you had the choice. And that is the point you are leaving out. If you think you don’t, then show me where 1 and 2 differ, and where you use the fact that you originally had two envelopes in your calculations. (Or will you say again that 1 is not TEP and therefore it does not interest you? If you do that then I know you are not able to show the difference.)

kkwan - 17 July 2012 07:19 PM

All your questions refer to two specific quantities which are assumed to be known, whether it is green/red paper or (X, 2X) but this is not so explicit in the TEP.

BS. X is unknown. And Y too. You are just refusing again, because the answers would show you are wrong. Answer the questions, kkwan, or admit you were wrong.

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 Posted: 18 July 2012 10:38 AM [ Ignore ]   [ # 1406 ]
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So is this the new Free Will thread then?  Will this one also spin endlessly in meaningless circles until it too collapses under its own pointless weight?

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 Posted: 18 July 2012 10:57 AM [ Ignore ]   [ # 1407 ]
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Isn’t it beautiful? Nearly 100 pages! Go now for the 200!

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 Posted: 18 July 2012 12:20 PM [ Ignore ]   [ # 1408 ]
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Dead Monky - 18 July 2012 10:38 AM

So is this the new Free Will thread then?  Will this one also spin endlessly in meaningless circles until it too collapses under its own pointless weight?

Belief in (libertarian) free will is one of the most challenging problems we face today.

The two envelope problem lags behind a little

Stephen

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 Posted: 19 July 2012 01:00 PM [ Ignore ]   [ # 1409 ]
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GdB - 16 July 2012 10:53 PM

No. Free will still does better.

I just refuse to give up when I see that people make logical errors, especially when they seem too stubborn to acknowledge that. Call it a personality problem if you wish…

In the case of free will there are no logical errors.

Assuming determinism you would need a different past in order to behave differently. As the past is out of your control it follows that you would need something out of your control to be different in order to behave differently. It is merely your good or bad luck whether it was or it wasn’t.

This is what is meant by we don’t have libertarian free will.

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 Posted: 19 July 2012 03:54 PM [ Ignore ]   [ # 1410 ]
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kkwan,
There are two equally possible events i.e. either (X, 2X) or (X, 1/2X) are in the two envelopes. In each mutually exclusive event, there are two amounts.

Allow me a little observation. There are always only two amounts. (X, 2X).  No matter which we pick this TEP holds true and when we pick we pick either the smallest or the largets. In order to arrive at a second TEP we must first switch the envelopes in order to create an appearance of (X, 1/2X), by making X into 1/2 X and making 2X into X. But that makes absolutely no difference to the outcome. We either pick the envelope with the largest or the envelope with the smallest amount. By switching we do not realize a potential gain other than a 50/50 chance of losing or winning.  Relative gain cannot enter this equation.. The TEP is and always will be X, 2X.

I mentioned Schrodinger before but that was dismissed as the amounts are not imaginary and cannot be both amounts at the same time.

[ Edited: 19 July 2012 04:18 PM by Write4U ]
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