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The two envelopes problem
 Posted: 15 November 2012 03:28 AM [ Ignore ]   [ # 1576 ]
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GdB - 12 November 2012 12:09 AM

Yes, I am totally ignorant about the exact amounts. But I am not about the following:

1. The two envelopes are closed, so the amounts cannot change.
2. Corollary 1: the total amount does not change.
3. Corollary 2: the amounts do not change dependent on which envelope I take first. So also the total amount will stay the same.
4. The amount of one envelope is twice the amount of the other.
5. Corollary: the amount of one envelope is half of the other.

Which of my presuppositions is wrong?

You switch in your argument between two possibilities: (X,2X) and (X,X/2) for the same X are different amounts. So you are contravening rule 1.

All your presuppositions are not wrong. However, both (X, 2X) or (X, 1/2X) fit your presuppositions and as such we cannot assume (X, 2X) and discard/ignore (X, 1/2X) as we do not know with certainty what are actually in the two envelopes.

Thus, if (X, 2X) are in the two envelopes, the total amount is 3X or if (X, 1/2X) are in the two envelopes, the total amount is 3/2X.

This is consistent with denoting the amount in the selected envelope as A and as A could be the smaller or the larger amount, either (A, 2A) or (A, 1/2A) are in the two envelopes. The total amounts are either 3A or 3/A.

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 Posted: 15 November 2012 04:52 AM [ Ignore ]   [ # 1577 ]
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kkwan - 15 November 2012 03:28 AM
GdB - 12 November 2012 12:09 AM

Yes, I am totally ignorant about the exact amounts. But I am not about the following:

1. The two envelopes are closed, so the amounts cannot change.
2. Corollary 1: the total amount does not change.
3. Corollary 2: the amounts do not change dependent on which envelope I take first. So also the total amount will stay the same.
4. The amount of one envelope is twice the amount of the other.
5. Corollary: the amount of one envelope is half of the other.

Which of my presuppositions is wrong?

You switch in your argument between two possibilities: (X,2X) and (X,X/2) for the same X are different amounts. So you are contravening rule 1.

All your presuppositions are not wrong. However, both (X, 2X) or (X, 1/2X) fit your presuppositions and as such we cannot assume (X, 2X) and discard/ignore (X, 1/2X) as we do not know with certainty what are actually in the two envelopes.

Thus, if (X, 2X) are in the two envelopes, the total amount is 3X or if (X, 1/2X) are in the two envelopes, the total amount is 3/2X.

This is consistent with denoting the amount in the selected envelope as A and as A could be the smaller or the larger amount, either (A, 2A) or (A, 1/2A) are in the two envelopes. The total amounts are either 3A or 3/A.

If my presuppositions are correct, then it follows that your derivation is not correct: dependent on the amount in the first envelope you look at different amounts.That contradicts presupposition 1.
Imagine I put \$10 and \$20 in the envelopes. Then if you pick the \$10 first you suppose there are (\$5,\$10) or (\$10,\$20) in the envelopes; if you pick \$20 first then you suppose there are (\$10,\$20) or (\$20,\$40) in the envelopes. Your (total) amount(s) are not fixed. In reality they are.

Let’s do it another way: the total amount in the two envelopes is an unknown constant. ‘Constant’ means: whatever envelope you choose, this makes no difference for the total amount. Call this amount T. Then the two amounts are (T/3,2T/3). If you pick T/3 first, and you switch you gain T/3. If you pick 2T/3 first and you switch you loose T/3. Loss and gain are the same, so there is no advantage in switching. This is keeping the total amount the same. You do not keep it the same. You are muddling up ‘being unknown’ and ‘not being constant’.

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 Posted: 15 November 2012 05:37 AM [ Ignore ]   [ # 1578 ]
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StephenLawrence - 12 November 2012 01:09 AM

The assumption in question is rock solid Kkwan. It’s simply that there are two amounts in the envelopes one double the other.

From that assumption we can make this more concrete to think about by imagining any two amounts we like.

We can do that because it makes no diference, the result is the same which ever two amounts we think about.

You are not facing up to this. Changing your mind is a good thing! (sometimes)

That one envelope contains twice the amount in the other is given in the TEP. This requirement is satisfied by two distinct possibilities: either (X, 2X) or (X, 1/2X), not one, for any finite value of X.

However, we cannot assume it is only just any (X, 2X) as (X, 1/2X) is equally possible because we don’t know what are actually in the two envelopes. Without that knowledge, we cannot make any assumption at all wrt what amounts are in the two envelopes.

Hence, analyzing the TEP by assuming any (X, 2X) as equivalent to any (X, 1/2X), thus we need only consider any (X, 2X), is incomplete, misleading and begs the question.

Thus, to be complete wrt the TEP, we do not assume anything to not mislead and not beg the question, we must consider ALL the possible situations for any finite value of X, i.e. it is either (X, 2X) or (X, 1/2X).

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 Posted: 15 November 2012 06:09 AM [ Ignore ]   [ # 1579 ]
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GdB - 15 November 2012 04:52 AM

If my presuppositions are correct, then it follows that your derivation is not correct: dependent on the amount in the first envelope you look at different amounts.That contradicts presupposition 1.
Imagine I put \$10 and \$20 in the envelopes. Then if you pick the \$10 first you suppose there are (\$5,\$10) or (\$10,\$20) in the envelopes; if you pick \$20 first then you suppose there are (\$10,\$20) or (\$20,\$40) in the envelopes. Your (total) amount(s) are not fixed. In reality they are.

Wrt your propositions, if the denoted amount A in the selected envelope is:

1. the smaller amount, the amount in the other envelope is 2A

2. the larger amount, the amount in the other envelope is 1/2A.

1 and 2 refer to different mutually exclusive situations and in the context of each situation does not contradict any of your presuppositions.

If you did put (\$10, \$20) into the two envelopes, obviously there cannot be \$5 or \$40 in the two envelopes.

However, in the TEP, you did not do that and neither do you know what are actually in the two envelopes. As such, if you only imagine (\$10, \$20) are in the envelopes, it is possible that (\$10, \$5) are in the envelopes.

Let’s do it another way: the total amount in the two envelopes is an unknown constant. ‘Constant’ means: whatever envelope you choose, this makes no difference for the total amount. Call this amount T. Then the two amounts are (T/3,2T/3). If you pick T/3 first, and you switch you gain T/3. If you pick 2T/3 first and you switch you loose T/3. Loss and gain are the same, so there is no advantage in switching. This is keeping the total amount the same. You do not keep it the same. You are muddling up ‘being unknown’ and ‘not being constant’.

The total amount T is fixed, but it could be either 3X or 3/2X.

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 Posted: 15 November 2012 06:12 AM [ Ignore ]   [ # 1580 ]
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kkwan - 15 November 2012 05:37 AM
StephenLawrence - 12 November 2012 01:09 AM

The assumption in question is rock solid Kkwan. It’s simply that there are two amounts in the envelopes one double the other.

From that assumption we can make this more concrete to think about by imagining any two amounts we like.

We can do that because it makes no diference, the result is the same which ever two amounts we think about.

You are not facing up to this. Changing your mind is a good thing! (sometimes)

That one envelope contains twice the amount in the other is given in the TEP. This requirement is satisfied by two distinct possibilities: either (X, 2X) or (X, 1/2X), not one, for any finite value of X.

However, we cannot assume it is only just any (X, 2X) as (X, 1/2X) is equally possible because we don’t know what are actually in the two envelopes. Without that knowledge, we cannot make any assumption at all wrt what amounts are in the two envelopes.

Hence, analyzing the TEP by assuming any (X, 2X) as equivalent to any (X, 1/2X), thus we need only consider any (X, 2X), is incomplete, misleading and begs the question.

That’s wrong.  X is not a variable, 2X is not a variable. (X, 2X) is not a variable, IOW,  (X, 2X) = (X, 1/2X) and that doesn’t work.

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 Posted: 15 November 2012 07:30 AM [ Ignore ]   [ # 1581 ]
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kkwan - 15 November 2012 06:09 AM

1 and 2 refer to different mutually exclusive situations and in the context of each situation does not contradict any of your presuppositions.

Then you must treat them as mutually exclusive, but you don’t. See here where I do that. The result is that switching does not help.

kkwan - 15 November 2012 06:09 AM

If you did put (\$10, \$20) into the two envelopes, obviously there cannot be \$5 or \$40 in the two envelopes.

However, in the TEP, you did not do that and neither do you know what are actually in the two envelopes. As such, if you only imagine (\$10, \$20) are in the envelopes, it is possible that (\$10, \$5) are in the envelopes.

This is too stupid, kkwan. Let’s face it, you do not understand mathematics, how it must be applied. Either you suppose there are only two fixed amounts, or you calculate with two possibilities, but then you must do this consistently as in the above referenced posting.

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 Posted: 15 November 2012 09:34 PM [ Ignore ]   [ # 1582 ]
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Write4U - 15 November 2012 06:12 AM

That’s wrong.  X is not a variable, 2X is not a variable. (X, 2X) is not a variable, IOW,  (X, 2X) = (X, 1/2X) and that doesn’t work.

X is a variable, a symbol that represents a quantity in a mathematical expression.

If we assign a quantity of 10 for X,  then (X, 2X) is (10, 20) and (X, 1/2X) is (10, 5).

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 Posted: 15 November 2012 09:51 PM [ Ignore ]   [ # 1583 ]
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GdB - 15 November 2012 07:30 AM

Then you must treat them as mutually exclusive, but you don’t. See here where I do that. The result is that switching does not help.

You should treat both mutually exclusive events together and not separately as both events are equally possible because you don’t know what are actually in the two envelopes.

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 Posted: 15 November 2012 11:37 PM [ Ignore ]   [ # 1584 ]
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kkwan - 15 November 2012 09:51 PM

You should treat both mutually exclusive events together and not separately as both events are equally possible because you don’t know what are actually in the two envelopes.

Taking together means the total is 3X/2 and 3X at the same time. What a bogus, kkwan.

You do as if the second envelopes magically can change its contents: in reality its contents is determined by your choice of envelope: if you have two envelopes, one with \$X, the other with \$Y, then if you choose the one with \$X the other one has \$Y. If you choose the one with \$Y, then the other one contains \$X. That is the fact you wipe under the carpet. For any pair of amounts (must not even be TEP), possible loss or gain are the same: \$X - \$Y. Whatever other condition I apply, like TEP where \$Y = 2 x \$X, this fact is not changed in any way.

So tell me plainly: do you agree that for any two amounts (X,Y), the possible gain or loss by switching them is the same, namely X - Y?

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 Posted: 16 November 2012 12:04 AM [ Ignore ]   [ # 1585 ]
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GdB - 15 November 2012 07:30 AM

Then you must treat them as mutually exclusive, but you don’t. See here where I do that. The result is that switching does not help.

But this ceases to work if you open your envelope. And you say opening your envelope makes no difference.

So you aren’t making any sense.

What is needed to stay coherent and solve the puzzle is:

A) A (sensible) equation that shows the expected gain is equal to the expected loss with your envelope open.

Or

B) An explanation for why there isn’t one.

Or

C) An explanation for why in fact opening the envelope does make a difference.

Or

D) A change of mind over the position we are in when both envelopes are closed.

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 Posted: 16 November 2012 01:47 AM [ Ignore ]   [ # 1586 ]
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StephenLawrence - 16 November 2012 12:04 AM
GdB - 15 November 2012 07:30 AM

Then you must treat them as mutually exclusive, but you don’t. See here where I do that. The result is that switching does not help.

But this ceases to work if you open your envelope. And you say opening your envelope makes no difference.

Why? I start with “you open the envelope and you see the amount \$X”.

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 Posted: 17 November 2012 11:28 PM [ Ignore ]   [ # 1587 ]
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GdB - 15 November 2012 11:37 PM

Taking together means the total is 3X/2 and 3X at the same time. What a bogus, kkwan.

Not so. It means that the total amount is either 3X or 3/2X, not 3X and 3/2X. This is because you don’t know what are actually the amounts in the two envelopes.

You do as if the second envelopes magically can change its contents: in reality its contents is determined by your choice of envelope: if you have two envelopes, one with \$X, the other with \$Y, then if you choose the one with \$X the other one has \$Y. If you choose the one with \$Y, then the other one contains \$X. That is the fact you wipe under the carpet. For any pair of amounts (must not even be TEP), possible loss or gain are the same: \$X - \$Y. Whatever other condition I apply, like TEP where \$Y = 2 x \$X, this fact is not changed in any way.

So tell me plainly: do you agree that for any two amounts (X,Y), the possible gain or loss by switching them is the same, namely X - Y?

In the TEP, what is X and Y is ambiguous as we have two possible situations:

1. Y is 2X, hence (X, 2X) are in the two envelopes

2. Y is 1/2X, hence (X, 1/2X) are in the two envelopes.

In reality, as you don’t know which situation is actual, so you must consider both possible situations: that either (X, 2X) or (X, 1/2X) are in the two envelopes.

There are no facts to consider wrt the amounts in the two envelopes because you don’t know what are actually in the two envelopes. It is that simple.

Thus, if you select an envelope, all you can say is to denote the amount in it as A and if A is either the smaller or the larger amount, then either (A, 2A) or (A, 1/2A) are in the two envelopes.

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 Posted: 18 November 2012 12:56 AM [ Ignore ]   [ # 1588 ]
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kkwan - 17 November 2012 11:28 PM

Not so. It means that the total amount is either 3X or 3/2X, not 3X and 3/2X. This is because you don’t know what are actually the amounts in the two envelopes.

BS. If you treat them separately, then do so:
Either your potential gain/loss is X when you describe the total amount as 3X, or it is X/2 when you describe the total amount as 3X/2. You do not treat the two descriptions separately, but as if they are one.

kkwan - 17 November 2012 11:28 PM

In the TEP, what is X and Y is ambiguous as we have two possible situations:

1. Y is 2X, hence (X, 2X) are in the two envelopes

2. Y is 1/2X, hence (X, 1/2X) are in the two envelopes.

In reality, as you don’t know which situation is actual, so you must consider both possible situations: that either (X, 2X) or (X, 1/2X) are in the two envelopes.

There are no facts to consider wrt the amounts in the two envelopes because you don’t know what are actually in the two envelopes. It is that simple.

Thus, if you select an envelope, all you can say is to denote the amount in it as A and if A is either the smaller or the larger amount, then either (A, 2A) or (A, 1/2A) are in the two envelopes.

As I said: you do not use the knowledge you have, namely that the amounts are fixed. Show me where you use this information in your calculation.

And you did not answer my question: do you agree that for any two amounts (X,Y), the possible gain or loss by switching them is the same, namely X - Y?

[ Edited: 18 November 2012 10:40 AM by GdB ]
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 Posted: 18 November 2012 05:44 PM [ Ignore ]   [ # 1589 ]
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Kkwan has literally been repeating the same “your envelope is X and the other is either 2X or 1/2X with 50-50 probability” nonsense for almost a year now, despite having been shown time and time again that mathematics doesn’t work that way when dealing with unknown probability distributions.  Isn’t it time for everyone to move on?

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 Posted: 18 November 2012 09:37 PM [ Ignore ]   [ # 1590 ]
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No! Mingy.  The thread must go on.  If Kkwan were not caught up in this nightmarish perpetuity, he might be using his superior intellect to take over the world.  That must not happen.

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