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The two envelopes problem
Posted: 08 December 2012 03:33 AM   [ Ignore ]   [ # 1771 ]
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kkwan - 08 December 2012 12:58 AM

You have not considered the following problem wrt the TEP.

The problem is, you don’t know whether either (X, 2X) or (X, 1/2X) are actually in the two envelopes.

Therefore I derived that under both descriptions switching does not help.

kkwan - 08 December 2012 12:58 AM

You cannot assume you are in situation 1 or 2 because you don’t know.

Agreed?

I agree that I don’t know. But it does not matter, as for any two amounts, my possible gain and possible loss are the same when I switch. And you agreed with that: (X - Y) = -1 * (Y - X), for every X and Y.

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Posted: 08 December 2012 09:32 AM   [ Ignore ]   [ # 1772 ]
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GdB - 08 December 2012 03:33 AM

Therefore I derived that under both descriptions switching does not help.

By doing that you have assumed that you know one of the situations is actual which is not justified as you don’t know which situation is actual. Thus, the two situations must be considered together and not separately as you don’t know which situation is actual.

I agree that I don’t know. But it does not matter, as for any two amounts, my possible gain and possible loss are the same when I switch. And you agreed with that: (X - Y) = -1 * (Y - X), for every X and Y.

That is true if and only if you have definite information that one of the situations is actual which you don’t have, in the context of the TEP.

In the context of the TEP, there are two and only two distinct possible situations (X, 2X) or (X, 1/2X) for any finite value of X but you don’t know which situation is actual. Thus, you must consider ALL the possible situations together and not separately.

With the two possible situations considered together, the potential gain/loss is not equal (with A as the denoted amount in the selected envelope):

1. If A is the smaller amount, then 2A is in the other envelope. On switching, gain is A

2. If A is the larger amount, then 1/2A is in the other envelope. On switching, loss is 1/2A.

For any finite value of A, the potential gain, A is twice the potential loss, 1/2A.

Another approach to the TEP

In the TEP, that one envelope contains twice the amount in the other (which incidentally, is actually equivalent to double or half), can be interpreted as:

1. Self-referential wrt in any one specific distinct situation. So, with (A, 2A), 2A is double of A and A is half of 2A whereas with (A, 1/2A), A is double of 1/2A and 1/2A is half of A. This is useful in deducing that either (A, 2A) or (A, 1/2A) for any finite value of A, satisfy the TEP, but it has it’s limitations.

However, with two envelopes in play, double or half on switching should now be interpreted as:

2.  2A or 1/2A with A as the common starting reference amount.

2 is not self-referential as A is only the common starting reference amount for comparing gain/loss between two and only two distinct possible situations (A, 2A) and (A, 1/2A) for any finite value of A.

In comparing gain/loss wrt the above two distinct possible situations it is necessary to have a common starting reference amount as in 2 to be coherent whereas 1, being self-referential, is inapplicable because it has no conceptual room for comparing two distinct situations.

[ Edited: 08 December 2012 09:37 AM by kkwan ]
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Posted: 08 December 2012 09:59 AM   [ Ignore ]   [ # 1773 ]
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kkwan - 08 December 2012 09:32 AM

By doing that you have assumed that you know one of the situations is actual which is not justified as you don’t know which situation is actual. Thus, the two situations must be considered together and not separately as you don’t know which situation is actual.

The italic sentence is nonsense. Yes, I know that one of the situations is actual, which means I do not know which one. You are completely lost, kkwan. In the context of TEP it just means you picked the biggest amount first, or the smallest: that are the two situations.

kkwan - 08 December 2012 09:32 AM

That is true if and only if you have definite information that one of the situations is actual which you don’t have, in the context of the TEP.

Yes, one knows it is one of the situations: one has the biggest or the smallest amount. There is no other situation.

kkwan - 08 December 2012 09:32 AM

In the context of the TEP, there are two and only two distinct possible situations (X, 2X) or (X, 1/2X) for any finite value of X but you don’t know which situation is actual. Thus, you must consider ALL the possible situations together and not separately.

In the italic sentence you are falling back on $Schrödinger money again. X is two amounts at the same time, so is the amount in the other envelope. So X (or A or whatever) may refer to the envelope of your choice, but not to a well defined amount. So it is not a constant, and you cannot use it in a derivation as you do. So the resto of your posting is just invalid.

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Posted: 08 December 2012 10:49 AM   [ Ignore ]   [ # 1774 ]
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GdB - 08 December 2012 09:59 AM

The italic sentence is nonsense. Yes, I know that one of the situations is actual, which means I do not know which one. You are completely lost, kkwan. In the context of TEP it just means you picked the biggest amount first, or the smallest: that are the two situations.

You don’t know which situation is actual unless you opened both envelopes which you cannot in the context of the TEP. Neither can you specifically say that the selected envelope contains the smaller or the larger amount.

Yes, one knows it is one of the situations: one has the biggest or the smallest amount. There is no other situation.

Not necessarily so.

In the italic sentence you are falling back on $Schrödinger money again. X is two amounts at the same time, so is the amount in the other envelope. So X (or A or whatever) may refer to the envelope of your choice, but not to a well defined amount. So it is not a constant, and you cannot use it in a derivation as you do. So the resto of your posting is just invalid.

Not so.

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Posted: 08 December 2012 11:30 AM   [ Ignore ]   [ # 1775 ]
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kkwan - 08 December 2012 10:49 AM

You don’t know which situation is actual unless you opened both envelopes which you cannot in the context of the TEP. Neither can you specifically say that the selected envelope contains the smaller or the larger amount.

Right. You do not know. But you also don’t have to know. In both situations, you do not gain on average by switching.

kkwan - 08 December 2012 10:49 AM

Not necessarily so.

Not so.

Seems you run out of arguments.

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Posted: 09 December 2012 01:30 AM   [ Ignore ]   [ # 1776 ]
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Kkwan,

kkwan - 08 December 2012 01:30 AM
StephenLawrence - 02 December 2012 07:20 AM

Dunno quite what you are doing above but it’s wrong.

Simpler to write it out like this:

1. $20 is selected gain on switching is $20

2. $20 is selected loss on switching is $10

3. $10 is selected gain on switching is $10

4. $40 is selected loss on switching is $20

This is equivalent to what I wrote.

OK, so we can write it like this:

1. Gain $20 with 1/4 probability

2. Lose $10 with 1/4 probability

3. Gain $10 with 1/4 probability

4. Lose $20 with 1/4 probability

We can all see that these cancel out and so switching makes no difference.

If you don’t get this bit right we can’t move on to the next stage of the solution.

Stephen

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Posted: 09 December 2012 03:36 AM   [ Ignore ]   [ # 1777 ]
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GdB - 08 December 2012 11:30 AM

Right. You do not know. But you also don’t have to know. In both situations, you do not gain on average by switching.

Without the knowledge of what are the actual amounts in the two envelopes (when there are two possible situations which are not equivalent), you cannot correctly determine gain/loss by switching (with each situation considered separately) because it is begging the question, which is fallacious reasoning.

You can only do that correctly (without begging the question) by considering both situations jointly.

So, your approach to the TEP is fundamentally flawed, incomplete and misleading.

Hence, any “resolution” you proposed is not the resolution to the TEP per se.

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Posted: 09 December 2012 05:38 AM   [ Ignore ]   [ # 1778 ]
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StephenLawrence - 09 December 2012 01:30 AM

OK, so we can write it like this:

1. Gain $20 with 1/4 probability

2. Lose $10 with 1/4 probability

3. Gain $10 with 1/4 probability

4. Lose $20 with 1/4 probability

We can all see that these cancel out and so switching makes no difference.

If you don’t get this bit right we can’t move on to the next stage of the solution.

Not quite so, as by doing that, it is equivalent to having four envelopes in two groups instead of two envelopes.

Now, with group 1 ($20, $40) and group 2 ($20, $10):

Selecting $20 from group 1 and switching is your 1.

Selecting $20 from group 2 and switching is your 2.

Selecting $10 from group 2 and switching is your 3

Selecting $40 from group 1 and switching is your 4.

But, that is not the TEP per se.

In the TEP, either ($20, $40) or ($20, $!0) are in the two envelopes:

1. If $20 is selected, it could be from either of the two possible situations and the probabilities are 1/2 and 1/2.

2. If $10 is selected, it must be from ($20, $10) as then ($20, $40) does not exist.

3. If $40 is selected, it must be from ($20, $40) as then ($20, $10) does not exist.

However, (with total ignorance in the context of the TEP) all we can deduce is 1. If we can deduce either 2 or 3, then we are begging the question and which also contravenes the TEP because either 2 or 3 refer to one and only one situation.

Furthermore, if we are considering two either/or situations, we cannot also consider any one of the two situations separately as well because it is a contradiction and incoherent.

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Posted: 09 December 2012 06:55 AM   [ Ignore ]   [ # 1779 ]
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kkwan - 09 December 2012 05:38 AM

But, that is not the TEP per se.

I said that in the first place.

The point is to gain agreement on my example and then for me to show it’s relevance to the TEP in stages.

So can we gain agreement on my example before we move on?

Stephen

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Posted: 09 December 2012 07:57 AM   [ Ignore ]   [ # 1780 ]
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kkwan - 09 December 2012 03:36 AM

Without the knowledge of what are the actual amounts in the two envelopes (when there are two possible situations which are not equivalent), you cannot correctly determine gain/loss by switching (with each situation considered separately) because it is begging the question, which is fallacious reasoning.

Right. I cannot determine the gain or loss. But I can derive that at least they are equal: X - Y = -1 * (Y - X) for every two amounts. I have no idea how much X - Y is. But I know that its absolute value is the same as the absolute value of Y - X.

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Posted: 09 December 2012 04:09 PM   [ Ignore ]   [ # 1781 ]
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StephenLawrence - 09 December 2012 06:55 AM

I said that in the first place.

The point is to gain agreement on my example and then for me to show it’s relevance to the TEP in stages.

So can we gain agreement on my example before we move on?

We can, with the caveat and proviso that as it is non-TEP, you will have to explicitly show and explain why your non-TEP example is relevant to the TEP at all without either begging the question, misinterpreting or any other contradictions wrt the TEP, otherwise it will be irrelevant, unacceptable, unconvincing, incoherent and ultimately, fatal.

We should be mindful of the pitfalls expressed in the above paragraph before moving on..

[ Edited: 09 December 2012 05:41 PM by kkwan ]
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Posted: 09 December 2012 05:27 PM   [ Ignore ]   [ # 1782 ]
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GdB - 09 December 2012 07:57 AM

Right. I cannot determine the gain or loss. But I can derive that at least they are equal: X - Y = -1 * (Y - X) for every two amounts. I have no idea how much X - Y is. But I know that its absolute value is the same as the absolute value of Y - X.

Notwithstanding that, but the tautology is inapplicable to the TEP as it is because fundamentally, the TEP is not about one possible situation but two possible situations whereby you don’t know which situation is actual.

If and only if you know what is actually in the two envelopes which implies there is one and only one situation but which begs the question and contravenes the basic nature of the TEP, then and only then, will the above tautology be applicable.

So, we are trapped in the paradoxical symmetry inherent in the TEP whereby the one and only one way forward without breaking the symmetry, begging the question and contravening the basic nature of the TEP, is to consider both situations jointly.

So, gain/loss by switching can only be determined in the context of the two possible situations but not in the context of any one situation.

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Posted: 09 December 2012 11:08 PM   [ Ignore ]   [ # 1783 ]
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Let’s get it straight:

I give you two envelopes, with two different amounts. I do not give you any more information. You may pick one envelope, but before you open it, I say to you that you can switch. Now you know that for every two amounts X -Y = -1 * (Y - X). So you know switching does not give you advantage. Agree?

Now I say to you that the difference between the amounts equals the smaller of the two amounts.

At that moment you immediately realise you should switch, isn’t it? Explain, kkwan, why, even if X -Y = -1 * (Y - X) is true for all X and Y, now that Y - X = X it is not true anymore…

[ Edited: 09 December 2012 11:54 PM by GdB ]
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Posted: 09 December 2012 11:59 PM   [ Ignore ]   [ # 1784 ]
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kkwan - 09 December 2012 04:09 PM

We can, with the caveat and proviso that as it is non-TEP, you will have to explicitly show and explain why your non-TEP example is relevant to the TEP at all without either begging the question, misinterpreting or any other contradictions wrt the TEP, otherwise it will be irrelevant, unacceptable, unconvincing, incoherent and ultimately, fatal.

We should be mindful of the pitfalls expressed in the above paragraph before moving on..

grin  OK.

So 2. The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2. is not true in my example, agreed?

Stephen

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Posted: 10 December 2012 12:29 PM   [ Ignore ]   [ # 1785 ]
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GdB - 09 December 2012 11:08 PM

Let’s get it straight:
I give you two envelopes, with two different amounts. I do not give you any more information. You may pick one envelope, but before you open it, I say to you that you can switch. Now you know that for every two amounts X -Y = -1 * (Y - X). So you know switching does not give you advantage. Agree?

Now I say to you that the difference between the amounts equals the smaller of the two amounts.

At that moment you immediately realise you should switch, isn’t it? Explain, kkwan, why, even if X -Y = -1 * (Y - X) is true for all X and Y, now that Y - X = X it is not true anymore…

If there is one and only one possibility wrt what are the amounts in the two envelopes, then there is no advantage in switching as the potential gain is equal to the potential loss (in amount) irrespective of the difference of the amounts in the two envelopes.

However, in the TEP, there are two and only two possible situations wrt what amounts are in the envelopes, but we don’t know which situation is actual. As such, we cannot apply the above as we must consider both situations jointly.

So, by denoting the amount in the selected envelope as A:

1. If A is the smaller amount, then 2A is in the other envelope. On switching, gain is A.

2. If A is the larger amount, then 1/2A is in the other envelope. On switching, loss is 1/2A.

Now, the potential gain, A is twice the potential loss, 1/2A (in amount).

So, it is beneficial to switch.

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