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The two envelopes problem
Posted: 28 December 2012 11:40 PM   [ Ignore ]   [ # 1921 ]
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kkwan - 28 December 2012 07:11 PM

Of course, the entity who selected what amounts to put into the envelopes knows, but it will not inform us to keep us totally ignorant and clueless.

It will inform us because it tells us what would happen if we switched.

The entity could select at random from the two possible pairs, perhaps by tossing an unbiased coin, with a random number generator (depending on whether the number generated is even or odd) or with any other similar method.

OK good. Take the simplest case, selecting from two possible pairs with an unbiased coin.

Now what are the two pairs?

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Posted: 29 December 2012 03:44 AM   [ Ignore ]   [ # 1922 ]
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kkwan - 27 December 2012 11:50 AM

You do the calculation of your ridiculous non-TEP.

I did already, but I’ll copy/paste it for you:

Situation:
You can pick an envelope with an for you unknown amount. After your pick, a second envelope is filled with another amount that is half or twice the amount you have in your envelope, with a chance of 50% for each. You are allowed to switch. Should you?

Solution:
1. Call the amount on the envelope you have A
2. Then the other envelope might contain 2A with a chance of 1/2.
3. Or the other envelope might contain A/2 with a chance of 1/2.
4. Now the expected value expressed in terms the amount you have of switching is:
    c0f75c1a69af64e06f77ce0ec051c958.png
5. This is more than the expected value of you envelope which is of course A.
6. So you should switch.

Here it is correct to use ‘the amount you picked first’ as a constant, because the amount in the second envelope is a function of the amount in your first envelope: it is defined in its terms, as twice or half of that amount.

So your task kkwan, for this situation, to derive the formula that helps to decide if you should switch or not. Criticising the situation, as you did before, or showing the differences with TEP, as you did before, does not make sense. Make you own derivation, and with that show me why mine is not valid. If you cannot then explain how it is possible that, even that this situation differs with TEP, the same derivation holds.

kkwan - 27 December 2012 11:50 AM

(X, 2X) is equivalent to (2X, X) as the two indistinguishable envelopes can be switched around and also we don’t know which is which, which means your a) and b) (for Y = 2X) is trivially tautological. LOL

The LOL is not at its place. Every true sentence of axiomatic mathematics is tautological. ‘Trivial’ is every true proposition that immediately can be seen to follow from the axioms. I showed that only for Y = 2X we confirm to TEP, which says there are only two amounts. For Y = X we have 3 possible amounts, except when X = 0, which is explicitly excluded in TEP. You split up TEP in two inconsistent situations when Y = X. In each description on its own it is clear that switching does not help. Only by joining them, by stating that Y = X, you get at your answer, but it does not conform to TEP which says there are only two amounts.

kkwan - 27 December 2012 11:50 AM

No joining? Sure? So then in (X,2X) I can gain or loose X, and in (X,X/2) I can gain or loose X/2. So in both separate events there is no advantage in switching. So under both descriptions, there is no reason to switch. Agree? In the end, I do not know if I took the smallest or the biggest amount first. Now kkwan, why is my derivation not valid? In your answer you are not allowed to join both situations: they cannot be the case at the same time. I only do not know which event is the case. But I know that in both cases loss and gain exactly cancel.

With A denoted as the amount in the selected envelope, either (A, 2A) or (A, 1/2A) are in the two envelopes. The two events: either (A, 2A) or (A, 1/2A) are mutually exclusive.

1. In (A, 2A) by switching, gain is A

2. In (A, 1/2A) by switching, loss is 1/2A.

Where is the “joining”?

The joining is in the fact that you say that A in (A,2A) and in (A,A/2) are the same amount. Say I give an envelope to you: I filled it it with $5 or $10, depending of the tossing of a fair coin. You denote this amount as A. Now I ask you to calculate, say $50 minus the amount in your closed envelope. Is this one single value? Is it $45 or $40? Now I fill one envelope with $5, and the other with $10. I give you the envelopes, and ask you to calculate the difference between the two. What is the difference? Well, you don’t know: it could be $10 - $5 = $5, or $5 - $10 = -$5. But if I ask you to say what the absolute value is, you know it exactly: it is $5.

Now of course $5 and $10 is a TEP pair, and above is valid for every TEP pair. And don’t come with ‘you did not put the amounts in the envelopes’. For TEP it doesn’t matter how the amounts got into the envelopes. It must just be guaranteed that the description ‘one envelope contains twice the amount of the other’ is true. TEP does not say anyting about how the values got in the envelopes, and it simply does not matter.

kkwan - 27 December 2012 11:50 AM

So two amounts, kkwan: (5,10) or (10,20). If it is (5,10) I possibly loose or gain 5, if it is (10,20) I possibly loose or gain 10. Beware: do not join both situations!

With 10 as the selected amount:

1. If we are (10, 20) by switching, gain is 10

2. If we are in (10, 5) by switching, loss is 5

Just wrong. With (5,10) in the envelopes I will have 5 or 10 in my envelope. With (10,20) I will have 10 or 20 in my envelope. You leave out two possibilities.

kkwan - 27 December 2012 11:50 AM

Considering jointly is not synonymous to “join both” (whatever it means).

What is the difference then? Explain formally and with an example.

[ Edited: 29 December 2012 06:00 AM by GdB ]
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Posted: 29 December 2012 04:03 AM   [ Ignore ]   [ # 1923 ]
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GdB - 29 December 2012 03:44 AM

Here is is correct to use ‘the amount you picked first’ as a constant, because the amount in the second envelope is a function of the amount in your first envelope: it is defined in its terms, as twice or half of that amount.

Yes, I’ve found the puzzle interesting because what one needs to get is even if we look we still can’t use the amount we picked first as a constant in the TEP.

A ‘common sense’ view to some one who doesn’t know a lot about this (like me) is once you know something it’s fixed. And that is why I think the puzzle is difficult in a way the Monty Hall problem isn’t. Of course we don’t look in the TEP but we naturally think about what if we did.

Stephen

[ Edited: 29 December 2012 04:06 AM by StephenLawrence ]
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Posted: 29 December 2012 04:39 AM   [ Ignore ]   [ # 1924 ]
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StephenLawrence - 29 December 2012 04:03 AM

Yes, I’ve found the puzzle interesting because what one needs to get is even if we look we still can’t use the amount we picked first as a constant in the TEP.

Don’t forget that TromboneAndrew already said this in the first reaction on kkwan’s OP… The reason this thread has become so long is that kkwan’s ego is now dependent on it. Believe me, he knows he is wrong, he just cannot afford to admit it. He never gives straight answers on my questions, because he knows the answers would show him wrong.

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Posted: 29 December 2012 05:16 AM   [ Ignore ]   [ # 1925 ]
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GdB - 29 December 2012 04:39 AM
StephenLawrence - 29 December 2012 04:03 AM

Yes, I’ve found the puzzle interesting because what one needs to get is even if we look we still can’t use the amount we picked first as a constant in the TEP.

Don’t forget that TromboneAndrew already said this in the first reaction on kkwan’s OP…

Yes, as I say it’s still difficult to understand, for a number of us, at least, because we open our envelope and there is A. That’s the mind trick.

Stephen

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Posted: 29 December 2012 09:28 PM   [ Ignore ]   [ # 1926 ]
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StephenLawrence - 28 December 2012 11:40 PM

It will inform us because it tells us what would happen if we switched.

Of course, the information of what are the amounts in the two envelopes is crucial, but in the context of the TEP as it is, that information is unavailable.

However, we can deduce either (X, 2X) or (X, 1/2X) are in the two envelopes as we don’t know which pair is actual.

As such, neither can we assume (X, 2X) nor can we assume (X, 1/2X) are in the two envelopes because both assumptions beg the question.

Thus, we can only say either (X, 2X) or (X, 1/2X) are in the two envelopes without begging the question.

This is the approach of the argument to switch whereby with A denoted as the amount in the selected envelope, as A is either the smaller or the larger amount, then either (A, 2A) or (A, 1/2A) are in the two envelopes.

This is consistent with the form of what we deduced: that either (X, 2X) or (X, 1/2X) are in the two envelopes.

OK good. Take the simplest case, selecting from two possible pairs with an unbiased coin.

Now what are the two pairs?

The two pairs are either (X, 2X) or (X, 1/2X) because we don’t know which pair is actual.

With total ignorance, we cannot be more specific without begging the question.

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Posted: 29 December 2012 10:30 PM   [ Ignore ]   [ # 1927 ]
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GdB - 29 December 2012 03:44 AM

I did already, but I’ll copy/paste it for you:

Situation:
You can pick an envelope with an for you unknown amount. After your pick, a second envelope is filled with another amount that is half or twice the amount you have in your envelope, with a chance of 50% for each. You are allowed to switch. Should you?

Solution:
1. Call the amount on the envelope you have A
2. Then the other envelope might contain 2A with a chance of 1/2.
3. Or the other envelope might contain A/2 with a chance of 1/2.
4. Now the expected value expressed in terms the amount you have of switching is:
    c0f75c1a69af64e06f77ce0ec051c958.png
5. This is more than the expected value of you envelope which is of course A.
6. So you should switch.

Here it is correct to use ‘the amount you picked first’ as a constant, because the amount in the second envelope is a function of the amount in your first envelope: it is defined in its terms, as twice or half of that amount.

So your task kkwan, for this situation, to derive the formula that helps to decide if you should switch or not. Criticising the situation, as you did before, or showing the differences with TEP, as you did before, does not make sense. Make you own derivation, and with that show me why mine is not valid. If you cannot then explain how it is possible that, even that this situation differs with TEP, the same derivation holds.

Your non-TEP is not relevant as it is not the TEP per se. Please confine yourself to discussing only the TEP.

In the context of the TEP, A is a constant and the smaller amount in (A, 2A) or the larger amount in (A, 1/2A) without contradiction as we don’t know whether either (A, 2A) or (A, 1/2A) is actual.

The LOL is not at its place. Every true sentence of axiomatic mathematics is tautological. ‘Trivial’ is every true proposition that immediately can be seen to follow from the axioms. I showed that only for Y = 2X we confirm to TEP, which says there are only two amounts. For Y = X we have 3 possible amounts, except when X = 0, which is explicitly excluded in TEP. You split up TEP in two inconsistent situations when Y = X. In each description on its own it is clear that switching does not help. Only by joining them, by stating that Y = X, you get at your answer, but it does not conform to TEP which says there are only two amounts.

Your (X, 2X) is equivalent to (2X, X) which is of the form X = X, thus it is trivially tautological and inane.

The joining is in the fact that you say that A in (A,2A) and in (A,A/2) are the same amount. Say I give an envelope to you: I filled it it with $5 or $10, depending of the tossing of a fair coin. You denote this amount as A. Now I ask you to calculate, say $50 minus the amount in your closed envelope. Is this one single value? Is it $45 or $40? Now I fill one envelope with $5, and the other with $10. I give you the envelopes, and ask you to calculate the difference between the two. What is the difference? Well, you don’t know: it could be $10 - $5 = $5, or $5 - $10 = -$5. But if I ask you to say what the absolute value is, you know it exactly: it is $5.

Now of course $5 and $10 is a TEP pair, and above is valid for every TEP pair. And don’t come with ‘you did not put the amounts in the envelopes’. For TEP it doesn’t matter how the amounts got into the envelopes. It must just be guaranteed that the description ‘one envelope contains twice the amount of the other’ is true. TEP does not say anyting about how the values got in the envelopes, and it simply does not matter.

You have conveniently forgotten that in the context of the TEP, you are totally ignorant of:

1. The amount in the selected envelope

2. The amounts in the two envelopes.

As such, if A is denoted as the amount in the selected envelope:

1. A could be the smaller amount, then (A, 2A) are in the two envelopes.

2. A could be the larger amount, then (A, 1/2A) are in the two envelopes.

There is no “joining” of situations 1 and 2 as they are mutually exclusive.

Just wrong. With (5,10) in the envelopes I will have 5 or 10 in my envelope. With (10,20) I will have 10 or 20 in my envelope. You leave out two possibilities.

That is begging the question because you don’t know whether (10, 20) or (10, 5) are in the two envelopes and thus cannot specifically assume (!0, 20) or (10, 5) are in the two envelopes.

What is the difference then? Explain formally and with an example.

Considering both events jointly means considering both events together as we don’t know which event is actual.

Thus, with A denoted as the amount in the selected envelope, as A could be either the smaller or the larger amount, then either (A, 2A) or (A, 1/2A) are in the two envelopes.

As we don’t know which event is actual, gain/loss by switching is determined by considering:

1. If A is the smaller amount, then (A, 2A) are in the two envelopes and gain is A.

2. If A is the larger amount, then (A, 1/2A) are in the two envelopes and loss is 1/2A.

So, in that sense, both events are considered jointly and not separately as you do.

The onus is on you to explain what is “join both” because you wrote that.

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Posted: 29 December 2012 10:51 PM   [ Ignore ]   [ # 1928 ]
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GdB - 29 December 2012 04:39 AM

Don’t forget that TromboneAndrew already said this in the first reaction on kkwan’s OP… The reason this thread has become so long is that kkwan’s ego is now dependent on it. Believe me, he knows he is wrong, he just cannot afford to admit it. He never gives straight answers on my questions, because he knows the answers would show him wrong.

Charming. What has my “ego” got to do with the TEP?

GdB, I don’t believe you because you beg the question, mislead and incompletely describes the TEP as it is for your convenience in promoting your “resolution” of the TEP. 

As such, your purported “resolution” is not the resolution of the TEP at all. 

This much I know.

grin

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Posted: 30 December 2012 01:29 AM   [ Ignore ]   [ # 1929 ]
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kkwan - 29 December 2012 10:30 PM

Your non-TEP is not relevant as it is not the TEP per se. Please confine yourself to discussing only the TEP.

You see the point. That’s why you not answer on it. As a TEP-interested guy, you should wonder how another situation leads to the same formula. The only way to avoid the impression that you know that you are wrong is by explaining how the same formula is valid for another situation.

kkwan - 29 December 2012 10:30 PM

Your (X, 2X) is equivalent to (2X, X) which is of the form X = X, thus it is trivially tautological and inane.

Well, yes, of course. It seems you did not get it: simple true statements of mathematics are trivial tautologies. Now the simple trivial tautology you do not conform to, is that if you have any two amounts, and you switch one amount for the other, the loss or gain are equal, as you switch back again.

For the rest you make the same error again and again.

For two amounts, call them X and Y, and you pick one of them, you can only pick X or Y.
Agree?

Now if you pick X first, and change to Y you gain Y - X. Agree?
If you pick Y first, and switch to X, you gain X - Y. Agree?
Now we know it is trivially true that (Y - X) = -1 * (X - Y), (in other words |Y - X| = |X - Y|). So my possible gain and loss are the same.

Now in TEP we can describe the amounts with a:Y = 2X or with b:Y = X/2.
a: X - 2X = -X; 2X -X = X —> So my possible gain and loss are the same.
b: X - X/2 = X/2; X/2 - X = -X/2 —> So my possible gain and loss are the same.

What you do is taking half of both descriptions and join them:

c: 2X - X = X; X/2 - X = -X/2. So you join the two possibilities wrongly.

And now kkwan, tell me why in my non-TEP situation the same formula holds as in TEP. Avoiding to answer is interpreted as you seeing that you are wrong.

[ Edited: 30 December 2012 08:33 AM by GdB ]
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Posted: 30 December 2012 01:31 AM   [ Ignore ]   [ # 1930 ]
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kkwan - 29 December 2012 10:51 PM

GdB, I don’t believe you because you beg the question, mislead and incompletely describes the TEP as it is for your convenience in promoting your “resolution” of the TEP. 

My resolution amounts to the simple observation that you cannot get rich by switching envelopes. Why is this not possible kkwan? Why are you not rich already? Might there be an error in your argument?

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Posted: 30 December 2012 02:49 AM   [ Ignore ]   [ # 1931 ]
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kkwan - 29 December 2012 09:28 PM


The two pairs are either (X, 2X) or (X, 1/2X) because we don’t know which pair is actual.

No they are not. We can check and see. Say the two pairs are 10,20 and 70,140.

Stephen

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Posted: 30 December 2012 02:51 AM   [ Ignore ]   [ # 1932 ]
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kkwan - 29 December 2012 09:28 PM

Of course, the information of what are the amounts in the two envelopes is crucial, but in the context of the TEP as it is, that information is unavailable.

No Kkwan we can fill in numbers to see what would happen in reality if we switched as a general strategy.

To not do so and try to solve the puzzle is to be stupid.

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Posted: 30 December 2012 01:16 PM   [ Ignore ]   [ # 1933 ]
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GdB - 30 December 2012 01:29 AM

You see the point. That’s why you not answer on it. As a TEP-interested guy, you should wonder how another situation leads to the same formula. The only way to avoid the impression that you know that you are wrong is by explaining how the same formula is valid for another situation.

No, I don’t see the point or the rationale of you deliberately introducing a fixed and manipulatable non-TEP situation whereby there is no initial choice and no safeguard against manipulation as the amount in the other envelope is put in after the initial “selection” unlike the TEP whereby there is an initial choice in selecting an envelope from two indistinguishable envelopes, also the amounts have been put into the two envelopes before selection and thus there is no possibility of manipulation.

That the argument to switch is also applicable to your non-TEP does not in any way invalidate it in the context of the TEP. It is analogous to comparing apples with oranges and declaring that as both are eatable fruit, eating apples is synonymous to eating oranges and to imply when we eat apples, we are eating oranges instead, which is misleading, unjustified and preposterous..

So, the impression that I am wrong is irrelevant and unjustified as you have not directly shown the flaw in the argument to switch, in the context of the TEP.

Your non-TEP is a red herring to deflect the focus on the TEP as it is because you have no direct means of showing why the argument to switch is invalid and must resort to this ridiculous
method.

Well, yes, of course. It seems you did not get it: simple true statements of mathematics are trivial tautologies. Now the simple trivial tautology you do not conform to, is that if you have any two amounts, and you switch one amount for the other, the loss or gain are equal, as you switch back again.

However, your true statement that (X, 2X) = (2X, X) is obvious, trivially true and vacuous.

For the rest you make the same error again and again.

For two amounts, call them X and Y, and you pick one of them, you can only pick X or Y.
Agree?

Now if you pick X first, and change to Y you gain Y - X. Agree?
If you pick Y first, and switch to X, you gain X - Y. Agree?
Now we know it is trivially true that (Y - X) = -1 * (X - Y), (in other words |Y - X| = |X - Y|). So my possible gain and loss are the same.

Now in TEP we can describe the amounts with a:Y = 2X or with b:Y = X/2.
a: X - 2X = -X; 2X -X = X —> So my possible gain and loss are the same.
b: X - X/2 = X/2; X/2 - X = -X/2 —> So my possible gain and loss are the same.

What you do is taking half of both descriptions and join them:

c: 2X - X = X; X/2 - X = -X/2. So you join the two possibilities wrongly.

And now kkwan, tell me why in my non-TEP situation the same formula holds as in TEP. Avoiding to answer is interpreted as you seeing that you are wrong.

How many times must I reiterate that Y is superfluous?

And how many times must I reiterate that it is begging the question to assume you know what are the amounts in the two envelopes as either (X, 2X) or (X, 1/2X) are in the two envelopes and you don’t know which event is actual?

You can interpret it in any way you like, but it is irrelevant as explained above.

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Posted: 30 December 2012 01:25 PM   [ Ignore ]   [ # 1934 ]
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GdB - 30 December 2012 01:31 AM

My resolution amounts to the simple observation that you cannot get rich by switching envelopes. Why is this not possible kkwan? Why are you not rich already? Might there be an error in your argument?

Your resolution begs the question and as such is not the resolution of the TEP.

It is only possible if and only if you know what are the amounts in the two envelopes which you don’t, in the context of the TEP.

Whether I am rich or not has nothing to do with my argument per se.

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Posted: 30 December 2012 01:35 PM   [ Ignore ]   [ # 1935 ]
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StephenLawrence - 30 December 2012 02:49 AM

No they are not. We can check and see. Say the two pairs are 10,20 and 70,140.

Your two pairs are of the form (X, 2X) whereby X is 10 and 70 respectively.

There should also be the form (X, 1/2X) whereby they are (10, 5) and (70, 35) respectively.

So, (10, 20) or (10, 5) are two pairs and (70, 140) or (70, 35) are another two pairs for the same value of X.

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