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The two envelopes problem
 Posted: 30 December 2012 01:45 PM [ Ignore ]   [ # 1936 ]
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StephenLawrence - 30 December 2012 02:51 AM

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.

Of course we can, but generally it is not necessary as we can deduce either (X, 2X) or (X, 1/2X) are in the two envelopes for any finite value of X.

That is sufficient for us to proceed as in the argument to switch by denoting the amount in the selected envelope as A. We can fill in numbers at any time.

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 Posted: 31 December 2012 12:53 AM [ Ignore ]   [ # 1937 ]
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kkwan - 30 December 2012 01:16 PM

No, I don’t see the point or the rationale of you deliberately introducing a fixed and manipulatable non-TEP situation whereby (...)

The rationale is that the formula is exactly the same, where the situations where we apply them to are totally different. That is something to be explained.

Now from following passage, it is clear you cannot explain it:

kkwan - 30 December 2012 01:16 PM

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..

But it is good to see that you admit that my formula is correct for my situation. Why didn’t you say that simply before? Why all the fuss?

Now your comparison with apple and oranges is plainly stupid. You simply show you have no explanation why in my counter example the same formula holds as in TEP.

However I can explain why your formula is not valid in TEP: you do not use the fact that there are only two amounts where you pick from.

I do not know if I can put it more basically than:

1. If you have picked the envelope with one of the amounts, then the other remains, and this is the only one you can switch to.

One step concreter:
2a. If you have picked the envelope with the smallest amount, the other one is the envelope with the biggest amount.
or
2b. If you have picked the envelope with the biggest amount, the other one is the envelope with the smallest amount.

From 2a. and b. follows:
3. The difference between the biggest and the smallest amount is always the same, so your possible gain and loss when switching are also always the same.

Even more concrete:

Call the smallest amount X (that does not mean I know what the amount is!), then the biggest amount is 2X.

Now I can rephrase 2:
4a. If you have picked the envelope with X, then the other contains 2X.
4b. If you have picked the envelope with 2X, the other one contains X.

So by switching you can loose or gain X.

If you want you can call the smallest amount X/2, then the biggest amount is X:

5a. If you have picked the envelope with X/2, then the other contains X.
5b. If you have picked the envelope with X, the other one contains X/2.

So by switching you can loose or gain X/2.

So kkwan: either you show me you use 1. in your derivation, or you must show us that 1. is not valid in TEP.

In my ‘counter example’ (other situation, but your formula applies), 1. does not apply because for the other amount there are two possibilities.

kkwan - 30 December 2012 01:16 PM

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.

Yes I have. You do not use 1:

If you have picked the envelope with one of the amounts, then the other remains, and this is the only one you can switch to.

kkwan - 30 December 2012 01:16 PM

How many times must I reiterate that Y is superfluous?

And how many times I must repeat that for any two amounts, there is one possible value for their difference, and this is your possible gain or loss?

kkwan - 30 December 2012 01:16 PM

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?

And how many times I must say that you can only describe TEP as either (X,2X) or as (X,X/2), but not as both at the same time?

But kkwan, have you thought at combining the two descriptions in another way?:

Call the amount in the other envelope X. Then you have X/2 or 2X in your envelope. If you have X/2 now, then you gain X/2 by switching. If you have 2X now, then you loose X by switching. So you should not switch.

Explain what is wrong with this derivation, kwann.

[ Edited: 01 January 2013 09:15 AM by GdB ]
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 Posted: 31 December 2012 01:44 AM [ Ignore ]   [ # 1938 ]
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kkwan - 30 December 2012 01:45 PM
StephenLawrence - 30 December 2012 02:51 AM

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.

Of course we can, but generally it is not necessary…..............

It is necessary because otherwise you won’t see what is wrong.

Right, now you said that the entity could have picked the pair we are playing with from two pairs. Can you fill numbers in please. I suggest you have some separation between the pairs like 20,40 and 70,140 for instance, but it’s up to you.

[ Edited: 31 December 2012 04:06 AM by StephenLawrence ]
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 Posted: 02 January 2013 01:50 PM [ Ignore ]   [ # 1939 ]
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GdB - 31 December 2012 12:53 AM

The rationale is that the formula is exactly the same, where the situations where we apply them to are totally different. That is something to be explained.

That does not mean the argument to switch is invalid for the TEP. It is just that your contrived non-TEP, with it’s entirely initial situation has subsequent similar features to the TEP and that is all there is to it.

But it is good to see that you admit that my formula is correct for my situation. Why didn’t you say that simply before? Why all the fuss?

Now your comparison with apple and oranges is plainly stupid. You simply show you have no explanation why in my counter example the same formula holds as in TEP.

However I can explain why your formula is not valid in TEP: you do not use the fact that there are only two amounts where you pick from.

I do not know if I can put it more basically than:

1. If you have picked the envelope with one of the amounts, then the other remains, and this is the only one you can switch to.

One step concreter:
2a. If you have picked the envelope with the smallest amount, the other one is the envelope with the biggest amount.
or
2b. If you have picked the envelope with the biggest amount, the other one is the envelope with the smallest amount.

From 2a. and b. follows:
3. The difference between the biggest and the smallest amount is always the same, so your possible gain and loss when switching are also always the same.

Even more concrete:

Call the smallest amount X (that does not mean I know what the amount is!), then the biggest amount is 2X.

Now I can rephrase 2:
4a. If you have picked the envelope with X, then the other contains 2X.
4b. If you have picked the envelope with 2X, the other one contains X.

So by switching you can loose or gain X.

If you want you can call the smallest amount X/2, then the biggest amount is X:

5a. If you have picked the envelope with X/2, then the other contains X.
5b. If you have picked the envelope with X, the other one contains X/2.

So by switching you can loose or gain X/2.

So kkwan: either you show me you use 1. in your derivation, or you must show us that 1. is not valid in TEP.

In my ‘counter example’ (other situation, but your formula applies), 1. does not apply because for the other amount there are two possibilities.

Your explanation is fundamentally flawed because you assume (X, 2X) are in the two envelopes which is begging the question because you don’t know which event is actual in the context of the TEP.

The same applies to assuming (X, 1/2X) are in the two envelopes.

Yes I have. You do not use 1:

If you have picked the envelope with one of the amounts, then the other remains, and this is the only one you can switch to.

If we select an envelope, denote the amount in it as A and as it is either the smaller or the larger amount, then either (A, 2A) or (A, 1/2A) are in the two envelopes.

On switching, gain is A or loss is 1/2A.

The above is consistent with your 1 because (with A selected), then we switch to 2A in (A, 2A) and to 1/2A in (A, 1/2A).

And how many times I must say that you can only describe TEP as either (X,2X) or as (X,X/2), but not as both at the same time?

Not quite so.

That either (X, 2X) or (X, 1/2X) are mutually exclusive events means both events cannot occur at the same time.

However, as both events satisfy the requirement of the TEP and they are distinct events, then {either (X, 2X) or (X, 1/2X) are in the two envelopes}, is the complete description of the TEP.

But kkwan, have you thought at combining the two descriptions in another way?:

Call the amount in the other envelope X. Then you have X/2 or 2X in your envelope. If you have X/2 now, then you gain X/2 by switching. If you have 2X now, then you loose X by switching. So you should not switch.

Explain what is wrong with this derivation, kwann.

That is saying (2X, X) or (1/2X, X) are in the two envelopes and we select either 2X or 1/2X.

However, as we don’t know which event is actual:

1. We cannot assume we have selected 2X or 1/2X because that is begging the question.

2. We are totally ignorant of what we have selected. As such, we could have selected X.

Thus, all we can say, without begging the question and with total ignorance is:

1.We select an envelope and denote the amount in it as A

2. A is a constant in any particular trial of the TEP.

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 Posted: 02 January 2013 01:56 PM [ Ignore ]   [ # 1940 ]
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StephenLawrence - 31 December 2012 01:44 AM

It is necessary because otherwise you won’t see what is wrong.

Generally, with X as in (X, 2X) or (X, 1/2X), we can do that without numbers.

Right, now you said that the entity could have picked the pair we are playing with from two pairs. Can you fill numbers in please. I suggest you have some separation between the pairs like 20,40 and 70,140 for instance, but it’s up to you.

It is necessary and sufficient to refer to (X, 2X) or (X, 1/2X) and that is all there is to it.

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 Posted: 02 January 2013 02:08 PM [ Ignore ]   [ # 1941 ]
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GdB - 25 November 2012 11:15 AM
StephenLawrence - 25 November 2012 10:16 AM

Now I’m just waiting to test my answer on Kkwan, If I convince him that will be conclusive proof that I’m right.  :lol:

He will never admit that he was wrong. All these pages, all for nothing?? He will repeat his mantra again:

”(X,2X), (X,1/2X)”

You were right. :-)

Stephen

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 Posted: 03 January 2013 01:13 AM [ Ignore ]   [ # 1942 ]
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kkwan - 02 January 2013 01:50 PM

That does not mean the argument to switch is invalid for the TEP. It is just that your contrived non-TEP, with it’s entirely initial situation has subsequent similar features to the TEP and that is all there is to it.

No, no, kkwan. Why don’t you just admit that you cannot explain the fact that in my counter example the same formula is valid as you say is for TEP. A few postings above you have gone in great pain to declare what the differences between TEP and my counter example are. Now, how is it possible that with all these differences, the same formula applies?

kkwan - 02 January 2013 01:50 PM

Your explanation is fundamentally flawed because you assume (X, 2X) are in the two envelopes which is begging the question because you don’t know which event is actual in the context of the TEP.

The same applies to assuming (X, 1/2X) are in the two envelopes.

I assume there is (X,2X) in the envelopes or there is (X,X/2) in the envelopes. I consider both. It results in 4 possibilities (remember: <X,Y> means first pick is X, second is Y):

a. <X,2X>
b. <2X,X>
c. <X,X/2>
d. <X/2,X>

You happily leave out b. and d.

kkwan - 02 January 2013 01:50 PM

That either (X, 2X) or (X, 1/2X) are mutually exclusive events means both events cannot occur at the same time.

However, as both events satisfy the requirement of the TEP and they are distinct events, then {either (X, 2X) or (X, 1/2X) are in the two envelopes}, is the complete description of the TEP.

It is not complete: when you want to use two descriptions for one TEP situation, then you must also include b. and d.

kkwan - 02 January 2013 01:50 PM

That is saying (2X, X) or (1/2X, X) are in the two envelopes and we select either 2X or 1/2X.

However, as we don’t know which event is actual:

1. We cannot assume we have selected 2X or 1/2X because that is begging the question.

2. We are totally ignorant of what we have selected. As such, we could have selected X.

What a bullshit, kkwan. (2X, X) and (1/2X, X) perfectly describe the TEP: or do you think one of them does not conform to the description ‘one envelope contains twice the other’? So again, you leave out possibilities b. and d.

[ Edited: 03 January 2013 03:08 AM by GdB ]
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 Posted: 03 January 2013 09:51 AM [ Ignore ]   [ # 1943 ]
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GdB - 03 January 2013 01:13 AM

No, no, kkwan. Why don’t you just admit that you cannot explain the fact that in my counter example the same formula is valid as you say is for TEP. A few postings above you have gone in great pain to declare what the differences between TEP and my counter example are. Now, how is it possible that with all these differences, the same formula applies?

The essential differences are wrt to no initial selection as in the TEP and the amount in the other envelope is put in after the one envelope is “selected”.

Notwithstanding that, subsequently, (assuming there is no cheating wrt to putting the amount in the other envelope), i.e. the amount is selected at random, we have the following:-

The amount in the “selected” envelope is denoted as A.

1. If A is the smaller amount, the other envelope contains 2A

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

Thus, either (A, 2A) or (A, 1/2A) are in the two envelopes.

This is equivalent to the TEP and as such the argument to switch is also applicable which does not imply that the argument to switch in the TEP is invalid.

Actually, it does not matter whether there is any initial selection at all and whether the amount in the other envelope was put in before of after “selection” as long as there is no cheating in that the amount was selected at random to put into the other envelope.

In that sense, your non-TEP is essentially the TEP and as such, it is not surprising that the argument to switch is applicable.

I assume there is (X,2X) in the envelopes or there is (X,X/2) in the envelopes. I consider both. It results in 4 possibilities (remember: <X,Y> means first pick is X, second is Y):

a. <X,2X>
b. <2X,X>
c. <X,X/2>
d. <X/2,X>

You happily leave out b. and d.

a and b is equivalent, so is c and d, as events

And being totally ignorant, we cannot say whether we are in a, b, c or d without begging the question.

As such, we select an envelope and denote the amount in it as A. If we say A could be X, then A is the smaller amount in (X, 2X) or if we say A could be 2X, then A is the larger amount in (X, 2X).

However, by doing that we assume (X, 2X) are actually in the two envelopes (which is begging the question) because then and only then can we select as we do in the above.

We have the same problem wrt (X, 1/2X).

Thus, to not beg the question, we must say A is X in either (X, 2X) or (X, 1/2X) and as such, only a and c can be considered.

Hence, because b and d cannot be considered without begging the question, they can be ruled out in the context of the TEP as it is.

It is not complete: when you want to use two descriptions for one TEP situation, then you must also include b. and d.

It is, as a and b is equivalent, so is c and d, as events.

What a bullshit, kkwan. (2X, X) and (1/2X, X) perfectly describe the TEP: or do you think one of them does not conform to the description ‘one envelope contains twice the other’? So again, you leave out possibilities b. and d.

As (2X, X) is equivalent to (X, 2X) and (1/2X, X) is equivalent to (X, 1/2X) it is complete to describe the TEP as: either (X, 2X) or (X, 1/2X) in the two envelopes.

[ Edited: 03 January 2013 08:39 PM by kkwan ]
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 Posted: 04 January 2013 02:16 AM [ Ignore ]   [ # 1944 ]
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kkwan - 03 January 2013 09:51 AM

This is equivalent to the TEP and as such the argument to switch is also applicable which does not imply that the argument to switch in the TEP is invalid.

So here we have it: you say that these situations are the same:

1. You get an envelope with a certain amount, and are allowed to switch to another amount, which with a chance of a half, contains twice or half the amount as the amount in your envelope. You are offered to switch to the other envelope.
2. There are two envelopes with each an amount in it, one is twice the other. You select one envelope, but are then offered to switch.

Now I have a few questions:

I. In 2. I obviously start with two amounts. Express the possibilities (to switch or not to switch) in terms of the lowest amount (i.e. describe the lowest amount as ‘X’). Show your derivation.
II. Same with 1. Show your derivation.
III. What is the total amount in the two envelopes together, expressed in terms of the lowest amount in 1. and 2.? Show your derivation.

kkwan - 03 January 2013 09:51 AM

In that sense, your non-TEP is essentially the TEP and as such, it is not surprising that the argument to switch is applicable.

No, it isnt. The difference is the following: in 1. the first amount is fixed, and the other amount is defined by it. It is half or twice that amount. In 2, i.e. in TEP, the first amount is not fixed. The only way the second envelope contains twice the amount of the first, is that you picked from the two envelopes the one with the smallest amount. The other way round, the only way that the second envelope contains half of the first, is that you picked from the two envelopes the one with the biggest amount. However you do as if the amount in your envelope is the same in both situations independent from the fact if you picked the biggest or smallest amount first.

kkwan - 03 January 2013 09:51 AM

And being totally ignorant, we cannot say whether we are in a, b, c or d without begging the question.

As I am totally ignorant, I take all of a. to d. having the same chance. It follows that switching does not help.

kkwan - 03 January 2013 09:51 AM

However, by doing that we assume (X, 2X) are actually in the two envelopes (which is begging the question) because then and only then can we select as we do in the above.

We have the same problem wrt (X, 1/2X).

If there are two possibilities, then I look what what happens in each of them. If the result of both is that switching does not help, then switching does not help.

kkwan - 03 January 2013 09:51 AM

It is, as a and b is equivalent, so is c and d, as events.

As possible amounts in the envelopes you mean. Yes. But with each you should take into account that you could have picked the other value first, and these possibilities you do not take into account.

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 Posted: 04 January 2013 10:10 PM [ Ignore ]   [ # 1945 ]
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GdB - 04 January 2013 02:16 AM

1. You get an envelope with a certain amount, and are allowed to switch to another amount, which with a chance of a half, contains twice or half the amount as the amount in your envelope. You are offered to switch to the other envelope.
2. There are two envelopes with each an amount in it, one is twice the other. You select one envelope, but are then offered to switch.

1 and 2 are not exactly synonymous. In 1 we are given an envelope whereas in 2 we can select an envelope from two indistinguishable envelopes.

However, it does not matter (if we forgo choice initially) as we will end up with one envelope anyway whether in 1 or 2 and as we don’t know what amount is in that envelope irrespective of choice.

I. In 2. I obviously start with two amounts. Express the possibilities (to switch or not to switch) in terms of the lowest amount (i.e. describe the lowest amount as ‘X’). Show your derivation.
II. Same with 1. Show your derivation.
III. What is the total amount in the two envelopes together, expressed in terms of the lowest amount in 1. and 2.? Show your derivation.

Whether in 1 or 2, as we don’t know what is the amount in the given envelope in 1 or the selected envelope in 2, we denote the amount in that envelope as A and proceed as follows:-

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

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

Thus, either (A, 2A) or (A, 1/2A) are the amounts in the two envelopes.

No, it isnt. The difference is the following: in 1. the first amount is fixed, and the other amount is defined by it. It is half or twice that amount. In 2, i.e. in TEP, the first amount is not fixed. The only way the second envelope contains twice the amount of the first, is that you picked from the two envelopes the one with the smallest amount. The other way round, the only way that the second envelope contains half of the first, is that you picked from the two envelopes the one with the biggest amount. However you do as if the amount in your envelope is the same in both situations independent from the fact if you picked the biggest or smallest amount first.

Obviously, the amounts in either the given envelope in 1 or the selected envelope in 2 are fixed. However, we don’t know what the amounts are.

And with total ignorance, we don’t know whether the amount in the given or the selected envelope is the smaller or the larger amount.

Hence, we denote the amount in that envelope as A (which is a constant) and proceed as in the above.

As I am totally ignorant, I take all of a. to d. having the same chance. It follows that switching does not help.

Being totally ignorant, you can only consider a or c because b or d begs the question.

Also, a and b are equivalent, so is c and d, as events.

If there are two possibilities, then I look what what happens in each of them. If the result of both is that switching does not help, then switching does not help.

You can only consider possibilities which do not beg the question, in the context of the TEP.

As possible amounts in the envelopes you mean. Yes. But with each you should take into account that you could have picked the other value first, and these possibilities you do not take into account.

That we do not consider each event separately is because we don’t know which event is actual and as such, with A denoted as the amount in the selected envelope, we proceed as in the above.

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 Posted: 06 January 2013 08:21 AM [ Ignore ]   [ # 1946 ]
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I think I’ll stop here. For everyone it is clear that TEP and my counter example are different situations, the difference being that in TEP you initially have the choice between 2 amounts. The player knows these amounts are fixed, and that is enough. You do not use this fact in your formula: you do as if there is no dependency between the amount in your choice of envelope and the amount in the other envelope. You do as if when the other envelope has the smaller amount, or the other envelope has the bigger amount, your amount stays the same. It is so simple: if you first picked the smallest amount, the other envelope contains the remaining biggest amount; if you first picked the biggest amount, the other envelope contains the remaining smallest amount. You deny that in both situations the smallest amounts are the same, respectively the biggest amounts are the same.

You must stand the criticism that when you have two fixed amounts, the difference between them is also fixed. You do as if you can increase money by switching envelopes with unknown amounts. Do you really believe that?

You happily jump over the fact that if you have two amounts, as stated in TEP, that you have just one of the two possible fixed amounts, so that A cannot serve as a single constant. And if you take a constant for the other envelope, it is clear that according the same argument you should not switch. What if you would have chosen the other first: would it then be better not to switch? You would have called that amount A.

Also you make no difference between having two fixed but unknown amounts, and descriptions of them: all the following descriptions are full equivalent descriptions of TEP: (100A,50A), (3.14X, 6.28X), (0.001Y, 0.0005Y),(X,2X) and (X,X/2). In your calculation you just jump from one description to the other. If you stick to one singe description, and describe the different possibilities as ‘you picked the biggest amount first’ or the ‘you picked the smallest first’ then the logical absurdity evaporates: Then you see that with (50A,100A), (6.28X,3.14X), (0.0005Y,0.001Y), (X,2X), (X/2,X) the alternative that you picked the other envelope is fully described.

What also is very clear is that you always avoid direct answers to questions I ask. I must really press you to get some answer, and this answer shows that you are trying to crawl out under my arguments, instead of really coping with them. And you do not trust your own arguments when I am trying to bet with you: it is obvious you know that your argument is wrong.

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 Posted: 07 January 2013 07:57 AM [ Ignore ]   [ # 1947 ]
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GdB - 06 January 2013 08:21 AM

I think I’ll stop here. For everyone it is clear that TEP and my counter example are different situations, the difference being that in TEP you initially have the choice between 2 amounts. The player knows these amounts are fixed, and that is enough. You do not use this fact in your formula: you do as if there is no dependency between the amount in your choice of envelope and the amount in the other envelope. You do as if when the other envelope has the smaller amount, or the other envelope has the bigger amount, your amount stays the same. It is so simple: if you first picked the smallest amount, the other envelope contains the remaining biggest amount; if you first picked the biggest amount, the other envelope contains the remaining smallest amount. You deny that in both situations the smallest amounts are the same, respectively the biggest amounts are the same.

It is not so simple because we don’t know whether either (X, 2X) or (X, 1/2X) are in the two envelopes in your non-TEP or the TEP.

You must stand the criticism that when you have two fixed amounts, the difference between them is also fixed. You do as if you can increase money by switching envelopes with unknown amounts. Do you really believe that?

Because you are totally ignorant, you cannot assume there are only two fixed amounts because that is begging the question.

You happily jump over the fact that if you have two amounts, as stated in TEP, that you have just one of the two possible fixed amounts, so that A cannot serve as a single constant. And if you take a constant for the other envelope, it is clear that according the same argument you should not switch. What if you would have chosen the other first: would it then be better not to switch? You would have called that amount A.

The amount in the selected envelope (denoted as A), is clearly a constant in any particular trial of the TEP.

Also you make no difference between having two fixed but unknown amounts, and descriptions of them: all the following descriptions are full equivalent descriptions of TEP: (100A,50A), (3.14X, 6.28X), (0.001Y, 0.0005Y),(X,2X) and (X,X/2). In your calculation you just jump from one description to the other. If you stick to one singe description, and describe the different possibilities as ‘you picked the biggest amount first’ or the ‘you picked the smallest first’ then the logical absurdity evaporates: Then you see that with (50A,100A), (6.28X,3.14X), (0.0005Y,0.001Y), (X,2X), (X/2,X) the alternative that you picked the other envelope is fully described.

Because we are totally ignorant, either (X, 2X) or (X, 1/2X) are in the two envelopes and we must consider both mutually exclusive events together and not separately.

What also is very clear is that you always avoid direct answers to questions I ask. I must really press you to get some answer, and this answer shows that you are trying to crawl out under my arguments, instead of really coping with them. And you do not trust your own arguments when I am trying to bet with you: it is obvious you know that your argument is wrong.

You beg the question and as such, your arguments are not cogent.

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 Posted: 07 January 2013 09:06 AM [ Ignore ]   [ # 1948 ]
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kkwan - 07 January 2013 07:57 AM

The amount in the selected envelope (denoted as A), is clearly a constant in any particular trial of the TEP.

It apparently isn’t for reasons already given (over and over).

Please say why you disagree using a concrete example.

Stephen

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 Posted: 07 January 2013 10:12 AM [ Ignore ]   [ # 1949 ]
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StephenLawrence - 07 January 2013 09:06 AM

It apparently isn’t for reasons already given (over and over).

Please say why you disagree using a concrete example.

In any particular trial of the TEP, the amount in the selected envelope is a fixed amount and it cannot change within the context of that trial, i.e. it is a constant.

So, if we denote that amount as A, then A is a constant.

Concretely, if the amount in the selected envelope is \$20, it cannot be any other amount in the context of that particular trial of the TEP, i.e. it is a constant.

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 Posted: 07 January 2013 10:19 AM [ Ignore ]   [ # 1950 ]
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kkwan - 07 January 2013 10:12 AM
StephenLawrence - 07 January 2013 09:06 AM

It apparently isn’t for reasons already given (over and over).

Please say why you disagree using a concrete example.

In any particular trial of the TEP, the amount in the selected envelope is a fixed amount and it cannot change within the context of that trial, i.e. it is a constant.

So, if we denote that amount as A, then A is a constant.

Concretely, if the amount in the selected envelope is \$20, it cannot be any other amount in the context of that particular trial of the TEP, i.e. it is a constant.

There is no sense to this because to say we should switch means switching would work as a general principle.

If we check concrete examples we see that A is not a constant because the amount in our envelope could be other amounts and we need to factor that into our calculations. We also see this is true even if we look.

Stephen

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