No. You do not use all available information. The only information you use is that the other envelope contain half or twice the amount as your envelope. But you know more: that you could have taken the other envelope first, and you would be in exactly the same situation. And that is not the case in 2:

“The other envelope contains half or twice the amount as in the envelope I gave to you.”

Not so. What I wrote in post 1352 was:

2. If gain/loss is wrt the initial amounts in each situation, then on switching, gain/loss is 100%/50% and switching does makes a difference, as the potential gain is twice the potential loss.

If the other envelope with 2X was selected first, then on switching, loss/gain is 50%/100%. The potential loss is half the potential gain which is the same as in 2.

If in your formulation you would pick the envelope with 2X and you decide again to switch or not one should consider (2X,4X), and if you picked the one with X/2 one should consider (X/2,X/4) etc.

The point is, you don’t know whether the envelope you selected has the smaller or the larger amount. As such, A is denoted as the amount in the selected envelope. If A is the smaller amount, then (A, 2A) are in the two envelopes i.e. (X, 2X) and if A is the larger amount, then (A, 1/2A) are in the two envelopes i.e. (X, 1/2X).

4. If you have X, the other envelope contains 2X.
5. If you have 2X, the other envelope contains X.

That is assuming (X, 2X) is fixed as the amounts in the two envelopes, but in the TEP, either (X, 2X) or (X, 1/2X) are in the two envelopes.

As said, you are taking a changing reference in which you do as if X (in 4) is the same as 2X (in 5) which is of course not true.

What I wrote was wrt gain/loss which is not related to 4 or 5 at all.

4. If you have X, the other envelope contains 2X.
5. If you have 2X, the other envelope contains X.

That is assuming (X, 2X) is fixed as the amounts in the two envelopes, but in the TEP, either (X, 2X) or (X, 1/2X) are in the two envelopes.

Yes, of course the amounts are fixed. The envelopes are closed.

It is in non-TEP:
“The envelope you have has a certain amount. The other envelope has half or twice of that amount”.
Then you can say it is (X,X/2) or (X,2X). But not in TEP, as the amounts are fixed.

kkwan - 30 June 2012 07:41 AM

As said, you are taking a changing reference in which you do as if X (in 4) is the same as 2X (in 5) which is of course not true.

What I wrote was wrt gain/loss which is not related to 4 or 5 at all.

So what? That is the place where you make the error. Again, this is correct:
4. If you have X, the other envelope contains 2X.
5. If you have 2X, the other envelope contains X.

What you do is replace 2X with A in 5, so we get 5a:
5a: If you have A, then the other envelope contains A/2.
This one, taken on its own, is true. But then you replace X with A in 4:
4a. If you have A, then the other envelope contains 2A.
Which also is correct taken on its own.

But then you equate both A’s: in 4a, A = X, but in 5a A = 2X. Your two A’s are just not the same, so you cannot treat them as that as you do in your formula.

You are comparing the relative changes, not the absolute ones. What when a price increases from $50 to $100? That is an increase of 100%. What when the price decreases from $100 to $50? That is a decrease of 50%. so the decrease is less than the increase???? Or is the increase and decrease just the same, namely $50. Can you gain money by increasing and decreasing prices in such a way?

In each of the two different situations in the TEP, it is the relative change wrt to the initial amount which matters in the context of each situation. Your example of price increase/decrease is wrt to one common situation whereby the $50 change is wrt increase/decrease and as such, it is valid as a reference.

Sorry kkwan, I replaced 4 and 5 with propositions that contain the complete information you have, not 6 and 7. If you use the correct values in 6 and 7 you come at the correct answer for the expected value: (0.5 x 2A) + (0.5 x 0.5A) = X + (0.5 x 0.5 x 2X) = 1.5X. You are leaving out information that you have.

That is based on (A, 2A) fixed in the two envelopes. But, in the context of the TEP, either (A, 2A) or (A, 1/2A) are in the two envelopes and as such, steps 6 and 7 are valid.

Yes. But it also indicates that it is not valid with higher values. And, as correctly noticed in the article, some funny amounts as SFR 1234.01 also indicate that you have the smaller amount, as you cannot divide 0.1 by 2 (no, Swiss cheques have just two decimal places for the rappen) So you formula, even when it where valid (one can use it in some other situations than TEP), it is not valid universally in real situations.

Except for the limit of the uncle’s wealth and some funny amount which is not divisible by 2 to 2 decimal places, there is no reason not to have 50/50 chance for the amounts in the two envelopes. In reality, it is simple and practical to write cheques of twice or half the amount in integers rather than some ridiculous amount in integers and decimals.

Yes, of course the amounts are fixed. The envelopes are closed.

It is in non-TEP:
“The envelope you have has a certain amount. The other envelope has half or twice of that amount”.
Then you can say it is (X,X/2) or (X,2X). But not in TEP, as the amounts are fixed.

In the TEP, that one envelope contains twice the amount of the other, implies either (X, 2X) or (X, 1/2X) are in the envelopes. So, the amounts are not fixed as (X, 2X).

So what? That is the place where you make the error. Again, this is correct:
4. If you have X, the other envelope contains 2X.
5. If you have 2X, the other envelope contains X.

Not so, if either (X, 2X) or (X, 1/2X) are in the envelopes.

What you do is replace 2X with A in 5, so we get 5a:
5a: If you have A, then the other envelope contains A/2.
This one, taken on its own, is true. But then you replace X with A in 4:
4a. If you have A, then the other envelope contains 2A.
Which also is correct taken on its own.

But then you equate both A’s: in 4a, A = X, but in 5a A = 2X. Your two A’s are just not the same, so you cannot treat them as that as you do in your formula.

A, denoted as the amount in the selected envelope does not vary as either (A, 2A) or (A, 1/2A) are in the envelopes.

Kkwan, you have given no arguments against my position, even if it might seem so to you. You are just repeating (X,2X), (X, 1/2X), but at no place you show that my single (X,2X) is not a complete description of TEP:

4 If you have X, then the other envelope contains twice as much
5. If you have 2X, the other envelope contains half as much.

So I have this half/twice stuff in it. You with your (X,2X), (X, 1/2X) are introducing an extra possibility, i.e. an extra amount, that is just not there.

And where is the article where a mathematician says that your formula is valid for TEP? The Gill article doesn’t.

Kkwan, you have given no arguments against my position, even if it might seem so to you. You are just repeating (X,2X), (X, 1/2X), but at no place you show that my single (X,2X) is not a complete description of TEP:

4 If you have X, then the other envelope contains twice as much
5. If you have 2X, the other envelope contains half as much.

So I have this half/twice stuff in it. You with your (X,2X), (X, 1/2X) are introducing an extra possibility, i.e. an extra amount, that is just not there.

Your (X, 2X) is not the complete description of the TEP because you have omitted (X, 1/2X) which also satisfies “one envelope contains twice the amount of the other” as per the TEP. For any finite value of X, (X, 2X) is not equivalent to (X, 1/2X).

In your 4, if you have X, then (X, 2X) are in the envelopes.
In your 5, if you have 2X, then (2X, X) are in the envelopes.

Clearly, (X, 2X) is equivalent to (2X, X) for any finite value of X, which is circular and trivial. This is the consequence of fixing the amounts in the envelopes as (X, 2X) which is not what is specifically stated in the TEP. It is tantamount to simplifying the TEP to a one-sided fixed game which is a travesty.

OTOH, with either (X, 2X) or (X, 1/2X) as possible amounts in the two envelopes, is the complete description of the TEP.

And where is the article where a mathematician says that your formula is valid for TEP? The Gill article doesn’t.

You have ignored what I wrote in post 1364 whereby I quoted from Richard D Gill’s paper wrt this issue and he is a mathematician:

If the author only meant to write that since he knows almost nothing about X, it then follows that given A, is pretty certain to be very close to Bernoulli(1/2), we could not fault steps 6 and 7.

Steps 6 and 7 are:

6. Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2.

7. So the expected value of the money in the other envelope is

Thus, the formula for the expected value is valid with either (A, 2A) or (A, 1/2A) in the two envelopes is in concordance with the TEP per se and as such, is the complete description of the TEP.

Your (X, 2X) is not the complete description of the TEP because you have omitted (X, 1/2X) which also satisfies “one envelope contains twice the amount of the other” as per the TEP. For any finite value of X, (X, 2X) is not equivalent to (X, 1/2X).

If (X,2X) satisfies TEP, then does (X,2X) which if you want to write it otherwise can write as (Y,Y/2). Who says it must be for the same X?

kkwan - 02 July 2012 09:28 AM

This is the consequence of fixing the amounts in the envelopes as (X, 2X) which is not what is specifically stated in the TEP.

Yes, it is: the envelopes are closed.

kkwan - 02 July 2012 09:28 AM

You have ignored what I wrote in post 1364 whereby I quoted from Richard D Gill’s paper wrt this issue and he is a mathematician:

No I didn’t. Just fill in the correct amounts in 7 (the right X and 2X), and you get the correct answer. The problem is in the combination of 4 and 5, not in 6 and 7. I’ve shown that before.

If (X,2X) satisfies TEP, then does (X,2X) which if you want to write it otherwise can write as (Y,Y/2). Who says it must be for the same X?

Obviously (X, 2X) and (2X, X) satisfies the TEP. They are equivalent.

You did and to be consistent, you should stick to X. Then for any finite value of X, (X, 2X) is not equivalent to (X, 1/2X). Introducing (Y, 1/2Y) is superfluous if it is equivalent to (2X, X).

Yes, it is: the envelopes are closed.

It is fixed as what? (X, 2X) or (X, 1/2X)?

No I didn’t. Just fill in the correct amounts in 7 (the right X and 2X), and you get the correct answer. The problem is in the combination of 4 and 5, not in 6 and 7. I’ve shown that before.

That is because you have fixed (X, 2X) as the amounts in the two envelopes.

It is not so if either (X, 2X) or (X, 1/2X) are in the two envelopes.

There is no problem with steps 4 and 5 in the context of the following steps:

1. I denote by A the amount in my selected envelope.
2. The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2.
3. The other envelope may contain either 2A or A/2.
4. If A is the smaller amount the other envelope contains 2A.
5. If A is the larger amount the other envelope contains A/2.

with (X, 2X) not fixed as the amounts in the two envelopes

You choose. It doesn’t matter. But stick to your choice. In the first case you gain or loose X, in the second case you gain or loose X/2.

kkwan - 02 July 2012 10:43 AM

There is no problem with steps 4 and 5 in the context of the following steps:

1. I denote by A the amount in my selected envelope.
2. The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2.
3. The other envelope may contain either 2A or A/2.
4. If A is the smaller amount the other envelope contains 2A.
5. If A is the larger amount the other envelope contains A/2.

with (X, 2X) not fixed as the amounts in the two envelopes

Again: the amounts are fixed. That you don’t know what amounts they are does not matter. And the A’s in 4 and 5 are different amounts, expressed in money. In 4 it is X, in 5 it is 2X. Or if you want: in 4 it is X/2, in 5 it is X. (But an exclusive ‘or’, one or the other but not both!). So you cannot just add them up as if they are the same. A + A is not just 2A. It is: A(4) + A(5) = 3X. So expressed in amounts the gain and loss are the same. But that is what TEP is about.
Again, you leave out the information that you originally had the choice between the 2 envelopes, i.e. two amounts, not three. If you think you do not, then show me where you use it. In 3, 4 and 5 you are definitely neglecting it: you use only a generalisation of it.

PS Do you realise that the situation is symmetrical? Make the same argumentation, and start with denoting the amount in the other envelope as A. What then? Should you switch or not? Please make the calculation for us.

You choose. It doesn’t matter. But stick to your choice. In the first case you gain or loose X, in the second case you gain or loose X/2.

The point is, the player does not know and thus cannot determine whether (X, 2X) or (X, 1/2X) are in the two envelopes. As such, he proceeds with steps 1 to 5:

1. I denote by A the amount in my selected envelope.
2. The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2.
3. The other envelope may contain either 2A or A/2.
4. If A is the smaller amount the other envelope contains 2A.
5. If A is the larger amount the other envelope contains A/2.

If A is the smaller amount, on switching, gain is A.
If A is the larger amount, on switching, loss is 1/2A.

The potential gain (100%) is twice the potential loss (50%).

Again: the amounts are fixed. That you don’t know what amounts they are does not matter. And the A’s in 4 and 5 are different amounts, expressed in money. In 4 it is X, in 5 it is 2X. Or if you want: in 4 it is X/2, in 5 it is X. (But an exclusive ‘or’, one or the other but not both!). So you cannot just add them up as if they are the same. A + A is not just 2A. It is: A(4) + A(5) = 3X. So expressed in amounts the gain and loss are the same. But that is what TEP is about.
Again, you leave out the information that you originally had the choice between the 2 envelopes, i.e. two amounts, not three. If you think you do not, then show me where you use it. In 3, 4 and 5 you are definitely neglecting it: you use only a generalisation of it.

Again, as the player does not know whether (X, 2X) or (X, 1/2X) are in the two envelopes he proceeds with steps 1 to 5 as in the above.

PS Do you realise that the situation is symmetrical? Make the same argumentation, and start with denoting the amount in the other envelope as A. What then? Should you switch or not? Please make the calculation for us.

The situation is symmetrical because you have fixed the amounts as (X, 2X) in the two envelopes.

OTOH, if either (X, 2X) or (X, 1/2X) are in the two envelopes, the situation is not symmetrical as expressed in step 7:

7. So the expected value of the money in the other envelope is

The point is, the player does not know and thus cannot determine whether (X, 2X) or (X, 1/2X) are in the two envelopes. As such, he proceeds with steps 1 to 5:

BS. For every valid TEP pair (1,2)(0.01,0.02),(10,20) etc the description (X,2X) is valid and complete. Of course you can also describe it as (X,X/2), but not for the same X. But you treat X as if it is the same in both descriptions.

kkwan - 03 July 2012 10:57 PM

The situation is symmetrical because you have fixed the amounts as (X, 2X) in the two envelopes.

Yes, after the envelopes are closed the amounts are fixed.

Do you deny that the amounts in the envelopes cannot change anymore after they are closed? Do you deny that the player knows this? Or are you going to change TEP again with Schrödinger money, or computers that change the amount after you picked your envelope?

The only other condition, as formulated in TEP, is that one amount is twice the other. That can be described by (X,2X), or (0.111Y,0.222Y), (X/2,X) or (X,X/2), every description is fine. But you must stick to the description that you have chosen.

kkwan - 03 July 2012 10:57 PM

OTOH, if either (X, 2X) or (X, 1/2X) are in the two envelopes, the situation is not symmetrical as expressed in step 7:

Step 7 is much too late. It starts with 1: “Denote the amount in the other envelope as B”. Now make the argumentation, derive the expected amount of the envelope you have in terms of B.

And you did not answer my question: where do you use the information that there were two envelopes with just two amounts in your argumentation?

BS. For every valid TEP pair (1,2)(0.01,0.02),(10,20) etc the description (X,2X) is valid and complete. Of course you can also describe it as (X,X/2), but not for the same X. But you treat X as if it is the same in both descriptions.

(X, 2X) is not equivalent to (X, 1/2X) for any finite value of X. Clearly, X must be the same for both the descriptions (X, 2X) and (X, 1/2X), to be consistent.

As such, your description of (X, 2X) the other way round is merely (2X, X) and not (X, 1/2X).

Yes, after the envelopes are closed the amounts are fixed.

Fixed as what? (X, 2X) or (X, 1/2X)? The player does know what are actually the amounts in the two envelopes. He does not even know whether the amount in the selected envelope is either the smaller or the larger amount.

Do you deny that the amounts in the envelopes cannot change anymore after they are closed? Do you deny that the player knows this? Or are you going to change TEP again with Schrödinger money, or computers that change the amount after you picked your envelope?

The only other condition, as formulated in TEP, is that one amount is twice the other. That can be described by (X,2X), or (0.111Y,0.222Y), (X/2,X) or (X,X/2), every description is fine. But you must stick to the description that you have chosen.

The amounts in the two envelopes do not change, but the player does not know what are the amounts and whether the amount in the selected envelope is the smaller or the larger amount.

The amounts are either (X, 2X) or (X, 1/2X) in the two envelopes. You have omitted (X, 1/2X) in your description of the amounts in the two envelopes.

You have shifted the value of X to make (X, 2X) equivalent to (X, 1/2X) which is inconsistent as to be consistent (X, 2X) is equivalent to (2X, X).

Step 7 is much too late. It starts with 1: “Denote the amount in the other envelope as B”. Now make the argumentation, derive the expected amount of the envelope you have in terms of B.

No, it is not “much too late” (whatever it means). Step 7 is derived from steps 1 to 6.

But, step 1 is:

1. I denote by A the amount in my selected envelope.

This is not “It starts with 1: “Denote the amount in the other envelope as B”” which is what you wrote.

And you did not answer my question: where do you use the information that there were two envelopes with just two amounts in your argumentation?

That either (X, 2X) or (X, 1/2X) are in the two envelopes means there are only two envelopes with two amounts. It is just that there are two possibilities to consider instead of one.

(X, 2X) is not equivalent to (X, 1/2X) for any finite value of X. Clearly, X must be the same for both the descriptions (X, 2X) and (X, 1/2X), to be consistent.

Yep. Assume X=10, then (X,2X)=(10,20) and (X,X/2)=(10,5). But you know that only one of the two pairs is in the envelopes. If it is (10,20) you gain or loose 10, and if it is (10,5) you gain or loose 5. That you do not know which amounts are in the envelopes does not matter.

kkwan - 04 July 2012 07:28 PM

Fixed as what? (X, 2X) or (X, 1/2X)? The player does know what are actually the amounts in the two envelopes. He does not even know whether the amount in the selected envelope is either the smaller or the larger amount.

Fixed as two, just two, amounts. Do you think that choosing different descriptions changes the amounts in the envelopes?

kkwan - 04 July 2012 07:28 PM

The amounts in the two envelopes do not change, but the player does not know what are the amounts and whether the amount in the selected envelope is the smaller or the larger amount.

The amounts are either (X, 2X) or (X, 1/2X) in the two envelopes. You have omitted (X, 1/2X) in your description of the amounts in the two envelopes.

You have shifted the value of X to make (X, 2X) equivalent to (X, 1/2X) which is inconsistent as to be consistent (X, 2X) is equivalent to (2X, X).

What you do is shifting X. If the TEP pair is (10,20), then X is 10 or 20. That is not a ‘value’ you can calculate with.

kkwan - 04 July 2012 07:28 PM

But, step 1 is:

1. I denote by A the amount in my selected envelope.

This is not “It starts with 1: “Denote the amount in the other envelope as B”” which is what you wrote.

You seem to miss the understanding what I mean. I said: solve TEP by starting with the other envelope, i.e. start with “Denote the amount in the other envelope as B”. What result do you get? And if it is the same, how do you explain this? Or are you simply refusing again?

kkwan - 04 July 2012 07:28 PM

That either (X, 2X) or (X, 1/2X) are in the two envelopes means there are only two envelopes with two amounts. It is just that there are two possibilities to consider instead of one.

The essence of having the choice between two objects is that if you have picked one, the other one is fixed as being the other one, not another one.
Example: If I have a red piece of paper and a green one, then:
4. If I pick the red one, the other is green.
5. If I pick the green one, the other is red.

I am neglecting information when I would state:
5a. If I pick the green one, then the other is not green. (Oh, so it could be orange!).

In each of 4 and 5 there is only one other combination. 5a is true, but it does not contain all information you have.

So once you picked one envelope, there is only one possibility for the other one. That means you must only consider 2 possibilities:

4b. If you pick X, the other has 2X
5b. if you pick 2X, the other has X.

That means, with 4b and 5b we describe the two excluding possibilities, with each the only single possibility, given what you picked. And that is what you leave out. By equating X(4b) with 2X(5b) you create a possibility that simply does not exist.

Again, your formula applies when, after you picked your envelope, the amount of the other envelope is twice or half of the amount you picked with a chance of 1/2 each (that were your trials with Schrödinger money or with a computer that does not fix the amount of the second envelope). For short, you left out that your choice exclusively determines the amount in the other envelope, which is correctly, i.e. completely pictured in 4b and 5b.

Yep. Assume X=10, then (X,2X)=(10,20) and (X,X/2)=(10,5). But you know that only one of the two pairs is in the envelopes. If it is (10,20) you gain or loose 10, and if it is (10,5) you gain or loose 5. That you do not know which amounts are in the envelopes does not matter.

The problem is, neither you nor the player knows what is actually in the two envelopes. It could be either (X, 2X) or (X, 1/2X). As such, it is reasonable to proceed from steps 1 to 7.

Fixed as two, just two, amounts. Do you think that choosing different descriptions changes the amounts in the envelopes?

Which two amounts? Are they (X, 2X) or (X, 1/2X)? Neither you nor the player knows with certainty.

What you do is shifting X. If the TEP pair is (10,20), then X is 10 or 20. That is not a ‘value’ you can calculate with.

To be consistent, if X is 10 in (X, 2X), then it is also 10 in (2X, X) or (X, 1/2X).

You seem to miss the understanding what I mean. I said: solve TEP by starting with the other envelope, i.e. start with “Denote the amount in the other envelope as B”. What result do you get? And if it is the same, how do you explain this? Or are you simply refusing again?

That is not what you meant in your post 1376. What you are referring to is:

9. After the switch, I can denote that content by B and reason in exactly the same manner as above.
10. I will conclude that the most rational thing to do is to swap back again.
11. To be rational, I will thus end up swapping envelopes indefinitely.
12. As it seems more rational to open just any envelope than to swap indefinitely, we have a contradiction.

The puzzle: The puzzle is to find the flaw in the very compelling line of reasoning above.

So, where is the flaw?

The essence of having the choice between two objects is that if you have picked one, the other one is fixed as being the other one, not another one.
Example: If I have a red piece of paper and a green one, then:
4. If I pick the red one, the other is green.
5. If I pick the green one, the other is red.

I am neglecting information when I would state:
5a. If I pick the green one, then the other is not green. (Oh, so it could be orange!).

In each of 4 and 5 there is only one other combination. 5a is true, but it does not contain all information you have.

So once you picked one envelope, there is only one possibility for the other one. That means you must only consider 2 possibilities:

4b. If you pick X, the other has 2X
5b. if you pick 2X, the other has X.

That means, with 4b and 5b we describe the two excluding possibilities, with each the only single possibility, given what you picked. And that is what you leave out. By equating X(4b) with 2X(5b) you create a possibility that simply does not exist.

With two indistinguishable envelopes with two unknown amounts, you don’t know which is which. Neither do you know whether the selected envelope contains the smaller or the larger amount. As such, it is reasonable to proceed as follows:

1. I denote by A the amount in my selected envelope.
2. The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2.
3. The other envelope may contain either 2A or A/2.
4. If A is the smaller amount the other envelope contains 2A.
5. If A is the larger amount the other envelope contains A/2.

Steps 1 to 5 are valid in the context of the TEP.

Again, your formula applies when, after you picked your envelope, the amount of the other envelope is twice or half of the amount you picked with a chance of 1/2 each (that were your trials with Schrödinger money or with a computer that does not fix the amount of the second envelope). For short, you left out that your choice exclusively determines the amount in the other envelope, which is correctly, i.e. completely pictured in 4b and 5b.

Your 4b and 5b is wrt to fixing the amounts as (X, 2X) in the two envelopes.

If either (X, 2X) or (X, 1/2X) are in the two envelopes as per the TEP, there is no problem with:

4. If A is the smaller amount the other envelope contains 2A.
5. If A is the larger amount the other envelope contains A/2.

Fixed as two, just two, amounts. Do you think that choosing different descriptions changes the amounts in the envelopes?

Which two amounts? Are they (X, 2X) or (X, 1/2X)? Neither you nor the player knows with certainty.

So your answer is ‘yes’. You think that choosing different descriptions changes the amounts in the envelopes. In fact you have no idea what amounts are in the envelopes, so your saying there are two possibilities is silly. Is (44,22) the same as (X,X/2)? Or is it the same as (2X,X)? Is it the same as (22,44)? (X,2X)? Or (44X,22X)? Or (4X,2X)? (7X,3.5X)? What does it matter that you use two different descriptions for the same fact? As long as it expresses “one envelope contains twice the other” every description is fine. And even the description “one envelope contains half of the other” is the same, you must only ‘exchange’ one envelope with the other. But that is no problem, they are identical, remember.

kkwan - 06 July 2012 08:24 PM

What you do is shifting X. If the TEP pair is (10,20), then X is 10 or 20. That is not a ‘value’ you can calculate with.

To be consistent, if X is 10 in (X, 2X), then it is also 10 in (2X, X) or (X, 1/2X).

You do not deny what I am saying. Say there are (10,20) in the envelopes. What is then your A, denoting the envelope that you pick first? It is 10 or 20.

kkwan - 06 July 2012 08:24 PM

You seem to miss the understanding what I mean. I said: solve TEP by starting with the other envelope, i.e. start with “Denote the amount in the other envelope as B”. What result do you get? And if it is the same, how do you explain this? Or are you simply refusing again?

That is not what you meant in your post 1376.

No, that is exactly what I meant. And I say it again: start your argumentation with denoting the amount in the other envelope as B. It is obvious that the conclusion then is that you should stick to your envelope. You are avoiding to answer my question.

kkwan - 06 July 2012 08:24 PM

With two indistinguishable envelopes with two unknown amounts, you don’t know which is which. Neither do you know whether the selected envelope contains the smaller or the larger amount.

Until here it is correct. Notice the words I made bold. There are only two possibilities, and you shift that fact under the carpet. You are not using the complete information you have available.

kkwan - 06 July 2012 08:24 PM

Your 4b and 5b is wrt to fixing the amounts as (X, 2X) in the two envelopes.

I fix the fact given in the description of TEP: that there are only two amounts. That you have no idea what amounts does not matter, but there are only two amounts. You calculate with three. You leave out some information that you have.