It means that for any finite value of X, both (X, 2X) or (X, 1/2X) has defined distinct amounts.

For instance, if X is 10, then either (10, 20) or (10, 5) are in the two envelopes and they are distinct and not equivalent amounts.

This is true for any finite value of X.

Thus, (X, 2X) or (X, 1/2X) are distinct and not equivalent.

Since both (X, 2X) or (X, 1/2X) satisfy the requirement in the TEP that “one envelope contains twice the amount in the other” and they are distinct and not equivalent for any finite value of X, it follows that there are two and only two possibilities instead of one and only one possibility.

Therefore, with total ignorance of the actual amounts in the two envelopes, we must consider both possibilities instead of one.

The problem is ignorance of the amounts cannot make a difference because the puzzle is just the same whether we know the two amounts or not.

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.

But hey, that’s funny… what if somebody comes in after your first switch, and he is offered the same choice? It is still true that one envelope contains twice the amount of the other! Then he starts with B, and he concludes he should switch…

It is pointless to switch twice or any even number of times, but it is not so for switching once or any odd number of times.

The situations are symmetrical.

Hence, the rationale of steps 9, 10 and 11.

And is step 12 more rational when there is potential gain/loss (100%/50%) with switching once or any odd number of times whereas there is no gain/loss by just opening any envelope?

The two possibilities are (X, 2X) and (2X, X), but you are right you do not know which we are in.

Not so. The two possible events are either (X, 2X) or (X, 1/2X).

The rest is inconsistent representation that creates an incorrect EV for switching. Your representation creates an implicit and incorrect assumption that the total money in the envelopes is higher if you are holding the lower one (3X versus 3/2X). This implicit fact makes it so that you are biased toward assuming you are holding the lower one at the start, but when you consider it closely it doesn’t make any sense for the total money in the envelopes to change based on whether you are holding the lower. That assumption comes with defining X to be the amount of money you are holding, but that makes X incomparable between the two possibilities, and when you try to rectify the two possibilities into one outcome space via calculating the EV you treat the X as representing the same quantity. To get around that you pick a consistent value of X between the two possibilities that holds the total sum of money in the envelopes fixed. One consistent value is that there is a lower envelope and X is it.

If either (X, 2X) or (X, 1/2X) are in the two envelopes, then either 3X or 3/2X are the total amounts in the two envelopes which is consistent with denoting the amount in the selected envelope as A and A being either the smaller or the larger amount.

You can take this problem to an extreme by saying the possibilities are (50000X, 100000X) versus (X, 1/2X). Now such a person would be forced to assume they had the lower envelope because the lost possibility of getting 50000X greatly trumps avoiding losing the 1/2X for acting as if they had the larger envelope. See the problem?

Not so. Either (50000X, 100000X) or (50000X, 25000X) are in the two envelopes.

As the player does not know whether she has the smaller or the larger amount, she cannot assume she has the smaller amount and as such it is synonymous to the above.

We have had the argument of the changing total amount already a long time ago. kkwan has decided not to give in, that’s all. He knows he is wrong, but will never confess that. In fact the first response on kkwan’s OP already answered the whole thing.

With either (X, 2X) or (X, 1/2X) in the two envelopes, the total amounts are either 3X or 3/2X.

The problem is ignorance of the amounts cannot make a difference because the puzzle is just the same whether we know the two amounts or not.

So you’re conclusion doesn’t follow.

It does, because with total ignorance of the amounts in the two envelopes, we must consider both possibilities: either (X, 2X) or (X, 1/2X) and the total amounts as either 3X or 3/2X (which are not equivalent for any finite value of X).

We cannot assume (X, 2X) are in the two envelopes as GdB does which simplifies the TEP but is misleading and incomplete.

Thus, with (X, 2X) fixed in the two envelopes:

1. If the selected envelope contains X, the other envelope contains 2X.

2. If the selected envelope contains 2X, the other envelope contains X.

3. The total amount is fixed as 3X.

However, with either (X, 2X) or (X, 1/2X) in the two envelopes:

1. If the selected envelope contains X, the other envelope contains either 2X or 1/2X depending on whether X is the smaller or the larger amount.

It is pointless to switch twice or any even number of times, but it is not so for switching once or any odd number of times.

The situations are symmetrical.

Hence, the rationale of steps 9, 10 and 11.

And is step 12 more rational when there is potential gain/loss (100%/50%) with switching once or any odd number of times whereas there is no gain/loss by just opening any envelope?

This is just plain nonsense, kkwan. Steps 9, 10, 11 say your chance of gain increases on every switch. And my argument tells you that what is an even switch for one, can be an odd switch for somebody who comes in later, under the same valid description ‘one envelope contains twice the amount of the other’. You have not reacted on these arguments at all.

Concerning the other argument strain: the sum of the amounts in the envelopes does not change when you switch amounts. Agree, kkwan? It does not matter that you don’t know what the amount is, but you know it cannot change. Agree? If you do not agree with this, we all know you have left logical ground…

Call the total amount T, then with the two amounts calling X and Y we have:

X + Y = T

Agree?

Now in TEP I know a little more: one envelope contains twice the amount of the other.
Agree?

Now you say:

X + 2X = T, and
(X/2) + X = T for the same X.

Proof this can be the case for fixed values of X and T, and both unequal zero (is a TEP condition: ‘each of which contains a positive sum of money’).

X is the amount in his envelope. If it is $20 dollars the other could contain $40 or $10. So he is dealing with two different total amounts $30 and $60 and saying the total might be either of these. He is then saying he gains more by switching if the total is $60 than he loses if the total is $30.

The problem is ignorance of the amounts cannot make a difference because the puzzle is just the same whether we know the two amounts or not.

So you’re conclusion doesn’t follow.

It does, because with total ignorance of the amounts in the two envelopes, we must consider both possibilities:

The trouble is you need to say why we must because there is good reason to believe there is no need to. That reason is knowledge of the two amounts makes no difference.

So it is not true to say we must.

It’s still interesting to try to understand what is going wrong if we do though.

X is the amount in his envelope. If it is $20 dollars the other could contain $40 or $10. So he is dealing with two different total amounts $30 and $60 and saying the total might be either of these. He is then saying he gains more by switching if the total is $60 than he loses if the total is $30.

And? Does that fit that, whatever the total amount is, it does not change?

The only way out is denying that:

GdB - 16 October 2012 10:44 PM

Concerning the other argument strain: the sum of the amounts in the envelopes does not change when you switch amounts.

Say, before playing the game, I shuffle the envelopes in such a way that I really have no idea which is which. That means I switched them in ways I could not follow. During this shuffling, did the total amount change? Does it change when I choose one envelope? Does it change when I switch it once? So whatever envelope I choose, the total amount stays the same. And that kkwan cannot take into account when stating that the envelopes can contain (X,2X) or (X,X/2). I do not use any knowledge about the total amount, it can be anything, $30 or $60 or anything, but whatever envelope I choose, the total amount stays the same.

X is the amount in his envelope. If it is $20 dollars the other could contain $40 or $10. So he is dealing with two different total amounts $30 and $60 and saying the total might be either of these. He is then saying he gains more by switching if the total is $60 than he loses if the total is $30.

And? Does that fit that, whatever the total amount is, it does not change?

Does what fit what?

Of course Kkwan’s argument isn’t based on the amounts changing. The argument is based on it being true that he might be in either of these situations.

I really don’t know whether you’ve managed to say what is wrong with doing it Kkwan’s way.

Say I look in my envelope and have $20 dollars. What I need to know is why I should not switch.

I assume you mean the right thing, but saying it wrong: it is no use to switch, it doesn’t matter.

Yep, you can see why it looks like switching is advantage. It’s seeing why that is an illusion that solves the problem.

kkwan’s equation is meant to cover the situation without looking.

That is not important because looking makes no difference in my example.

The puzzle is looking makes no difference (at least in some situations) and in these situations it appears that if you look it’s advantages to switch and therefore it also is if you don’t look.

I don’t know if you’ve just side stepped the puzzle or if you have an answer to this.