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The two envelopes problem
 Posted: 16 September 2012 11:42 PM [ Ignore ]   [ # 1471 ]
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kkwan - 16 September 2012 07:54 PM

All your examples make the fundamental assumption that what are in the two envelopes are known which is not so in the context of the TEP.

From 3 onwards, the amounts are unknown. Your objection does not hold.

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 Posted: 16 September 2012 11:50 PM [ Ignore ]   [ # 1472 ]
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kkwan - 16 September 2012 07:31 PM

The amounts in the two envelopes are determined by the value of X.

What does that mean?

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 Posted: 22 September 2012 08:18 PM [ Ignore ]   [ # 1473 ]
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Just checking in to make sure that this thread is progressing as it should.  If any of you get tired of this thread, there’s a guy named Sisyphus who would like you to take over something he’s been working on.

Otherwise, carry on.

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 Posted: 02 October 2012 01:31 AM [ Ignore ]   [ # 1474 ]
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GdB - 16 September 2012 11:42 PM

From 3 onwards, the amounts are unknown. Your objection does not hold.

My objection is valid for all your examples.

In the context of the TEP, the amounts are unknown unlike your 1 and 2.

Your “unknown amounts” in 3 to 7 is based on the funfamental assumption that there is one and only one possibility whereas in the TEP, there are two and only two possibilities i.e. (X, 2X) or (X, 1/2X) for any finite value of X.

As such, your fundamental assumption is not justified unless you have direct acquaintance of the contents in the two envelopes which you don’t have, in the context of the TEP.

Thus, all you can say is either (X, 2X) or (X, 1/2X) are in the two envelopes because you are totally ignorant of the actual contents in the two envelopes.

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 Posted: 02 October 2012 01:48 AM [ Ignore ]   [ # 1475 ]
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So you deny that:

3. I have two envelopes, one with a piece of paper with the number X on it and one with a piece of paper with the number Y on it. I choose one. If I change, the chances of
X—> Y, so gaining Y - X,
Y—> X, so gaining X - Y, are equal.

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 Posted: 02 October 2012 02:18 AM [ Ignore ]   [ # 1476 ]
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StephenLawrence - 16 September 2012 11:50 PM

What does that mean?

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.

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 Posted: 02 October 2012 02:43 AM [ Ignore ]   [ # 1477 ]
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GdB - 02 October 2012 01:48 AM

So you deny that:

3. I have two envelopes, one with a piece of paper with the number X on it and one with a piece of paper with the number Y on it. I choose one. If I change, the chances of
X—> Y, so gaining Y - X,
Y—> X, so gaining X - Y, are equal.

That is because switching twice or an even number of times is equivalent to not switching at all.

However, if you switched once or an odd number of times, you could either gain or lose (if X is not equal to Y) as in the TEP.

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 Posted: 02 October 2012 02:45 AM [ Ignore ]   [ # 1478 ]
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TimB - 22 September 2012 08:18 PM

Just checking in to make sure that this thread is progressing as it should.  If any of you get tired of this thread, there’s a guy named Sisyphus who would like you to take over something he’s been working on.

Otherwise, carry on.

Here we have two envelopes, not one boulder.

[ Edited: 02 October 2012 03:02 AM by kkwan ]
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 Posted: 02 October 2012 04:11 AM [ Ignore ]   [ # 1479 ]
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kkwan - 02 October 2012 02:43 AM

That is because switching twice or an even number of times is equivalent to not switching at all.

And you think that in TEP after switching twice you have gained?

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 Posted: 07 October 2012 09:31 PM [ Ignore ]   [ # 1480 ]
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GdB - 02 October 2012 04:11 AM

And you think that in TEP after switching twice you have gained?

You have neither gained nor lost anything with switching twice because it is equivalent to not switching.

For instance, if you denote the amount in the selected envelope as A, switched and switched again, you now have the envelope with A.

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 Posted: 08 October 2012 12:21 AM [ Ignore ]   [ # 1481 ]
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kkwan - 07 October 2012 09:31 PM
GdB - 02 October 2012 04:11 AM

And you think that in TEP after switching twice you have gained?

You have neither gained nor lost anything with switching twice because it is equivalent to not switching.

Exactly.

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 Posted: 08 October 2012 06:09 PM [ Ignore ]   [ # 1482 ]
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GdB - 08 October 2012 12:21 AM

Exactly.

However, if you switch once or any odd number of times, there is potential gain/loss depending on whether the amount denoted as A in your selected envelope is the smaller/larger amount.

So, should you switch once or not switch at all (which is equivalent to switching twice)?

That is the crux of the TEP.

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 Posted: 08 October 2012 11:14 PM [ Ignore ]   [ # 1483 ]
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kkwan - 08 October 2012 06:09 PM

However, if you switch once or any odd number of times, there is potential gain/loss depending on whether the amount denoted as A in your selected envelope is the smaller/larger amount.

So, should you switch once or not switch at all (which is equivalent to switching twice)?

That is the crux of the TEP.

You already removed half of the TEP argument:

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…

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 Posted: 09 October 2012 07:53 PM [ Ignore ]   [ # 1484 ]
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kkwan - 02 October 2012 02:18 AM
StephenLawrence - 16 September 2012 11:50 PM

What does that mean?

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

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.

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?

[ Edited: 09 October 2012 09:25 PM by qutsemnie ]
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 Posted: 09 October 2012 10:33 PM [ Ignore ]   [ # 1485 ]
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qutsemnie,

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.

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