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The two envelopes problem
 Posted: 17 October 2012 06:29 PM [ Ignore ]   [ # 1501 ]
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GdB - 16 October 2012 10:23 PM

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.

Not so.

Steps 9, 10 and 11 just say that because the situations are symmetrical, we should switch indefinitely.

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.

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 Posted: 17 October 2012 06:48 PM [ Ignore ]   [ # 1502 ]
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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. 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…

Of course, the total amount does not change upon switching, but it is either 3X or 3/2X.

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

X + Y = T

T is not fixed as it could be either 3X or 3/2X.

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

So, either X + 2X = 3X or X + 1/2X = 3/2X

for any finite value of X.

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 Posted: 17 October 2012 07:03 PM [ Ignore ]   [ # 1503 ]
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StephenLawrence - 16 October 2012 11:16 PM

We know Kkwan doesn’t say that.

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.

Exactly, Stephen.

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 Posted: 17 October 2012 07:48 PM [ Ignore ]   [ # 1504 ]
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StephenLawrence - 16 October 2012 11:21 PM

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.

The fundamental nature of the TEP is not dependent on knowledge of the two amounts.

However, with total ignorance of the amounts in the two envelopes, we must consider ALL possible situations which are:

Situation 1

(X, 2X) are in the two envelopes

Situation 2

(X, 1/2X) are in the two envelopes

Why must we?

1. Situations 1 and 2 are not synonymous for any finite value of X.

2. The outcomes on switching once or any odd number of times are different.

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 Posted: 17 October 2012 10:51 PM [ Ignore ]   [ # 1505 ]
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kkwan - 17 October 2012 06:48 PM

T is not fixed as it could be either 3X or 3/2X.

T is fixed, it does not change from 3X to 3X/2.

kkwan - 17 October 2012 06:48 PM

Of course, the total amount does not change upon switching, but it is either 3X or 3/2X.

OK.

Assume the total is 3X.
Then, to conform to the TEP condition, your envelope contains X and 2X. You pick one of the envelopes. If you pick X first, then on switching you gain X. If you pick 2X first, then on switching you loose X. So switching will not help.

Now assume the total is 3X/2.
Then, to conform to the TEP condition, your envelope contains X and X/2. You pick one of the envelopes. If you pick X first, then on switching you loose X/2. If you pick X/2 first, then on switching you gain X/2. So switching will not help.

So in both situations, switching does not help. So if you take the total as an amount that does not change because of you switching envelopes, you see that switching does not give you any advantage. You are assuming 2 different totals, depending on which envelope you pick first.

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 Posted: 17 October 2012 10:54 PM [ Ignore ]   [ # 1506 ]
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kkwan - 17 October 2012 06:29 PM
GdB - 16 October 2012 10:23 PM

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.

Not so.

Steps 9, 10 and 11 just say that because the situations are symmetrical, we should switch indefinitely.

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.

10 says nothing about that you must switch an odd number of times. It says you should always swap.

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 Posted: 24 October 2012 06:15 AM [ Ignore ]   [ # 1507 ]
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GdB - 17 October 2012 10:51 PM

T is fixed, it does not change from 3X to 3X/2.

T does not change but it is not fixed as it is either 3X or 3/2X.

Assume the total is 3X.
Then, to conform to the TEP condition, your envelope contains X and 2X. You pick one of the envelopes. If you pick X first, then on switching you gain X. If you pick 2X first, then on switching you loose X. So switching will not help.

Now assume the total is 3X/2.
Then, to conform to the TEP condition, your envelope contains X and X/2. You pick one of the envelopes. If you pick X first, then on switching you loose X/2. If you pick X/2 first, then on switching you gain X/2. So switching will not help.

So in both situations, switching does not help. So if you take the total as an amount that does not change because of you switching envelopes, you see that switching does not give you any advantage. You are assuming 2 different totals, depending on which envelope you pick first.

You can neither assume the total amounts are 3X nor 3/2X as you don’t know.

All you can say is either (X, 2X) or (X, 1/2X) are in the envelopes i.e. either 3X or 3/2X are the total amounts.

As such, if the amount in the selected envelope is denoted as A and as A could be the smaller or the larger amount, therefore either (A, 2A) or (A, 1/2A) are in the two envelopes.

Hence, if you switched, you could either gain A or lose 1/2A.

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 Posted: 24 October 2012 06:54 AM [ Ignore ]   [ # 1508 ]
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GdB - 17 October 2012 10:54 PM

10 says nothing about that you must switch an odd number of times. It says you should always swap.

Neither does 10 say you must switch twice or any even number number of times. It only says you should switch back.

However, switching twice or any even number of times is equivalent to not switching at all.

So, if you do switch, you should switch once or any odd number of times.

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 Posted: 24 October 2012 07:26 AM [ Ignore ]   [ # 1509 ]
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kkwan - 24 October 2012 06:15 AM

You can neither assume the total amounts are 3X nor 3/2X as you don’t know.

All you can say is either (X, 2X) or (X, 1/2X) are in the envelopes i.e. either 3X or 3/2X are the total amounts.

Let me get this straight: you know either (X, 2X) or (X, X/2) is in the envelopes, but it is not allowed to treat both case separately?

I have proven that:
- If there is (X, 2X)  in the envelopes, then switching is no use
- If there is (X, X/2)  in the envelopes, then switching is no use

So yes, there is either (X, 2X) or (X, X/2) in the envelopes: and in both cases switching is no use.

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 Posted: 24 October 2012 07:28 AM [ Ignore ]   [ # 1510 ]
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kkwan - 24 October 2012 06:54 AM

So, if you do switch, you should switch once or any odd number of times.

And that is just what 10 does not say. 10 implies one should always swap. And of course that is irrational, as you are starting to see.

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 Posted: 24 October 2012 07:43 AM [ Ignore ]   [ # 1511 ]
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GdB - 24 October 2012 07:28 AM
kkwan - 24 October 2012 06:54 AM

So, if you do switch, you should switch once or any odd number of times.

And that is just what 10 does not say. 10 implies one should always swap. And of course that is irrational, as you are starting to see.

That is correct.

Stephen

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 Posted: 24 October 2012 07:47 AM [ Ignore ]   [ # 1512 ]
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kkwan - 17 October 2012 07:03 PM
StephenLawrence - 16 October 2012 11:16 PM

We know Kkwan doesn’t say that.

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.

Exactly, Stephen.

Right, so if the odds are 50/50 of you being in either of these two situations it seems right that you should switch.

So the problem is how to overcome this. I’m sure it isn’t right that you should switch but I’m not sure GdB and others have managed to solve the problem, as they suppose. (but perhaps they have and I’m just not getting it)

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 Posted: 24 October 2012 04:40 PM [ Ignore ]   [ # 1513 ]
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There has to be a fixed parameter that is valid

There is a 50/50 chance of picking,

system I
a = 10
b = 5

Or

system II
a = 5
b = 10

no change in outcome, either way

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 Posted: 24 October 2012 08:47 PM [ Ignore ]   [ # 1514 ]
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GdB - 24 October 2012 07:26 AM

Let me get this straight: you know either (X, 2X) or (X, X/2) is in the envelopes, but it is not allowed to treat both case separately?

You cannot treat both situations separately as both are equally possible.

I have proven that:
- If there is (X, 2X)  in the envelopes, then switching is no use
- If there is (X, X/2)  in the envelopes, then switching is no use

So yes, there is either (X, 2X) or (X, X/2) in the envelopes: and in both cases switching is no use.

Not so for the same rationale as above.

As such, if the amount in the selected envelope is denoted as A and as A could be the smaller or the larger amount, therefore either (A, 2A) or (A, 1/2A) are in the two envelopes.

Hence, if you switched, you could either gain A or lose 1/2A.

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 Posted: 24 October 2012 08:59 PM [ Ignore ]   [ # 1515 ]
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GdB - 24 October 2012 07:28 AM

And that is just what 10 does not say. 10 implies one should always swap. And of course that is irrational, as you are starting to see.

Since switching twice or any even number of times is equivalent to not switching, you should switch once or any odd number of times.

Of course, 11 is logically absurd.

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