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The two envelopes problem
Posted: 27 January 2013 12:30 AM   [ Ignore ]   [ # 1966 ]
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kkwan - 26 January 2013 02:37 PM
GdB - 15 January 2013 11:03 AM

To be short, kkwan, you are not able to give mathematical reasons for why I am wrong. You do not use your (pejorative) remarks in mathematical reasoning, that shows why I am wrong. You only state that I am wrong. For a ‘TEP specialist’ as you are it should be easy to derive from the TEP definition why my argument of making a countable set of all TEP games (not pairs!) is wrong. Why don’t just show us? Show us how assuming an upper limit influences the result of TEP games. Show us why, when I start with the smallest possible amount, this changes the outcome of TEP.

You can’t. You will avoid again, and give a load of derogatory sentences, without rigidly demonstrating why my argumentation is wrong. You have nothing else on offer, kkwan.

The TEP is not just a mathematical problem per se as the given information (that one envelope has twice the amount in the other) is insufficient to resolve the TEP mathematically as that information implies there are two and only two possibilities: either (X, 2X) or (X, 1/2X) are in the two envelopes and we are totally ignorant which is actual.

Thus, to try and resolve the TEP with one and only one possibility (X, 2X) or with limiting cases as you do, contravenes the fundamental nature of the TEP, begs the question and is clearly misguided and wrong.

QED

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Posted: 27 January 2013 03:00 AM   [ Ignore ]   [ # 1967 ]
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kkwan - 26 January 2013 02:59 PM
StephenLawrence - 15 January 2013 12:20 PM

The problem is whatever the TEP- universe, the result will be just the same. This is because for every possible gain there will always be the same possible loss with equal probability.

So if $10, $20 is one of the possible pairs you can gain $10 if you have the smaller amount and lose $10 if you have the larger amount.

However you set the TEP up you can’t avoid this because all you can do is add more and more possible pairs.

If you believe you can avoid this it’s up to you to give an example.

If ($10, $20) is one possible pair, then we must also consider ($10, $5) as the other possible pair.


As such, with A denoted as the amount in the selected envelope and as A is either the smaller or the larger amount:

1. If we are in (A, 2A), by switching gain is A

2. If we are in (A, 1/2A), by switching loss is 1/2A.

As the gain is twice the loss, it is beneficial to switch.

This is equivalent to A as $10 in either ($10, $20) or ($10, $5) for the same outcomes.

Your description is incomplete because it leaves out that A could be $5 and could be $20.

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Posted: 31 January 2013 04:40 AM   [ Ignore ]   [ # 1968 ]
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God, you just reminded me what OCD is like…  I’m not really great with mathematics, but the first issue I have is with line 8 and maybe the formula before it.

8. This is greater than A, so I gain on average by swapping.

How is it logical to consider the average of the two amounts?  If you only choose one envelope, and gain one amount, then isn’t the average irrelevant?

And it seems to me there is a certain factor of discomfort missing from the reasoning.  Even if the person choosing between the envelopes cannot acknowledge that he or she is stuck in a loop, wouldn’t a biological need of some kind force them to at some point make a choice.

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Posted: 05 February 2013 05:26 PM   [ Ignore ]   [ # 1969 ]
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GdB - 27 January 2013 12:30 AM

QED

LOL

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Posted: 05 February 2013 05:53 PM   [ Ignore ]   [ # 1970 ]
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StephenLawrence - 27 January 2013 03:00 AM

Your description is incomplete because it leaves out that A could be $5 and could be $20.

1. If A is $5, then either ($5, $10) or ($5, $2.5) are in the two envelopes.

2. If A is $20, then either ($20, $40) or ($20, $10) are in the two envelopes.

However, your query was ($10, $20) as one of the possible pairs and as such, ($10, $5) is the other possible pair with A as $10.

OTOH, if ($20, $10) is one of the possible pairs, then we have situation 2 with A as $20.

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Posted: 05 February 2013 06:23 PM   [ Ignore ]   [ # 1971 ]
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TheHatredOfRainbows - 31 January 2013 04:40 AM

God, you just reminded me what OCD is like…  I’m not really great with mathematics, but the first issue I have is with line 8 and maybe the formula before it.

8. This is greater than A, so I gain on average by swapping.

How is it logical to consider the average of the two amounts?  If you only choose one envelope, and gain one amount, then isn’t the average irrelevant?

And it seems to me there is a certain factor of discomfort missing from the reasoning.  Even if the person choosing between the envelopes cannot acknowledge that he or she is stuck in a loop, wouldn’t a biological need of some kind force them to at some point make a choice.

What appears initially to be elementary turns out not to be so? grin

Step 8 refers to the average of many repetitions of the TEP.

And what might that kind force be to get out of the endless loop?

OTOH, switching once or any odd number of times has a different outcome.

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Posted: 07 February 2013 06:38 AM   [ Ignore ]   [ # 1972 ]
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kkwan - 05 February 2013 05:53 PM
StephenLawrence - 27 January 2013 03:00 AM

Your description is incomplete because it leaves out that A could be $5 and could be $20.

1. If A is $5, then either ($5, $10) or ($5, $2.5) are in the two envelopes.

No, in the example this comes from the possible pairs are $5,$10 and $10, $20.

In that example A could be $5.

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Posted: 01 March 2013 08:09 PM   [ Ignore ]   [ # 1973 ]
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StephenLawrence - 07 February 2013 06:38 AM

No, in the example this comes from the possible pairs are $5,$10 and $10, $20.

In that example A could be $5.

Not quite so, as $5 is not in ($10, $20).

Remember, A is denoted as the amount in the selected envelope.

As such, A must be in both possible pairs.

Therefore, A cannot be $5 in your example because:

1. If A is $5, then either ($5, $10) or ($5, $2.5) are in the two envelopes.

2. If the two possible pairs are ($10, $20) or ($10, $5) then A is $10, not $5.

Generally, A must be in both (A, 2A) or (A, 1/2A) whereby A is the smaller amount in (A, 2A) or the larger amount in (A, 1/2A).

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Posted: 02 March 2013 09:43 AM   [ Ignore ]   [ # 1974 ]
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kkwan - 01 March 2013 08:09 PM

As such, A must be in both possible pairs.

A might be $5 and the two possible pairs might be $5,10 and $20,$40 so that is not true.

Also to save us wasting more time arguing over the above point let’s use an example in which there are 100 possible pairs. A can’t be in all of them.

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Posted: 02 March 2013 12:52 PM   [ Ignore ]   [ # 1975 ]
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StephenLawrence - 02 March 2013 09:43 AM

A might be $5 and the two possible pairs might be $5,10 and $20,$40 so that is not true.

With A denoted as the amount in the selected envelope:

1. If A is $5, then ($20, $40) is not a possible pair as $5 is not in ($20, $40).

2. If A is $5, then there are two and only two possible pairs i.e. ($5, $10) or ($5, $2.5).

Also to save us wasting more time arguing over the above point let’s use an example in which there are 100 possible pairs. A can’t be in all of them.

There are two and only two possible pairs for any finite value of A, i.e. (A, 2A) or (A, 1/2A).

For instance, if A is $10 then ($10, $20) or ($10, $5) are the two and only two possible pairs. There cannot be more pairs than that for any specific finite value of A.

So, if there are 100 possible pairs, there must be 50 distinct values of A of which for any specific value of A there are two and only two possible pairs (A, 2A) or (A, 1/2A).

.

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