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The two envelopes problem
Posted: 19 July 2012 11:16 PM   [ Ignore ]   [ # 1411 ]
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Write4U - 19 July 2012 03:54 PM

Allow me a little observation.

I am prepared to make an exception. OK,yes. wink

Write4U - 19 July 2012 03:54 PM

There are always only two amounts. (X, 2X).  No matter which we pick this TEP holds true and when we pick we pick either the smallest or the largets. In order to arrive at a second TEP we must first switch the envelopes in order to create an appearance of (X, 1/2X), by making X into 1/2 X and making 2X into X. But that makes absolutely no difference to the outcome. We either pick the envelope with the largest or the envelope with the smallest amount. By switching we do not realize a potential gain other than a 50/50 chance of losing or winning.  Relative gain cannot enter this equation.. The TEP is and always will be X, 2X.

You are perfectly right. Different descriptions will not make different outcomes, and in terms of ‘absolute money’ (which is the only thing we’re interested in, isn’t it?) there is no difference.

But I suppose the great kkwan will soon come showing that different descriptions will make a difference, and that going from 1 to 2 is an increase of 100%, but going from 2 to 1 is decrease of only 50%...

Write4U - 19 July 2012 03:54 PM

I mentioned Schrodinger before but that was dismissed as the amounts are not imaginary and cannot be both amounts at the same time.

Yep. I hope you have no problem with it?

PS I ordered me a T-Shirt with attached print.

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Posted: 22 July 2012 07:12 PM   [ Ignore ]   [ # 1412 ]
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GdB - 17 July 2012 10:44 PM

If you have two envelopes with two different amounts in front of you, you know there are only two amounts. If you describe them as (X, 2X) or (X, 1/2X) does not change that fact. ‘X’ is an unknown and could be any amount, so as Stephen already rightfully said, there is no reason to give (X,X/2) a special treatment, because it means nothing. Your juggling with (X, 2X) or (X, 1/2X) does not change the simple fact that there are two fixed amounts in two closed envelopes in front of you, one amount is the smallest and one is the biggest. You are not using that fact, i.e. you do not use all information available.

There are two distinct possible events, (X, 2X) or (X, 1/2X) which are not equivalent for any finite value of X, but you don’t know which event is actual. That is all the information provided in the TEP and both events satisfy it.

Compare:
1. I give you an envelope, and I have a second in my hand and say it contains twice or half the amount as in the envelope you have
with
2. I give you two envelopes, say that one contains twice the amount as the other and you may pick one envelope

You say these situations are equivalent: but in 1 the point is missing that you had the choice. And that is the point you are leaving out. If you think you don’t, then show me where 1 and 2 differ, and where you use the fact that you originally had two envelopes in your calculations. (Or will you say again that 1 is not TEP and therefore it does not interest you? If you do that then I know you are not able to show the difference.)

1 and 2 are equivalent irrespective of choice and both events, (X, 2X) or (X, 1/2X) satisfy them. Hence both possible events must be considered.

BS. X is unknown. And Y too. You are just refusing again, because the answers would show you are wrong. Answer the questions, kkwan, or admit you were wrong.

X is a variable, but (X, 2X) is considered as a known unique event if (X, 1/2X) is omitted.

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Posted: 22 July 2012 07:33 PM   [ Ignore ]   [ # 1413 ]
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Write4U - 19 July 2012 03:54 PM

Allow me a little observation. There are always only two amounts. (X, 2X).  No matter which we pick this TEP holds true and when we pick we pick either the smallest or the largets. In order to arrive at a second TEP we must first switch the envelopes in order to create an appearance of (X, 1/2X), by making X into 1/2 X and making 2X into X. But that makes absolutely no difference to the outcome. We either pick the envelope with the largest or the envelope with the smallest amount. By switching we do not realize a potential gain other than a 50/50 chance of losing or winning.  Relative gain cannot enter this equation.. The TEP is and always will be X, 2X.

There are two distinct possible events (X, 2X) or (X, 1/2X) which are not equivalent and both events satisfy the requirement of the TEP. However, we don’t know which event is actual.

As such, we cannot assume (X, 2X) must be in the two envelopes and omit (X, 1/2X).

I mentioned Schrodinger before but that was dismissed as the amounts are not imaginary and cannot be both amounts at the same time.

Precisely.

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Posted: 22 July 2012 11:10 PM   [ Ignore ]   [ # 1414 ]
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kkwan - 22 July 2012 07:12 PM

There are two distinct possible events, (X, 2X) or (X, 1/2X) which are not equivalent for any finite value of X, but you don’t know which event is actual. That is all the information provided in the TEP and both events satisfy it.

There are only 2 amounts, and if you describe them as (smallest,biggest), or (biggest,smallest) makes no difference. By calling the first amount X, you use the same X for ‘smallest amount’ and ‘biggest amount’, which in TEP is not the same amount per definition.

kkwan - 22 July 2012 07:12 PM

Compare:
1. I give you an envelope, and I have a second in my hand and say it contains twice or half the amount as in the envelope you have
with
2. I give you two envelopes, say that one contains twice the amount as the other and you may pick one envelope

You say these situations are equivalent: but in 1 the point is missing that you had the choice. And that is the point you are leaving out. If you think you don’t, then show me where 1 and 2 differ, and where you use the fact that you originally had two envelopes in your calculations. (Or will you say again that 1 is not TEP and therefore it does not interest you? If you do that then I know you are not able to show the difference.)

1 and 2 are equivalent irrespective of choice and both events, (X, 2X) or (X, 1/2X) satisfy them. Hence both possible events must be considered.

So 1 and 2 are the same? The same formula applies even if the situations differ? How do you explain this where there are only 2 amounts in TEP, but three possible amounts in situation 1?

kkwan - 22 July 2012 07:12 PM

BS. X is unknown. And Y too. You are just refusing again, because the answers would show you are wrong. Answer the questions, kkwan, or admit you were wrong.

X is a variable, but (X, 2X) is considered as a known unique event if (X, 1/2X) is omitted.

As you keep refusing to answer the questions, just one by one, I am sure that you realise that it will lead you to see that you are wrong. The only way to avoid this conclusion is by answering them, one by one.

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Posted: 25 July 2012 07:08 PM   [ Ignore ]   [ # 1415 ]
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GdB - 22 July 2012 11:10 PM

There are only 2 amounts, and if you describe them as (smallest,biggest), or (biggest,smallest) makes no difference. By calling the first amount X, you use the same X for ‘smallest amount’ and ‘biggest amount’, which in TEP is not the same amount per definition.

There are two amounts in each distinct possible event (X, 2X) or (X, 1/2X) and both events satisfy the requirement of the TEP.

So 1 and 2 are the same? The same formula applies even if the situations differ? How do you explain this where there are only 2 amounts in TEP, but three possible amounts in situation 1?

In 1, because you don’t know the amount in the given envelope, the amount in the other envelope could be twice or half of the amount in the given envelope to satisfy the requirement in the TEP “that one envelopes contains twice the amount in the other”.

In 2, “that one contains twice the amount as the other” is synonymous to 1.

Both 1 and 2 imply either (X, 2X) or (X, 1/2X) are in the two envelopes.

As you keep refusing to answer the questions, just one by one, I am sure that you realise that it will lead you to see that you are wrong. The only way to avoid this conclusion is by answering them, one by one.

All your questions assume the equivalent of (X, 2X) as known which is not so in the TEP as (X, 1/2X) is also possible.

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Posted: 25 July 2012 09:01 PM   [ Ignore ]   [ # 1416 ]
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Ye gods. O_o 1,415 posts discussing a trick question. The only correct solution is to grab both envelopes and run for it! :^D [/end thread]

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Posted: 26 July 2012 03:36 AM   [ Ignore ]   [ # 1417 ]
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kkwan - 25 July 2012 07:08 PM

There are two amounts in each distinct possible event (X, 2X) or (X, 1/2X) and both events satisfy the requirement of the TEP.

Exactly the answer to expect: from a difference in description you conclude to a difference in reality.

kkwan - 25 July 2012 07:08 PM

Both 1 and 2 imply either (X, 2X) or (X, 1/2X) are in the two envelopes.

But you left out that you had the choice between two amounts. You do not use all available information in TEP, and therefore you cannot distinguish between the 2 situations, even that they are different.

kkwan - 25 July 2012 07:08 PM

As you keep refusing to answer the questions, just one by one, I am sure that you realise that it will lead you to see that you are wrong. The only way to avoid this conclusion is by answering them, one by one.

All your questions assume the equivalent of (X, 2X) as known which is not so in the TEP as (X, 1/2X) is also possible.

These are the first 4 statements. Where do I assume there are only (X,2X) in it? Why can’t you say if you agree or not?

I have two envelopes, one with a green piece of paper, the other with a red one.

1. If I have the envelope with the green one the other contains the red one.
2. If I have the envelope with the red one the other contains the green one.

I have two envelopes with different amounts, call them X and Y:

3. If I have the envelope with X, then the other one contains Y.
4. If I have the envelope with Y, then the other one contains X.

And what is you problem when I explicitly state that there are (X,2X) in it? Does that make the questions unanswerable? I want to know your answer: do you agree with the following statements:

I have two envelopes with amounts X and 2X:

5. If I have the envelope with X, then the other one contains 2X.
6. If I have the envelope with 2X, then the other one contains X.

Edit: Layout and typos.

[ Edited: 26 July 2012 10:30 PM by GdB ]
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Posted: 26 July 2012 04:06 AM   [ Ignore ]   [ # 1418 ]
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kkwan,
There are two distinct possible events (X, 2X) or (X, 1/2X) which are not equivalent and both events satisfy the requirement of the TEP. However, we don’t know which event is actual.

As such, we cannot assume (X, 2X) must be in the two envelopes and omit (X, 1/2X).

Hummm, what if X is an indivisible number (prime number)?  2X is easy, but 1/2X may be problematic in theory. Thus a unique situation without a direct relationship can very well exist.

In this case the TEP of X, 2X holds for both situations under all tests. But the TEP for X, 1/2X does not necessarily quarantee an = exchange and is not allowed.

[ Edited: 26 July 2012 04:15 AM by Write4U ]
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Posted: 26 July 2012 04:20 AM   [ Ignore ]   [ # 1419 ]
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Write4U - 26 July 2012 04:06 AM

Hummm, what if X is an indivisible number (prime number)?  2X is easy, but 1/2X may be problematic in theory. Thus a unique situation without a direct relationship can very well exist.

Also an interesting objection. But here the great kkwan will introduce dollarcents or milli cents, or bank money… Even if you can’t put this in envelopes. And it is not just theory. If you allow dollars only, one can still play the TEP. But you have a problem with X/2 for every odd amount (you do not have to stick to prime numbers, there are not so many of them… tongue rolleye )

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Posted: 26 July 2012 04:36 AM   [ Ignore ]   [ # 1420 ]
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GdB - 26 July 2012 04:20 AM
Write4U - 26 July 2012 04:06 AM

Hummm, what if X is an indivisible number (prime number)?  2X is easy, but 1/2X may be problematic in theory. Thus a unique situation without a direct relationship can very well exist.

Also an interesting objection. But here the great kkwan will introduce dollarcents or milli cents, or bank money… Even if you can’t put this in envelopes. And it is not just theory. If you allow dollars only, one can still play the TEP. But you have a problem with X/2 for every odd amount (you do not have to stick to prime numbers, there are not so many of them… tongue rolleye )

Thanks GdB, you made me look up prime numbers… cheese
This may surprise you.
http://primes.utm.edu/mersenne/

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Posted: 26 July 2012 05:59 AM   [ Ignore ]   [ # 1421 ]
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The topic of prime numbers is the queen of number theory.

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Posted: 29 July 2012 06:25 AM   [ Ignore ]   [ # 1422 ]
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Grigory - 25 July 2012 09:01 PM

Ye gods. O_o 1,415 posts discussing a trick question. The only correct solution is to grab both envelopes and run for it! :^D [/end thread]

Cutting the Gordian Knot?  cheese

The TEP is about what is the best course of action to take in the face of total ignorance and uncertainty wrt what are actually in the two envelopes.

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Posted: 29 July 2012 07:08 AM   [ Ignore ]   [ # 1423 ]
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GdB - 26 July 2012 03:36 AM

Exactly the answer to expect: from a difference in description you conclude to a difference in reality.

Not so. (X, 2X) and (X, 1/2X) are distinct possible events, not “a difference in description”.

But you left out that you had the choice between two amounts. You do not use all available information in TEP, and therefore you cannot distinguish between the 2 situations, even that they are different.

Given a choice, you don’t know what is actually in the envelope you have chosen. Hence, 1 and 2 are synonymous notwithstanding choice.

These are the first 4 statements. Where do I assume there are only (X,2X) in it? Why can’t you say if you agree or not?

Are your statements actual facts or only merely assertions?

I have two envelopes, one with a green piece of paper, the other with a red one.

1. If I have the envelope with the green one the other contains the red one.
2. If I have the envelope with the red one the other contains the green one.

I have two envelopes with different amounts, call them X and Y:

3. If I have the envelope with X, then the other one contains Y.
4. If I have the envelope with Y, then the other one contains X.

In the TEP, it is not so obvious as in your examples. Either (X, 2X) or (X, 1/2X) are in the two envelopes but you don’t know which are actually in the two envelopes.

And what is you problem when I explicitly state that there are (X,2X) in it? Does that make the questions unanswerable? I want to know your answer: do you agree with the following statements:

I have two envelopes with amounts X and 2X:

5. If I have the envelope with X, then the other one contains 2X.
6. If I have the envelope with 2X, then the other one contains X.

How about (X, 1/2X) which you have ignored and omitted by asserting (X, 2X) are in the two envelopes without knowing what are actually in the two envelopes?

With that inherent ambiguity, if you select an envelope and it contains X, either the other envelope contains 2X or 1/2X.

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Posted: 29 July 2012 07:32 AM   [ Ignore ]   [ # 1424 ]
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Write4U - 26 July 2012 04:06 AM

Hummm, what if X is an indivisible number (prime number)?  2X is easy, but 1/2X may be problematic in theory. Thus a unique situation without a direct relationship can very well exist.

In this case the TEP of X, 2X holds for both situations under all tests. But the TEP for X, 1/2X does not necessarily quarantee an = exchange and is not allowed.

From the wiki on prime number

Among the numbers 1 to 6, the numbers 2, 3, and 5 are the prime numbers, while 1, 4, and 6 are not prime. 1 is excluded as a prime number, for reasons explained below. 2 is a prime number, since the only natural numbers dividing it are 1 and 2.

So, 2 is a prime number and what is the problem with 1/2X for any finite prime value of X?

For example, 929/2 = 464.5 which is a valid sum of money in dollars and cents.

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Posted: 29 July 2012 08:16 AM   [ Ignore ]   [ # 1425 ]
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kkwan - 29 July 2012 07:32 AM

For example, 929/2 = 464.5 which is a valid sum of money in dollars and cents.

Write4U, do you see? He does exactly as I predicted. He introduces dollar cents.

kkwan, suppose I restrict TEP to whole dollar amounts only. So I say to you that the envelopes contain just full dollars. Am I now restricted to even numbers?

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