116 of 132
116
The two envelopes problem
Posted: 28 November 2012 04:15 AM   [ Ignore ]   [ # 1726 ]
Sr. Member
RankRankRankRankRankRankRankRankRankRank
Total Posts:  1824
Joined  2007-10-28
GdB - 28 November 2012 12:34 AM

I will. No problem.  cool grin

But you only get your gain after 100 TEP plays where you won 1/4 more than the sum of your first picks. Otherwise you pay me, let’s say, 3/2 of the sum of your first picks.
And if you do not dare to do that: why do you know you will fail?

Where is it in the TEP that the player must pay you anything at all?  smile

The TEP is about double or half upon switching. That is all there is to it, GdB.

So, if I selected an envelope and switched, I either get double or half without paying anything at all.

Are you still willing to play the TEP with me?

 Signature 

I am, therefore I think.

Profile
 
 
Posted: 28 November 2012 04:25 AM   [ Ignore ]   [ # 1727 ]
Sr. Member
RankRankRankRankRankRankRankRankRankRank
Total Posts:  1824
Joined  2007-10-28
GdB - 28 November 2012 12:39 AM

Following I know:
1. The amounts are done in the envelopes, the envelopes are closed, so the amounts are fixed
2. One amount is twice the other.
3. I pick one of these envelopes, and no other.

You do not use 1 and 3. You make yourself more ignorant then you really are.

Total ignorance means that you don’t know the actual amounts in the two envelopes and neither do you know whether the amount in the selected envelope is the smaller or the larger amount.

So, even with 1, 2 and 3, you are still clueless and totally ignorant.

 Signature 

I am, therefore I think.

Profile
 
 
Posted: 28 November 2012 05:55 AM   [ Ignore ]   [ # 1728 ]
Sr. Member
Avatar
RankRankRankRankRankRankRankRankRankRank
Total Posts:  4377
Joined  2007-08-31
kkwan - 28 November 2012 04:05 AM

Either (X, 2X) or (X, 1/2X) are valid pairs as per the TEP. They are distinct pairs for any finite value of X.

There is just one pair, kkwan. You do not know what it is, but it is one pair. To say that you cannot describe it as (X,2X) and (X,X/2) for the same X has no meaning at all.

kkwan - 28 November 2012 04:05 AM

1. Show me where you use the fact that after the envelopes are closed, the amounts are fixed. You only say the amounts are fixed, but you do not use it in your calculation. That means you use an incomplete description of the TEP situation.

The amounts are fixed but you don’t know whether they are either (X, 2X) or (X, 1/2X). It is superfluous and misleading to introduce Y when X is sufficient in the context of the TEP. Occam’s razor.

I did not introduce X and Y here. What you state here is that you do not use your knowledge that there is just one fixed pair. That is wrong. Naming the amounts differently based on which (unknown) amount in the envelope of your first choice does not mean you can have different amounts. Not using the fact that there is only one fixed pair, independent of what you know about the factual amounts, is essential for the solution, and you do not use it.

kkwan - 28 November 2012 04:05 AM

2. Do you agree that for any two amounts (X,Y), the possible gain or loss by switching them is the same, namely X - Y?

OTOH, with X and Y, unless you know the relative values of X and Y which you don’t, X-Y or Y-X makes no sense.

This means you deny that you know that (X - Y) = -1 * (Y -X) for all possible finite values of X and Y.

Bye bye, kkwan.

kkwan - 28 November 2012 04:15 AM

Where is it in the TEP that the player must pay you anything at all?  smile

Nowhere. Let’s do it differently. Let’s play just with points, not with money. I generate a table of 100 randomised TEP numbers with a computer program, and then you pick one of the numbers and then switch. You count your points together, and I count the remaining points. According to you, you should have about a 25% higher score than the remaining points. Right? Shall we organise this? I create a zip file with a password with my numbers, then you send me a zip file with your password with your switched choices, and then we reveal the passwords to each other, and we can compare.

So I would send you some like this:

Envelope 1   Envelope2
34           68
4             2
132          66
etc

You send me a list:

Envelope 1    Envelope2
X
                X
                X
etc


Where the X represents the envelope you switched to.

Then we open the zip files, and we see how many points you have.
If you gained close to 25% compared to what is in the remaining envelopes you get $1000, if it is close to break even I get $1000. OK?

If you do not trust to do that, then obviously you know that you are wrong. And how do you know?

 Signature 

GdB

“The light is on, but there is nobody at home”

Profile
 
 
Posted: 28 November 2012 06:03 AM   [ Ignore ]   [ # 1729 ]
Sr. Member
Avatar
RankRankRankRankRankRankRankRankRankRank
Total Posts:  4377
Joined  2007-08-31
kkwan - 28 November 2012 04:25 AM
GdB - 28 November 2012 12:39 AM

Following I know:
1. The amounts are done in the envelopes, the envelopes are closed, so the amounts are fixed
2. One amount is twice the other.
3. I pick one of these envelopes, and no other.

You do not use 1 and 3. You make yourself more ignorant then you really are.

Total ignorance means that you don’t know the actual amounts in the two envelopes and neither do you know whether the amount in the selected envelope is the smaller or the larger amount.

So, even with 1, 2 and 3, you are still clueless and totally ignorant.

I am? I say this is enough to prove you are wrong.

4. From the two amounts being fixed it follows that the difference between the amounts is fixed.
5. That means when I switch I loose or gain the same fixed amount.
6. So switching does not help.

I do not even need 2, because 4 till 6 is valid for all pairs of amounts, so this includes TEP pairs.

You on the other side only use 2.

 Signature 

GdB

“The light is on, but there is nobody at home”

Profile
 
 
Posted: 28 November 2012 08:00 AM   [ Ignore ]   [ # 1730 ]
Sr. Member
RankRankRankRankRankRankRankRankRankRank
Total Posts:  5939
Joined  2006-12-20
GdB - 28 November 2012 02:15 AM
StephenLawrence - 28 November 2012 01:41 AM

Asserting this is no solution.

So you think you can deal with the problem by just leaving out some information you do have? It definitely belongs to the solution, as done here. How would you put in the fact that you started with just one and fixed TEP pair?

You see that the outcomes differ, depending on taking this into account or not.

The outcome doesn’t differ depending upon taking that into account.

In your last example in which I had $20, the outcome differed depending upon if I knew the probability distribution, as in both scenarios I should switch if I knew the probability distribution and in one I did and the other I didn’t.

Stephen

Profile
 
 
Posted: 28 November 2012 08:05 AM   [ Ignore ]   [ # 1731 ]
Sr. Member
RankRankRankRankRankRankRankRankRankRank
Total Posts:  5939
Joined  2006-12-20
kkwan - 28 November 2012 03:37 AM
StephenLawrence - 28 November 2012 12:13 AM

Hi Kkwan, are you open to a solution if it’s shown to you?

If 2. were true the rest follows but 2. is false and your reasoning over 2, is fallacious.

I think I’ve pinpointed the problem.

Of course. However, any proposed solution must explicitly show me why step 2 is false.

OK, good.

Stephen

Profile
 
 
Posted: 28 November 2012 08:07 AM   [ Ignore ]   [ # 1732 ]
Sr. Member
Avatar
RankRankRankRankRankRankRankRankRankRank
Total Posts:  4377
Joined  2007-08-31
StephenLawrence - 28 November 2012 08:00 AM

In your last example in which I had $20, the outcome differed depending upon if I knew the probability distribution, as in both scenarios I should switch if I knew the probability distribution and in one I did and the other I didn’t.

No idea what example you are talking about.

And on what does your ‘probability distribution’ depend?

 Signature 

GdB

“The light is on, but there is nobody at home”

Profile
 
 
Posted: 28 November 2012 08:17 AM   [ Ignore ]   [ # 1733 ]
Sr. Member
RankRankRankRankRankRankRankRankRankRank
Total Posts:  5939
Joined  2006-12-20
kkwan - 28 November 2012 03:37 AM

What you say is:

it is false that the amount in your envelope is the smaller or larger amount with 50/50 probability.

It is false if and only if you know the relative value of A which you don’t (with total ignorance) in the context of the TEP.

OK I’ll start with an example similar to the TEP in which this wouldn’t work and apply it to the TEP at a later stage. I’ll need to ask you questions to get there.

So imagine the TEP played with two pairs of envelopes $10 and $20 and $20 and $40. You have this information but don’t know which pair is in play nor which amount you have in your envelope.

The amount in your envelope is $20.

What should you do and why in this example?

Stephen

Profile
 
 
Posted: 28 November 2012 08:18 AM   [ Ignore ]   [ # 1734 ]
Sr. Member
RankRankRankRankRankRankRankRankRankRank
Total Posts:  5939
Joined  2006-12-20
GdB - 28 November 2012 08:07 AM
StephenLawrence - 28 November 2012 08:00 AM

In your last example in which I had $20, the outcome differed depending upon if I knew the probability distribution, as in both scenarios I should switch if I knew the probability distribution and in one I did and the other I didn’t.

No idea what example you are talking about.

rolleyes

Profile
 
 
Posted: 30 November 2012 05:12 AM   [ Ignore ]   [ # 1735 ]
Sr. Member
RankRankRankRankRankRankRankRankRankRank
Total Posts:  1824
Joined  2007-10-28
GdB - 28 November 2012 05:55 AM

There is just one pair, kkwan. You do not know what it is, but it is one pair. To say that you cannot describe it as (X,2X) and (X,X/2) for the same X has no meaning at all.

There is one pair, but it could be either (X, 2X) or (X, 1/2X).

I did not introduce X and Y here. What you state here is that you do not use your knowledge that there is just one fixed pair. That is wrong. Naming the amounts differently based on which (unknown) amount in the envelope of your first choice does not mean you can have different amounts. Not using the fact that there is only one fixed pair, independent of what you know about the factual amounts, is essential for the solution, and you do not use it.

Knowing that there is just one fixed pair does not tell you whether it is either (X, 2X) or (X, 1/2X).

This means you deny that you know that (X - Y) = -1 * (Y -X) for all possible finite values of X and Y.

Why introduce Y when either (X, 2X) or (X, 1/2X) is sufficient for the TEP? 

Nowhere. Let’s do it differently.

No, let’s play the TEP as it is.

 Signature 

I am, therefore I think.

Profile
 
 
Posted: 30 November 2012 05:32 AM   [ Ignore ]   [ # 1736 ]
Sr. Member
RankRankRankRankRankRankRankRankRankRank
Total Posts:  1824
Joined  2007-10-28
GdB - 28 November 2012 06:03 AM

I am? I say this is enough to prove you are wrong.

4. From the two amounts being fixed it follows that the difference between the amounts is fixed.
5. That means when I switch I loose or gain the same fixed amount.
6. So switching does not help.

I do not even need 2, because 4 till 6 is valid for all pairs of amounts, so this includes TEP pairs.

You on the other side only use 2.

The two amounts being fixed could be either (X, 2X) or (X, 1/2X). Being totally ignorant, you don’t know whether it is (X, 2X) or (X, 1/2X).

Your 4, 5 and 6 is based on the unjustified premise that you know whether (X, 2X) or (X, 1/2X) are in the two envelopes which you don’t, in the context of the TEP.

As such, as either (X, 2X) or (X, 1/2X) are in the two envelopes, all that you can say is to denote A as the amount in the selected envelope and as A is either the smaller or the larger amount, then the amounts in the two envelopes are either (A, 2A) or (A, 1/2A).

 Signature 

I am, therefore I think.

Profile
 
 
Posted: 30 November 2012 06:06 AM   [ Ignore ]   [ # 1737 ]
Sr. Member
RankRankRankRankRankRankRankRankRankRank
Total Posts:  1824
Joined  2007-10-28
StephenLawrence - 28 November 2012 08:17 AM

OK I’ll start with an example similar to the TEP in which this wouldn’t work and apply it to the TEP at a later stage. I’ll need to ask you questions to get there.

So imagine the TEP played with two pairs of envelopes $10 and $20 and $20 and $40. You have this information but don’t know which pair is in play nor which amount you have in your envelope.

The amount in your envelope is $20.

What should you do and why in this example?

Based on the information that either ($10, $20) or (20, $40) are in the two envelopes and with $20 in the selected envelope:

1. If $20 is the smaller amount, then ($20, $40) are in the two envelopes.

2. If $20 is the larger amount, then ($20, $10) are in the two envelopes.

In situation 1, on switching, gain is $20

In situation 2, on switching, loss is $10.

So, one should switch because the potential gain ($20) is twice the potential loss ($10).

 Signature 

I am, therefore I think.

Profile
 
 
Posted: 30 November 2012 06:10 AM   [ Ignore ]   [ # 1738 ]
Sr. Member
Avatar
RankRankRankRankRankRankRankRankRankRank
Total Posts:  4377
Joined  2007-08-31
kkwan - 30 November 2012 05:12 AM

This means you deny that you know that (X - Y) = -1 * (Y -X) for all possible finite values of X and Y.

Why introduce Y when either (X, 2X) or (X, 1/2X) is sufficient for the TEP? 

You don’t react on what I write: you deny that you know that (X - Y) = -1 * (Y -X) for all possible finite values of X and Y. If not, prove that to me.

kkwan - 30 November 2012 05:32 AM

Your 4, 5 and 6 is based on the unjustified premise that you know whether (X, 2X) or (X, 1/2X) are in the two envelopes which you don’t, in the context of the TEP.

BS. I claim I know nothing about the amounts, except that they are fixed from the moment they are in the closed envelopes.

4 follows from 1 and 3; 5 follows from 4; and 6 follows from 5. I have no premise about the ratio between the amounts in the envelopes at all.

 Signature 

GdB

“The light is on, but there is nobody at home”

Profile
 
 
Posted: 30 November 2012 09:48 AM   [ Ignore ]   [ # 1739 ]
Sr. Member
RankRankRankRankRankRankRankRankRankRank
Total Posts:  1824
Joined  2007-10-28
GdB - 30 November 2012 06:10 AM

You don’t react on what I write: you deny that you know that (X - Y) = -1 * (Y -X) for all possible finite values of X and Y. If not, prove that to me.

That is irrelevant in the context of the TEP with two and only two possibilities: (X, 2X) or (X, 1/2X) for any finite value of X and total ignorance.

BS. I claim I know nothing about the amounts, except that they are fixed from the moment they are in the closed envelopes.

4 follows from 1 and 3; 5 follows from 4; and 6 follows from 5. I have no premise about the ratio between the amounts in the envelopes at all.

They are fixed, but you don’t know what are their actual values and whether (X, 2X) or (X, 1/2X) are in the two envelopes because of total ignorance.

GdB, because of total ignorance, all possible situations must be considered together. Your 4,5, and 6 refer the universe of one and only one possible situation which is not the case in the TEP where there are two and only two possible situations for any finite value of X.

 Signature 

I am, therefore I think.

Profile
 
 
Posted: 30 November 2012 10:23 AM   [ Ignore ]   [ # 1740 ]
Sr. Member
Avatar
RankRankRankRankRankRankRankRankRankRank
Total Posts:  4377
Joined  2007-08-31
kkwan - 30 November 2012 09:48 AM
GdB - 30 November 2012 06:10 AM

You don’t react on what I write: you deny that you know that (X - Y) = -1 * (Y -X) for all possible finite values of X and Y. If not, prove that to me.

That is irrelevant in the context of the TEP with two and only two possibilities: (X, 2X) or (X, 1/2X) for any finite value of X and total ignorance.

I don’t want to know if you find it irrelevant, I want to know if you agree. I know you do not react because you know it shows that your argumentation is wrong.

kkwan - 30 November 2012 09:48 AM

BS. I claim I know nothing about the amounts, except that they are fixed from the moment they are in the closed envelopes.

4 follows from 1 and 3; 5 follows from 4; and 6 follows from 5. I have no premise about the ratio between the amounts in the envelopes at all.

They are fixed, but you don’t know what are their actual values and whether (X, 2X) or (X, 1/2X) are in the two envelopes because of total ignorance.

As said, I presuppose less than you. And still I can prove that your possible gain and loss are equal.

kkwan - 30 November 2012 09:48 AM

GdB, because of total ignorance, all possible situations must be considered together. Your 4,5, and 6 refer the universe of one and only one possible situation which is not the case in the TEP where there are two and only two possible situations for any finite value of X.

This is nonsense. I say that for every set of values (X,Y) it is valid that (X - Y) = -1 * (Y -X). This includes then also your (X, 2X) or (X, 1/2X).

[ Edited: 30 November 2012 10:29 AM by GdB ]
 Signature 

GdB

“The light is on, but there is nobody at home”

Profile
 
 
   
116 of 132
116