105 of 132
105
The two envelopes problem
Posted: 11 November 2012 07:03 PM   [ Ignore ]   [ # 1561 ]
Sr. Member
RankRankRankRankRankRankRankRankRankRank
Total Posts:  1884
Joined  2007-10-28
StephenLawrence - 04 November 2012 11:05 PM

And they must cancel each other out.

As both players are totally ignorant, from their respective perspectives, they are not wrong to expect potential gain/loss.

From our perspective, one of the player’s gain is the other player’s loss. That does not “cancel each other out”.

 Signature 

I am, therefore I think.

Profile
 
 
Posted: 11 November 2012 07:18 PM   [ Ignore ]   [ # 1562 ]
Sr. Member
RankRankRankRankRankRankRankRankRankRank
Total Posts:  1884
Joined  2007-10-28
StephenLawrence - 04 November 2012 11:08 PM

Kkwan the contradiction is there you can’t remove it as you are trying to do.

The reason is we also know that on the second go if you are asked if you want to switch the formula says you should switch too.

The same formula is saying you should and you shouldn’t

Step 10 recommends switching, but we can deduce switching twice or any even number of times is equivalent to not switching at all.

It does not specifically recommends switching any particular number of times.

Thus, step 10 is not contradictory per se because we can switch once or any odd number of times.

 Signature 

I am, therefore I think.

Profile
 
 
Posted: 11 November 2012 07:36 PM   [ Ignore ]   [ # 1563 ]
Sr. Member
RankRankRankRankRankRankRankRankRankRank
Total Posts:  1884
Joined  2007-10-28
StephenLawrence - 04 November 2012 11:18 PM

What we know is we either have $10 and $20 or $5 and $10, or another pair of numbers.

Because whatever two numbers we have makes no difference we can simply imagine any pair we like and work with that.

We don’t have to compare two different situations that we might be in at all.

I’m sure the only interesting bit is what goes wrong when we do the T.E.P that way and how should we work out what to do when we open our envelope.

For any pair of numbers, (X, 2X) there is also the possibility of (X, 1/2X) and they are not equivalent for any finite value of X. That is the complete TEP.

So, we must consider two and only two possible events: (X, 2X) or (X, 1/2X). If we only consider one and only one event, that is an incomplete description of the TEP.

Opening our selected envelope makes no difference wrt what to do as we don’t know whether it contains the smaller or the larger amount.

 Signature 

I am, therefore I think.

Profile
 
 
Posted: 11 November 2012 07:53 PM   [ Ignore ]   [ # 1564 ]
Sr. Member
RankRankRankRankRankRankRankRankRankRank
Total Posts:  1884
Joined  2007-10-28
GdB - 04 November 2012 11:39 PM

kkwan,
Choosing different descriptions for two unknown but fixed amounts is of no relevance for any calculation. But you do as if it makes a difference.

The amounts are fixed, but it could be either (X, 2X) or (X, 1/2X).

This is a contradiction kkwan: you use 3X when you choose the envelope with smallest amount first, and you use (3/2)X when you pick the envelope with the biggest amount first. For the same X these are different amounts.

Again: show me where you use that the total amount does not change. You haven’t done that yet.

There are two mutually exclusive events: either (X, 2X) or (X, 1/2X).

If it is (X, 2X), then 3X is the total amount in the two envelopes.

If it is (X, 1/2X), then 3/2X is the total amount in the two envelopes.

So, in each event, the total amount is fixed and does not change.

 Signature 

I am, therefore I think.

Profile
 
 
Posted: 11 November 2012 07:58 PM   [ Ignore ]   [ # 1565 ]
Sr. Member
RankRankRankRankRankRankRankRankRankRank
Total Posts:  1884
Joined  2007-10-28
StephenLawrence - 04 November 2012 11:42 PM

Kkwan only needs to say that he is comparing two different situations that he might be in, which he has agreed to many times.

So the total amount does not change.

Exactly. The total amounts refer to different distinct possible events.

 Signature 

I am, therefore I think.

Profile
 
 
Posted: 11 November 2012 08:43 PM   [ Ignore ]   [ # 1566 ]
Sr. Member
RankRankRankRankRankRankRankRankRankRank
Total Posts:  1884
Joined  2007-10-28
GdB - 05 November 2012 01:13 AM

Imagine we have two amounts, $5 and $10. (Total $15). Imagine I pick $10 first. Oh, but then I look at $10 and $20. (Total $30). Sorry, this is wrong as wrong can be. Depending on what is picked first, the amounts change.

Correct is: if I have $5 and $10 and I pick $5 then the other amount is $10. And if I pick the $10, then the other is $5. That is the fact kwann wipes under the carpet.

The crucial point is, you are totally ignorant wrt the amounts in the two envelopes. It could be either (X, 2X) or (X, 1/2X).

As such, if A is denoted as the amount in the selected envelope, then either (A, 2A) or (A, 1/2A) are in the two envelopes.

Thus, assuming ($5, $10) are in the envelopes is unwarranted because ($5, $2.5) is also possible per the TEP.

($5, $10) is a fact if and only if you know what are actually in the two envelopes which you do not, in the context of the TEP. Imagining it does not make it a fact.

To treat an unwarranted assumption as a fact is clearly wrong and misleading.  smile

 Signature 

I am, therefore I think.

Profile
 
 
Posted: 12 November 2012 12:09 AM   [ Ignore ]   [ # 1567 ]
Sr. Member
Avatar
RankRankRankRankRankRankRankRankRankRank
Total Posts:  4455
Joined  2007-08-31
kkwan - 11 November 2012 08:43 PM

To treat an unwarranted assumption as a fact is clearly wrong and misleading.  smile

Yes, I am totally ignorant about the exact amounts. But I am not about the following:

1. The two envelopes are closed, so the amounts cannot change.
2. Corollary 1: the total amount does not change.
3. Corollary 2: the amounts do not change dependent on which envelope I take first. So also the total amount will stay the same.
4. The amount of one envelope is twice the amount of the other.
5. Corollary: the amount of one envelope is half of the other.

Which of my presuppositions is wrong?

You switch in your argument between two possibilities: (X,2X) and (X,X/2) for the same X are different amounts. So you are contravening rule 1.

 Signature 

GdB

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

Profile
 
 
Posted: 12 November 2012 01:05 AM   [ Ignore ]   [ # 1568 ]
Sr. Member
RankRankRankRankRankRankRankRankRankRank
Total Posts:  5944
Joined  2006-12-20

GdB,

The problem is if we open our envelope we see that it does at least look reasonable to do the sum Kkwan’s way. Also it looks like opening the envelope makes no difference.

This is the illusion we are trying to overcome.

You are not solving the problem you are side stepping it.

Do it with the envelope open for goodness sake.

Stephen

Profile
 
 
Posted: 12 November 2012 01:09 AM   [ Ignore ]   [ # 1569 ]
Sr. Member
RankRankRankRankRankRankRankRankRankRank
Total Posts:  5944
Joined  2006-12-20
kkwan - 11 November 2012 08:43 PM

To treat an unwarranted assumption as a fact is clearly wrong and misleading.  smile

The assumption in question is rock solid Kkwan. It’s simply that there are two amounts in the envelopes one double the other.

From that assumption we can make this more concrete to think about by imagining any two amounts we like.

We can do that because it makes no diference, the result is the same which ever two amounts we think about.

You are not facing up to this. Changing your mind is a good thing! (sometimes)

Stephen

Profile
 
 
Posted: 12 November 2012 03:30 AM   [ Ignore ]   [ # 1570 ]
Sr. Member
Avatar
RankRankRankRankRankRankRankRankRankRank
Total Posts:  4455
Joined  2007-08-31
StephenLawrence - 12 November 2012 01:05 AM

The problem is if we open our envelope we see that it does at least look reasonable to do the sum Kkwan’s way. Also it looks like opening the envelope makes no difference.

kkwan refuses to confess that his solution is wrong for closed envelopes. That’s what I am not side stepping. The open envelope is your personal problem…

 Signature 

GdB

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

Profile
 
 
Posted: 12 November 2012 04:04 AM   [ Ignore ]   [ # 1571 ]
Sr. Member
RankRankRankRankRankRankRankRankRankRank
Total Posts:  5944
Joined  2006-12-20
GdB - 12 November 2012 03:30 AM
StephenLawrence - 12 November 2012 01:05 AM

The problem is if we open our envelope we see that it does at least look reasonable to do the sum Kkwan’s way. Also it looks like opening the envelope makes no difference.

kkwan refuses to confess that his solution is wrong for closed envelopes. That’s what I am not side stepping. The open envelope is your personal problem…

It is not my personal problem.

As I said the illusion is created by the appearance that opening the envelope makes no difference so we might as well treat the puzzle as if our envelope is open.

Kkwan treats the puzzle as if he knows what is in his envelope, and to solve the puzzle you need to say what is wrong with treating it like that i.e why doesn’t that work?

Stephen

Profile
 
 
Posted: 12 November 2012 05:21 AM   [ Ignore ]   [ # 1572 ]
Sr. Member
Avatar
RankRankRankRankRankRankRankRankRankRank
Total Posts:  4455
Joined  2007-08-31

OK, let’s try.

You pick an envelope, and it contains $X.
Now you have two possibilities: you picked from the TEP-pair (X,2X) or from (X,X/2).
Let’s make the assumption that the chances of both pair are equal. (In reality, there are limits to this assumption: when X is smaller than the smallest possible coin, or 2X is more than the play master can afford.)
a. Now the chance that you picked X with those two pairs is 1/2.
b. The chance that you picked 2X is 1/4.
c. The chance that you picked X/2 is 1/4.
So the expected value when not switching is X/2 + 2X/4 + (X/2)/4 = 9X/8.

Now if you switch it looks as follows:
i. in case b., i.e. you got 2X first, if you switch you get X, so the expected value for this one is: X/4 = 2X/8.
ii. in case c,, you got X/2 first, if you switch you get X, so the expected value for this one is: X/4 = 2X/8.
iii. In case a, the other envelope can contain X/2 or 2X, both with a chance of 1/2. Multiplied with the chance you got X (1/2), we get:
(1/2)(X/4 + 2X/2) = X/8 + X/2 = 5X/8
Add the expected values of i, ii, and iii:
2X/8 + 2X/8 + 5X/8 = 9X/8.

So the expected values for switching and not switching are the same. (This is just in other words what Mingy Jongo derived a long time ago).

The problem with kkwan’s solution that he does not fix the amounts. Of course, if I pick, say $10, then the other envelope can contain $5 or $20. But I must take into account that the amounts are constants. So I must calculate the expected values for both pairs separately, and then add them according their chances. In stating that the chances for $5 and $20 are equal, I do not take into account that the chance for the amount that I picked is twice as big as the other two (1/2 against 1/4 for $5 and $10 respectively).

The simpler approach we also discussed earlier: As long as you do not look, you know that your possible gain and loss are the same. Now if you look, that fact does not change.

 Signature 

GdB

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

Profile
 
 
Posted: 12 November 2012 07:12 AM   [ Ignore ]   [ # 1573 ]
Sr. Member
RankRankRankRankRankRankRankRankRankRank
Total Posts:  5944
Joined  2006-12-20
GdB - 12 November 2012 05:21 AM

OK, let’s try.

Thank you. I expect to find myself looking through it carefully.

Stephen

Profile
 
 
Posted: 12 November 2012 01:25 PM   [ Ignore ]   [ # 1574 ]
Sr. Member
RankRankRankRankRankRankRankRankRankRank
Total Posts:  5944
Joined  2006-12-20
GdB - 12 November 2012 05:21 AM

The simpler approach we also discussed earlier: As long as you do not look, you know that your possible gain and loss are the same. Now if you look, that fact does not change.

I don’t think this can be coherent, something must be wrong. I’ll try to explain.

Of course, if I pick, say $10, then the other envelope can contain $5 or $20. But I must take into account that the amounts are constants. So I must calculate the expected values for both pairs separately, and then add them according their chances. In stating that the chances for $5 and $20 are equal, I do not take into account that the chance for the amount that I picked is twice as big as the other two (1/2 against 1/4 for $5 and $10 respectively)

Is it possible to make sense of this? I don’t think so. We agree that there can be $5 or $20 in the other envelope. And we agree that we can’t know that the probability is 50/50.

But one way to put the problem is if we don’t know that the probability is 50/50 what do we know it is?!

You can answer with we just dunno but that doesn’t fit with your idea that we do know that our expected gain is equal to our expected loss.

We can answer that the probability of there being $20 in the other envelope is 1 in 3 and the chance of there being $5 in the other envelope is 2 in 3 which would mean our expected gain is equal to our expected loss and so would fit with what we are supposed to know but obviously that isn’t the solution.

What else?

Well, what you are trying to do just doesn’t coherently fit, if I haven’t managed to show that above I can do it another way.

You pick an envelope, and it contains $X.

Say I pick $10.

Now you have two possibilities: you picked from the TEP-pair (X,2X) or from (X,X/2).

OK.

Let’s make the assumption that the chances of both pair are equal.

OK so with this information I can set the situation up this way. I have two pairs of envelopes one pair contains $10 and $20 and the other contains $5 and $10. Now I toss a coin to discard one of the pairs. I tell you all this. There seems to be nothing wrong with this as I’ve set it up so the chances of both pairs are equal.

Now you pick an envelope and you don’t look. What should you do? Your sums work, switching makes no difference because if you have $5 you would gain $5  if you have $10 you might gain $10 and you might lose $5 and if you have $20 you will lose $10 This all cancels out.

But the above fits with it being right to switch if you open the envelope and find $10 because you can gain $10 or lose $5 with an equal chance, so it isn’t true that the rest isn’t taken account of, it’s that there is no possible way to take account of it as it becomes unapplicable once you discover you have $10.

Stephen

[ Edited: 12 November 2012 01:37 PM by StephenLawrence ]
Profile
 
 
Posted: 13 November 2012 12:29 AM   [ Ignore ]   [ # 1575 ]
Sr. Member
Avatar
RankRankRankRankRankRankRankRankRankRank
Total Posts:  4455
Joined  2007-08-31

Stephen,

The problem with all this is that you fall for your intuitions, leaving out everything you know about the history how you got into the situation.

The description of TEP is not: here you have an amount $X, and the other envelope contains half or twice of the amount with equal chances. It is a true description of your situation, but it is not a complete description. It is the same disease that kkwan suffers from. Therefore I introduced, much earlier in the thread, other situations where our intuitions go astray because of incomplete descriptions: the Monty Hall problem and the diagnostics information. In the Monty Hall problem, when I leave out my knowledge of the history of the two boxes that are left, I am sure the other box contains the price or not. But that does not mean that my chance for the price is 50%. We know it is 66%. Same with the diagnostic test: its reliability of 99% does not mean that when the test is positive, that my chance of being ill is 99%. I did not take into account how many people really have the illness, and so the number of false positives is much larger than the correct positives.

So when looking at TEP, you must take the real history in account, which is that you have two envelopes, you had a free choice which one you picked first, then opened it, and then may choose if you take the other or not. In none of you argumentations you do that. But I did. Your examples are simplifications of the situation, and invalid simplifications. You must show me where I make an error in my derivation, where I take all possible information in account.

 Signature 

GdB

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

Profile
 
 
   
105 of 132
105