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The two envelopes problem
 Posted: 07 July 2012 11:00 PM [ Ignore ]   [ # 1381 ]
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GdB - 28 June 2012 11:28 PM
StephenLawrence - 28 June 2012 10:40 PM

Yes, and I think that solving the puzzle includes answering why there isn’t. That’s what most interests me.

(the quote refers to the TEP in which we see our amount)

It is in fact quite simple, but, I must confess, a little counter intuitive.
Take the following: there are two envelopes, I say that one envelope contains \$10 more than the other. Now you open your envelope. Whatever the amount you see, you have no clue if you should switch or not. Right?

I do have a clue, I know switching makes no difference because I might gain or lose £10 with 50/50 probability so switching makes no difference.

Now one step further: I say that one envelope contains an unknown amount more than the other. Now you open your envelope. Whatever the amount you see, you have no clue if you should switch or not. Right?

I could gain or lose that amount with 50/50 probability and so switching makes no difference.

Now one step further: I say that one envelope contains one amount, the other a certain amount more: the difference between the amounts is the same as the contents of one of the envelopes. Now you open your envelope. Whatever the amount you see, you have no clue if you should switch or not. Right?

I open the envelope and I see \$20, from this I know that the other contains \$10 or \$40. If it’s true that switching makes no difference, like in the first two examples, there should be a sum I can do with these numbers that agrees with that, like in the first two examples.

Why isn’t there?

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 Posted: 08 July 2012 10:54 PM [ Ignore ]   [ # 1382 ]
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StephenLawrence - 07 July 2012 11:00 PM

If it’s true that switching makes no difference, like in the first two examples, there should be a sum I can do with these numbers that agrees with that, like in the first two examples.

Why isn’t there?

I don’t understand what you mean. What are the sums in the first 2 examples, and what do you expect from a sum in the third example?

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 Posted: 08 July 2012 11:07 PM [ Ignore ]   [ # 1383 ]
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GdB - 06 July 2012 11:51 PM

So your answer is ‘yes’. You think that choosing different descriptions changes the amounts in the envelopes. In fact you have no idea what amounts are in the envelopes, so your saying there are two possibilities is silly. Is (44,22) the same as (X,X/2)? Or is it the same as (2X,X)? Is it the same as (22,44)? (X,2X)? Or (44X,22X)? Or (4X,2X)? (7X,3.5X)? What does it matter that you use two different descriptions for the same fact? As long as it expresses “one envelope contains twice the other” every description is fine. And even the description “one envelope contains half of the other” is the same, you must only ‘exchange’ one envelope with the other. But that is no problem, they are identical, remember.

Precisely because you do not know what amounts are in the two envelopes, the two possibilities must be considered and it matters whether (X, 2X) or (X, 1/2X) are in the two envelopes because they are not equivalent for any finite value of X.

You do not deny what I am saying. Say there are (10,20) in the envelopes. What is then your A, denoting the envelope that you pick first? It is 10 or 20.

You cannot assume (10, 20) are in the two envelopes because you don’t know it is so. It could be (10, 5) as per the TEP.

As such, if A is 10 in the selected envelope:

If A is the smaller amount, (10, 20) are in the two envelopes.
If A is the larger amount, (10, 5) are in the two envelopes.

No, that is exactly what I meant. And I say it again: start your argumentation with denoting the amount in the other envelope as B. It is obvious that the conclusion then is that you should stick to your envelope. You are avoiding to answer my question.

That would be:

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.

Whereby the amount in the other envelope is denoted by B.

Until here it is correct. Notice the words I made bold. There are only two possibilities, and you shift that fact under the carpet. You are not using the complete information you have available.

There are two possibilities, either (X, 2X) or (X, 1/2X).

I fix the fact given in the description of TEP: that there are only two amounts. That you have no idea what amounts does not matter, but there are only two amounts. You calculate with three. You leave out some information that you have.

There are two amounts but they are either (X, 2X) or (X, 1/2X) and they are both possible events in the context of the TEP.

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 Posted: 08 July 2012 11:32 PM [ Ignore ]   [ # 1384 ]
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kkwan - 08 July 2012 11:07 PM

You cannot assume (10, 20) are in the two envelopes because you don’t know it is so. It could be (10, 5) as per the TEP.

Rubbish. You could also assume there are (3,6), (231,462) or whatever in the envelopes. Why should I not need to consider those? If I say: “Assume there are 10 and 20 in the envelopes”, and you say there could be 5 in one of the envelopes, you do not follow the assumption.

It is so simple: For every TEP pair there is exactly one unique value of X for which (X,2X) is a complete and exact description of that TEP pair. Do you deny that?
Of course, there is also a unique value of X for which (X,X/2) is a complete and exact description of that same TEP pair. But it is another value of X then above, because we talk about the same TEP pair. But you treat those X’s as the same.

kkwan - 08 July 2012 11:07 PM

There are two amounts but they are either (X, 2X) or (X, 1/2X) and they are both possible events in the context of the TEP.

Correctly formulated:

There are two amounts and these can be described by (X, 2X) or (X, 1/2X). But because there are only two amounts, the X in these description do not refer to the same amount, if both descriptions describe the same TEP pair.

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 Posted: 09 July 2012 12:18 PM [ Ignore ]   [ # 1385 ]
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GdB - 08 July 2012 10:54 PM
StephenLawrence - 07 July 2012 11:00 PM

If it’s true that switching makes no difference, like in the first two examples, there should be a sum I can do with these numbers that agrees with that, like in the first two examples.

Why isn’t there?

I don’t understand what you mean.

Obviously if it’s true that switching makes no difference it should be possible to express this mathematically.

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 Posted: 09 July 2012 01:50 PM [ Ignore ]   [ # 1386 ]
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StephenLawrence - 09 July 2012 12:18 PM

Obviously if it’s true that switching makes no difference it should be possible to express this mathematically.

That is still not clear to me. What is the math you see in the first 2 examples, but not in the third?

BTW, you have been away for a long time. I posed a lot more questions and deliberations for you: see here and the second half of this posting.

I have no idea what you expect more. The point is that kkwan (and probably you) are not using all information available. Saying that the second envelope contains half or twice the amount is only is a generalisation of the fact that you started with exactly these two envelopes, and you you chose one of them, and the remaining is the other one. That is the main argument I make in my story line in my first link above. Please read it carefully.

See also here, second half, where I start once again with green and red pieces of paper.

Or do you agree with all these? Then still I ask: what exactly are you then still looking for?

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 Posted: 09 July 2012 11:21 PM [ Ignore ]   [ # 1387 ]
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GdB - 09 July 2012 01:50 PM

That is still not clear to me. What is the math you see in the first 2 examples, but not in the third?

GdB,

It can’t be clearer so I’m not playing ball. If it’s true that switching makes no difference in the third example it should be possible to mathematically express that.

The key question is why isn’t it?

Stephen

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 Posted: 10 July 2012 09:29 AM [ Ignore ]   [ # 1388 ]
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Step 1:
If you have two amounts, X and Y, whereby Y > X, then their difference is Y - X. So if you chose X first, then by switching you gain Y - X. If you chose Y first, you loose Y - X by switching.

Right?

Step 2:
Say you open one envelope, it contains \$40. If \$40 is the smallest amount, then you gain Y - 40. If it is the biggest amount, then you loose 40 - X. So what is the difference: (Y - 40) + (40 - X) = Y - X. Hey, isn’t that interesting? That I have \$40 in my envelope does not change anything in my decision! I did not get any useful information from it!

Right?

Step 3:
Now I take TEP: In TEP I know Y = 2X. So if you chose X first, then by switching you gain X. If you chose 2X first, you loose X by changing. Now you see \$40 in your envelope. What is the difference between the two amounts? Again, as above: (Y - 40) + (40 - X) = (2X - 40) + (40 - X) = X. Again, you have no help for your decision! Knowing the amount makes no difference.

kkwan is denying that we should write 4 and 5:

4. If A is the smaller amount the other envelope contains 2A.
5. If A is the larger amount the other envelope contains A/2.

As:
4a. If you choose X, the smaller amount, then the other envelope contains 2X.
5a. If you choose 2X, the larger amount, then the other envelope contains X.

So he suggests that there are 3 amounts, where there are only 2, which is correctly pictured in 4a and 5a. He wipes the relationship between the two amounts under the carpet, which simply is: if you pick one envelope, say with X, then the other envelope contains the other amount, Y. And if you choose Y, then the other contains X.
And above you see that your knowledge of one of the amounts does not help you. Formulas enough?

[ Edited: 10 July 2012 11:00 PM by GdB ]
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 Posted: 10 July 2012 08:19 PM [ Ignore ]   [ # 1389 ]
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GdB - 08 July 2012 11:32 PM

Rubbish. You could also assume there are (3,6), (231,462) or whatever in the envelopes. Why should I not need to consider those? If I say: “Assume there are 10 and 20 in the envelopes”, and you say there could be 5 in one of the envelopes, you do not follow the assumption.

All you can say is either (X, 2X) or (X, 1/2X) are in the two envelopes. If X is 10, then either (10, 20) or (10, 5) are in the two envelopes.

It is so simple: For every TEP pair there is exactly one unique value of X for which (X,2X) is a complete and exact description of that TEP pair. Do you deny that?
Of course, there is also a unique value of X for which (X,X/2) is a complete and exact description of that same TEP pair. But it is another value of X then above, because we talk about the same TEP pair. But you treat those X’s as the same.

For any finite value of X, by considering both (X, 2X) and (X, 1/2X), is the complete description of the TEP.

There are two amounts and these can be described by (X, 2X) or (X, 1/2X). But because there are only two amounts, the X in these description do not refer to the same amount, if both descriptions describe the same TEP pair.

For any finite value of X, (X, 2X) and (X, 1/2X) is not equivalent but they are possible events in the context of the TEP. Hence, both events must be considered in a complete analysis of the TEP.

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 Posted: 10 July 2012 10:26 PM [ Ignore ]   [ # 1390 ]
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GdB - 10 July 2012 09:29 AM

Step 1:
If you have two amounts, X and Y, whereby Y > X, then their difference is Y - X. So if you chose X first, then by switching you gain Y - X. If you chose Y first, you loose Y - X by switching.

Right?

Yep so we know switching makes no difference

Step 2:
Say you open one envelope, it contains \$40. If \$40 is the smallest amount, then you gain Y - 40.

Which is \$80 - \$40 = \$40

If it is the biggest amount, then you loose 40 - X. So what is the difference:

Which is \$40 - \$20 = \$20

So you see you can gain \$40 or lose \$20

Now, if it’s true that switching makes no difference what equation can we do with this potential gain of \$40 or loss of \$20 to express that?

(Y - 40) - (40 - X) = Y - X.

Nope.

Hey, isn’t that interesting? That I have \$40 in my envelope does not change anything in my decision! I did not get any useful information from it!

Right?

Fact is the information you get is that there is either \$20 or \$80 in the other envelope and so you will gain \$40 or lose \$20 by switching.

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 Posted: 10 July 2012 10:41 PM [ Ignore ]   [ # 1391 ]
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kkwan,

You must consider one TEP pair, and not just a ‘denotation of the amount you pick first’. By taking only one amount fixed, you consider two TEP pairs, but there is only one, the one in front of you. By considering only one amount you deprive yourself of the knowledge that, “of one TEP pair, if you pick the smallest amount, the other one is the biggest; if you pick the biggest one first, then the other one is the smallest”. This is what you deny. That if you pick one envelope there is only one other one, that is there lying in front of you. This is the knowledge you have, but you do not use in your ‘derivation’ of TEP.

Tell me if you agree wit the numbered statements:

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.

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.

I have two envelopes according to TEP:
7. If I have the envelope with the smallest amount, then the other one contains the biggest.
8. If I have the envelope with the biggest amount, then the other one contains the smallest.

Or are you going to refuse again, and give some avoiding answer?

[ Edited: 11 July 2012 03:29 AM by GdB ]
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 Posted: 10 July 2012 10:44 PM [ Ignore ]   [ # 1392 ]
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StephenLawrence - 10 July 2012 10:26 PM

(Y - 40) - (40 - X) = Y - X.

Nope.

Sorry that had to be a plus. Mixing up gains and losses…

Correction:
(Y - 40) + (40 - X) = Y - X.

I’ll correct it in my original posting.

[ Edited: 10 July 2012 10:59 PM by GdB ]
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 Posted: 10 July 2012 11:03 PM [ Ignore ]   [ # 1393 ]
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StephenLawrence - 10 July 2012 10:26 PM

Step 2:
Say you open one envelope, it contains \$40. If \$40 is the smallest amount, then you gain Y - 40.

Which is \$80 - \$40 = \$40

Step 2 is not TEP yet. Tell me why the argumentation in step 2 is wrong.

Then you can go on and tell me why in step 3, which is TEP, the argumentation in step 2 is not valid anymore.

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 Posted: 10 July 2012 11:50 PM [ Ignore ]   [ # 1394 ]
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kkwan - 10 July 2012 08:19 PM

For any finite value of X, (X, 2X) and (X, 1/2X) is not equivalent but they are possible events in the context of the TEP. Hence, both events must be considered in a complete analysis of the TEP.

It isn’t true that you must consider both these possibilities.

You know you have a pair, whatever that is, and you can just consider what would be the case if you had the larger or the smaller amount of a pair.

There is nothing wrong with doing it that way. The puzzle is why doesn’t your way work as well.

Stephen

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 Posted: 11 July 2012 12:33 AM [ Ignore ]   [ # 1395 ]
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StephenLawrence - 10 July 2012 11:50 PM

You know you have a pair, whatever that is, and you can just consider what would be the case if you had the larger or the smaller amount of a pair.

There is nothing wrong with doing it that way. The puzzle is why doesn’t your way work as well.

I do not understand why you are still asking this. kkwan’s way does not work while he does not use the known fact you mention above.

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