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The two envelopes problem
Posted: 24 October 2012 09:15 PM   [ Ignore ]   [ # 1516 ]
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StephenLawrence - 24 October 2012 07:47 AM

Right, so if the odds are 50/50 of you being in either of these two situations it seems right that you should switch.

Yes, once or any odd number of times.

So the problem is how to overcome this. I’m sure it isn’t right that you should switch but I’m not sure GdB and others have managed to solve the problem, as they suppose. (but perhaps they have and I’m just not getting it)

Why is it not right to switch? The potential gain/loss is 100%/50%.

What GdB and others have tried to “solved the problem” is to consider only one possibility instead of both possibilities together, which is the complete TEP.

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Posted: 24 October 2012 09:29 PM   [ Ignore ]   [ # 1517 ]
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Write4U - 24 October 2012 04:40 PM

There is a 50/50 chance of picking,

system I
a = 10
b = 5

Or

system II
a = 5
b = 10

no change in outcome, either way

Not so. There is a 50/50 chance that either (X, 2X) or (X, 1/2X) are in the two envelopes.

So, if you select the envelope with X, switch, you either gain X or lose 1/2X.

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Posted: 24 October 2012 10:04 PM   [ Ignore ]   [ # 1518 ]
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kkwan - 24 October 2012 08:47 PM

As such, if the amount in the selected envelope is denoted as A and as A could be the smaller or the larger amount, therefore either (A, 2A) or (A, 1/2A) are in the two envelopes.

Hence, if you switched, you could either gain A or lose 1/2A.

You keep repeating you mantra. It doesn’t help. Dependent on the amount of the first choice, you define different totals. But it is obvious that the total amount of the two envelopes is not dependent on which envelope you choose first.

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Posted: 24 October 2012 10:10 PM   [ Ignore ]   [ # 1519 ]
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kkwan - 24 October 2012 08:59 PM
GdB - 24 October 2012 07:28 AM

And that is just what 10 does not say. 10 implies one should always swap. And of course that is irrational, as you are starting to see.

Since switching twice or any even number of times is equivalent to not switching, you should switch once or any odd number of times.

Of course, 11 is logically absurd.

Right.

And now, after you you made your first choice, somebody else comes into the room. He is said, one envelope contains twice the amount of the other. (That is still true!). He picks the envelope you did not choose first. Should he swap to increase his chance for an higher amount? And you? Should you still swap?

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Posted: 24 October 2012 11:04 PM   [ Ignore ]   [ # 1520 ]
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kkwan - 24 October 2012 09:15 PM
StephenLawrence - 24 October 2012 07:47 AM

Right, so if the odds are 50/50 of you being in either of these two situations it seems right that you should switch.

Yes, once or any odd number of times.

You’ve made this bit up. The formula doesn’t say that at all. It’s just you’ve reasoned that if you switch twice you are back to square 1.

Why is it not right to switch? The potential gain/loss is 100%/50%.

Because it can’t be right. Imagine two players in the game both having one of the envelopes. It cannot be beneficial for both of them to switch.

What GdB and others have tried to “solved the problem” is to consider only one possibility instead of both possibilities together, which is the complete TEP.

There is nothing wrong with doing that.

But it doesn’t say what is wrong with your way.

Stephen

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Posted: 24 October 2012 11:26 PM   [ Ignore ]   [ # 1521 ]
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kkwan - 24 October 2012 09:29 PM
Write4U - 24 October 2012 04:40 PM

There is a 50/50 chance of picking,

system I
a = 10
b = 5

Or

system II
a = 5
b = 10

no change in outcome, either way

Not so. There is a 50/50 chance that either (X, 2X) or (X, 1/2X) are in the two envelopes.

So, if you select the envelope with X, switch, you either gain X or lose 1/2X.

I understand what you are saying but, IMO the relative equivalencies should be expressed as,

where X = 5 :  (X, 2X)  or (a = 5), (b = 10)

next equation,

where X = 10 :  (X, X/2) or (a = 10), (b = 5)

X is the variable and becomes either the greater or the smaller of any 2 fixed amounts (5 or 10) actually physically present in the envelopes.

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Posted: 26 October 2012 08:22 PM   [ Ignore ]   [ # 1522 ]
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GdB - 24 October 2012 10:04 PM

You keep repeating you mantra. It doesn’t help. Dependent on the amount of the first choice, you define different totals. But it is obvious that the total amount of the two envelopes is not dependent on which envelope you choose first.

It is not my “mantra”. That is from the argument to switch. It is reasonable not to assume what are the total amounts in the two envelopes as you don’t know. Neither can you determine whether the amount in the selected envelope is the smaller or the larger amount. As such, the amount in your selected envelope is denoted as A and as A could be the smaller or the larger amount, the amounts in the two envelopes are either (A, 2A) or (A, 1/2A) which is consistent with the requirement of the TEP that one envelope contains twice the amount in the other.

Of course, the total amounts in the two envelopes is not dependent on which envelope you selected, but as you don’t know the total amounts, it could be either (X, 2X) or (X, 1/2X) in the two envelopes notwithstanding whether you selected any envelope at all.

In other words, there is an inherent ambiguity in the TEP which cannot be resolved or ignored.

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Posted: 26 October 2012 08:29 PM   [ Ignore ]   [ # 1523 ]
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GdB - 24 October 2012 10:10 PM

And now, after you you made your first choice, somebody else comes into the room. He is said, one envelope contains twice the amount of the other. (That is still true!). He picks the envelope you did not choose first. Should he swap to increase his chance for an higher amount? And you? Should you still swap?

Because the situations are symmetrical, both should switch. However, they should switch once or any odd number of times.

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Posted: 26 October 2012 08:52 PM   [ Ignore ]   [ # 1524 ]
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StephenLawrence - 24 October 2012 11:04 PM

You’ve made this bit up. The formula doesn’t say that at all. It’s just you’ve reasoned that if you switch twice you are back to square 1.

Switching twice or any even number of times is equivalent to not switching at all.

Thus, 10 can be interpreted as switching once or any odd number of times.

Because it can’t be right. Imagine two players in the game both having one of the envelopes. It cannot be beneficial for both of them to switch.

Why not? Both could gain/lose 100%/50% from their respective perspectives.

There is nothing wrong with doing that.

But it doesn’t say what is wrong with your way.

There is, because of the assumption of only one possibility whereas the TEP has two instead. By doing that it is misleading and incomplete and thus it is not the solution of the TEP per se.

There is an inherent ambiguity/symmetry in the TEP which makes it intractable.

[ Edited: 27 October 2012 05:57 AM by kkwan ]
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Posted: 26 October 2012 09:10 PM   [ Ignore ]   [ # 1525 ]
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Write4U - 24 October 2012 11:26 PM

I understand what you are saying but, IMO the relative equivalencies should be expressed as,

where X = 5 :  (X, 2X)  or (a = 5), (b = 10)

next equation,

where X = 10 :  (X, X/2) or (a = 10), (b = 5)

X is the variable and becomes either the greater or the smaller of any 2 fixed amounts (5 or 10) actually physically present in the envelopes.

Not so, because for any finite value of X, (X, 2X) and (X, 1/2X) are not equivalent.

So, for any finite value of X, either (X, 2X) or (X, 1/2X) are in the two envelopes.

If X = 10, either (10, 20 ) or (10, 5) are in the two envelopes.

If X = 5, either (5, 10) or (5, 5/2) are in the two envelopes.

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Posted: 26 October 2012 11:07 PM   [ Ignore ]   [ # 1526 ]
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kkwan - 26 October 2012 09:10 PM
Write4U - 24 October 2012 11:26 PM

I understand what you are saying but, IMO the relative equivalencies should be expressed as,

where X = 5 :  (X, 2X)  or (a = 5), (b = 10)

next equation,

where X = 10 :  (X, X/2) or (a = 10), (b = 5)

X is the variable and becomes either the greater or the smaller of any 2 fixed amounts (5 or 10) actually physically present in the envelopes.

Not so, because for any finite value of X, (X, 2X) and (X, 1/2X) are not equivalent.

So, for any finite value of X, either (X, 2X) or (X, 1/2X) are in the two envelopes.

If X = 10, either (10, 20 ) or (10, 5) are in the two envelopes.

If X = 5, either (5, 10) or (5, 5/2) are in the two envelopes.

This is the angle I am exploring.  IMO, it is not an algebraic equation where x can be assigned an arbitrary (finite) value. It is already unalterably fixed by the amount contained in the other envelope.

X cannot be given a fixed (finite) value, because the values are fixed in both envelopes. Trying to solve the equation by assigning a fixed value for X in algebraic terms always yields one incorrect answer in reality. Thus it must be logically false.

One can say that theoretically the equation holds true, but in reality it does not hold true for one set. Therefore no logical reason can be applied to switch or not to switch.

[ Edited: 27 October 2012 12:23 AM by Write4U ]
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Posted: 26 October 2012 11:28 PM   [ Ignore ]   [ # 1527 ]
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kkwan - 26 October 2012 08:22 PM

It is reasonable not to assume what are the total amounts in the two envelopes as you don’t know. Neither can you determine whether the amount in the selected envelope is the smaller or the larger amount.

You do not know the total amount, that is true. But it is also true that it will not be different dependent on your choice of the first envelope. Whatever the total amount, it does not change when you pick another envelope first. In your argumentation you suppose that if you pick the smallest amount, the total amount has become half of the amount it would have been if you would have chosen the biggest amount first.

kkwan - 26 October 2012 08:29 PM
GdB - 24 October 2012 10:10 PM

And now, after you you made your first choice, somebody else comes into the room. He is said, one envelope contains twice the amount of the other. (That is still true!). He picks the envelope you did not choose first. Should he swap to increase his chance for an higher amount? And you? Should you still swap?

Because the situations are symmetrical, both should switch. However, they should switch once or any odd number of times.

But we definitely know that on switching one will loose and the other will gain. And we know that the amount one looses is the same as the other gains. And even worse: both player know that too. So what ground do they have for switching?

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Posted: 27 October 2012 06:18 AM   [ Ignore ]   [ # 1528 ]
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Write4U - 26 October 2012 11:07 PM

This is the angle I am exploring.  IMO, it is not an algebraic equation where x can be assigned an arbitrary (finite) value. It is already unalterably fixed by the amount contained in the other envelope.

X cannot be given a fixed (finite) value, because the values are fixed in both envelopes. Trying to solve the equation by assigning a fixed value for X in algebraic terms always yields one incorrect answer in reality. Thus it must be logically false.

One can say that theoretically the equation holds true, but in reality it does not hold true for one set. Therefore no logical reason can be applied to switch or not to switch.

(X, 2X) or (X, 1/2X) are algebraic expressions which are determined by the value of X.

For any finite value of X, they are distinct and not equivalent.

In reality (as we don’t know what are actually in the two envelopes), it is rational and valid to consider both possibilities:
 
Either (X, 2X) or (X, 1/2X) are in the two envelopes.

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Posted: 27 October 2012 06:40 AM   [ Ignore ]   [ # 1529 ]
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GdB - 26 October 2012 11:28 PM

In your argumentation you suppose that if you pick the smallest amount, the total amount has become half of the amount it would have been if you would have chosen the biggest amount first.

If the amount in the selected envelope is denoted as A (as we don’t know whether it is the smaller or the larger amount):

1. If A is the smaller amount, then the amount in the other envelope is 2A.

2. If A is the larger amount, then the amount in the other envelope is 1/2A.

Thus, the total amounts in the two envelopes are either 3A or 3/2A.

But we definitely know that on switching one will loose and the other will gain. And we know that the amount one looses is the same as the other gains. And even worse: both player know that too. So what ground do they have for switching?

We or the players don’t know who will gain or lose as we/they don’t know what are actually the amounts in the two envelopes.

All we/they can say is that potential gain/loss is 100%/50% on switching once or any odd number of times.

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Posted: 27 October 2012 06:58 AM   [ Ignore ]   [ # 1530 ]
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kkwan - 27 October 2012 06:40 AM

Thus, the total amounts in the two envelopes are either 3A or 3/2A.

So according to you the total amount changes dependent on the choice of the first envelope.

kkwan - 27 October 2012 06:40 AM

We or the players don’t know who will gain or lose as we/they don’t know what are actually the amounts in the two envelopes.

All we/they can say is that potential gain/loss is 100%/50% on switching once or any odd number of times.

So you say by switching envelopes the two player’s gain and loss are not the same.

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