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The two envelopes problem
Posted: 26 December 2012 06:41 AM   [ Ignore ]   [ # 1906 ]
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GdB - 26 December 2012 05:18 AM

Yes, I deny that. You can describe the two possibilities of TEP as (remember, <X,Y> means X is your first pick, Y your second):

1. <X,2X> and <2X,X>
or as:
2. <X,2X> and <Y, Y/2>, with the addition that Y = 2X (or X = Y/2).

Taking X = Y is wrong, it introduces 3 amounts.

Describing the TEP with X and Y is superfluous as X is necessary and sufficient.

I assume your argument is that I must consider the mutually exclusive events jointly?

In any mutually exclusive event if there are two amounts, there are only two amounts.

Yes. And the smaller amount is a different amount than the bigger amount, but you do as if they are the same.

There is no contradiction if A is the smaller amount in (A, 2A) and the larger amount in (A, 1/2A).

A mathematical exercise is fundamentally flawed? I ask you to calculate the chances of gaining by switching in the non-TEP situation. Your refusing to do it shows that you see your error, but you just cannot admit it. C’mon kwann, it is an easy task. Just show me how your solution for the non-TEP situation differs from TEP. It is not necessary to show me the differences between the situations, I am fully aware of them. You should give a clear argument, instead of throwing words as “fundamentally flawed”, “misleading” and whatever.

You do it.

No kkwan, you only showed that the situations differ. You did not show why my solution is wrong. You cannot, because it is not wrong. Show me, mathematically, that it is wrong.

Not so.

No. TEP informs you that you have two amounts, one twice as big as the other. Or, what is equivalent, one amount is half of the other.
If you take the smallest amount as reference, then you have (X,2X), if you take the biggest as reference then you have (X,X/2). But the references differ: X in (X,2X) is not the same as X in (X,X/2). However, in both mutually exclusive descriptions (they are not different situations), you can pick the biggest or the smallest amount first:

1. <X,2X> or <2X,X>:—> your possible gain and loss are X
2. <X,X/2> or <X/2,X>:—> your possible gain and loss are X/2

You mixup description and reality, maybe this is what Scott is at.

LOL

It seems you don’t get you even don’t know you have 10.

In the context of the TEP, if the selected amount is 10, then either (10, 20) or (10, 5) are in the two envelopes. It is that simple.

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Posted: 26 December 2012 07:03 AM   [ Ignore ]   [ # 1907 ]
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kkwan - 26 December 2012 06:41 AM

Describing the TEP with X and Y is superfluous as X is necessary and sufficient.

Yes, it is. But you do it, with the addition that Y = X.

kkwan - 26 December 2012 06:41 AM

In any mutually exclusive event if there are two amounts, there are only two amounts.

Yes, and in both mutually exclusive events possible gain and loss are the same.

kkwan - 26 December 2012 06:41 AM

There is no contradiction if A is the smaller amount in (A, 2A) and the larger amount in (A, 1/2A).

Except when you take A as a single and fixed value. Then you ‘join the mutually exclusive events’, and create a situation with three amounts.

kkwan - 26 December 2012 06:41 AM

You do it.

I proved my point already here. You only reacted by pointing out differences between TEP and my non-TEP. You were not able to show why my derivation was wrong.

kkwan - 26 December 2012 06:41 AM

Not so.

You are running out of arguments, kkwan.

kkwan - 26 December 2012 06:41 AM

LOL

Yes, definitely, you are running out of arguments.

kkwan - 26 December 2012 06:41 AM

In the context of the TEP, if the selected amount is 10, then either (10, 20) or (10, 5) are in the two envelopes. It is that simple.

With 10 and 20 in the envelopes you pick 10 first, or 20. It is that simple.

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Posted: 26 December 2012 09:12 AM   [ Ignore ]   [ # 1908 ]
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kkwan - 26 December 2012 05:02 AM

Either pair could be in the two envelopes but we don’t know which pair.

What are the amounts in the envelopes?

Some entity put either of the two pairs in the two envelopes randomly..

So the entity selected from two pairs just like in my example?

Stephen

[ Edited: 27 December 2012 03:26 AM by StephenLawrence ]
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Posted: 27 December 2012 07:38 AM   [ Ignore ]   [ # 1909 ]
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GdB - 26 December 2012 07:03 AM

Yes, it is. But you do it, with the addition that Y = X.

If Y is superfluous, then we need not introduce Y at all.

So, if either (X, 2X) or (X, 1/2X) are in the two envelopes with no Y, how can Y=X?

Yes, and in both mutually exclusive events possible gain and loss are the same.

That is true if and only if you know which event is actual which you don’t in the context of the TEP.

As such, when an envelope is selected and the amount in it is denoted as A, if it is the smaller amount, then (A, 2A) are in the two envelopes and if it is the larger amount, then (A, 1/2A) are in the two envelopes.

Except when you take A as a single and fixed value. Then you ‘join the mutually exclusive events’, and create a situation with three amounts.

A is a constant and it can be the smaller amount in (A, 2A) and the larger amount in (A, 1/2A). There is no “joining of the events” and neither are there “three amounts” in each of the two mutually exclusive events.

With 10 and 20 in the envelopes you pick 10 first, or 20. It is that simple.

It is, but in the context of the TEP either (10, 20) or (10, 5) are in the two envelopes and not exclusively only (10, 20) because you don’t know which event is actual. As such, you cannot assume (10, 20) are in the two envelopes because that is begging the question.

It is not so simple.

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Posted: 27 December 2012 08:07 AM   [ Ignore ]   [ # 1910 ]
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kkwan - 27 December 2012 07:38 AM

It is, but in the context of the TEP either (10, 20) or (10, 5) are in the two envelopes and not exclusively only (10, 20) because you don’t know which event is actual. As such, you cannot assume (10, 20) are in the two envelopes because that is begging the question.

It is not so simple.

You can either simplify and consider one possible pair as GdB does. Or consider all the possible pairs.

What doesn’t make sense is to just consider two possible pairs. It would if you knew what was in your envelope and knew the probability of having either pair. Since you don’t know these two things you can’t get this specific because it doesn’t work. We can tell it doesn’t work by considering what would happen if you played the game over and over. Apart from seeing that switching makes no difference, we also see that we would not keep getting the same two pairs and you wouldn’t keep selecting $10.

If you want to reduce all these possible games into one, A would be all the numbers you would get divided by the number of games. And what is in the other envelope would be all those numbers divided by the number of games.

Stephen

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Posted: 27 December 2012 08:41 AM   [ Ignore ]   [ # 1911 ]
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StephenLawrence - 26 December 2012 09:12 AM

What are the amounts in the envelopes?

Generally, either (X, 2X) or (X, 1/2X) are in the two envelopes, but we don’t know which pair is actual in the context of the TEP.

So the entity selected from two pairs just like in my example?

Not quite so.

In your example, the setup is such that both pairs (X, 2X) and (X, 1/2X) coexist at the same time, so there was no selection by the entity at all, with the setup as it is. We then select an envelope with an amount from one of the two pairs. Thus, we can select any one of the four amounts in the two pairs.

In the TEP, the setup is such that one of the two pairs (X, 2X) or (X, 1/2X) was selected by the entity randomly and then we select an envelope with an amount from the selected pair. However, we don’t know which pair was selected by the entity.

So, if we select an amount, we cannot determine from which pair it came from and as such, it is prudent and rational to denote the amount as A as in step 1 and deduce if A is either the smaller or the larger amount, then either (A, 2A) or (A, 1/2A) are in the two envelopes which is consistent with the setup of the TEP without begging the question.

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Posted: 27 December 2012 09:03 AM   [ Ignore ]   [ # 1912 ]
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To start with, kkwan, again you do not do the calculation for my non-TEP situation. if you are not able to do this, then it is clear that you see that this the situation where your formula fits, and not for TEP.

kkwan - 27 December 2012 07:38 AM

If Y is superfluous, then we need not introduce Y at all.

It is not quite superfluous. It makes a point clear. TEP fits two descriptions:

a) One envelope contains twice the amount of the other.
b) One envelope contains half of the other.

Now, as the envelopes are indistinguishable, we don’t know what the ‘one’ is and the ‘other’. So the descriptions can be formalised by:

a) (X,2X)
b) (Y,Y/2)

Now the question is: what is Y expressed in terms of X? It follows from the fact that we have only two envelopes with only two amounts.
This is solved by Y = 2X:
Then b) changes into:

(2X,X),

i.e. the same two envelopes with the same two amounts, but switched. And that is exactly what the TEP is about: two amounts.

You say Y = X, so b) becomes:

(X,X/2)

However there is no meaningful value (except 0) for which (X,2X) can be transformed to (X,X/2): except you introduce a third amount, which is not according TEP.

However 3 amounts fit to my non-TEP situation.

kkwan - 27 December 2012 07:38 AM

As such, when an envelope is selected and the amount in it is denoted as A, if it is the smaller amount, then (A, 2A) are in the two envelopes and if it is the larger amount, then (A, 1/2A) are in the two envelopes.

If you have only two amounts, which TEP clearly states, then A being the smallest or the biggest value means that it has different values in (A,2A) and (A,A/2), which means you cannot treat them as the same value, as you do.

kkwan - 27 December 2012 07:38 AM

A is a constant and it can be the smaller amount in (A, 2A) and the larger amount in (A, 1/2A). There is no “joining of the events” and neither are there “three amounts” in each of the two mutually exclusive events.

No joining? Sure? So then in (X,2X) I can gain or loose X, and in (X,X/2) I can gain or loose X/2. So in both separate events there is no advantage in switching. So under both descriptions, there is no reason to switch. Agree? In the end, I do not know if I took the smallest or the biggest amount first. Now kkwan, why is my derivation not valid? In your answer you are not allowed to join both situations: they cannot be the case at the same time. I only do not know which event is the case. But I know that in both cases loss and gain exactly cancel.

kkwan - 27 December 2012 07:38 AM

It is, but in the context of the TEP either (10, 20) or (10, 5) are in the two envelopes and not exclusively only (10, 20) because you don’t know which event is actual. As such, you cannot assume (10, 20) are in the two envelopes because that is begging the question.

So two amounts, kkwan: (5,10) or (10,20). If it is (5,10) I possibly loose or gain 5, if it is (10,20) I possibly loose or gain 10. Beware: do not join both situations!

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Posted: 27 December 2012 09:54 AM   [ Ignore ]   [ # 1913 ]
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StephenLawrence - 27 December 2012 08:07 AM

You can either simplify and consider one possible pair as GdB does. Or consider all the possible pairs.

If we consider one possible pair as GdB does, it is simpler but it is incomplete and begs the question because there are two possible pairs but we don’t know which pair is actual.

If we consider all the possible pairs, then it is either (X, 2X) or (X, 1/2) which is what the argument to switch does and that does not beg the question.

What doesn’t make sense is to just consider two possible pairs. It would if you knew what was in your envelope and knew the probability of having either pair. Since you don’t know these two things you can’t get this specific because it doesn’t work. We can tell it doesn’t work by considering what would happen if you played the game over and over. Apart from seeing that switching makes no difference, we also see that we would not keep getting the same two pairs and you wouldn’t keep selecting $10.

The TEP is about what to do if we are totally ignorant of the amount in the selected envelope and what are the amounts in the two envelopes. What we can deduce is either (X, 2X) or (X, 1/2X) are in the two envelopes.

It does not matter what is in the selected envelope but we know the probability of having either pair is 1/2 as there are only two pairs.

As such, it is prudent and rational to denote the amount in the selected envelope as A (as a constant) and deduce if it is either the smaller or the larger amount, then either (A, 2A) or (A, 1/2A) are in the two envelopes which is consistent with the above.

There are only two pairs to consider, so why should we not keep getting the same two pairs?

We can’t tell what works because we are totally ignorant. We only have the argument to switch as a guide in the face of uncertainty which recommends it is beneficial to switch.

A is the amount in the selected envelope irrespective of how many times we play the TEP.

If you want to reduce all these possible games into one, A would be all the numbers you would get divided by the number of games. And what is in the other envelope would be all those numbers divided by the number of games.

A, denoted as the amount in the selected envelope is a constant in the TEP. If there are 100 repetitions of the TEP, A is (100A divided by 100) which is A.

What is in the other envelope is either 2A or 1/2A. With 100 repetitions of the TEP, it is either (200A divided by 100) which is 2A or (50A divided by 100) which is 1/2A.

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Posted: 27 December 2012 09:56 AM   [ Ignore ]   [ # 1914 ]
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kkwan - 27 December 2012 08:41 AM
StephenLawrence - 26 December 2012 09:12 AM

What are the amounts in the envelopes?

Generally, either (X, 2X) or (X, 1/2X) are in the two envelopes, but we don’t know which pair is actual in the context of the TEP.

As you know I’m asking for a concrete example. So I’m asking which pair is actual.

So the entity selected from two pairs just like in my example?

Not quite so.

Then how did the entity select the pair?

Stephen

[ Edited: 27 December 2012 10:04 AM by StephenLawrence ]
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Posted: 27 December 2012 10:10 AM   [ Ignore ]   [ # 1915 ]
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kkwan - 27 December 2012 09:54 AM


There are only two pairs to consider, so why should we not keep getting the same two pairs?

There is only one pair, once the money is in the envelopes!

But if we consider there are two pairs then we would still not get the same amount in our envelope each time, which is why switching wouldn’t work.

Stephen

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Posted: 27 December 2012 11:50 AM   [ Ignore ]   [ # 1916 ]
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GdB - 27 December 2012 09:03 AM

To start with, kkwan, again you do not do the calculation for my non-TEP situation. if you are not able to do this, then it is clear that you see that this the situation where your formula fits, and not for TEP.

You do the calculation of your ridiculous non-TEP.

It is not quite superfluous. It makes a point clear. TEP fits two descriptions:

a) One envelope contains twice the amount of the other.
b) One envelope contains half of the other.

Now, as the envelopes are indistinguishable, we don’t know what the ‘one’ is and the ‘other’. So the descriptions can be formalised by:

a) (X,2X)
b) (Y,Y/2)

Now the question is: what is Y expressed in terms of X? It follows from the fact that we have only two envelopes with only two amounts.
This is solved by Y = 2X:
Then b) changes into:

(2X,X),

i.e. the same two envelopes with the same two amounts, but switched. And that is exactly what the TEP is about: two amounts.

You say Y = X, so b) becomes:

(X,X/2)

However there is no meaningful value (except 0) for which (X,2X) can be transformed to (X,X/2): except you introduce a third amount, which is not according TEP.

However 3 amounts fit to my non-TEP situation.

(X, 2X) is equivalent to (2X, X) as the two indistinguishable envelopes can be switched around and also we don’t know which is which, which means your a) and b) (for Y = 2X) is trivially tautological. LOL

OTOH, if Y = X, then b) is (X, 1/2X) which is distinct from a) is (X, 2X).

We are not transforming (X, 2X). We are merely evaluating b) for Y = X as you do with Y = 2X.

There is no third amount as (X, 2X) and (X, 1/2X) are mutually exclusive events.

Y is superfluous as the given relation in the TEP “that one envelope contains twice the amount in the other envelope” is completely described by X as either (X, 2X) or (X, 1/2X).

Occam’s razor. grin

If you have only two amounts, which TEP clearly states, then A being the smallest or the biggest value means that it has different values in (A,2A) and (A,A/2), which means you cannot treat them as the same value, as you do.

There are two amounts does not rule out two amounts in each of two mutually exclusive events.

A is a constant which means it has the same value in either (A, 2A) or (A, 1/2A). So, A is the smaller amount in (A, 2A) and the larger amount in (A, 1/2A).

No joining? Sure? So then in (X,2X) I can gain or loose X, and in (X,X/2) I can gain or loose X/2. So in both separate events there is no advantage in switching. So under both descriptions, there is no reason to switch. Agree? In the end, I do not know if I took the smallest or the biggest amount first. Now kkwan, why is my derivation not valid? In your answer you are not allowed to join both situations: they cannot be the case at the same time. I only do not know which event is the case. But I know that in both cases loss and gain exactly cancel.

With A denoted as the amount in the selected envelope, either (A, 2A) or (A, 1/2A) are in the two envelopes. The two events: either (A, 2A) or (A, 1/2A) are mutually exclusive.

1. In (A, 2A) by switching, gain is A

2. In (A, 1/2A) by switching, loss is 1/2A.

Where is the “joining”?

So two amounts, kkwan: (5,10) or (10,20). If it is (5,10) I possibly loose or gain 5, if it is (10,20) I possibly loose or gain 10. Beware: do not join both situations!

With 10 as the selected amount:

1. If we are (10, 20) by switching, gain is 10

2. If we are in (10, 5) by switching, loss is 5

However, we don’t know whether we are in 1 or 2, so we cannot assume we are in 1 or 2 as you do, which is begging the question.

So, we must consider both 1 and 2 jointly without begging the question and not separately as you do in calculating gain/loss which is begging the question and clearly wrong.

Considering jointly is not synonymous to “join both” (whatever it means).

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Posted: 27 December 2012 12:04 PM   [ Ignore ]   [ # 1917 ]
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StephenLawrence - 27 December 2012 09:56 AM

As you know I’m asking for a concrete example. So I’m asking which pair is actual.

We don’t know which pair is actual in the context of the TEP.

So, there is no cast in stone concreteness. We can assign values to the pairs but that does not tell us which pair is actual at all.

Then how did the entity select the pair?

Randomly, as the probability of either pair is 1/2.

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Posted: 27 December 2012 12:24 PM   [ Ignore ]   [ # 1918 ]
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StephenLawrence - 27 December 2012 10:10 AM

There is only one pair, once the money is in the envelopes!

But if we consider there are two pairs then we would still not get the same amount in our envelope each time, which is why switching wouldn’t work

The problem is, we don’t know which pair is actual.

That is obvious in your non-TEP example where the two pairs coexist, but in the TEP, it is not so obvious as the two pairs do not coexist.

But, switching works half the time in your non-TEP example when $20 is selected.

OTOH, the argument to switch in the TEP says switching works all the time if the selected amount is denoted as A (which is a constant).

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Posted: 27 December 2012 12:40 PM   [ Ignore ]   [ # 1919 ]
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kkwan - 27 December 2012 12:04 PM
StephenLawrence - 27 December 2012 09:56 AM

As you know I’m asking for a concrete example. So I’m asking which pair is actual.

We don’t know which pair is actual in the context of the TEP.

So, there is no cast in stone concreteness. We can assign values to the pairs but that does not tell us which pair is actual at all.

Kkwan you can take a God like view and say what is in the envelopes. Or you can have the entity know. Or have it all recorded on film to look at later.

If you won’t you’re just deliberately avoiding finding anything out.

Then how did the entity select the pair?

Randomly, as the probability of either pair is 1/2.

That’s only true if he selected at random from two possible pairs.

And answer with a little detail for goodness sake. Did he toss a coin or what? How many numbers did he select the two number from?

This is like extracting teeth. smile

Stephen

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Posted: 28 December 2012 07:11 PM   [ Ignore ]   [ # 1920 ]
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StephenLawrence - 27 December 2012 12:40 PM

Kkwan you can take a God like view and say what is in the envelopes. Or you can have the entity know. Or have it all recorded on film to look at later.

If you won’t you’re just deliberately avoiding finding anything out.


If we are omniscient, opened both envelopes, have X-ray vision or peeked (in recording the whole selection process by the entity on video or film) then and only then, will we know what are the amounts in the two envelopes. However, as none of the options are possible or allowed in the TEP, we are shackled to ignorance forever.

Of course, the entity who selected what amounts to put into the envelopes knows, but it will not inform us to keep us totally ignorant and clueless.

That’s only true if he selected at random from two possible pairs.

And answer with a little detail for goodness sake. Did he toss a coin or what? How many numbers did he select the two number from?

This is like extracting teeth. smile

The entity could select at random from the two possible pairs, perhaps by tossing an unbiased coin, with a random number generator (depending on whether the number generated is even or odd) or with any other similar method.

It could select the two numbers from a finite number of numbers with the proviso that for any finite number X, (X, 2X) and (X, 1/2X) are the two possible pairs.

There is a maximum amount of 32 teeth to extract, in any adult person.  grin

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