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The two envelopes problem
 Posted: 10 December 2012 01:04 PM [ Ignore ]   [ # 1786 ]
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So you say:

- For every two amounts it is true that X -Y = -1 * (Y -X), i.e. where you have no idea what X and Y are
- But not for its subset where you do not know if Y = 2X or X = 2Y

Wow, that is interesting. Something is true for all elements of a set, but not for a subset of that set. Mathematics was wrong all the time…

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 Posted: 10 December 2012 01:48 PM [ Ignore ]   [ # 1787 ]
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StephenLawrence - 09 December 2012 11:59 PM

So 2. The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2. is not true in my example, agreed?

Yes, as it is non-TEP with four envelopes in two existing groups of two envelopes.

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 Posted: 10 December 2012 02:11 PM [ Ignore ]   [ # 1788 ]
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GdB - 10 December 2012 01:04 PM

So you say:

- For every two amounts it is true that X -Y = -1 * (Y -X), i.e. where you have no idea what X and Y are
- But not for its subset where you do not know if Y = 2X or X = 2Y

Wow, that is interesting. Something is true for all elements of a set, but not for a subset of that set. Mathematics was wrong all the time…

Not so, GdB.

This was what I wrote in post 1785:

If there is one and only one possibility wrt what are the amounts in the two envelopes, then there is no advantage in switching as the potential gain is equal to the potential loss (in amount) irrespective of the difference of the amounts in the two envelopes.

However, in the TEP, there are two and only two possible situations wrt what amounts are in the envelopes, but we don’t know which situation is actual. As such, we cannot apply the above as we must consider both situations jointly.

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 Posted: 11 December 2012 04:57 AM [ Ignore ]   [ # 1789 ]
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kkwan - 10 December 2012 02:11 PM

If there is one and only one possibility wrt what are the amounts in the two envelopes, then there is no advantage in switching as the potential gain is equal to the potential loss (in amount) irrespective of the difference of the amounts in the two envelopes.

However, in the TEP, there are two and only two possible situations wrt what amounts are in the envelopes, but we don’t know which situation is actual. As such, we cannot apply the above as we must consider both situations jointly.

I did not react because it just makes no sense. In TEP in the first place you have two amounts. Now for any combination of amounts, it is true that (X -Y) = -1 * (Y -X). Now , without the condition ‘one envelope contains twice the amount of the other’ there are more possibilities than with the condition. The TEP pairs are a special, but true subset of all combinations, i.e. there are less possibilities then for the set described as ‘one envelope contains one amount, and the other envelope contains another amount’.

So what we have is a statement that is true for all possible combinations of amounts, but you say this statement does not apply to a true subset of these. That is impossible.

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 Posted: 12 December 2012 12:19 AM [ Ignore ]   [ # 1790 ]
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GdB - 11 December 2012 04:57 AM

I did not react because it just makes no sense. In TEP in the first place you have two amounts. Now for any combination of amounts, it is true that (X -Y) = -1 * (Y -X). Now , without the condition ‘one envelope contains twice the amount of the other’ there are more possibilities than with the condition. The TEP pairs are a special, but true subset of all combinations, i.e. there are less possibilities then for the set described as ‘one envelope contains one amount, and the other envelope contains another amount’.

So what we have is a statement that is true for all possible combinations of amounts, but you say this statement does not apply to a true subset of these. That is impossible.

Not quite so.

What you are proposing is to consider one and only one possible situation and that implies you know what are actually in the two envelopes which begs the question and also contravenes the basic nature of the TEP.

In the TEP, there are two and only two possible situations, either (X, 2X) or (X, 1/2X) for any finite value of X, but you don’t know which situation is actual.

As such, we must consider and compare both possible situations jointly in order to determine gain/loss by switching without begging the question and contravening the basic nature of the TEP.

It is feasible and possible to do that. Is that so difficult to understand?

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 Posted: 12 December 2012 12:36 AM [ Ignore ]   [ # 1791 ]
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kkwan - 12 December 2012 12:19 AM

It is feasible and possible to do that. Is that so difficult to understand?

It is not feasible, and it is impossible to understand, because it is wrong.

The TEP clearly says: there are two envelopes with each one single amount in it. It is not ‘there is one envelope with a fixed amount, and the other varies’. You pick one of two fixed amounts, that is TEP. For the amounts there are infinite possibilities, not two. But in these infinite possibilities two subsets can be distinguished, dependent on your choice: those where you pick the smallest first, and those where you pick the biggest amount first. This is the correct way of formulating your possibilities: the two possibilities are you pick X first and the other contains 2X, or you pick the envelope with 2X first and the other contains X.

Is that so difficult to understand?

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 Posted: 12 December 2012 02:10 AM [ Ignore ]   [ # 1792 ]
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kkwan - 10 December 2012 01:48 PM
StephenLawrence - 09 December 2012 11:59 PM

So 2. The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2. is not true in my example, agreed?

Yes, as it is non-TEP with four envelopes in two existing groups of two envelopes.

OK

I think we can see that if the pair is picked from finite options 2. is always false as all cases will be essentially the same as my example. Do you think that is right?

Now imagine you look in the envelope and see that you have \$20. What should you do? I’ll give the answer to save time, you should switch because you know the probability is 50/50.

Now let’s change it a bit. There are two pairs of envelopes one of which has been selected at random. You know that in each pair one envelope contains half the other. You don’t know what the pairs are though. In fact they are as in my previous example
(\$10, \$20)  (\$20 \$40)

You open your envelope and see you have \$20. What should you do? (I think the answer to this is revealing)

Stephen

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 Posted: 12 December 2012 09:57 PM [ Ignore ]   [ # 1793 ]
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GdB - 12 December 2012 12:36 AM
kkwan - 12 December 2012 12:19 AM

It is feasible and possible to do that. Is that so difficult to understand?

It is not feasible, and it is impossible to understand, because it is wrong.

Not quite so.

What is so infeasible and impossible to understand (that we must consider the two possible situations jointly instead of the one and only one situation because we don’t know which situation is actual)?

The TEP clearly says: there are two envelopes with each one single amount in it. It is not ‘there is one envelope with a fixed amount, and the other varies’. You pick one of two fixed amounts, that is TEP. For the amounts there are infinite possibilities, not two. But in these infinite possibilities two subsets can be distinguished, dependent on your choice: those where you pick the smallest first, and those where you pick the biggest amount first. This is the correct way of formulating your possibilities: the two possibilities are you pick X first and the other contains 2X, or you pick the envelope with 2X first and the other contains X.

No, your approach is flawed and incorrect because it fundamentally begs the question, misleads, is incomplete and contravenes the basic nature of the TEP whereby for any finite value of X, there are two and only two possible situations: either (X, 2X) or (X, 1/2X), not one and only one situation (X, 2X) because we don’t know which situation is actual.

As such, with A is denoted as the amount in the selected amount:

1. If A is the smaller amount, then 2A is in the other envelope, thus (A, 2A) are in the two envelopes

2. If A is the larger amount, then 1/2A is in the other envelope, thus (A, 1/2A) are in the two envelopes.

This is consistent with the basic nature of the TEP whereas your interpretation of the TEP, is not.

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 Posted: 12 December 2012 11:07 PM [ Ignore ]   [ # 1794 ]
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kkwan - 12 December 2012 09:57 PM

No, your approach is flawed and incorrect because it fundamentally begs the question, misleads, is incomplete and contravenes the basic nature of the TEP whereby for any finite value of X, there are two and only two possible situations: either (X, 2X) or (X, 1/2X), not one and only one situation (X, 2X) because we don’t know which situation is actual.

Sorry, the description of TEP is ‘One envelope contains twice as much as the other’. This is fully described by having a pair (X,2X). The two possibilities are then that you pick the smallest first, and then switch to the biggest, or the other way round. One can describe the first with X—> 2X, and the second as 2X—> X. What you do, is divide the second by 2, and you get X—> X/2, with other words, you changed the amounts. What you should do, if you want, is write the second situation as Y—> Y/2, but you must add to it that Y = 2X.

Your solution is invalid, because you are changing the value of ‘X’, i.e. X is not a constant. The amount you pick first is the smallest or the biggest, and therefore is not a constant. You do not use the fact that there are two, and only two, unknown, but fixed amounts, and your solution does not obey the simple rule that for all pairs of amounts (X - Y) = -1 * (Y - X).

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 Posted: 13 December 2012 08:05 AM [ Ignore ]   [ # 1795 ]
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StephenLawrence - 12 December 2012 02:10 AM

I think we can see that if the pair is picked from finite options 2. is always false as all cases will be essentially the same as my example. Do you think that is right?

Not necessarily so, as “all cases” in the TEP is not the same as that in your non-TEP example.

Your non-TEP example is equivalent to four envelopes in two existing groups fo two envelopes.

So, we have Group 1 (\$20, \$40) and group 2 (\$20, \$10).

With A as the amount in the selected envelope, A must be from either one of the two groups:

1. If A is the smaller amount, it could be from group 1 (\$20) or from group 2 (\$10).

2. If A is the larger amount, it could be from group 1 (\$40) or from group 2 (\$20).

So, A varies in value in both 1 and 2 depending on whether A is the smaller or the larger amount.

With A varying in values in both 1 and 2, there is no specific A to assign probabilities of A as either the smaller or the larger amount as in the TEP.

Thus, which A is the reference A to assign probability? Is it A (\$20), A (\$10), A (\$40) or A (\$20)?

With no reference A, it seems no probabilities can be determined at all.

However, as we can compare A to the other amount in any one group, the probability is 1/2 that A is either the smaller or the larger amount because there are four possible values of A of which two are the smaller and two are the larger amount.

So, step 2 is true even in the context of your non-TEP example.

OTOH, in the context of the TEP, Step 2 refer to A (denoted as the amount in the selected envelope) which must be of a constant value and that is the reference A but A (notwithstanding it is of a constant value) could be either the smaller or the larger amount as in either (A, 2A) or (A, 1/2A) but not as variable value A as in the above.

In the TEP, with two and only two mutually exclusive possible events of either (A, 2A) or (A, 1/2A) for any finite value of A, step 2 is valid as the probability that A is the smaller amount is 1/2 and that it is the larger amount is also 1/2, is clearly true.

So, the probability of what A is (as either the smaller or the larger amount) is 1/2 + 1/2 = 1, which is also obviously true as there are two and only two mutually exclusive possible events for any finite value of A.

OTOH, your non-TEP example is about mutually inclusive events as both groups 1 and 2 do exist at the same time. As such, step 2 does not necessarily apply to your example as step 2 is specific to the TEP whereby the events are mutually exclusive.

So, you are comparing the validity of Step 2 in the context of two entirely different games i.e. the TEP and your non-TEP which is inappropriate as it is like comparing apples and oranges.

Notwithstanding that, as I have shown above, step 2 is true even in the context of your non-TEP example.

Now imagine you look in the envelope and see that you have \$20. What should you do? I’ll give the answer to save time, you should switch because you know the probability is 50/50.

In the TEP, yes, but in the context of your non-TEP example, it is not so simple.

Now let’s change it a bit. There are two pairs of envelopes one of which has been selected at random. You know that in each pair one envelope contains half the other. You don’t know what the pairs are though. In fact they are as in my previous example
(\$10, \$20)  (\$20 \$40)

You open your envelope and see you have \$20. What should you do? (I think the answer to this is revealing)

With mutually inclusive events, as in your non-TEP example:

1. If \$20 is selected from group 1 (\$20, \$40), by switching, gain is \$20.

2. If \$20 is selected from group 2 (\$20, \$10), by switching, loss is \$10.

The probability of 1, P (1) =1/2 and the probability of 2, P (2) = 1/2.

And the probability of 1 and 2, P (1 and 2) = 2/4 = 1/2.

For mutually inclusive events, P (1 or 2) = P (1) + P (2) - P (1 and 2)

So, the probability of 1 or 2, P (1 or 2) = 1/2 + 1/2 - 1/2 = 1/2

As we don’t know whether we are in (1 or 2), we should consider the probability of 1 or 2 which is 1/2. We cannot consider the separate probabilities of 1 or 2 because we don’t know whether we are in 1 or 2.

This is not the same as in the TEP (with mutually exclusive events) whereby the probability of gain/loss is simply 1/2.

So, the benefit of switching is not so simple as in the TEP.

However, as the probability of 1 or 2 is 1/2, it is beneficial to switch.

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 Posted: 13 December 2012 08:42 AM [ Ignore ]   [ # 1796 ]
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GdB - 12 December 2012 11:07 PM

Sorry, the description of TEP is ‘One envelope contains twice as much as the other’. This is fully described by having a pair (X,2X). The two possibilities are then that you pick the smallest first, and then switch to the biggest, or the other way round. One can describe the first with X—> 2X, and the second as 2X—> X. What you do, is divide the second by 2, and you get X—> X/2, with other words, you changed the amounts. What you should do, if you want, is write the second situation as Y—> Y/2, but you must add to it that Y = 2X.

Not so. The TEP is complete if and only if for any finite value of X, either (X, 2X) or (X, 1/2X) are in the two envelopes.

Your solution is invalid, because you are changing the value of ‘X’, i.e. X is not a constant. The amount you pick first is the smallest or the biggest, and therefore is not a constant. You do not use the fact that there are two, and only two, unknown, but fixed amounts, and your solution does not obey the simple rule that for all pairs of amounts (X - Y) = -1 * (Y - X).

X is a variable, a symbol that represents a quantity in a mathematical expression. As such, X can be assigned any finite value without contradiction. There is no requirement in the TEP that X must be a constant. However, if a value is assigned to X, it becomes a constant.

Your simple rule is tautological which is applicable if and only if you know what are the actual amounts in the two envelopes which you don’t, in the context of the TEP.

If you assume you know, it begs the question and contravenes the basic nature of the TEP. So, fallacious reasoning and contravening the basic nature of the TEP render your “solution” as untenable.

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 Posted: 13 December 2012 09:24 AM [ Ignore ]   [ # 1797 ]
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kkwan - 13 December 2012 08:42 AM

Not so. The TEP is complete if and only if for any finite value of X, either (X, 2X) or (X, 1/2X) are in the two envelopes.

Let us say you are given two indistinguishable envelopes, each of which contains a positive sum of money. One envelope contains twice as much as the other. You may pick one envelope and keep whatever amount it contains. You pick one envelope at random but before you open it you are offered the possibility to take the other envelope instead.

The question is if it is a good strategy to switch. All your X,2X,X/2 is part of your argument, not of the way the TEP is posed.

kkwan - 13 December 2012 08:42 AM

X is a variable, a symbol that represents a quantity in a mathematical expression. As such, X can be assigned any finite value without contradiction. There is no requirement in the TEP that X must be a constant. However, if a value is assigned to X, it becomes a constant.

There is no requirement for X in TEP at all. Using of constants is a way the puzzle can be solved. But you do not use X as a constant.
Example: I say X - X <> 0. A mathematician says that cannot be: for every number, when you subtract the same number you get zero. But then I say “Yes, but I do not know what X is, it can be anything, so the first can be different from the second X”.

kkwan - 13 December 2012 08:42 AM

Your simple rule is tautological which is applicable if and only if you know what are the actual amounts in the two envelopes which you don’t, in the context of the TEP.

My rule is valid for all pair of numbers, known and unknown equally, as is the rule X - X = 0 for all possible numbers, known or unknown. You are saying that when you have two unknown amounts, that when you switch your possible gain and loss are not the same. That is only true when you change the amounts during the switching.

[ Edited: 13 December 2012 09:27 AM by GdB ]
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 Posted: 13 December 2012 12:09 PM [ Ignore ]   [ # 1798 ]
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kkwan - 13 December 2012 08:05 AM
StephenLawrence - 12 December 2012 02:10 AM

I think we can see that if the pair is picked from finite options 2. is always false as all cases will be essentially the same as my example. Do you think that is right?

Not necessarily so, as “all cases” in the TEP is not the same as that in your non-TEP example.

Uh Uh, note I said in all cases in which the pair you are playing with is selected from finite options. So it will always be like my example, I think you’ll find.

Thus, which A is the reference A to assign probability? Is it A (\$20), A (\$10), A (\$40) or A (\$20)?

A is the amount in your envelope what ever it is. What else could A be?

With no reference A, it seems no probabilities can be determined at all.

Looks like you’ve stumbled upon the solution to me.

So, step 2 is true even in the context of your non-TEP example.

No, we’ve already agreed step 2. is false in my set up.

So, you are comparing the validity of Step 2 in the context of two entirely different games i.e. the TEP and your non-TEP which is inappropriate as it is like comparing apples and oranges.

The difference is my envelopes are picked from finite options and in yours the options are infinite, that’s all, I think. And I’ve made no comparison to the TEP, I’m just working through these examples. Applying the information to the TEP comes shortly.

Now let’s change it a bit. There are two pairs of envelopes one of which has been selected at random. You know that in each pair one envelope contains half the other. You don’t know what the pairs are though. In fact they are as in my previous example
(\$10, \$20)  (\$20 \$40)

You open your envelope and see you have \$20. What should you do? (I think the answer to this is revealing)

With mutually inclusive events, as in your non-TEP example:

1. If \$20 is selected from group 1 (\$20, \$40), by switching, gain is \$20.

2. If \$20 is selected from group 2 (\$20, \$10), by switching, loss is \$10.

The probability of 1, P (1) =1/2 and the probability of 2, P (2) = 1/2.

And the probability of 1 and 2, P (1 and 2) = 2/4 = 1/2.

For mutually inclusive events, P (1 or 2) = P (1) + P (2) - P (1 and 2)

So, the probability of 1 or 2, P (1 or 2) = 1/2 + 1/2 - 1/2 = 1/2

As we don’t know whether we are in (1 or 2), we should consider the probability of 1 or 2 which is 1/2. We cannot consider the separate probabilities of 1 or 2 because we don’t know whether we are in 1 or 2.

This is not the same as in the TEP (with mutually exclusive events) whereby the probability of gain/loss is simply 1/2.

So, the benefit of switching is not so simple as in the TEP.

However, as the probability of 1 or 2 is 1/2, it is beneficial to switch.

That is the wrong answer. You’re forgetting you don’t know what the two pairs are.

Stephen

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 Posted: 13 December 2012 09:12 PM [ Ignore ]   [ # 1799 ]
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GdB - 13 December 2012 09:24 AM

Let us say you are given two indistinguishable envelopes, each of which contains a positive sum of money. One envelope contains twice as much as the other. You may pick one envelope and keep whatever amount it contains. You pick one envelope at random but before you open it you are offered the possibility to take the other envelope instead.

That one envelope contains twice as much as the other is satisfied by two and only two possibilities for any finite value of X, either (X, 2X) or (X, 1/2X).

This is implicit in the TEP which you have ignored and as such your interpretation and description of the TEP is incomplete, misleading and begs the question.

The question is if it is a good strategy to switch. All your X,2X,X/2 is part of your argument, not of the way the TEP is posed.

It is the correct way (without begging the question) and it is beneficial to switch.

There is no requirement for X in TEP at all. Using of constants is a way the puzzle can be solved. But you do not use X as a constant.
Example: I say X - X <> 0. A mathematician says that cannot be: for every number, when you subtract the same number you get zero. But then I say “Yes, but I do not know what X is, it can be anything, so the first can be different from the second X”.

Not so. X is a variable, but it becomes a constant when a value is assigned to it.

My rule is valid for all pair of numbers, known and unknown equally, as is the rule X - X = 0 for all possible numbers, known or unknown. You are saying that when you have two unknown amounts, that when you switch your possible gain and loss are not the same. That is only true when you change the amounts during the switching.

Your rule is valid if there is one and only one possible situation, but in the TEP as it is, there are two and only two possible situations but as we don’t know which situation is actual, we must consider the two situations jointly. In such an environment, your rule is inappropriate and thus, is inapplicable.

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 Posted: 13 December 2012 10:21 PM [ Ignore ]   [ # 1800 ]
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StephenLawrence - 13 December 2012 12:09 PM

Uh Uh, note I said in all cases in which the pair you are playing with is selected from finite options. So it will always be like my example, I think you’ll find.

A is the amount in your envelope what ever it is. What else could A be?

Yes, but in your non-TEP example, there is no fixed value of A which can be used as a reference.

Looks like you’ve stumbled upon the solution to me.

Yes.

No, we’ve already agreed step 2. is false in my set up.

I did, without fully analyzing your set up. My apologies for not taking the time and effort to do that.

Now, I disagree as we can compare A to the other amount in any one group, the probability is 1/2 that A is either the smaller or the larger amount because there are four possible values of A of which two are the smaller and two are the larger amount.

The difference is my envelopes are picked from finite options and in yours the options are infinite, that’s all, I think. And I’ve made no comparison to the TEP, I’m just working through these examples. Applying the information to the TEP comes shortly.

How does finite options differ and why do you think my options are infinite? Please explain.

That is the wrong answer. You’re forgetting you don’t know what the two pairs are.

I did not forget that, but it does not matter.

We know that “in each pair one envelope contains half the other”.

Let the two pairs be in two and only two mutually inclusive groups for any finite value of X.

Group 1 (X, 2X) and group 2 (X, 1/2X):

1. If X is selected from group 1 (X, 2X), by switching, gain is X

2. If X is selected from group 2 (X, 1/2X), by switching, loss is 1/2X

The probability of 1 is 1/2 and the probability of 2 is 1/2.

The probability of 1 and 2 is 2/4 =1/2.

So, the probability of 1 or 2 is 1/2 + 1/2 - 1/2 = 1/2.

As we don’t know whether we are in situation 1 or 2, we should consider the probability of 1 or 2 which is 1/2.

Hence, it is beneficial to switch.

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