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The two envelopes problem
 Posted: 23 December 2012 12:54 PM [ Ignore ]   [ # 1846 ]
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kkwan - 23 December 2012 09:49 AM

3. Thus, in step 2, the probability of A as the smaller is 1/2 and as the larger amount is also 1/2 is clearly true as the probability of A (as either the smaller or the larger amount) is 1.

Nope.

In my example if A is \$10 the probability it is the smaller amount is not 1/2 it is 1.

Stephen

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 Posted: 24 December 2012 07:36 AM [ Ignore ]   [ # 1847 ]
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kkwan - 23 December 2012 10:53 AM
GdB - 22 December 2012 04:11 AM

- You pick the smallest of the two amounts first,
or
- You pick the biggest of the same two amounts first

Do you deny this?

We select either the smaller or the larger amount X in either (X, 2X) or (X, 1/2X) but not in exclusively (X, 2X).

It is ‘yes’ or ‘no’ kkwan.

kkwan - 23 December 2012 10:53 AM

Not so. Either (X, 2X) or (X, 1/2X) can occur at the same time. Thus, at any one time, there are two and only two possible amounts: either (X, 2X) or (X, 1/2X), not three possible amounts.

That will not do: you say there are two possibilities, (X,2X) and (X,X/2), so for the same X we have three possible amounts. Of course not at the same time, but there are are three possibilities. But I know there are only two.

kkwan - 23 December 2012 10:53 AM

What you are proposing are two possibilities in one event {(X, 2X) which is equivalent to (2X, X)}.

There are only two amounts, and given the TEP, there are two possibilities: I pick the smallest amount first, or the second. Do you deny this?

kkwan - 23 December 2012 10:53 AM

My reasoning fits the TEP, not your non-TEP.

Proof? What formula would be valid? Oh, I know again, you will say again that you are interested in TEP only. I have shown that the argument fits the non-TEP situation. Now show me where I am wrong, kkwan. Here it is again, in short:

Situation:
You can pick an envelope with an for you unknown amount. After your pick, a second envelope is filled with another amount that is half or twice the amount you have in your envelope, with a chance of 50% of both. You are allowed to switch. Should you?

Solution:
1. Call the amount on the envelope you have A
2. Then the other envelope might contain 2A with a chance of 1/2.
3. Or the other envelope might contain A/2 with a chance of 1/2.
4. Now the expected value expressed in terms the amount you have of switching is:

5. This is more than the expected value of you envelope which is of course A.
6. So you should switch.

Two questions:
a. Where is the error?
b. When it has no errors, why are the solutions of this situation and of TEP the same, where at the same time the situations do differ?

kkwan - 23 December 2012 10:53 AM

That one envelope contains twice as much as the other is satisfied by two and only two possible distinct events either (X, 2X) or (X, 1/2X) for any finite value of X and not by exclusively one (X, 2X) event.

Do you deny that?

Yes of course: the descriptions ‘one envelope contains twice the amount of the other’ is equivalent to ‘one envelope contains half the amount of the other’. Agree? Don’t forget: the envelopes are identical, and ‘one’ or the ‘other’ are no fix references!
So your descriptions (X,2X) and (X,X/2) are the same; but not for the same X of course. By doing as if both X’es represent the same amount your ‘derivation’ is wrong.

kkwan - 23 December 2012 10:53 AM

It has meaning.

No. You do it just again. What the hell does ‘considering both mutually exclusive events jointly’ mean???

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 Posted: 24 December 2012 12:31 PM [ Ignore ]   [ # 1848 ]
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kkwan - 23 December 2012 08:53 AM

It is a logical paradox. Historically, it is of the form: “I am lying”.

This is the liar paradox. It is self-referential which cannot be resolved with classical two value logic.

...

In philosophy and logic, the liar paradox or liar’s paradox (pseudomenon in Ancient Greek) is the statement “this sentence is false.” Trying to assign to this statement a classical binary truth value leads to a contradiction (see paradox).

If “this sentence is false” is true, then the sentence is false, which would in turn mean that it is actually true, but this would mean that it is false, and so on ad infinitum.

Similarly, if “this sentence is false” is false, then the sentence is true, which would in turn mean that it is actually false, but this would mean that it is true, and so on ad infinitum.

Logical structure of the liar paradox:

“A = ‘A = false’”

This is an equation from which the truth value of A = “this statement is false” could hopefully be obtained. In the boolean domain “A = false” is equivalent to “not A” and therefore the equation is not solvable.

I brought this up because it is commonly assumed that this is an unresolvable paradox as you’ve stated. The problem is is that it is only an apparent one because you (and most others) interpret the statement’s meaning of the sentence as a real claim of ironic truth when the sentence’s statement is merely a set of symbols that, in this case, only an English reader provides meaning to. I am capable of printing the statement, “I always lie.” But to presume that its meaning can ever be an absolute truth is just a game of pretending for the sake of entertainment.
Godel’s Theorem is wrong and set back science in an unfortunate way. Statements or theories cannot make a logical claim of truth or falsity regarding its very own statements. It’s a rhetorical trick.

Likewise, you are mixing up semantic meaning and the statements in your TEP argument. The claimed formula is NOT a possible nor reasonable expression because it tricks you into thinking that what is in the envelopes means something to determining the probability. Just because they are quantities of something that can be algebraically expressed as something relative to one another does not mean they have any relevance to the numbers for the formula required to determine the probability of receiving any choice. When you state a probability, it’s not reasonable to interpret the particular or specific meaning of the content the probability refers to but the general arbitrary class which represents separate individual possibilities. You can say:

1/2(the time you will choose one particular selection of two) + 1/2(the time you will choose the other particular selection of two) = 1(of the whole set of both possibilities)

If I pick one and am told that I may opt to select the other one, then the odds are reset to the beginning.

Envelope є {A(1), A(2)}

1/2 [chance of A(1)] + 1/2 [chance of A(2)] = 1[chances of getting A(1) or A(2)]

By the way, ‘classical’ logic isn’t necessarily a static convention of the old days. When people dis it, from the context of their writings, they appear not to have studied it. Rather, they have reinvented the wheel because they usually impose things on it that was never meant to be. For instance, the whole of Fuzzy Logic is a recreation of Caltech on what they presumed classical logic lacked. But reading their arguments for what traditional arguments lacked proved that they were focused in the wrong area. They picked on deductive reasoning as being incapable of supplying ‘degrees’ of things. But traditional logic uses “inductive” reasoning for that, not deductive! I think of ‘classical’ logic as the whole set of all logic while the newbies have just created “alternative” logics as if they were significantly groundbreaking and different. Perhaps its to gain political recognition for schools. Who knows.

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 Posted: 24 December 2012 01:05 PM [ Ignore ]   [ # 1849 ]
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The Monty Hall problem seems similar but isn’t. Kwan, perhaps you can check that out and see how Vos Savant resolved it and try to compare for argument’s sake.

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 Posted: 24 December 2012 09:15 PM [ Ignore ]   [ # 1850 ]
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Scott Mayers - 24 December 2012 12:31 PM

I brought this up because it is commonly assumed that this is an unresolvable paradox as you’ve stated. The problem is is that it is only an apparent one because you (and most others) interpret the statement’s meaning of the sentence as a real claim of ironic truth when the sentence’s statement is merely a set of symbols that, in this case, only an English reader provides meaning to. I am capable of printing the statement, “I always lie.” But to presume that its meaning can ever be an absolute truth is just a game of pretending for the sake of entertainment.

This is nonsense, Scott. Sentences can express facts, and facts are language independent. The sentence “This sentence is false” is a real paradox: it represents a fact that is true when it is false, and false when it is true.

Scott Mayers - 24 December 2012 12:31 PM

Godel’s Theorem is wrong and set back science in an unfortunate way. Statements or theories cannot make a logical claim of truth or falsity regarding its very own statements. It’s a rhetorical trick.

All right! The in mathematics general accepted Theorem of Gödel is wrong, and you have proof of that? The statement “This sentence contains one verb” is not a valid claim? Grammatics cannot be true or false? And in what ways did Gödel set back science?

To be very clear: TEP is not a paradox at all. kkwan’s argument is just plain wrong, and you do not have to use metamathematical arguments about paradoxes to solve it.

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 Posted: 24 December 2012 09:44 PM [ Ignore ]   [ # 1851 ]
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StephenLawrence - 23 December 2012 12:54 PM

In my example if A is \$10 the probability it is the smaller amount is not 1/2 it is 1.

What I wrote in post 1844 (which you quoted) was in the context of the TEP where step 2 was proposed and whereby it is clearly true.

In your example, if A is \$10, it must be from group 2 (\$20, \$10) where \$10 is the smaller amount but as you selected \$10 from the two possible selections of either \$10 or \$20 in group 2, the probability of selecting \$10 is 1/2.

It depends on how you interpret step 2 (in the context of your example) keeping in mind that step 2 was proposed specifically in the context of the TEP which is not synonymous to your example:

1. The probability that \$10 is the smaller amount in group 2 is 1.

2. The probability of selecting the smaller amount in group 2 is 1/2.

Step 2:

2. The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2.

I interpret the phrase “the probability that A is the smaller amount is 1/2” as in 2 whereby we don’t know whether A is \$10 or \$20 whereas you interpret it as in 1 whereby you assume you know A is \$10 (which begs the question) as you don’t.

This is the crucial difference wrt how you and I interpret your non-TEP example as well as the TEP.

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 Posted: 25 December 2012 12:48 AM [ Ignore ]   [ # 1852 ]
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GdB - 24 December 2012 07:36 AM

It is ‘yes’ or ‘no’ kkwan.

Yes, in the context that either (A, 2A) or (A, 1/2A) are in the two envelopes with A in the selected envelope.

That will not do: you say there are two possibilities, (X,2X) and (X,X/2), so for the same X we have three possible amounts. Of course not at the same time, but there are are three possibilities. But I know there are only two.

Not so. There are two possible amounts in each mutually exclusive event: either (X, 2X) or (X, 1/2X) for the same finite value of X and not three possible amounts amounts and that is all there is to it.

Counting possible amounts as you do is improper, inappropriate and incoherent in the context of the above two mutually exclusive events.

There are only two amounts, and given the TEP, there are two possibilities: I pick the smallest amount first, or the second. Do you deny this?

That is incomplete and misleading. There are two amounts in two and only two distinct mutually exclusive possible events either (X, 2X) or (X, 1/2X) for any finite value of X.

Proof? What formula would be valid? Oh, I know again, you will say again that you are interested in TEP only. I have shown that the argument fits the non-TEP situation. Now show me where I am wrong, kkwan. Here it is again, in short:

Situation:
You can pick an envelope with an for you unknown amount. After your pick, a second envelope is filled with another amount that is half or twice the amount you have in your envelope, with a chance of 50% of both. You are allowed to switch. Should you?

The following are the flaws/errors in your proposed non-TEP example:

1. Your non-TEP example has no selection per se as there only one envelope to pick. Hence, “selection” makes no sense.

2. In the TEP (whereby both indistinguishable envelopes are filled before selection) and subsequently we select one, but in your non-TEP example the second envelope is filled only after “selection”, which is preposterous.

3. As such, if A is the amount in your envelope, the other envelope could contain either 2A or 1/2A after “selection”, but not before.

4. Hence, your non-TEP can be manipulated after “selection” (which is no selection at all as stated in 1).

5. So, what safeguard is there (before the game begins) to ensure that the other envelope is not always filled with 2A or 1/2A after “selection” which means that the game is manipulated or fixed?

6. Without such a preexisting safeguard, your non-TEP example is fundamentally flawed as an unbiased game.

Solution:
1. Call the amount on the envelope you have A
2. Then the other envelope might contain 2A with a chance of 1/2.
3. Or the other envelope might contain A/2 with a chance of 1/2.
4. Now the expected value expressed in terms the amount you have of switching is:

5. This is more than the expected value of you envelope which is of course A.
6. So you should switch.

Two questions:
a. Where is the error?
b. When it has no errors, why are the solutions of this situation and of TEP the same, where at the same time the situations do differ?

The culminating fatal flaws/errors are from 1 to 6 as stated above. As such, the above analysis is likely to be invalid as steps 2 and 3 could be false in your non-TEP example.

OTOH, the TEP has no such potential flaws/errors and as all the steps 1 to 8 are valid, therefore the argument to switch is tenable.

Yes of course: the descriptions ‘one envelope contains twice the amount of the other’ is equivalent to ‘one envelope contains half the amount of the other’. Agree? Don’t forget: the envelopes are identical, and ‘one’ or the ‘other’ are no fix references!
So your descriptions (X,2X) and (X,X/2) are the same; but not for the same X of course. By doing as if both X’es represent the same amount your ‘derivation’ is wrong.

Not so, as either (X, 2X) or (X, 1/2X) are distinct possible events for any same finite value of X.

For instance, if X is 10, are either (10, 20) or (10, 5) not distinct possible events which satisfy the requirement of the TEP?

No. You do it just again. What the hell does ‘considering both mutually exclusive events jointly’ mean???

It means that we must consider the both mutually exclusive events together: either (X, 2X) or (X, 1/2X) when a selection is made, as we don’t know which event is actual.

[ Edited: 25 December 2012 03:03 AM by kkwan ]
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 Posted: 25 December 2012 01:09 AM [ Ignore ]   [ # 1853 ]
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kkwan,

It’s good that you are questioning my interpretation of 2. because it’s this that my solution depends upon. I think we can see that 2. should be interpreted as I am interpreting it and so there is the solution you are looking for.

This is what we need to focus on.

kkwan - 24 December 2012 09:44 PM

In your example, if A is \$10, it must be from group 2 (\$20, \$10) where \$10 is the smaller amount but as you selected \$10 from the two possible selections of either \$10 or \$20 in group 2, the probability of selecting \$10 is 1/2.

OK, since you are fixing the pair the probability of selecting \$10 is 1/2. (I’d say it’s 1/4 because you might have been playing with the other pair.)

But A means the amount in our envelope and since that is what A means the correct way to interpret 2. in this case, is if A is \$10 the probability of it being the smaller amount is 1. we can test this out and see in my example and we can move to the TEP once we’ve done that.

Stephen

[ Edited: 25 December 2012 01:11 AM by StephenLawrence ]
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 Posted: 25 December 2012 01:59 AM [ Ignore ]   [ # 1854 ]
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kkwan - 25 December 2012 12:48 AM

Yes, in the context that either (A, 2A) or (A, 1/2A) are in the two envelopes with A in the selected envelope.

My question concerns TEP, and not your “(A, 2A) or (A, 1/2A)” stuff.

So you said “yes” on:

- You pick the smallest of the two amounts first,
or
- You pick the biggest of the same two amounts first

Do you deny this?

What else do I need to show that your argumentations are absurd?

kkwan - 25 December 2012 12:48 AM

Not so. There are two possible amounts in each mutually exclusive event: either (X, 2X) or (X, 1/2X) for the same finite value of X and not three possible amounts amounts and that is all there is to it.

Counting possible amounts as you do is improper, inappropriate and incoherent in the context of the above two mutually exclusive events.

Sorry kkwan, you say there are two possible events, both with two possible values, from which the alternatives have one value in common (X), that makes three values. You can bend your meaningless “considering them jointly and separately” as you want, it does not help. If I treat the situations separately (and show you switching brings nothing), then you say I must consider them jointly. But if I take them jointly (3 amounts) you say I must consider them separately.

kkwan - 25 December 2012 12:48 AM

There are only two amounts, and given the TEP, there are two possibilities: I pick the smallest amount first, or the second. Do you deny this?

That is incomplete and misleading. There are two amounts in two and only two distinct mutually exclusive possible events either (X, 2X) or (X, 1/2X) for any finite value of X.

So again you deny the obvious: that in TEP I pick the smallest amount first, or the biggest.

Your long exposé about my non-TEP example is absurd. I give you a situation, and ask you to evaluate the solution. Instead you say of the situation that it is “fundamentally flawed”. I ask you what you should do in such a situation. Would it be better to switch? I gave you a derivation for it, that turns out to exactly the same as the one you use in TEP. Now you must tell where the error is. I know the situation is different, and therefore I want to know from you where the error in my derivation in this situation is. Different situations, but same derivation. You should be able to show where the error in my derivation is. If you can’t, it shows that the derivation belongs to such a situation, and not to TEP.

kkwan - 25 December 2012 12:48 AM

The culminating fatal flaws/errors are from 1 to 6 as stated above.

So this is a meaningless reaction, you cannot criticise the situation: I want to know how you argue for switching or not.

kkwan - 25 December 2012 12:48 AM

As such, the above analysis is likely to be invalid as steps 2 and 3 could be false in your non-TEP example.

Ah, I describe:

After your pick, a second envelope is filled with another amount that is half or twice the amount you have in your envelope, with a chance of 50% of both.

And then you say these are invalid:

2. Then the other envelope might contain 2A with a chance of 1/2.
3. Or the other envelope might contain A/2 with a chance of 1/2.

Explain, kkwan.

kkwan - 25 December 2012 12:48 AM

For instance, if X is 10, are either (10, 20) or (10, 5) not distinct possible events which satisfy the requirement of the TEP?

You take the value in your first picked envelope as reference. That is wrong. You should use the two original values as reference. So if there are (10,20) in the envelopes then your pick is 10 or 20. Possible gain or loss is 10. Or there are (10,5) (why not (5,10), I thought that was the same?) in the envelopes, and then possible gain or loss are 5. Sorry kkwan, if you consider the two possibilities mutually exclusive you see there is no reason to switch. If you take them together you have 3 amounts to consider, which is not according to TEP. But it is according to my non-TEP situation.

[ Edited: 25 December 2012 02:08 AM by GdB ]
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 Posted: 25 December 2012 02:31 AM [ Ignore ]   [ # 1855 ]
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Scott Mayers - 24 December 2012 12:31 PM

I brought this up because it is commonly assumed that this is an unresolvable paradox as you’ve stated. The problem is is that it is only an apparent one because you (and most others) interpret the statement’s meaning of the sentence as a real claim of ironic truth when the sentence’s statement is merely a set of symbols that, in this case, only an English reader provides meaning to. I am capable of printing the statement, “I always lie.” But to presume that its meaning can ever be an absolute truth is just a game of pretending for the sake of entertainment.

That the liar paradox is an unresolvable logical paradox with classical two value logic and not just merely an assumption, is clearly shown in the wiki:

“A = ‘A = false’”

This is an equation from which the truth value of A = “this statement is false” could hopefully be obtained. In the boolean domain “A = false” is equivalent to “not A” and therefore the equation is not solvable.

It is neither “merely a set of symbols”, an apparent paradox nor is it “just a game of pretending for the sake of entertainment”.

Godel’s Theorem is wrong and set back science in an unfortunate way. Statements or theories cannot make a logical claim of truth or falsity regarding its very own statements. It’s a rhetorical trick.

Godel will turn in his grave if you assert that his Theorems are “wrong and set back science in an unfortunate way” without tenable proof or justifiable reasons and that “it’s a rhetorical trick”, by fiat.

Likewise, you are mixing up semantic meaning and the statements in your TEP argument. The claimed formula is NOT a possible nor reasonable expression because it tricks you into thinking that what is in the envelopes means something to determining the probability. Just because they are quantities of something that can be algebraically expressed as something relative to one another does not mean they have any relevance to the numbers for the formula required to determine the probability of receiving any choice. When you state a probability, it’s not reasonable to interpret the particular or specific meaning of the content the probability refers to but the general arbitrary class which represents separate individual possibilities.

Because of total ignorance, all you can say is to denote the amount in the selected envelope as A and as A is either the smaller or the larger amount, then either (A, 2A) or (A, 1/2A) are in the two envelopes.

This analysis is possible and reasonable and there is no mind trick or any “mixing up of semantic meaning” per se.

1/2(the time you will choose one particular selection of two) + 1/2(the time you will choose the other particular selection of two) = 1(of the whole set of both possibilities)

The problem is, we don’t know whether either (A, 2A) or (A, 1/2A) are in the two envelopes at any one time.

So, with the selected amount denoted as A and by switching:

1. If (A, 2A) are in the two envelopes, we get 2A.

2. If (A, 1/2A) are in the two envelopes, we get 1/2A.

With many repetitions of the TEP, half the time (A, 2A) or (A, 1/2A) are in the two envelopes.

So, with many repetitions of the TEP and by switching, half the time we get 2A and half the time we get 1/2A.

Hence:

8. This is greater than A, so I gain on average by swapping.

This is the rationale for switching.

If I pick one and am told that I may opt to select the other one, then the odds are reset to the beginning.

Envelope є {A(1), A(2)}

1/2 [chance of A(1)] + 1/2 [chance of A(2)] = 1[chances of getting A(1) or A(2)]

Notwithstanding that, with total ignorance and with either (A, 2A) or (A, 1/2A) in the two envelopes, the above reasoning is valid as there are two possible events to consider, not one.

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 Posted: 25 December 2012 03:01 AM [ Ignore ]   [ # 1856 ]
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StephenLawrence - 25 December 2012 01:09 AM

kkwan,

It’s good that you are questioning my interpretation of 2. because it’s this that my solution depends upon. I think we can see that 2. should be interpreted as I am interpreting it and so there is the solution you are looking for.

This is what we need to focus on.

Your interpretation is correct if and only if we definitely know \$10 is in the envelope which we don’t, in the context of your non-TEP example.

OTOH, my interpretation is correct if we don’t know whether \$10 or \$20 is in the envelope which we don’t and therefore we cannot assume \$10 is in the selected envelope.

Not so. Your solution is valid only in the context of your non-TEP example whereby both groups are mutually inclusive.

OK, since you are fixing the pair the probability of selecting \$10 is 1/2. (I’d say it’s 1/4 because you might have been playing with the other pair.)

It cannot be 1/4 as the probability of selecting \$10 from group 1 (\$20, \$40) is 0.

But A means the amount in our envelope and since that is what A means the correct way to interpret 2. in this case, is if A is \$10 the probability of it being the smaller amount is 1. we can test this out and see in my example and we can move to the TEP once we’ve done that.

Not necessarily so.

You are working on the assumption that A is \$10 which is unjustified and begging the question as A could be either \$10 or \$20 because we don’t know for sure what A is.

All we can deduce is if A is selected from group 2 (\$20, \$10) it is either \$20 or \$10.

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 Posted: 25 December 2012 03:07 AM [ Ignore ]   [ # 1857 ]
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kkwan - 25 December 2012 03:01 AM

OTOH, my interpretation is correct if we don’t know whether \$10 or \$20 is in the envelope which we don’t and therefore we cannot assume \$10 is in the selected envelope.

No kkwan. In my example we don’t know that \$10 is in our envelope. If it is the probability of it being the smaller amount is 1.

Not so. Your solution is valid only in the context of your non-TEP example whereby both groups are mutually inclusive.

Well, start with my non TEP example and move from there in time. In my example 2. should be interpreted as I say and we know this because we know switching makes no difference.

Not necessarily so.

You are working on the assumption that A is \$10 which is unjustified and begging the question as A could be either \$10 or \$20 because we don’t know for sure what A is.

All we can deduce is if A is selected from group 2 (\$20, \$10) it is either \$20 or \$10.

I’m not working on the assumption that A is \$10. I’m working on the assumption that A is the amount in our envelope. If it is \$10 the probability of it being the smaller amount is 1.

Stephen

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 Posted: 25 December 2012 03:19 AM [ Ignore ]   [ # 1858 ]
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kkwan - 25 December 2012 03:01 AM

Your interpretation is correct if and only if we definitely know \$10 is in the envelope which we don’t, in the context of your non-TEP example.

Should we switch in my game? We agree no. Why not? Because when we get \$40 we will lose \$20. The probability of \$40 being the smaller amount is zero.

We do not need to know what is in our envelope to know this.

This is a black and white issue. If you won’t budge you won’t get to the solution. Do you really want to get to the solution, if there is one?

Stephen

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 Posted: 25 December 2012 03:42 AM [ Ignore ]   [ # 1859 ]
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GdB - 25 December 2012 01:59 AM

My question concerns TEP, and not your “(A, 2A) or (A, 1/2A)” stuff.

So you said “yes” on:

- You pick the smallest of the two amounts first,
or
- You pick the biggest of the same two amounts first

Do you deny this?

What else do I need to show that your argumentations are absurd?

That either (A, 2A) or (A, 1/2A) are in the two envelopes is the complete decription of the TEP which you deny.

Sorry kkwan, you say there are two possible events, both with two possible values, from which the alternatives have one value in common (X), that makes three values. You can bend your meaningless “considering them jointly and separately” as you want, it does not help. If I treat the situations separately (and show you switching brings nothing), then you say I must consider them jointly. But if I take them jointly (3 amounts) you say I must consider them separately.

Counting amounts from two mutually exclusive events is clearly wrong.

So again you deny the obvious: that in TEP I pick the smallest amount first, or the biggest.

I don’t deny that with either (A, 2A) or (A, 1/2A0 in the two envelopes, if A is the selected amount, then A is either the smaller or the larger amount.

Your long exposé about my non-TEP example is absurd. I give you a situation, and ask you to evaluate the solution. Instead you say of the situation that it is “fundamentally flawed”. I ask you what you should do in such a situation. Would it be better to switch? I gave you a derivation for it, that turns out to exactly the same as the one you use in TEP. Now you must tell where the error is. I know the situation is different, and therefore I want to know from you where the error in my derivation in this situation is. Different situations, but same derivation. You should be able to show where the error in my derivation is. If you can’t, it shows that the derivation belongs to such a situation, and not to TEP.

It is fundamentally flawed and has no relevance to the TEP at all.

So this is a meaningless reaction, you cannot criticise the situation: I want to know how you argue for switching or not.

Ah, I describe:

After your pick, a second envelope is filled with another amount that is half or twice the amount you have in your envelope, with a chance of 50% of both.

And then you say these are invalid:

2. Then the other envelope might contain 2A with a chance of 1/2.
3. Or the other envelope might contain A/2 with a chance of 1/2.

Explain, kkwan.

I have explained in post 1852. Please refer to it.

You take the value in your first picked envelope as reference. That is wrong. You should use the two original values as reference. So if there are (10,20) in the envelopes then your pick is 10 or 20. Possible gain or loss is 10. Or there are (10,5) (why not (5,10), I thought that was the same?) in the envelopes, and then possible gain or loss are 5. Sorry kkwan, if you consider the two possibilities mutually exclusive you see there is no reason to switch. If you take them together you have 3 amounts to consider, which is not according to TEP. But it is according to my non-TEP situation.

Of course, with total ignorance, the amount A in the selected envelope should be the reference.

You cannot assume that (10, 20) are in the two envelopes as it could be (10, 5) because you don’t know which event is actual. That is begging the question. It is that simple, but you don’t seem to get it.

There are no three amounts to consider as there are only two amounts in each mutually exclusive event. Counting amounts in this context as you do, is clearly wrong and misleading.

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 Posted: 25 December 2012 04:18 AM [ Ignore ]   [ # 1860 ]
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StephenLawrence - 25 December 2012 03:07 AM

No kkwan. In my example we don’t know that \$10 is in our envelope. If it is the probability of it being the smaller amount is 1.

Then the probability of A as the smaller amount is 1/2.

Well, start with my non TEP example and move from there in time. In my example 2. should be interpreted as I say and we know this because we know switching makes no difference.

I’m not working on the assumption that A is \$10. I’m working on the assumption that A is the amount in our envelope. If it is \$10 the probability of it being the smaller amount is 1.

If you don’t know what A is, then all you can say is that A is either \$20 or \$10 from group 2 (\$20, \$10) and the probability of selecting either is 1/2.

Hence, the probability of selecting \$10 (which is the smaller amount), is 1/2.

However, you cannot say “If it is \$10…...............” without implying A is \$10 and thus begging the question.

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