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The two envelopes problem
Posted: 18 December 2012 12:51 AM   [ Ignore ]   [ # 1831 ]
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GdB - 17 December 2012 04:32 AM

kkwan, stop this bullshit. The TEP is clearly formulated:

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. Should you switch?

Full stop. Everything else is your addition. So answer the above question in the above described situation.

That “one envelope contains twice as much as the other” does not mean only (X, 2X) but either (X, 2X) or (X, 1/2X) which are distinct events for any finite value of X.

You have misinterpreted the TEP as it is, for your convenience.

TEP is not defined as:

You are given one envelope. Now you are offered a second envelope, which can contain half or twice the amount in your envelope. Should you switch?

But that is the way you redefine TEP. You still owe me to show how you treat TEP and the above situation differently. Show me the calculations of both, and explain the difference to us. You are not able to do this, kkwan. For you, both situations are the same. But in the second is missing that you originally had the choice between the two envelopes, and that is exactly the fact you do not use in your ‘derivation’.

That is your perception of what you “think” is my “redefinition” of the TEP which is not so.

What a nonsense. How can be guaranteed that the amounts in the envelopes conform to ‘one amount is twice the other’, if there is not somebody, or some automaton programmed to put the amounts in the envelopes? For you there is no difference in how it is done. The only important thing is that you do not know what the factual amounts are.

Some entity put the amounts in the two envelopes (not you) and the amounts are either (X, 2X) or (X, 1/2X).

That is not the TEP as it is. Maybe I should point you to your first posting again, where you clearly state (well, copy from Wikipedia…):

The switching argument: Now suppose you reason as follows:

1. I denote by A the amount in my selected envelope.
2. The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2.
...

Bold by me: everything following is part of the reasoning, and not of the setup.

C’mon kkwan, I want to win $1000, why don’t you want to bet on it? Why are you so sure about yourself, and still do not take the chance to earn an easy $1000?

That was from the argument to switch in the wiki on the TEP, not from me, but if you were to follow through steps 3, 4 and 5 instead of selectively quoting only steps 1 and 2 for your convenience:

3. The other envelope may contain either 2A or A/2.
4. If A is the smaller amount the other envelope contains 2A.
5. If A is the larger amount the other envelope contains A/2.

3, 4 and 5 taken together mean either (A, 2A) or (A, 1/2A) are in the two envelopes which is synonymous and equivalent to: either (X, 2X) or (X, 1/2X) are in the two envelopes.

This is the complete description the TEP and it is the TEP as it is.

GdB, we are here to discuss the TEP, not to gamble. If you have any cogent arguments as to why the argument to switch is flawed thereby solving the puzzle and do not deride, insult or challenge me to gamble etc., I will be willing to consider them on their merits without prejudice.

The puzzle: The puzzle is to find the flaw in the very compelling line of reasoning above.

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Posted: 18 December 2012 02:31 AM   [ Ignore ]   [ # 1832 ]
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kkwan - 18 December 2012 12:51 AM

That “one envelope contains twice as much as the other” does not mean only (X, 2X) but either (X, 2X) or (X, 1/2X) which are distinct events for any finite value of X.

For one and the same X this means you are considering 3 amounts; but there are only two. This is correctly represented by:

1. You have the envelope with X, and you change to the envelope with 2X
2. You have the envelope with 2X, and you switch to the envelope with X,

For any constant value of X. This is the only correct way to represent that there are
a. only 2 amounts,
b. with the two possibilities that the second amount is twice the amount of my first pick (1), or that it is half of it (2).

Nowhere in your ‘reasoning’ you use the fact that there are only two amounts. By representing the situations as in 1. and 2. I do. And do not forget: I have no idea what X is.

You have misinterpreted the TEP as it is, for your convenience.

kkwan - 18 December 2012 12:51 AM

That is your perception of what you “think” is my “redefinition” of the TEP which is not so.

You are not able to prove that: so you assertion is hollow.

kkwan - 18 December 2012 12:51 AM

Some entity put the amounts in the two envelopes (not you) and the amounts are either (X, 2X) or (X, 1/2X).

So there are three possible amounts, where the definition of TEP clearly states there are only two amounts.

kkwan - 18 December 2012 12:51 AM

That was from the argument to switch in the wiki on the TEP, not from me, but if you were to follow through steps 3, 4 and 5 instead of selectively quoting only steps 1 and 2 for your convenience:

Again you are avoiding the point I am making: that the points 1 and 2 etc. belong to the (wrong) reasoning, and not to the definition of the TEP.

kkwan - 18 December 2012 12:51 AM

GdB, we are here to discuss the TEP, not to gamble. If you have any cogent arguments as to why the argument to switch is flawed thereby solving the puzzle and do not deride, insult or challenge me to gamble etc., I will be willing to consider them on their merits without prejudice.

It is clear that you do not trust your own argument. For you it should not be a gamble, because you should be sure that you win the bet. Obviously you are not.

[ Edited: 18 December 2012 04:29 AM by GdB ]
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Posted: 18 December 2012 05:58 AM   [ Ignore ]   [ # 1833 ]
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GdB - 18 December 2012 02:31 AM

For one and the same X this means you are considering 3 amounts; but there are only two. This is correctly represented by:

1. You have the envelope with X, and you change to the envelope with 2X
2. You have the envelope with 2X, and you switch to the envelope with X,

For any constant value of X. This is the only correct way to represent that there are
a. only 2 amounts,
b. with the two possibilities that the second amount is twice the amount of my first pick (1), or that it is half of it (2).

Nowhere in your ‘reasoning’ you use the fact that there are only two amounts. By representing the situations as in 1. and 2. I do. And do not forget: I have no idea what X is.

You have misinterpreted the TEP as it is, for your convenience.

Not so.

For any same finite value of X, there are two and only two possible mutually exclusive distinct events: either (X, 2X) or (X, 1/2X) and not exclusively one (X, 2X).

There are no 3 amounts in any of the two possible events. There are only 2 amounts in each mutually exclusive event because both events cannot occur at the same time.

OTOH, if the two events are mutually inclusive as in Stephen’s non-TEP example, then there are 3 amounts as both events can occur at the same time.

Counting amounts as you do is improper in the context of mutually exclusive events.

Your interpretation of the TEP as exclusively (X, 2X) is perversely misinterpreting the TEP, misleading, incomplete and begging the question because you have ignored the other possible event (X, 1/2X).

You are not able to prove that: so you assertion is hollow.

There is nothing to prove.

So there are three possible amounts, where the definition of TEP clearly states there are only two amounts.

There are no three possible amounts. There are only two amounts in each of the two mutually exclusive possible events as explained above.

Counting amounts as you do is improper in the context of mutually exclusive events.

Again you are avoiding the point I am making: that the points 1 and 2 etc. belong to the (wrong) reasoning, and not to the definition of the TEP.

How and why did you come to the assertion that 1 and 2 is the “wrong reasoning”?

Please explain.

It is clear that you do not trust your own argument. For you it should not be a gamble, because you should be sure that you win the bet. Obviously you are not.

LOL

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Posted: 18 December 2012 08:19 AM   [ Ignore ]   [ # 1834 ]
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I don’t get why this issue isn’t closed already. Kwan, is there any evidence that anyone could possibly put forward that could convince you of an error in your ways? If so, what kind of proof would satisfy you?

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Posted: 18 December 2012 08:43 AM   [ Ignore ]   [ # 1835 ]
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kkwan - 18 December 2012 05:58 AM

For any same finite value of X, there are two and only two possible mutually exclusive distinct events: either (X, 2X) or (X, 1/2X) and not exclusively one (X, 2X).

Given that the envelopes contain only two amounts, there are indeed two possibilities: if you pick X first and then the other contains 2X, if you pick 2X first, then the other contains X. In words the second one reads: the second envelope contains half of the envelope you picked first, and this corresponds two your (X,X/2), but of course not for the same X. That is what you do: your ‘two X’es’ in (X,2X) and (X,X/2) have not the same value.

kkwan - 18 December 2012 05:58 AM

There are no 3 amounts in any of the two possible events. There are only 2 amounts in each mutually exclusive event because both events cannot occur at the same time.

Sorry, when there are two mutually exclusive events, as in (X,2X) and (X,X/2) there are tree possible amounts.

kkwan - 18 December 2012 05:58 AM

Your interpretation of the TEP as exclusively (X, 2X) is perversely misinterpreting the TEP, misleading, incomplete and begging the question because you have ignored the other possible event (X, 1/2X).

It is not exclusively (X, 2X), it is also (2X,X), in which the first amount is your amount of the first pick.

kkwan - 18 December 2012 05:58 AM

There is nothing to prove.

OK. Here is the situation:

You are given one envelope. Now you are offered a second envelope, which can contain half or twice the amount in your envelope. Should you switch?

The switching argument: Now suppose you reason as follows:

  1. I denote by A the amount in my selected envelope.
  2. The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2.
  3. The other envelope may contain either 2A or A/2.
  4. If A is the smaller amount the other envelope contains 2A.
  5. If A is the larger amount the other envelope contains A/2.
  6. Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2.
  7. So the expected value of the money in the other envelope is

      c0f75c1a69af64e06f77ce0ec051c958.png

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

C’mon kkwan, how is it possible that those reasonings are the same, but the starting situation is not?

(Swapping back of course brings nothing, because the amount in the first envelope has not changed.)

kkwan - 18 December 2012 05:58 AM

How and why did you come to the assertion that 1 and 2 is the “wrong reasoning”?

My statement is that 1 and 2 do not belong to the description of TEP anymore. You are avoiding the point again again.

So again: this is TEP:

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 rest belongs to your reasoning, and not to the definition of TEP.

kkwan - 18 December 2012 05:58 AM

Counting amounts as you do is improper in the context of mutually exclusive events.

What does mutually exclusive mean, kkwan?
If the envelopes contain (X,2X) then I can gain or loss X, so they cancel out. When they contain (X,X/2) then the possible gain or loss is X/2, that also cancel each other out. Now that was mutually exclusive. But you mingle them, i.e. you do not treat them as ‘mutually exclusive’.

[ Edited: 19 December 2012 12:18 AM by GdB ]
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Posted: 21 December 2012 04:15 AM   [ Ignore ]   [ # 1836 ]
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Scott Mayers - 18 December 2012 08:19 AM

I don’t get why this issue isn’t closed already. Kwan, is there any evidence that anyone could possibly put forward that could convince you of an error in your ways? If so, what kind of proof would satisfy you?


There is no evidence that anyone can provide or the necessity of any proof whatsoever.

What is required is to convincingly show me the flaw in the argument to switch without begging the question by assuming that there is one and only possible event, (X, 2X) whereas there are two and only two possible events, either (X, 2X) or (X, 1/2X) inherently in the context of the TEP, which the argument to switch correctly addressed, without begging the question.

With total ignorance of the amount in the selected envelope and the actual amounts in the two envelopes and with the inherent symmetry in the TEP, it appears there can be no solution of the puzzle without begging the question and thereby breaking it’s inherent symmetry, which is fatal.

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Posted: 21 December 2012 06:02 AM   [ Ignore ]   [ # 1837 ]
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kkwan - 21 December 2012 04:15 AM

There is no evidence that anyone can provide or the necessity of any proof whatsoever.

 

Well, at least you proved something!

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Posted: 21 December 2012 08:10 AM   [ Ignore ]   [ # 1838 ]
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GdB - 18 December 2012 08:43 AM

Given that the envelopes contain only two amounts, there are indeed two possibilities: if you pick X first and then the other contains 2X, if you pick 2X first, then the other contains X. In words the second one reads: the second envelope contains half of the envelope you picked first, and this corresponds two your (X,X/2), but of course not for the same X. That is what you do: your ‘two X’es’ in (X,2X) and (X,X/2) have not the same value.

Not so. There are two possible events: either (X, 2X) or (X, 1/2X) not two possible selections: either X or 2X in one possible event (X, 2X).

By assuming that only (X, 2X) are in the two envelopes, you are begging the question as either (X, 2X) or (X, 1/2X) are in the two envelopes and your don’t know which event is actual.

Sorry, when there are two mutually exclusive events, as in (X,2X) and (X,X/2) there are tree possible amounts.

Not so.

In each mutually exclusive event, there are only two amounts. As such, there are no three possible amounts per se as both events cannot occur at the same time.

It is thus wrong to enumerate possible amounts as you do. You can only do that if (X, 2X) and (X, 1/2X) are mutually inclusive as in Stephen’s non-TEP example.

There are two amounts in each of the two mutually exclusive events is the correct description.

It is not exclusively (X, 2X), it is also (2X,X), in which the first amount is your amount of the first pick.

As (X, 2X) is equivalent to (2X, X), there is one and only one exclusive event.

OK. Here is the situation:

You are given one envelope. Now you are offered a second envelope, which can contain half or twice the amount in your envelope. Should you switch?

This is not exactly like the TEP, whereby there are two indistinguishable envelopes but one envelope contains twice as much as the other and you select an envelope. That is the starting situation.

The switching argument: Now suppose you reason as follows:

  1. I denote by A the amount in my selected envelope.
  2. The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2.
  3. The other envelope may contain either 2A or A/2.
  4. If A is the smaller amount the other envelope contains 2A.
  5. If A is the larger amount the other envelope contains A/2.
  6. Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2.
  7. So the expected value of the money in the other envelope is

      c0f75c1a69af64e06f77ce0ec051c958.png

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

C’mon kkwan, how is it possible that those reasonings are the same, but the starting situation is not?

The starting situation (as stated above) is on selecting an envelope and then proceeding to reason from step 1 to 8.

(Swapping back of course brings nothing, because the amount in the first envelope has not changed.)

Switching once or any odd number of times has potential gain or loss.

My statement is that 1 and 2 do not belong to the description of TEP anymore. You are avoiding the point again again.

1 is merely denoting A as the amount in the selected envelope.

2 is merely stating that as A is either the smaller or the larger amount, the probability of A being either the smaller or the larger amount, is 1/2 which is true.

1 is about denoting the selected amount and that is all there is to it.

2 is consistent with the complete description of the TEP as it is with either (X, 2X) or (X, 1/2X) in the two envelopes.

So, your contention that “1 and 2 do not belong to the description of TEP anymore” is unjustified.

 

So again: this is TEP:

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.

There is nothing in the TEP which prohibit either (X, 2X) or (X, 1/2X) to be in the two envelopes as both distinct events satisfy the requirement of the TEP. This is inherent in the TEP which you have ignored, for your convenience. You have thus incompletely described the TEP to be only (X, 2X) which is misleading and begs the question as it could be (X, 1/2X).

What does mutually exclusive mean, kkwan?
If the envelopes contain (X,2X) then I can gain or loss X, so they cancel out. When they contain (X,X/2) then the possible gain or loss is X/2, that also cancel each other out. Now that was mutually exclusive. But you mingle them, i.e. you do not treat them as ‘mutually exclusive’.

Mutually exclusive mean either (X, 2X) or (X, 1/2X) can occur at the same time but as you don’t know which is the actual event, you must consider the two possible events jointly and not separately as by doing so, you beg the question.

Considering both possible events jointly does not imply mingling them at all.

Thus, with A denoted as the amount in the selected envelope and as 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 two events being mutually exclusive.

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Posted: 21 December 2012 08:36 AM   [ Ignore ]   [ # 1839 ]
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Scott Mayers - 21 December 2012 06:02 AM

Well, at least you proved something!

That I cannot be convinced so easily, nevertheless I can be convinced if cogent arguments are presented? 

grin

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Posted: 21 December 2012 12:40 PM   [ Ignore ]   [ # 1840 ]
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Kwan,

“I always lie.”

What is wrong with this statement? Is this a real paradox?

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Posted: 22 December 2012 12:28 AM   [ Ignore ]   [ # 1841 ]
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kkwan - 21 December 2012 04:15 AM

What is required is to convincingly show me the flaw in the argument to switch without begging the question by assuming that there is one and only possible event, (X, 2X) whereas there are two and only two possible events, either (X, 2X) or (X, 1/2X) inherently in the context of the TEP, which the argument to switch correctly addressed, without begging the question.

I think I’ve already done that Kkwan.


The problem is you cannot fix A to mean the amount in your envelope as it actually is. This is because you don’t know what the probability is of having half or double of that amount in the other envelope. A needs to be the amount in your envelope whatever it might be. So there are not two possible events but many more. This is even if you look and see what you have. It’s still true that if you played the game over and over you would get different amounts and you need to take that into account. This is because to say we should switch is to say switching would work as a general strategy

If you think of all the possible combinations that could come up you see that when you switch you will double half of the amount that you will halve. So for example, in my game you double £10 and £20 = £30 but you also halve £20 and £40 = £60

From there we can simplify and just consider one pair, as playing the game with one pair is essentially the same as playing with numerous pairs, as on switching you double half the amount that you halve.

Stephen

[ Edited: 22 December 2012 01:15 AM by StephenLawrence ]
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Posted: 22 December 2012 04:11 AM   [ Ignore ]   [ # 1842 ]
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kkwan - 21 December 2012 08:10 AM

By assuming that only (X, 2X) are in the two envelopes, you are begging the question as either (X, 2X) or (X, 1/2X) are in the two envelopes and your don’t know which event is actual.

There are only two possibilities:

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

Do you deny this?

kkwan - 21 December 2012 08:10 AM

In each mutually exclusive event, there are only two amounts. As such, there are no three possible amounts per se as both events cannot occur at the same time.

But you consider them at the same time. You assume both (X,2X) and (X,X/2) can be the case, which means you assume both at the same time, and for the same X this means there are three possible amounts, which contradicts TEP which clearly says there are only two amounts.

kkwan - 21 December 2012 08:10 AM

It is not exclusively (X, 2X), it is also (2X,X), in which the first amount is your amount of the first pick.

As (X, 2X) is equivalent to (2X, X), there is one and only one exclusive event.

You leave out that I said you should consider the order. First picking the smallest or first picking the biggest amount are two different events. But I will introduce a new notation: <A,B> means first A, then B.

So the two possibilities in TEP are:

- <X,2X>
- <2X,X>

In this way I stick to the demand of TEP there are only two amounts. The second can be expressed by:

- <Y, Y/2> whereby Y = 2X.

You do as if X = Y, which is wrong.

kkwan - 21 December 2012 08:10 AM

This is not exactly like the TEP, whereby there are two indistinguishable envelopes but one envelope contains twice as much as the other and you select an envelope. That is the starting situation.

I know it is not TEP. But the way you reason fits to this situation, and not to TEP. You are not able to proof me wrong.

kkwan - 21 December 2012 08:10 AM

The starting situation (as stated above) is on selecting an envelope and then proceeding to reason from step 1 to 8.

Yes, and? Is it wrong or not for the non-TEP situation? You keep avoiding to answer.

kkwan - 21 December 2012 08:10 AM

My statement is that 1 and 2 do not belong to the description of TEP anymore. You are avoiding the point again again.

So, your contention that “1 and 2 do not belong to the description of TEP anymore” is unjustified.

Another point of avoiding: I state that this is not anymore TEP, that it belongs to your argument. TEP does not say that for the same X the other amount can be 2X or X/2. That belongs to your argumentation. So saying this belongs to the essence of TEP is simply wrong. This, and only this is TEP:

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.

Do you deny that this, and only this is the full description of TEP?

kkwan - 21 December 2012 08:10 AM

Mutually exclusive mean either (X, 2X) or (X, 1/2X) can occur at the same time but as you don’t know which is the actual event, you must consider the two possible events jointly and not separately as by doing so, you beg the question.

You sentence has no meaning: we must consider mutually exclusive events jointly and not separately?

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Posted: 23 December 2012 08:53 AM   [ Ignore ]   [ # 1843 ]
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Scott Mayers - 21 December 2012 12:40 PM

Kwan,

“I always lie.”

What is wrong with this statement? Is this a real paradox?

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.

http://en.wikipedia.org/wiki/Liar_paradox

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.

Applications of the liar paradox:

Gödel’s incompleteness theorems are two fundamental theorems of mathematical logic which state inherent limitations of all but the most trivial axiomatic systems for mathematics. The theorems were proven by Kurt Gödel in 1931, and are important in the philosophy of mathematics. Roughly speaking, in proving the first incompleteness theorem, Gödel used a modified version of the liar paradox, replacing “this sentence is false” with “this sentence is not provable”, called the “Gödel sentence G”. Thus for a theory “T”, “G” is true, but not provable in “T”. The analysis of the truth and provability of “G” is a formalized version of the analysis of the truth of the liar sentence.

However, the TEP is not synonymous to the liar paradox.

The TEP is paradoxical in that it’s inherent symmetry rule out the tenability of any proposed “resolution” which has one and only one possible event because:

1. It breaks it’s inherent symmetry and thereby contravenes the TEP as it is with two and only two possible events (for any finite value of X) instead of one and only one possible event.

2. It begs the question because with total ignorance, we don’t know which event is actual and as such, is fallacious and fatal.

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Posted: 23 December 2012 09:49 AM   [ Ignore ]   [ # 1844 ]
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StephenLawrence - 22 December 2012 12:28 AM

The problem is you cannot fix A to mean the amount in your envelope as it actually is. This is because you don’t know what the probability is of having half or double of that amount in the other envelope. A needs to be the amount in your envelope whatever it might be. So there are not two possible events but many more. This is even if you look and see what you have. It’s still true that if you played the game over and over you would get different amounts and you need to take that into account. This is because to say we should switch is to say switching would work as a general strategy

In the context of the TEP:

1. In step 1, the amount in the selected envelope is denoted as A.

2. A can be any amount but it cannot vary i.e. it is a constant. However (with A as a constant) that it can be either the smaller or the larger amount are the two and only two possibilities of A.

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.

If you think of all the possible combinations that could come up you see that when you switch you will double half of the amount that you will halve. So for example, in my game you double £10 and £20 = £30 but you also halve £20 and £40 = £60

That is so in your non-TEP example with the two mutually inclusive events, but not so in the TEP as it is with two mutually exclusive possible distinct events and total ignorance.

From there we can simplify and just consider one pair, as playing the game with one pair is essentially the same as playing with numerous pairs, as on switching you double half the amount that you halve.

By just considering only one pair in the context of the TEP instead considering the two possible pairs jointly, you beg the question and break it’s inherent symmetry, which is fallacious and fatal.

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Posted: 23 December 2012 10:53 AM   [ Ignore ]   [ # 1845 ]
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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).

But you consider them at the same time. You assume both (X,2X) and (X,X/2) can be the case, which means you assume both at the same time, and for the same X this means there are three possible amounts, which contradicts TEP which clearly says there are only two amounts.

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.

There are three possible amounts if and only if (X, 2X) and (X, 1/2X) can occur at the same time as in Stephen’s non-TEP example or there is only one event (X, 2X, 1/2X).

You leave out that I said you should consider the order. First picking the smallest or first picking the biggest amount are two different events. But I will introduce a new notation: <A,B> means first A, then B.

So the two possibilities in TEP are:

- <X,2X>
- <2X,X>

In this way I stick to the demand of TEP there are only two amounts. The second can be expressed by:

- <Y, Y/2> whereby Y = 2X.

You do as if X = Y, which is wrong.

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

I know it is not TEP. But the way you reason fits to this situation, and not to TEP. You are not able to proof me wrong.

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

Yes, and? Is it wrong or not for the non-TEP situation? You keep avoiding to answer.

It is correct for the TEP as it is with either (X, 2X) or (X, 1/2X) in the two envelopes.

Another point of avoiding: I state that this is not anymore TEP, that it belongs to your argument. TEP does not say that for the same X the other amount can be 2X or X/2. That belongs to your argumentation. So saying this belongs to the essence of TEP is simply wrong. This, and only this is TEP:

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.

Do you deny that this, and only this is the full description of TEP?

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?

You sentence has no meaning: we must consider mutually exclusive events jointly and not separately?

It has meaning.

You must consider both mutually exclusive events jointly if you don’t know which event is actual which you don’t, in the context of the TEP.

Thus, if you consider each event separately, you are begging the question in the context of the TEP.

You can only do that if and only if you know which event is actual which you don’t, in the context of the TEP.

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