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The two envelopes problem
 Posted: 25 December 2012 04:51 AM [ Ignore ]   [ # 1861 ]
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kkwan - 25 December 2012 04:18 AM
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.

A = the amount in your envelope whatever that happens to be. So if the amount in your envelope happens to be \$10 the probability of A being the smaller amount is 1. This is merely a matter of fact, so there is no point in arguing over it.

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.

Well,if 2. were interpreted as you are doing we should switch and we know that switching makes no difference. Interpreting 2. as I’m doing gives the right answer, so we know that is right.

We can move to the TEP because we also should intepret 2. as I’m doing in the TEP. There is no reason to interpret 2. differently in the TEP.  This we can discuss but we need to get everything straight in my simplified version first.

.

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

Yes I can Kkwan. I’m not implying A is \$10 at all. Say A happens to be \$20. Should we switch? No because if A happened to be \$40 we would lose \$20.This is true regardless of what amount we happen to have.

The thing is if you played the game over and over you wouldn’t get \$20 every time, you would get \$20 1/2 the time and you need to take that into account. You need to know what would happen if you got \$40 what would happen if you got \$10 etc and what you see is switching every time gets you no where.

And all this can be demonstrated beyond doubt in my example.

Stephen

[ Edited: 25 December 2012 05:00 AM by StephenLawrence ]
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 Posted: 25 December 2012 05:03 AM [ Ignore ]   [ # 1862 ]
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StephenLawrence - 25 December 2012 03:19 AM

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?

Switching is not beneficial in your non-TEP example because both group 1 (\$20, \$40) and group 2 (\$20, \$10) can coexist at the same time.

There are 3 possible amounts to select from and as such, the total gain equals the total loss by switching because the game is so designed to be so, irrespective of the amount in the selected envelope.

However, the TEP is not synonymous to your non-TEP example. In the TEP, either (\$20, \$40) or (\$20, \$10) can exist at the same time.

As such, you will have to show me why the solution to your non-TEP example is relevant and applicable to the TEP without begging the question and breaking the inherent symmetry of the TEP.

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 Posted: 25 December 2012 05:17 AM [ Ignore ]   [ # 1863 ]
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kkwan - 25 December 2012 05:03 AM

Switching is not beneficial in your non-TEP example because both group 1 (\$20, \$40) and group 2 (\$20, \$10) can coexist at the same time.

They can’t (meaning the pair I select from can be either or not both). And this is not the crucial point. The crucial point is 2. is false in my example. You need to accept that before we can apply it to the TEP

There are 3 possible amounts to select from and as such, the total gain equals the total loss by switching because the game is so designed to be so, irrespective of the amount in the selected envelope.

However the two envelopes got to be filled with the amounts they are, you’ll find it’s the same. If we take into account the finite possibilities they will always cancel out, in just the same manor. If not design a way of selecting the pair in which this isn’t true.

You can’t do it. You can’t do it because once we take all the possible pairs into account you’ll have double the amount in total on the goes you halve your money by switching and visa versa.

However, the TEP is not synonymous to your non-TEP example. In the TEP, either (\$20, \$40) or (\$20, \$10) can exist at the same time.

Just as in my example

Stephen

[ Edited: 25 December 2012 05:27 AM by StephenLawrence ]
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 Posted: 25 December 2012 06:28 AM [ Ignore ]   [ # 1864 ]
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GdB - 24 December 2012 09:15 PM
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.

The expression of a statement differs from the meaning or interpretation derived from the statement. They belong in extended or different logical universes. Therefore, they are allowed to appear to contradict but actually do not! For example, You can be here and not here given that the extended logical universe is a set of different times. This is why Gödel was wrong too for which I will present formally elsewhere. My point is to suggest that what appears paradoxical to many, like Kwan, is not being capable of understanding this type of error in reasoning. He thinks that the contents of the envelope belongs to the same universe as the probability measure when they are not. So he and those who think that it is a paradox is due to using the meaning of the contents of the envelope to determine an external statement that is irrelevant.

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 Posted: 25 December 2012 06:34 AM [ Ignore ]   [ # 1865 ]
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StephenLawrence - 25 December 2012 04:51 AM

A = the amount in your envelope whatever that happens to be. So if the amount in your envelope happens to be \$10 the probability of A being the smaller amount is 1. This is merely a matter of fact, so there is no point in arguing over it.

It is not a matter of fact that A is \$10 is in the selected envelope because we don’t know and cannot verify it to be so.

If and only if we can do that (which we cannot) then and only then, the probability of A being the smaller amount is 1, is merely a matter of fact.

Well,if 2. were interpreted as you are doing we should switch and we know that switching makes no difference. Interpreting 2. as I’m doing gives the right answer, so we know that is right.

Not necessarily so, as in the context of your non-TEP example (with the two mutually inclusive events and 3 possible amounts) is so designed to make switching not to be beneficial irrespective of whether step 2 is true or false.

We can move to the TEP because we also should intepret 2. as I’m doing in the TEP. There is no reason to interpret 2. differently in the TEP.  This we can discuss but we need to get everything straight in my simplified version first.

Not so. Step 2 (in the context of the TEP) cannot be interpreted as you do (in the context of your non-TEP example) as the TEP with the two mutually exclusive events of either (A, 2A) or (A, 1/2A), is fundamentally a different animal.

Yes I can Kkwan. I’m not implying A is \$10 at all. Say A happens to be \$20. Should we switch? No because if A happened to be \$40 we would lose \$20.This is true regardless of what amount we happen to have.

If you are not implying A is \$10, then you must say A is either \$10 or \$20.

Of course, because your non-TEP example (with it’s inherent dissymmetry) is so designed that whatever amount A is, switching is not beneficial. That whatever the selected amount is, the overall outcome is zero by switching is indicative of a fixed game with no degree of freedom.

However, the TEP with it’s inherent symmetry of either (A, 2A) or (A, 1/2A) in the two envelopes, is not such a fixed game per se.

It is futile to export the solution of your fixed game and import it into the unfixed environment of the TEP and expect it to be the solution of the TEP as well.

The thing is if you played the game over and over you wouldn’t get \$20 every time, you would get \$20 1/2 the time and you need to take that into account. You need to know what would happen if you got \$40 what would happen if you got \$10 etc and what you see is switching every time gets you no where.

And all this can be demonstrated beyond doubt in my example.

It is so because your non-TEP example was designed to be so.

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 Posted: 25 December 2012 06:54 AM [ Ignore ]   [ # 1866 ]
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kkwan - 25 December 2012 06:34 AM

It is not a matter of fact that A is \$10 is in the selected envelope because we don’t know and cannot verify it to be so.

This is just wrong. If there is \$10 in the envelope that is true regardless of whether we know it or not.

If and only if we can do that (which we cannot) then and only then, the probability of A being the smaller amount is 1, is merely a matter of fact.

No, if \$10 is in the envelope the probability of it being the smaller is 1.

Not necessarily so, as in the context of your non-TEP example (with the two mutually inclusive events and 3 possible amounts) is so designed to make switching not to be beneficial irrespective of whether step 2 is true or false.

Not at all. If step 2. were true we should switch. It’s because step 2. is false that the reasoning that says we should switch fails.

Not so. Step 2 (in the context of the TEP) cannot be interpreted as you do

Yes it can and must be. Otherwise we can’t apply the formula as a general strategy. You won’t get the same amount each time.

If you are not implying A is \$10, then you must say A is either \$10 or \$20.

No. A is the amount in your envelope whatever it happens to be. A might be \$10, it might be \$20 and it might be \$40 in my example.

Of course, because your non-TEP example (with it’s inherent dissymmetry) is so designed that whatever amount A is, switch is not beneficial

No. If A is \$10 switching is beneficial. If A is \$20 switching is beneficial. If A is \$40 switching is not beneficial

It is so because your non-TEP example was designed to be so.

No. The amounts in the two envelopes have to be selected some how. Any way of selecting them will give the same result.

Stephen

[ Edited: 25 December 2012 07:06 AM by StephenLawrence ]
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 Posted: 25 December 2012 07:27 AM [ Ignore ]   [ # 1867 ]
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kkwan - 25 December 2012 02:31 AM

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.

This is kinda my point. And since you do not know what’s in the envelopes, you cannot assign numerical meaning to them with respect to the probability equation!

A probability equation of this type must represent a balance of all the odds possible. This equation appears to suggest that if you add up parts of something whole, you get something greater! You cannot exceed 100% here. It’s as if I said, 50% of the time, I’ll get heads and 50% of the time I’ll get tails. Therefore, I get 120% odds getting heads. How the hell does this make sense to you? The probability equation must show that the parts add up to the whole, represented by 1 (or 100%). It’s like an exam you take in school. You may know more than what is given. But it is irrelevant. The maximum mark you can achieve is 100%.
Instead of a coin toss as either heads or tails, you can assign values to them if you choose just like the envelope problem and it won’t change the probabilities. Assign heads as A and tails as 2A and you have the envelope problem exactly.

or

50%(of a head) + 50%(of a tail) = 50%(of a head-tail combination*)

What it should be:

Because we want to keep track of which ones are uniquely used up, we may write it like this:

but then what is in the braces are only reminders of what each possibility is. They are not meant to represent part of an algebraic formula such that heads are multiplied by 1/2 as well as tails. All that the TEP did was to assign numerical relative values to the heads and tails set. You see numbers within brackets next to the probability and you transferred the expression to mean what you’ve traditionally seen with an algebra statement illegally. It’s like a comedian’s joke: “How did they find out that the girl on jaws had dandruff?” “They found her head and shoulders on the beach.” We find it funny because we transfer a term in one context to another.

[* corrected “multiple” to “combination”]

[ Edited: 25 December 2012 07:31 AM by Scott Mayers ]
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 Posted: 25 December 2012 08:10 AM [ Ignore ]   [ # 1868 ]
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StephenLawrence - 25 December 2012 05:17 AM

They can’t (meaning the pair I select from can be either or not both). And this is not the crucial point. The crucial point is 2. is false in my example. You need to accept that before we can apply it to the TEP

It does not matter from which group you select the envelope. It is a fact that both groups can coexist at the same time for mutually inclusive events. The crucial point is, you can select an envelope from either of the two groups at any one time which you cannot do, in the context of the TEP.

The crucial point is not whether step 2 is true or false in your non-TEP example (which I interpret as true) because it is designed to make switching not beneficial irrespective of whether step 2 is true or false

The crucial point is, is step 2 is true in the context of the TEP?

However the two envelopes got to be filled with the amounts they are, you’ll find it’s the same. If we take into account the finite possibilities they will always cancel out, in just the same manor. If not design a way of selecting the pair in which this isn’t true.

You can’t do it. You can’t do it because once we take all the possible pairs into account you’ll have double the amount in total on the goes you halve your money by switching and visa versa.

That is only in the context of your non-TEP example.

Just as in my example

Not so.

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 Posted: 25 December 2012 08:24 AM [ Ignore ]   [ # 1869 ]
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kkwan - 25 December 2012 08:10 AM

It does not matter from which group you select the envelope. It is a fact that both groups can coexist at the same time for mutually inclusive events. The crucial point is, you can select an envelope from either of the two groups at any one time which you cannot do, in the context of the TEP.

No. The pair you play with is selected from two pairs. However you do it, you will be in the same position which is that you have selected an envelope from a pair and you could have selected an envelope from a number of different pairs.

Kkwan if you are going to dispute something like this you have to give a concrete example. and you can’t. You should not be so stubborn if you can’t. You can’t give a concrete example because you are wrong.

The crucial point is, is step 2 is true in the context of the TEP?

It’s false in my example and my example is just a mini TEP.

However the two envelopes got to be filled with the amounts they are, you’ll find it’s the same. If we take into account the finite possibilities they will always cancel out, in just the same manor. If not design a way of selecting the pair in which this isn’t true.

You can’t do it. You can’t do it because once we take all the possible pairs into account you’ll have double the amount in total on the goes you halve your money by switching and visa versa.

That is only in the context of your non-TEP example.

Then give a concrete TEP example. Explain how the amounts in the two envelopes are selected and from what numbers and we’ll see. You need to ‘put up’ at this stage if you don’t want us to think the worst. The worst being that you are a bullshitter, who is not interested in finding out the truth.

Just as in my example

Not so.

Then give your own example and we can see!

Stephen

[ Edited: 25 December 2012 08:32 AM by StephenLawrence ]
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 Posted: 25 December 2012 09:23 AM [ Ignore ]   [ # 1870 ]
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StephenLawrence - 25 December 2012 06:54 AM

This is just wrong. If there is \$10 in the envelope that is true regardless of whether we know it or not.

Yes, if and only if, but in the context of total ignorance, we cannot determine that with certainty.

No, if \$10 is in the envelope the probability of it being the smaller is 1.

If and only if, but it is only a fact upon verification which is not possible in the context of your non-TEP example or the TEP.

Not at all. If step 2. were true we should switch. It’s because step 2. is false that the reasoning that says we should switch fails.

In the context of your non-TEP example, even if step 2 were true, switching is not beneficial because it was so designed that the loss on selecting \$20 (from group 2) or \$40 (from group 1) and switching nullifies the gain of selecting \$20 (from group 1) or \$10 (from group 2) and switching.

Yes it can and must be. Otherwise we can’t apply the formula as a general strategy. You won’t get the same amount each time.

Whether the same formula can be applied in the context of the TEP is not clearly shown as yet.

No. A is the amount in your envelope whatever it happens to be. A might be \$10, it might be \$20 and it might be \$40 in my example.

That is only in the context of your non-TEP example where A is not a constant.

No. If A is \$10 switching is beneficial. If A is \$20 switching is beneficial. If A is \$40 switching is not beneficial

Specifically, if A is \$20 from group 1 (\$20, \$40) switching is beneficial but if A is \$20 from group 2 (\$20, \$10) switching is not beneficial.

However, over many repetitions of your non-TEP example, switching is not beneficial because of it’s
inherent dissymmetry with the 3 possible selections and their respective values which are designed to be so.

No. The amounts in the two envelopes have to be selected some how. Any way of selecting them will give the same result.

It is obviously designed to be so. Thus, your non-TEP example is designed to be a zero-sum game with the “solution” of no gain or loss by switching.

http://en.wikipedia.org/wiki/Zero–sum_game

If the total gains of the participants are added up, and the total losses are subtracted, they will sum to zero.

OTOH, from the wiki on the TEP:

http://en.wikipedia.org/wiki/Two_envelopes_problem#A_new_variant

This problem can be considered from the point of view of game theory, where we make the game a two-person zero-sum game with outcomes win or lose, depending on whether the player ends up with the higher or lower amount of money. The organiser chooses the joint distribution of the amounts of money in both envelopes, and the player chooses the distribution of Z. The game does not have a “solution” (or saddle point) in the sense of game theory. This is an infinite game and von Neumann’s minimax theorem does not apply

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 Posted: 25 December 2012 10:16 AM [ Ignore ]   [ # 1871 ]
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StephenLawrence - 25 December 2012 08:24 AM

No. The pair you play with is selected from two pairs. However you do it, you will be in the same position which is that you have selected an envelope from a pair and you could have selected an envelope from a number of different pairs.

Not so. The envelope and the amount in it that you select is from two groups of two pairs.

Kkwan if you are going to dispute something like this you have to give a concrete example. and you can’t. You should not be so stubborn if you can’t. You can’t give a concrete example because you are wrong.

1. If \$20 is selected, it can be either from group 1 (\$20, \$40) or from group 2 (\$20, \$10)

2. If \$10 is selected, it must be from group 2 (\$20, \$10).

3. IF \$40 is selected, it must be from group 1 (\$20, \$40).

So, whichever amount you select, it is from either group 1 or group 2 which do coexist at the same time i.e. the two events are mutually inclusive.

This is not so in the context of the TEP where at one time, there is only one pair, not two pairs as in your non-TEP example.

It’s false in my example and my example is just a mini TEP.

My interpretation is, step 2 is true in your non-TEP example but it is not the TEP per se.

Then give a concrete TEP example. Explain how the amounts in the two envelopes are selected and from what numbers and we’ll see. You need to ‘put up’ at this stage if you don’t want us to think the worst. The worst being that you are a bullshitter, who is not interested in finding out the truth.

In the TEP, (with total ignorance) we denote the amount in the selected envelope as A and as A is either the smaller or the larger amount, therefore either (A, 2A) or (A, 1/2A) are in the two envelopes.

Let A be \$20. Then either (\$20, \$40) or (\$20, \$10) are in the two envelopes which is similar to the values in your non-TEP example:

1. If A (\$20) is the smaller amount, then (\$20, \$40) are in the two envelopes and gain by switching is \$20.

2. If A (\$20) is the larger, then (\$20, \$10) are in the two envelopes and loss by switching is \$10.

The benefit of switching is clearly stated in steps 7 and 8.

Then give your own example and we can see!

In your non-TEP example, both (\$20, \$40) and (\$20, \$10) can coexist at the same time.

In the TEP, either (\$20, \$40) or (\$20, \$10) can occur at the same time.

So, your non-TEP example is not synonymous to the TEP.

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 Posted: 25 December 2012 10:27 AM [ Ignore ]   [ # 1872 ]
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kkwan - 25 December 2012 09:23 AM

In the context of your non-TEP example, even if step 2 were true, switching is not beneficial because it was so designed that the loss on selecting \$20 (from group 2) or \$40 (from group 1) and switching nullifies the gain of selecting \$20 (from group 1) or \$10 (from group 2) and switching.

Step 2. can’t be true in my example without opening the envelope.

Whether the same formula can be applied in the context of the TEP is not clearly shown as yet.

Why not?

That is only in the context of your non-TEP example where A is not a constant.

We know A can’t be a constant because we know if we played the game over and over we would not continually get the same amount in our envelope. As I’ve explained switching needs to work as a general strategy.

Specifically, if A is \$20 from group 1 (\$20, \$40) switching is beneficial but if A is \$20 from group 2 (\$20, \$10) switching is not beneficial.

Yep but we don’t know which group we have and the probability of having either is 1/2. So we’ll gain \$20 half of the time and lose \$10 half of the time when we have \$20 in our envelope.

However, over many repetitions of your non-TEP example, switching is not beneficial because of it’s
inherent dissymmetry with the 3 possible selections and their respective values which are designed to be so.

They are not designed to be so, it just is so. The same goes for any way you care to pick the amounts in the two envelopes. The challenge is for you to find some way that contradicts this.

You can’t because there is no such way.

If you think you can then actually do it. Use numbers and say how they get selected.

And if you don’t…....................

Stephen

[ Edited: 25 December 2012 10:37 AM by StephenLawrence ]
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 Posted: 25 December 2012 10:35 AM [ Ignore ]   [ # 1873 ]
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kkwan - 25 December 2012 10:16 AM

Let A be \$20. Then either (\$20, \$40) or (\$20, \$10) are in the two envelopes which is similar to the values in your non-TEP example:

1. If A (\$20) is the smaller amount, then (\$20, \$40) are in the two envelopes and gain by switching is \$20.

2. If A (\$20) is the larger, then (\$20, \$10) are in the two envelopes and loss by switching is \$10.

The benefit of switching is clearly stated in steps 7 and 8.

A can’t be \$20 because A must refer to the amount in your envelope what ever it is. So A might be \$20, might be \$40 and might be \$10

There is no such thing as what A is, only what A might be. We know this because if we played over and over we wouldn’t get the same amount each time.

Stephen

[ Edited: 25 December 2012 10:38 AM by StephenLawrence ]
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 Posted: 25 December 2012 10:52 AM [ Ignore ]   [ # 1874 ]
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Scott Mayers - 25 December 2012 07:27 AM

This is kinda my point. And since you do not know what’s in the envelopes, you cannot assign numerical meaning to them with respect to the probability equation!

However, as “one envelope contains twice the amount in the other” is given in the TEP, and without knowing what amounts are in the two envelopes, we can deduce either (X, 2X) or (X, 1/2X) are in the two envelopes for any finite value of X as both distinct events satisfy the given requirement of the TEP.

There are two indistinguishable envelopes, so the probability of selecting either one is 1/2.

So, with an envelope selected, we can proceed from step 1 to 8 without contradiction.

A probability equation of this type must represent a balance of all the odds possible. This equation appears to suggest that if you add up parts of something whole, you get something greater! You cannot exceed 100% here. It’s as if I said, 50% of the time, I’ll get heads and 50% of the time I’ll get tails. Therefore, I get 120% odds getting heads. How the hell does this make sense to you? The probability equation must show that the parts add up to the whole, represented by 1 (or 100%). It’s like an exam you take in school. You may know more than what is given. But it is irrelevant. The maximum mark you can achieve is 100%.

None of the equations I presented or those from steps 1 to 8 indicate that the total probability is more than 1.

Instead of a coin toss as either heads or tails, you can assign values to them if you choose just like the envelope problem and it won’t change the probabilities. Assign heads as A and tails as 2A and you have the envelope problem exactly.

Following your analogy, it should be heads (A, 2A) or Tails (A, 1/2A) wrt the TEP.

or

50%(of a head) + 50%(of a tail) = 50%(of a head-tail combination*)

Not so.

The probability of heads (A, 2A) is 1/2.

The probability of tails (A, 1/2A) is 1/2.

The probability of either heads (A, 2A) or tails (A, 1/2A) is 1/2 + 1/2 = 1 because the two events are mutually exclusive.

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 Posted: 25 December 2012 10:59 AM [ Ignore ]   [ # 1875 ]
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We are not in a position to fix A Kkwan. And this we can check and see through examples.

A does not = the amount in our envelope. It is a mind trick to think it does.

Stephen

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