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The two envelopes problem
Posted: 07 January 2013 08:40 PM   [ Ignore ]   [ # 1951 ]
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StephenLawrence - 07 January 2013 10:19 AM

There is no sense to this because to say we should switch means switching would work as a general principle.

That the amount in the selected envelope is a constant is clearly true (irrespective of what is the amount) in the context of any particular trial of the TEP.

In the context of the TEP, switching is beneficial as we are totally ignorant of:

1. The amount in the selected envelope.

2. Whether the amount in the selected envelope is the smaller or the larger amount.

3. The amounts in the two envelopes which imply they are either (X, 2X) or (X, 1/2X).

If we check concrete examples we see that A is not a constant because the amount in our envelope could be other amounts and we need to factor that into our calculations. We also see this is true even if we look.

In any particular trial of the TEP, A (denoted as the amount in the selected envelope) is a constant:

1. Thus, in a particular trial of the TEP, if we opened the selected envelope and find $20, then A is a constant as $20 in this trial.

2. In another trial of the TEP, if we opened the selected envelope and find $10, then A is a constant as $10 in the context of this and only this trial.

1 and 2 are independent trials which means they are not related at all.

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Posted: 09 January 2013 05:15 AM   [ Ignore ]   [ # 1952 ]
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Okay, of course I am here again. So much about free will…

I thought of another way of approaching the whole thing, more ilustrative in ‘small TEP- and TEP-like universes’. In small universes we can make complete overviews of all possibilities.

Starting point is to realise what ‘expected value’ means. In case of TEP it is the average of all possible gains and losses. Even kkwan must admit, that even when it would be wise to switch, it increases the chances on a higher amount, it does not increase the amount itself. Half of the times you loose, half of the times you gain. I assume so far there is no problem. So we should take into account the statistical character of the expected value, and we can do this easily in a limited set of possibilities.

So I define my TEP-universe as:
1. We use cents only
2. The ‘play master’ has only 18 cents.
As additional condition we add that the player of course does not know about 2: he has no idea that there is such a limit.

Now we have two ways of making the complete set of possibilities:
a. according to ‘my method’: for every TEP pair, the opposite pair exists too, i.e. if (X,Y) is a TEP pair then (Y,X) is a TEP pair too.
b. according to kkwan’s method: for every amount as first choice, the other envelope contains half or twice the amount of the first.

Then we add up all the first choices and all the second choices: if the second choices add up to a higher amount, then it is advantageous to switch.

So here it is:

.
        
my method               method kkwan

        First   Second          First   Second
        choice  choice          choice  choice
        1       2               2       1
        2       1               2       4
        2       4               4       8
        4       2               4       2
        3       6               6       3
        6       3               6       12
        4       8
        8       4
        5       10
        10      5
        6       12
        12      6

Total   63      63              24      30 

Why is kkwan’s list so much shorter? Well obvious: the pair (1,2) does not work, because we then would need (1, 0.5) too, and there are no half dollar cents. Same for every other odd amount. Some other pairs are missing too, e.g. (8,4), because the pair to complete the possibilities would be (8,16), but the play master does not have 24 cents, he’s got only 18.
Now it is obvious that in the first list the total of all first choices and all second choices is always the same: it follows immediately from the rule that for every TEP pair (X,Y) (Y,X) is a TEP pair too. Every amount as first choice also appears as second choice.

Now in the ‘kkwan listing’ both columns are not the same: their ratio is 30/24 = 5/4. Exactly what kkwan’s formula says, so it means the pairs in kkwan’s row fit exactly to his argument. But it is obvious that kkwan’s list is not a complete list of all possible TEP pairs that we can create when we have 18 cents.

Now if you want you can try with every limited bound of available money. You will see that for every limit, the relation between the first and the second choice is 5/4. But it is also obvious that in kwann’s listing possible TEP pairs are left out. So kkwan’s formulation artificially makes only a reduced set of possible TEP pairs.

Edit:
For fun, I made an example with excel, maximum amount $150. Now half dollars are allowed of course, and as you see, that is not what makes the difference. It is the leaving out of the higher pairs that would pass the limit of total $150. In the bitmap you see the end of the excel sheet.

[ Edited: 09 January 2013 05:56 AM by GdB ]
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Posted: 12 January 2013 07:49 AM   [ Ignore ]   [ # 1953 ]
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GdB - 09 January 2013 05:15 AM

Okay, of course I am here again. So much about free will…

Is it about the urge to engage and to convince?

Starting point is to realise what ‘expected value’ means. In case of TEP it is the average of all possible gains and losses. Even kkwan must admit, that even when it would be wise to switch, it increases the chances on a higher amount, it does not increase the amount itself. Half of the times you loose, half of the times you gain. I assume so far there is no problem. So we should take into account the statistical character of the expected value, and we can do this easily in a limited set of possibilities.

In the TEP as it is, as the potential gain (A) is twice the potential loss (1/2A) by switching, the probability of gain/loss is 1/2 and the expected value is 5/4A, thus it is beneficial to switch.

So I define my TEP-universe as:
1. We use cents only
2. The ‘play master’ has only 18 cents.
As additional condition we add that the player of course does not know about 2: he has no idea that there is such a limit.

Your TEP-universe is too narrow and restrictive and as such, does not accurately model the TEP at all. In the TEP, it is not so narrow and there are no such restrictions wrt money as cents and the total amount being only 18 cents.

Now we have two ways of making the complete set of possibilities:
a. according to ‘my method’: for every TEP pair, the opposite pair exists too, i.e. if (X,Y) is a TEP pair then (Y,X) is a TEP pair too.
b. according to kkwan’s method: for every amount as first choice, the other envelope contains half or twice the amount of the first.

The fundamental issue with “your method” is that it begs the question. (X, Y) is equivalent to (Y, X) as one and only one possible event (wrt what are the amounts in the two envelopes) whereas either (X, 2X) or (X, 1/2X) are in the two envelopes are two possible distinct events, but we don’t know which event is actual.

Now it is obvious that in the first list the total of all first choices and all second choices is always the same: it follows immediately from the rule that for every TEP pair (X,Y) (Y,X) is a TEP pair too. Every amount as first choice also appears as second choice.

Your (X, Y) and (Y, X) describe only one and only one possible event, not two distinct possible events: as either (X, 2X) or (X, 1/2X) for any finite value of X, in the context of the TEP.

Now in the ‘kkwan listing’ both columns are not the same: their ratio is 30/24 = 5/4. Exactly what kkwan’s formula says, so it means the pairs in kkwan’s row fit exactly to his argument. But it is obvious that kkwan’s list is not a complete list of all possible TEP pairs that we can create when we have 18 cents.

Your “complete list” is based on your too narrow and restrictive “model” of the TEP, the unwarranted assumption that (X, Y) and (Y, X) are two distinct possible events (which is not so) and which begs the question, in the context of the TEP.

Now if you want you can try with every limited bound of available money. You will see that for every limit, the relation between the first and the second choice is 5/4. But it is also obvious that in kwann’s listing possible TEP pairs are left out. So kkwan’s formulation artificially makes only a reduced set of possible TEP pairs.

Not so, as your “possible TEP pairs” is based on the fallacious reasoning of:

1. Treating (X, Y) and (Y, X) as two distinct possible events whereas they actually describe one and only one possible event, in the context of the TEP.

2. As we are totally ignorant of which of the two possible events: either (X, 2X) or (X, 1/2X) is actual, to assume only one and only one event, is begging the question.

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Posted: 12 January 2013 08:21 AM   [ Ignore ]   [ # 1954 ]
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kkwan - 12 January 2013 07:49 AM

Your TEP-universe is too narrow and restrictive and as such, does not accurately model the TEP at all. In the TEP, it is not so narrow and there are no such restrictions wrt money as cents and the total amount being only 18 cents.

Explain how my example is not an accurate example of TEP. Show to us how alleviating the maximum amount changes the result.

kkwan - 12 January 2013 07:49 AM

(X, Y) is equivalent to (Y, X) as one and only one possible event (wrt what are the amounts in the two envelopes)

So having 4 as first choice and 8 as the second is the same as having 8 first and 4 as the second? In the first case you gain 4 cents, in the second you loose.

So tell me, kkwan: how many possibilities are there in a TEP game, when there are only 18 cents available? To be sure: order matters. Having 1 first with 2 as alternative, is another game then having 2 first, and then 1. The question is simply how many courses are possible with 18 cents in TEP games. Answer that question, kkwan.

It is so clearly obvious that with your algorithm you leave out possibilities that really exist: first 12, an then 6 is possible with 18 cents, but you cannot add it in your algorithm, because it would imply the existence of first 12 and then 24, it passes the limit of 18 cents.

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Posted: 12 January 2013 09:28 AM   [ Ignore ]   [ # 1955 ]
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Oh, and another interesting way to show my point.

Show me the beginning of the countable set of all possible TEP games. When you start with cents, you are sure you are able to make a list of all possible TEP sets. You can do this by basing the list on the available amount.

See attached bitmap. It is easy to do with Excel. If you know Excel, then use it, as you can use your descriptions (F5=B5/3, F6=F5/2, F7=F5*2),

Reading example: with 6 cents, I can make <2,4> and <4,2>, you can make <2,1> and <2,4>.
Is your list complete? Does <3,6> not belong to the set of TEP pairs, i.e. is it not possible that there is 3 cents in your first envelope and 6 in the second? Beware: having 6 first and then switch to 3 is not the same game!

In other words, start from the lowest possible amount, and show all possible ways that we can go through TEP games with all possible TEP pairs.

If you think I cheated then show the beginning of your countable set of TEP games (not TEP pairs!), and tell me why your version is correct.

Or are you starting to avoid again? At least I now alleviated the limit for you, and showed you for every total amount the possible TEP games according to my algorithm, and according to yours.

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Posted: 13 January 2013 09:35 AM   [ Ignore ]   [ # 1956 ]
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GdB - 12 January 2013 08:21 AM

Explain how my example is not an accurate example of TEP. Show to us how alleviating the maximum amount changes the result.

Your example is not an accurate description of the TEP because:

1. It’s ambit is too narrow by restricting money to cents only.

2. The total amount is only 18 cents whereas it is not specific what the total amount is, in the TEP.

3. There is no “play master” as such, in the context of the TEP.

4. Your additional condition is ludicrous.

5. Fundamentally, your example with (X, Y) and (Y, X) is fatally flawed because it begs the question as (X, Y) or (Y, X) refer to one and only one possible event whereas there are two and only two possible events: either (X, 2X) or (X, 1/2X) and we don’t know which event is actual.

So having 4 as first choice and 8 as the second is the same as having 8 first and 4 as the second? In the first case you gain 4 cents, in the second you loose.

(4, 8) or (8, 4) describes one and only one event, i.e. there are 4 and 8 cents in the two envelopes.
As such, if we can either select 4 or 8 cents, that implies that we know specifically that the above event is actual which is begging the question.

So tell me, kkwan: how many possibilities are there in a TEP game, when there are only 18 cents available? To be sure: order matters. Having 1 first with 2 as alternative, is another game then having 2 first, and then 1. The question is simply how many courses are possible with 18 cents in TEP games. Answer that question, kkwan.

If order matters, then you are begging the question because being totally ignorant implies we don’t know what is the amount we have selected and whether it is the smaller or the larger amount.

It is so clearly obvious that with your algorithm you leave out possibilities that really exist: first 12, an then 6 is possible with 18 cents, but you cannot add it in your algorithm, because it would imply the existence of first 12 and then 24, it passes the limit of 18 cents.

The total amount of 18 cents is an unjustified assumption and contravenes the TEP as explained above.

Thus, your example is not the TEP per se.

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Posted: 13 January 2013 09:45 AM   [ Ignore ]   [ # 1957 ]
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GdB - 12 January 2013 09:28 AM

Show me the beginning of the countable set of all possible TEP games. When you start with cents, you are sure you are able to make a list of all possible TEP sets. You can do this by basing the list on the available amount.

When you start with cents, it is not the TEP per se.

So, it is pointless and futile to make a list as you do.

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Posted: 13 January 2013 11:28 PM   [ Ignore ]   [ # 1958 ]
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kkwan - 13 January 2013 09:35 AM
GdB - 12 January 2013 08:21 AM

Explain how my example is not an accurate example of TEP. Show to us how alleviating the maximum amount changes the result.

Your example is not an accurate description of the TEP because:

1. It’s ambit is too narrow by restricting money to cents only.

2. The total amount is only 18 cents whereas it is not specific what the total amount is, in the TEP.

3. There is no “play master” as such, in the context of the TEP.

4. Your additional condition is ludicrous.

5. Fundamentally, your example with (X, Y) and (Y, X) is fatally flawed because it begs the question as (X, Y) or (Y, X) refer to one and only one possible event whereas there are two and only two possible events: either (X, 2X) or (X, 1/2X) and we don’t know which event is actual.

Ah, avoiding again!

You gave not a single mathematical ground how alleviating the maximum amount changes the result.

1. is bullshit. if you think one can only play TEP when there is no smallest unit of money then I am done with you.
3. It is not relevant how the amounts got into the envelopes. Assuming any special procedure, (and with special money (see 1)!) is begging the question. TEP states nothing about the way the two envelopes are prepared so it is irrelevant. Also in your argumentation you never use some fact about how the amounts got into the envelopes. So any criticism based on the question how the envelopes were prepared is irrelevant.
4. My additional condition is that the player does not know the play master has a limited amount of money, and that the player does not know what the limit is. It is not ludicrous, it is even superfluous. But I just wanted to make sure you see that the player is totally ignorant about the limited amount of money, and so also about any value he might have chosen.
5. And your (X, 2X) or (X, 1/2X) is the best example of begging the question: with the same two envelopes, with the same two amounts, we don’t know if <X,2X> or <2X,X> is actual (remember the meaning of <a,b>!).

And 2. is the point I asked you to explain, and I say it explicitly here, mathematically. Why does a limited amount of available money (18 cents in my example) gives a wrong set of possible courses in TEP plays? Give one or two examples, and then a rigid proof. (But I am sure you will avoid this again).

kkwan - 13 January 2013 09:35 AM

(4, 8) or (8, 4) describes one and only one event, i.e. there are 4 and 8 cents in the two envelopes.
As such, if we can either select 4 or 8 cents, that implies that we know specifically that the above event is actual which is begging the question.

You are happily avoiding the fact that I do not just describe TEP pairs, but possible courses that TEP games can have. Above my columns were very clear ‘first choice, second choice’. If you describe the ‘round’ first pick 4 and then 8, or first picking 8 and then 4 as the same round, i.e. gaining 4 cents and loosing 4 cents as the same event, then I am done with you.

[ Edited: 14 January 2013 05:16 AM by GdB ]
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Posted: 14 January 2013 12:03 AM   [ Ignore ]   [ # 1959 ]
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kkwan - 13 January 2013 09:45 AM
GdB - 12 January 2013 09:28 AM

Show me the beginning of the countable set of all possible TEP games. When you start with cents, you are sure you are able to make a list of all possible TEP sets. You can do this by basing the list on the available amount.

When you start with cents, it is not the TEP per se.

See 1. above.

So you deny that if you really play TEP, there is a smallest total amount you can start with?

1 cent: no TEP possible, only way to split the amount is (0,1) which violates 2 conditions of TEP (0 not allowed, one envelope twice the other is also impossible)
2 cents: only way to split up without 0 is (1,1), which is not a TEP pair
3 cents: now we can split up according to (1,2). So there are two possible ways a TEP round can take place: I pick 1 first and then 2, or I pick 2 first and then 1.
4 cents: (1,3), (2,2) are not TEP pairs
5 cents: (1,4) and (2,3) are not TEP pairs
6 cents: now we can split up according to (2,4). So there are two possible ways a TEP round can take place: I pick 2 first and then 4, or I pick 4 first and then 2.
etc.
So based on cents, I can produce every pair of TEP amounts in cents, and therefore also in dollars.

Now you proof that this algorithm does not produce a list with all possible TEP pairs, with all possible rounds that one can play with these pairs.

Secondly you can do the same with your algorithm. If you do it correctly it should produce the ratio of 5/4 as derived by your formula. If not, your algorithm is false, or your formula.

Another question: assume you know that my amount of money is limited, and I play TEP with you: is my formula valid or yours? Why? Show me how your knowledge about the limit influences your strategy.

And an additional question: take a snapshot of the earth at a moment. In this snapshot, is the amount of money limited or unlimited? What does that mean for two guys who are playing TEP at the moment of the snapshot?

[ Edited: 14 January 2013 11:52 PM by GdB ]
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Posted: 15 January 2013 09:13 AM   [ Ignore ]   [ # 1960 ]
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GdB - 13 January 2013 11:28 PM

You gave not a single mathematical ground how alleviating the maximum amount changes the result.

1. is bullshit. if you think one can only play TEP when there is no smallest unit of money then I am done with you.
3. It is not relevant how the amounts got into the envelopes. Assuming any special procedure, (and with special money (see 1)!) is begging the question. TEP states nothing about the way the two envelopes are prepared so it is irrelevant. Also in your argumentation you never use some fact about how the amounts got into the envelopes. So any criticism based on the question how the envelopes were prepared is irrelevant.
4. My additional condition is that the player does not know the play master has a limited amount of money, and that the player does not know what the limit is. It is not ludicrous, it is even superfluous. But I just wanted to make sure you see that the player is totally ignorant about the limited amount of money, and so also about any value he might have chosen.
5. And your (X, 2X) or (X, 1/2X) is the best example of begging the question: with the same two envelopes, with the same two amounts, we don’t know if <X,2X> or <2X,X> is actual (remember the meaning of <a,b>!).

And 2. is the point I asked you to explain, and I say it explicitly here, mathematically. Why does a limited amount of available money (18 cents in my example) gives a wrong set of possible courses in TEP plays? Give one or two examples, and then a rigid proof. (But I am sure you will avoid this again).

Your example is clearly a travesty of the TEP as the TEP does not specify any limited amount of money such as 18 cents and neither is money restricted to cents. As such, it is not the TEP per se.

The onus is on you to show convincingly that it is synonymous to the TEP as it is your example.

That either (X, 2X) or (X, 1/2X) are in the two envelopes does not beg the question as that implies we don’t know what are the actual amounts in the two envelopes as per the TEP.

You are happily avoiding the fact that I do not just describe TEP pairs, but possible courses that TEP games can have. Above my columns were very clear ‘first choice, second choice’. If you describe the ‘round’ first pick 4 and then 8, or first picking 8 and then 4 as the same round, i.e. gaining 4 cents and loosing 4 cents as the same event, then I am done with you.

It does not matter whether it is (4, 8) or (8, 4) as both describe one and only one event and that begs the question, which is fatal.

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Posted: 15 January 2013 10:19 AM   [ Ignore ]   [ # 1961 ]
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GdB - 14 January 2013 12:03 AM

See 1. above.

So you deny that if you really play TEP, there is a smallest total amount you can start with?

1 cent: no TEP possible, only way to split the amount is (0,1) which violates 2 conditions of TEP (0 not allowed, one envelope twice the other is also impossible)
2 cents: only way to split up without 0 is (1,1), which is not a TEP pair
3 cents: now we can split up according to (1,2). So there are two possible ways a TEP round can take place: I pick 1 first and then 2, or I pick 2 first and then 1.
4 cents: (1,3), (2,2) are not TEP pairs
5 cents: (1,4) and (2,3) are not TEP pairs
6 cents: now we can split up according to (2,4). So there are two possible ways a TEP round can take place: I pick 2 first and then 4, or I pick 4 first and then 2.
etc.
So based on cents, I can produce every pair of TEP amounts in cents, and therefore also in dollars.

Now you proof that this algorithm does not produce a list with all possible TEP pairs, with all possible rounds that one can play with these pairs.

Secondly you can do the same with your algorithm. If you do it correctly it should produce the ratio of 5/4 as derived by your formula. If not, your algorithm is false, or your formula.

Another question: assume you know that my amount of money is limited, and I play TEP with you: is my formula valid or yours? Why? Show me how your knowledge about the limit influences your strategy.

And an additional question: take a snapshot of the earth at a moment. In this snapshot, is the amount of money limited or unlimited? What does that mean for two guys who are playing TEP at the moment of the snapshot?

You are going off on a tangent and writing gibberish:

1. By insisting on only cents for money and limiting the total amount to only 18 cents in your example which is not so, in the context of the TEP. Hence, your ridiculous example is merely a travesty of the TEP and as such, should be discarded ASAP.

2. Making the unwarranted assumption that the amount of money is limited in the context of the TEP whereas there should be no such assumption except that money is discrete and not infinite.

3. What has “a snapshot of the earth at a moment”, “the amount of money limited or unlimited” and “what does that mean for two guys who are playing TEP at the moment of the snapshot” got to do with the TEP?

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Posted: 15 January 2013 11:03 AM   [ Ignore ]   [ # 1962 ]
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To be short, kkwan, you are not able to give mathematical reasons for why I am wrong. You do not use your (pejorative) remarks in mathematical reasoning, that shows why I am wrong. You only state that I am wrong. For a ‘TEP specialist’ as you are it should be easy to derive from the TEP definition why my argument of making a countable set of all TEP games (not pairs!) is wrong. Why don’t just show us? Show us how assuming an upper limit influences the result of TEP games. Show us why, when I start with the smallest possible amount, this changes the outcome of TEP.

You can’t. You will avoid again, and give a load of derogatory sentences, without rigidly demonstrating why my argumentation is wrong. You have nothing else on offer, kkwan.

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Posted: 15 January 2013 12:20 PM   [ Ignore ]   [ # 1963 ]
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Kkwan,

kkwan - 12 January 2013 07:49 AM

Your TEP-universe is too narrow and restrictive and as such, does not accurately model the TEP at all. In the TEP, it is not so narrow and there are no such restrictions wrt money as cents and the total amount being only 18 cents.

The problem is whatever the TEP- universe, the result will be just the same. This is because for every possible gain there will always be the same possible loss with equal probability.

So if $10, $20 is one of the possible pairs you can gain $10 if you have the smaller amount and lose $10 if you have the larger amount.

However you set the TEP up you can’t avoid this because all you can do is add more and more possible pairs.

If you believe you can avoid this it’s up to you to give an example.

Stephen

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Posted: 26 January 2013 02:37 PM   [ Ignore ]   [ # 1964 ]
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GdB - 15 January 2013 11:03 AM

To be short, kkwan, you are not able to give mathematical reasons for why I am wrong. You do not use your (pejorative) remarks in mathematical reasoning, that shows why I am wrong. You only state that I am wrong. For a ‘TEP specialist’ as you are it should be easy to derive from the TEP definition why my argument of making a countable set of all TEP games (not pairs!) is wrong. Why don’t just show us? Show us how assuming an upper limit influences the result of TEP games. Show us why, when I start with the smallest possible amount, this changes the outcome of TEP.

You can’t. You will avoid again, and give a load of derogatory sentences, without rigidly demonstrating why my argumentation is wrong. You have nothing else on offer, kkwan.

The TEP is not just a mathematical problem per se as the given information (that one envelope has twice the amount in the other) is insufficient to resolve the TEP mathematically as that information implies there are two and only two possibilities: either (X, 2X) or (X, 1/2X) are in the two envelopes and we are totally ignorant which is actual.

Thus, to try and resolve the TEP with one and only one possibility (X, 2X) or with limiting cases as you do, contravenes the fundamental nature of the TEP, begs the question and is clearly misguided and wrong.

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Posted: 26 January 2013 02:59 PM   [ Ignore ]   [ # 1965 ]
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StephenLawrence - 15 January 2013 12:20 PM

The problem is whatever the TEP- universe, the result will be just the same. This is because for every possible gain there will always be the same possible loss with equal probability.

So if $10, $20 is one of the possible pairs you can gain $10 if you have the smaller amount and lose $10 if you have the larger amount.

However you set the TEP up you can’t avoid this because all you can do is add more and more possible pairs.

If you believe you can avoid this it’s up to you to give an example.

If ($10, $20) is one possible pair, then we must also consider ($10, $5) as the other possible pair.

As such, with A denoted as the amount in the selected envelope and as A is either the smaller or the larger amount:

1. If we are in (A, 2A), by switching gain is A

2. If we are in (A, 1/2A), by switching loss is 1/2A.

As the gain is twice the loss, it is beneficial to switch.

This is equivalent to A as $10 in either ($10, $20) or ($10, $5) for the same outcomes.

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