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The two envelopes problem
Posted: 06 September 2012 12:47 AM   [ Ignore ]   [ # 1456 ]
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kkwan - 05 September 2012 07:01 PM

There are 2 amounts in each of the 2 possible events. You are conflating amounts with events.

Sorry kkwan, there is only one event: there are only two amounts. From the position of the player however there are endless possible events (not just two). And the endless possibilities are perfectly described by (X,2X). Also by (X,X/2), but they describe the same sets. We had that already.

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Posted: 07 September 2012 07:40 PM   [ Ignore ]   [ # 1457 ]
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GdB - 06 September 2012 12:47 AM

Sorry kkwan, there is only one event: there are only two amounts. From the position of the player however there are endless possible events (not just two). And the endless possibilities are perfectly described by (X,2X). Also by (X,X/2), but they describe the same sets. We had that already.

We are considering any specific finite value of X and not all the possible values of X. So, the issue of endless possibilities is not relevant.

There are two possible events (for any specific finite value of X) and they are (X, 2X) or (X, 1/2X).

From the perspective of the player (for any specific finite value of X), there are two and only two possible events, not endless possible events.

And as such, they do not describe the same sets as they are not equivalent in value.

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Posted: 08 September 2012 02:59 AM   [ Ignore ]   [ # 1458 ]
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kkwan - 07 September 2012 07:40 PM

We are considering any specific finite value of X and not all the possible values of X. So, the issue of endless possibilities is not relevant.

Why are we considering any specific finite value of X and not all the possible values of X? As long as the player has no idea about the amounts any pair of amounts is possible, as long as one amount is twice the other. Or said otherwise, one amount is half of the other. Only for you these two descriptions are not the same… I think you should ask your friends if these two description are equivalent or not, and see what they say…

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Posted: 09 September 2012 07:40 PM   [ Ignore ]   [ # 1459 ]
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GdB - 08 September 2012 02:59 AM

Why are we considering any specific finite value of X and not all the possible values of X? As long as the player has no idea about the amounts any pair of amounts is possible, as long as one amount is twice the other. Or said otherwise, one amount is half of the other. Only for you these two descriptions are not the same… I think you should ask your friends if these two description are equivalent or not, and see what they say…

We are only considering any specific finite value of X and not all the possible value of X because it defines a particular situation whereby X is also defined.

X is a variable and can be any finite value but for any specific finite value of X, (X, 2X) and (X, 1/2X) are not equivalent.

So, if X is 100, (X, 2X) is (100, 200) and (X, 1/2X) is (100, 50) and they are not equivalent.

This is true for any specific finite possible value of X. 

Thus, generally, for any specific finite possible value of X, (X, 2X) and (X, 1/2X) are the two and only two possible events and they are not equivalent.

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Posted: 09 September 2012 11:31 PM   [ Ignore ]   [ # 1460 ]
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No kkwan, we are considering two unknown amounts, from which we only know that one amount is half the other, or the other amount is twice the first - which is exactly the same.

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Posted: 11 September 2012 08:56 AM   [ Ignore ]   [ # 1461 ]
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kkwan - 04 January 2012 08:27 PM

What is it? From the wiki

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.
  9. After the switch, I can denote that content by B and reason in exactly the same manner as above.
  10. I will conclude that the most rational thing to do is to swap back again.
  11. To be rational, I will thus end up swapping envelopes indefinitely.
  12. As it seems more rational to open just any envelope than to swap indefinitely, we have a contradiction.

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

...
The conflict between human reasoning, logic and rationality. To switch or not to switch? LOL

I am thrown off by this ‘paradox’ as being anything real except for a misapplication of forcing meaning into something that shouldn’t be there. The formula is illegitimate on logical grounds. The probability of the terms are being assumed capable of being formulated within the same formula just because they represent numbers. Let me give an example:

If you were to think of the probabilities of heads(h) or tails(t) in one toss, you may say that you have 1/2 probability of a head turning up and 1/2 of a tail turning up. All probabilities added up are defined as 1. So, you might illustrate it like this:

A:

    1/2(H) + 1/2(T) = 1

In symbols for probabilities, we do not think of ‘H’ or ‘T’ as variables meant to represent a shared meaning with the formula. Otherwise, we could then do this:

B:

  (H + T)/2 = 1
            H + T = 2
                  T = 2 - H

This is absurd and means nothing because heads nor tails represent quantities. But even if we assigned quantitative meaning to them, they are still irrelevant to the probability formula. So let heads represent a quantity such that tails represents twice its value. Let H be represented by A, a variable quantity of money, and T be represented by 2A, twice that amount. Then,


C:

  T + H = total amount of money in play
        =  A + 2A = 3A


The heads or tails formulations in A are not substitutable with values from C because probability fractions don’t represent sums of different units of things, but rather of a common thing:

D:

        1/2 P + 1/2 P = 1(P),    where P = All unique possibilities of a set {P(1), P(2)}

So there is no such paradox.

[ Edited: 11 September 2012 08:59 AM by Scott Mayers ]
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Posted: 11 September 2012 11:01 AM   [ Ignore ]   [ # 1462 ]
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I think you make it a bit too complicated, but the gust of your argument is correct.

I gave a similar argumentation with a red piece of paper in one envelope and a green piece of paper in the other, one representing the smaller amount, the other the bigger. But kkwan has decided that his argumentation is correct, and just will blow some more smoke again.

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Posted: 11 September 2012 10:28 PM   [ Ignore ]   [ # 1463 ]
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GdB - 09 September 2012 11:31 PM

No kkwan, we are considering two unknown amounts, from which we only know that one amount is half the other, or the other amount is twice the first - which is exactly the same.

There are two unknown amounts but we don’t know what they are with certainty because “one amount is half the other/one amount is twice the other” is ambiguous and is satisfied by both (X, 2X) and (X, 1/2X) which are not the same or equivalent for any specific finite value of X.

So, there are two possible events (X, 2X) or (X, 1/2X) and not one (X, 2X), for any specific finite value of X.

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Posted: 11 September 2012 10:51 PM   [ Ignore ]   [ # 1464 ]
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Scott Mayers - 11 September 2012 08:56 AM

I am thrown off by this ‘paradox’ as being anything real except for a misapplication of forcing meaning into something that shouldn’t be there. The formula is illegitimate on logical grounds. The probability of the terms are being assumed capable of being formulated within the same formula just because they represent numbers. Let me give an example:

The formula in step 7 is logically valid as the player is totally ignorant as expressed in steps 1 to 6.

If you were to think of the probabilities of heads(h) or tails(t) in one toss, you may say that you have 1/2 probability of a head turning up and 1/2 of a tail turning up. All probabilities added up are defined as 1.

Think of the events (X, 2X) or (X, 1/2X) as analogous to head or tail.

The probability of either event is 1/2.

The two probabilities add up to 1, which means either (X, 2X) or (X, 1/2X) are in the two envelopes similar to either head or tail in any toss of an unbiased coin.

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Posted: 11 September 2012 11:19 PM   [ Ignore ]   [ # 1465 ]
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kkwan - 11 September 2012 10:28 PM

So, there are two possible events (X, 2X) or (X, 1/2X) and not one (X, 2X), for any specific finite value of X.

So what? I am interested in the amounts, not in X.

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Posted: 11 September 2012 11:20 PM   [ Ignore ]   [ # 1466 ]
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kkwan - 11 September 2012 10:51 PM

Think of the events (X, 2X) or (X, 1/2X) as analogous to head or tail.

Both events are the same. Where the hell do you get the idea that you should have the same X??? Why should you interpret the sentence “One envelope contains twice as much as the other” in such a way? On this point Scott is completely right. There are 2 amounts, they can be expressed as (X,2X), (X,X/2), (1001X,2002X), (0.01A,0.02A), all equally valid. And all leading to the same conclusion.

Let’s do it once again:

1. I have two envelopes, one with a red piece of paper and one with a green piece of paper. I choose one. If I change, the chances of
red—> green,
green—> red are equal.
2. I have two envelopes, one with a piece of paper with 1000 on it and one with a piece of paper with 2000 on it. I choose one. If I change, the chances of
1000—> 2000, so gaining 1000,
2000—> 1000, so loosing 1000, are equal.
3. I have two envelopes, one with a piece of paper with the number X on it and one with a piece of paper with the number Y on it. I choose one. If I change, the chances of
X—> Y, so gaining Y - X,
Y—> X, so gaining X - Y, are equal.
4. I have two envelopes, one with a piece of paper with the number X on it and one with a piece of paper with the number 2X on it. I choose one. If I change, the chances of
X—> 2X, so gaining 2X - X = X,
2X—> X, so ‘gaining’ X - 2X = -X, are equal.
And special for you kkwan:
5. I have two envelopes, one with a piece of paper with the number X on it and one with a piece of paper with the number X/2 on it. I choose one. If I change, the chances of
X—> X/2, so ‘gaining’ X/2 - X = -X/2,
X/2—> X, so gaining X - X/2 = X/2, are equal.
6.  I have two envelopes, one with an amount in it and one with twice the amount in it. I choose one. If I change, the chances of
amount—> twice the amount, so gaining the smallest amount,
twice the amount —> the smallest amount, so loosing the smallest amount are equal.
And special for you kkwan:
7.  I have two envelopes, one with an amount in it and one with half the amount in it. I choose one. If I change, the chances of
amount—> half the amount, so loosing the smallest amount,
half the amount —> the biggest amount, so gaining the smallest amount are equal.

Now you tell me for every step why it would be invalid.

[ Edited: 11 September 2012 11:54 PM by GdB ]
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Posted: 12 September 2012 09:25 AM   [ Ignore ]   [ # 1467 ]
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kkwan - 11 September 2012 10:51 PM

Think of the events (X, 2X) or (X, 1/2X) as analogous to head or tail.

The probability of either event is 1/2.

The two probabilities add up to 1, which means either (X, 2X) or (X, 1/2X) are in the two envelopes similar to either head or tail in any toss of an unbiased coin.

This is precisely my point. But what you are not following is that the values or any meaning given to heads or tails is irrelevant to the formulation of a probability equation. It just happens to be arbitrary that the meanings in this particular example have numerical values and have relational values that can create equations with each other: that is, what the envelope contains. But, just as they could be represented as heads or tails (non-numerical values),the envelopes could contain colored cards, as Gdb gives as one example in number one above. So you can only make a separate statement or sentence regarding those values if you so choose, as I showed in C in my previous post. But that equation or its terms have no connection whatsoever to the probability equation itself.

The illegal parts are the “1/2 times A” and “1/2 times A/2” because while the fractions represent the probability of getting A or A/2, the probability statement cannot interpret the meaning of what’s in the envelope. To rephrase Gdb’s color example, let’s assume that one envelope has a red card and the other has a green card. So that formula should go according to the same assumptions set up by this presumed paradox would be:

    1/2(red) + 1/2(green) = 1/2(red + green) = 1/2 yellow?  [Is this a contradiction too? Should it equal 1? ... or white?]

The envelopes could have contained differing items too, like a dollar bill and a blue card. So the formula would be:

    1/2($1) + 1/2(blue card) = 1/2($1 + blue card) ??

Do you get it now?

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Posted: 16 September 2012 07:31 PM   [ Ignore ]   [ # 1468 ]
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GdB - 11 September 2012 11:19 PM

So what? I am interested in the amounts, not in X.

The amounts in the two envelopes are determined by the value of X.

So, for any finite value of X, (X, 2X) and (X, 1/2X) are the two and only two possible events and they are not equivalent.

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Posted: 16 September 2012 07:54 PM   [ Ignore ]   [ # 1469 ]
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GdB - 11 September 2012 11:20 PM

Both events are the same. Where the hell do you get the idea that you should have the same X??? Why should you interpret the sentence “One envelope contains twice as much as the other” in such a way? On this point Scott is completely right. There are 2 amounts, they can be expressed as (X,2X), (X,X/2), (1001X,2002X), (0.01A,0.02A), all equally valid. And all leading to the same conclusion.

(X, 2X) or (X, 1/2X) are distinct possible events for any specific finite value of X.

1. I have two envelopes, one with a red piece of paper and one with a green piece of paper. I choose one. If I change, the chances of
red—> green,
green—> red are equal.
2. I have two envelopes, one with a piece of paper with 1000 on it and one with a piece of paper with 2000 on it. I choose one. If I change, the chances of
1000—> 2000, so gaining 1000,
2000—> 1000, so loosing 1000, are equal.
3. I have two envelopes, one with a piece of paper with the number X on it and one with a piece of paper with the number Y on it. I choose one. If I change, the chances of
X—> Y, so gaining Y - X,
Y—> X, so gaining X - Y, are equal.
4. I have two envelopes, one with a piece of paper with the number X on it and one with a piece of paper with the number 2X on it. I choose one. If I change, the chances of
X—> 2X, so gaining 2X - X = X,
2X—> X, so ‘gaining’ X - 2X = -X, are equal.
And special for you kkwan:
5. I have two envelopes, one with a piece of paper with the number X on it and one with a piece of paper with the number X/2 on it. I choose one. If I change, the chances of
X—> X/2, so ‘gaining’ X/2 - X = -X/2,
X/2—> X, so gaining X - X/2 = X/2, are equal.
6.  I have two envelopes, one with an amount in it and one with twice the amount in it. I choose one. If I change, the chances of
amount—> twice the amount, so gaining the smallest amount,
twice the amount —> the smallest amount, so loosing the smallest amount are equal.
And special for you kkwan:
7.  I have two envelopes, one with an amount in it and one with half the amount in it. I choose one. If I change, the chances of
amount—> half the amount, so loosing the smallest amount,
half the amount —> the biggest amount, so gaining the smallest amount are equal.

All your examples make the fundamental assumption that what are in the two envelopes are known which is not so in the context of the TEP.

As such, your fundamental assumption is not justified unless you have direct acquaintance of the contents in the two envelopes which you don’t have, in the context of the TEP.

Thus, all you can say is either (X, 2X) or (X, 1/2X) are in the two envelopes because you are totally ignorant of the actual contents in the two envelopes.

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Posted: 16 September 2012 08:50 PM   [ Ignore ]   [ # 1470 ]
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Scott Mayers - 12 September 2012 09:25 AM

This is precisely my point. But what you are not following is that the values or any meaning given to heads or tails is irrelevant to the formulation of a probability equation. It just happens to be arbitrary that the meanings in this particular example have numerical values and have relational values that can create equations with each other: that is, what the envelope contains. But, just as they could be represented as heads or tails (non-numerical values),the envelopes could contain colored cards, as Gdb gives as one example in number one above. So you can only make a separate statement or sentence regarding those values if you so choose, as I showed in C in my previous post. But that equation or its terms have no connection whatsoever to the probability equation itself.

As head or tail are two possible mutually exclusive events, so are (X, 2X) or (X, 1/2X).

The illegal parts are the “1/2 times A” and “1/2 times A/2” because while the fractions represent the probability of getting A or A/2, the probability statement cannot interpret the meaning of what’s in the envelope. To rephrase Gdb’s color example, let’s assume that one envelope has a red card and the other has a green card. So that formula should go according to the same assumptions set up by this presumed paradox would be:

    1/2(red) + 1/2(green) = 1/2(red + green) = 1/2 yellow?  [Is this a contradiction too? Should it equal 1? ... or white?]

The envelopes could have contained differing items too, like a dollar bill and a blue card. So the formula would be:

    1/2($1) + 1/2(blue card) = 1/2($1 + blue card) ??

The fundamental assumption of GdB that “one envelope has a red card and the other has a green card” is unjustified because we don’t know what are actually in the two envelopes in the context of the TEP.

As such, we must be circumspect.

The information provided in the TEP implies either (X, 2X) or (X, 1/2X) are the two and only two distinct possible events for any finite value of X

Thus, all we can say is either (X, 2X) or (X, 1/2X) are in the two envelopes as we are totally ignorant of the actual contents in the two envelopes.

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