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The two envelopes problem
 Posted: 21 January 2012 12:21 PM [ Ignore ]   [ # 166 ]
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Axegrrl - 21 January 2012 12:53 AM

Someone on another board looked at this and said:

this is a misapplication of expected value i think. it’s not a continuous function in this case.

It need not be so.

From the wiki on continous function

In mathematics, a continuous function is a function for which, intuitively, “small” changes in the input result in “small” changes in the output. Otherwise, a function is said to be “discontinuous”.

Examples:

As an example, consider the function h(t), which describes the height of a growing flower at time t. This function is continuous. In fact, a dictum of classical physics states that in nature everything is continuous. By contrast, if M(t) denotes the amount of money in a bank account at time t, then the function jumps whenever money is deposited or withdrawn, so the function M(t) is discontinuous.

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 Posted: 21 January 2012 12:45 PM [ Ignore ]   [ # 167 ]
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Axegrrl - 21 January 2012 01:01 AM

Someone else commented:

the problem is in the incorrect use of “expected value.”  Another way of looking at is that the expected value is not the value that one would expect is in the other envelope, it’s simply a fancy statistical term for the average of the amounts in the two envelopes.

Of course, it is “not the value that one would expect is in the other envelope”, but neither is it “a fancy statistical term for the average of the amounts in the two envelopes”.

From the wiki on the expected value

More informally, it can be interpreted as the long-run average of the results of many independent repetitions of an experiment (e.g. a dice roll).

However, the writer has to show incorrect use of expected value explicitly, otherwise it is only an unsupported assertion.

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 Posted: 21 January 2012 01:23 PM [ Ignore ]   [ # 168 ]
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GdB - 21 January 2012 03:34 AM

Assume one envelope has A dollars in it, the other 2A.

But if A is an unknown quantity, how can you assume one envelope has A dollars, the other 2A?

By doing so, you have considered A as a known quantity.

If you are shown the amount in the envelope you have selected, you still don’t know whether the other has twice or half the amount.

In this case, A is a known quantity, but you don’t know whether it is the larger or the smaller amount.

So, it is more appropriate to:

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.

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 Posted: 22 January 2012 12:39 AM [ Ignore ]   [ # 169 ]
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kkwan - 21 January 2012 01:23 PM
GdB - 21 January 2012 03:34 AM

Assume one envelope has A dollars in it, the other 2A.

But if A is an unknown quantity, how can you assume one envelope has A dollars, the other 2A?

What a bullshit, kkwan. This is what defines the whole situation:

One envelope contains twice as much as the other.

The amounts are already put into the envelopes, they do not change as in my adaptation of the game. To say that one envelope has an (unknown) amount of A and the other 2A, and I don’t know which amount is in what envelope, is a perfect translation of the ‘problem’. You may also express it as A and 0.5A, but you cannot intermingle them as you do. In your presentation the total amount in the two envelopes ‘magically’ changes from 1.5A to 3A or the other way round.

You may think, but you do not think correctly.

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 Posted: 22 January 2012 09:13 AM [ Ignore ]   [ # 170 ]
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GdB - 22 January 2012 12:39 AM

One envelope contains twice as much as the other.

The amounts are already put into the envelopes, they do not change as in my adaptation of the game. To say that one envelope has an (unknown) amount of A and the other 2A, and I don’t know which amount is in what envelope, is a perfect translation of the ‘problem’. You may also express it as A and 0.5A, but you cannot intermingle them as you do. In your presentation the total amount in the two envelopes ‘magically’ changes from 1.5A to 3A or the other way round.

That one envelope contains twice as much as the other is the only information known to the player, but what is A is not known. In other words, A could be any amount but unknown. Neither is A known as the smaller or the larger amount.

What you are saying is tantamount to assuming the player could know them as A and 2A which she does not at all times.

Thus, from the perspective of the player:

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.

Even if A is known to the player, the above steps are just as valid.

It is as simple as that, so why can you not accept that?

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 Posted: 22 January 2012 10:09 AM [ Ignore ]   [ # 171 ]
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Yes, A is unknown, I’ve no idea what A is, except that it is bigger than 0. That is said in the description. So why do you think I put more information in my description?

Please also give one single example for amounts in the enveloped that fit the description of the problem and in which I am not able to find a description in terms of an amount A and an amount 2A (e.g. 4\$ and 8\$), and you do not know which envelope you choose, the bigger amount or the smaller amount. If there is none, then my translation is correct.

You make a wrong translation of a problem into mathematical language. I notice you are just pasting in the same texts again and again. You are not reasoning, you are just ‘kkwanning’ again.

It seems you have the same problems with the concept of ‘unknown’ and the concept of ‘endless’ in mathematics.

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 Posted: 22 January 2012 10:51 PM [ Ignore ]   [ # 172 ]
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GdB - 22 January 2012 10:09 AM

Yes, A is unknown, I’ve no idea what A is, except that it is bigger than 0. That is said in the description. So why do you think I put more information in my description?

If A is the amount in the envelope you selected, A is unknown and you do not know whether A has the smaller or the larger amount, that is all there is to it.

However, your description in post 170:

To say that one envelope has an (unknown) amount of A and the other 2A, and I don’t know which amount is in what envelope, is a perfect translation of the ‘problem’.

How do you know that the other envelope has 2A if A is unknown and you don’t know whether A is the smaller or the larger amount? It is equally possible that the other envelope has A/2.

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 Posted: 23 January 2012 12:39 AM [ Ignore ]   [ # 173 ]
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kkwan - 22 January 2012 10:51 PM

How do you know that the other envelope has 2A if A is unknown and you don’t know whether A is the smaller or the larger amount? It is equally possible that the other envelope has A/2.

I don’t know. But it is equivalent to say there are 2 envelopes, one with an amount of A\$ and another with 2A\$, and I do not know which one I picked first.

Say I picked the envelope with 2A\$ first (still not knowing what A is). Then I can introduce a new variable B defined by B=2A. Filling this in I get B\$ in the first envelope and (B/2)\$ in the second (still not knowing what A is, and therefore not knowing what B is). And that exactly covers the situation in which I picked the envelope with the biggest amount first.

So if you you want it this way, then describe it as such. Call the lower amount A, the higher amount B:

4b If you pick the lower amount A, then the other envelope contains 2A
5b If you pick the higher amount B, then the other envelope contains B/2

Now according to the problem as stated, B = 2A, so we can replace 5b with 5b`:

5b` If you pick the higher amount 2A then the other envelope contains A

So a slight detour to get at my 4a and 5a:

4a If you pick the lower amount A, then the other envelope contains 2A
5a If you pick the higher amount 2A, then the other envelope contains A

Which lead to the correct solution that switching is no use.

[ Edited: 23 January 2012 01:33 AM by GdB ]
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 Posted: 23 January 2012 12:50 AM [ Ignore ]   [ # 174 ]
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“sigh of relief”......

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 Posted: 23 January 2012 01:56 AM [ Ignore ]   [ # 175 ]
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Write4U - 23 January 2012 12:50 AM

“sigh of relief”......

kkwan will produce some smoke again…

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 Posted: 23 January 2012 02:53 AM [ Ignore ]   [ # 176 ]
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Write4U - 23 January 2012 12:50 AM

“sigh of relief”

`
*ditto-ing Write4U’s relief*

You nailed it GdB :)  hopefully, kwan will be able to understand why….....

`

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 Posted: 23 January 2012 06:18 AM [ Ignore ]   [ # 177 ]
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I haven’t read all 12 pages of this post so excuse me if someone has already said this but the problem with the explanation is that when you describe 2A and A/2 you are really describing two different A’s. A does not =A in this equation.  They are not the same. If one envelope has \$1 and the other has \$2 then the only way you could be trading for the A/2 envelope is if A=\$2, and the only way you could be trading for the 2A envelope is if A=\$1.

Am I missing something here. Doesn’t that pretty much clear up the paradox?

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 Posted: 23 January 2012 07:52 AM [ Ignore ]   [ # 178 ]
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Hi macgyver,

I think we are all already convinced of this since the first page. Only kkwan says he does not agree…

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 Posted: 23 January 2012 08:29 AM [ Ignore ]   [ # 179 ]
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GdB - 23 January 2012 12:39 AM
kkwan - 22 January 2012 10:51 PM

How do you know that the other envelope has 2A if A is unknown and you don’t know whether A is the smaller or the larger amount? It is equally possible that the other envelope has A/2.

I don’t know. But it is equivalent to say there are 2 envelopes, one with an amount of A\$ and another with 2A\$, and I do not know which one I picked first.

To say that “there are 2 envelopes, one with an amount of A\$ and another with 2A\$”
is equivalent to say “there are 2 envelopes, one with half the amount of the other”

And as such, why introduce the term A at all?

However, if A is denoted as the amount in the envelope you selected, and if you don’t know the value of A and you don’t know whether it is the larger or the smaller amount, what you can say is that as the value of A is unknown and whether it is the smaller or the larger amount is also unknown, there are two possible situations:

1. If A is the smaller amount, then the other envelope has 2A
2. If A is the larger amount, then the other envelope has A/2

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 Posted: 23 January 2012 08:48 AM [ Ignore ]   [ # 180 ]
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Poor kkwan. Don’t you understand mathematics, or do you have problems with your ego that you cannot admit that you are wrong?

Here an exact derivation.

The smaller amount we call A, the bigger amount B.

1. If you pick A, the other envelope contains B.
2. If you pick B the other envelope contains A.

Agree?

Now, it is given that B = 2A.

So we can substitute and express the conditions above as follows:

1a. If you pick A, then the other envelope contains 2A.
2a. If you pick the 2A, then the other envelope contains A.

Agree?

3. If I switch in situation 1a, I gain A.
4. If I switch in situation 2a, I lose A.

Agree?

As the chance of both situations is equal (1/2), my expected gain is A - A = 0 by switching.

Agree?

To make it complete:

5. If I do not switch in 1a, I lose A
6. If I do not switch in 2a, I gain A

Agree?

As the chance of both situations is equal (1/2), my expected gain is -A + A = 0 by not switching. So to switch or not makes no difference.

Agree?

You can repeat the whole stuff by turning B = 2A into A = B/2.

Edit: small edits for clarity.

[ Edited: 23 January 2012 09:11 AM by GdB ]
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