Search

 126 of 132 « First Prev 124 125 126 127 128 Next Last »
The two envelopes problem
 Posted: 25 December 2012 11:18 AM [ Ignore ]   [ # 1876 ]
Sr. Member
Total Posts:  2012
Joined  2007-10-28
StephenLawrence - 25 December 2012 10:27 AM

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

Why not?

There are 4 possible amounts in your example of which 2 are the smaller and 2 are the larger amount.

Thus, without opening any envelope, step 2 is true as the probability of A being either the smaller or the larger amount is 2/4 = 1/2.

Why not?

You will have to show me why.

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.

That is obvious in the setup of your example, but A is a constant in the TEP.

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.

But, as you pointed out earlier, we only get \$20 half of the time. So, we only gain/lose half of half of the time which is 1/4 of the time.

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…....................

Not in the context of your example which is a zero-sum fixed game with no degree of freedom.

However, the TEP is not like that.

Signature

I am, therefore I think.

 Profile

 Posted: 25 December 2012 11:35 AM [ Ignore ]   [ # 1877 ]
Sr. Member
Total Posts:  2012
Joined  2007-10-28
StephenLawrence - 25 December 2012 10:35 AM

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.

That is only in the context of your non-TEP example whereby A is not a constant as it could be \$20, \$40 or \$10.

In the context of the TEP, A, denoted as the amount in the selected envelope is a constant i.e. it does not vary in value anywhere in the TEP.

As such, we can assign A a value of \$20 without contradiction.

Thus, with A as \$20, then either (\$20, \$40) or (\$20, \$10) are in the two envelopes.

Signature

I am, therefore I think.

 Profile

 Posted: 25 December 2012 11:41 AM [ Ignore ]   [ # 1878 ]
Sr. Member
Total Posts:  6851
Joined  2006-12-20
kkwan - 25 December 2012 11:35 AM
StephenLawrence - 25 December 2012 10:35 AM

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.

That is only in the context of your non-TEP example whereby A is not a constant as it could be \$20, \$40 or \$10.

Well, good we agree on that.

In the context of the TEP, A, denoted as the amount in the selected envelope is a constant i.e. it does not vary in value anywhere in the TEP.

Fact is A would vary if you played the TEP over and over. And that’s what we are interested in because the claim is switching would work as a general strategy.

Stephen

 Profile

 Posted: 25 December 2012 11:48 AM [ Ignore ]   [ # 1879 ]
Sr. Member
Total Posts:  6851
Joined  2006-12-20

This is what happens when you decide not to celebrate Christmas.

Stephen

 Profile

 Posted: 25 December 2012 12:46 PM [ Ignore ]   [ # 1880 ]
Sr. Member
Total Posts:  2012
Joined  2007-10-28
StephenLawrence - 25 December 2012 10:59 AM

We are not in a position to fix A Kkwan. And this we can check and see through examples.

We are, in the context of the TEP with A (denoted as the amount in the selected envelope in step 1), it is implicit and mandatory that A is a constant in order to proceed to step 2 etc.

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

Then what is A in the context of the TEP and can you provide cogent arguments to support your contention that it is a mind trick?

Signature

I am, therefore I think.

 Profile

 Posted: 25 December 2012 12:51 PM [ Ignore ]   [ # 1881 ]
Sr. Member
Total Posts:  2012
Joined  2007-10-28
StephenLawrence - 25 December 2012 11:41 AM

Fact is A would vary if you played the TEP over and over. And that’s what we are interested in because the claim is switching would work as a general strategy.

If A varies, the consequences are dire.

Signature

I am, therefore I think.

 Profile

 Posted: 25 December 2012 12:59 PM [ Ignore ]   [ # 1882 ]
Sr. Member
Total Posts:  6851
Joined  2006-12-20
kkwan - 25 December 2012 12:46 PM
StephenLawrence - 25 December 2012 10:59 AM

We are not in a position to fix A Kkwan. And this we can check and see through examples.

We are, in the context of the TEP with A (denoted as the amount in the selected envelope in step 1), it is implicit and mandatory that A is a constant in order to proceed to step 2 etc.

That must mean the amount in the selected envelope whatever it is. so A isn’t a fixed amount.

Then what is A in the context of the TEP and can you provide cogent arguments to support your contention that it is a mind trick?

I’ve said what A is. And I’ve provided cogent arguments as to why it is a mind trick. In my example A is not \$20 even if \$20 is in your envelope!

I’ve said this also must be the case in the TEP because switching needs to work as a general strategy. I’ve said that however the pair is selected the position will be the same.

Now it’s up to you to come up with a way of selecting the pair that would change this.

really Kkwan at this point (much earlier actually) you should put up or shut up.

Stephen

 Profile

 Posted: 25 December 2012 01:00 PM [ Ignore ]   [ # 1883 ]
Sr. Member
Total Posts:  6851
Joined  2006-12-20
kkwan - 25 December 2012 12:51 PM
StephenLawrence - 25 December 2012 11:41 AM

Fact is A would vary if you played the TEP over and over. And that’s what we are interested in because the claim is switching would work as a general strategy.

If A varies, the consequences are dire.

A does vary. The consequence have nothing to do with the truth of the matter.

Stephen

 Profile

 Posted: 25 December 2012 01:01 PM [ Ignore ]   [ # 1884 ]
Sr. Member
Total Posts:  2012
Joined  2007-10-28
StephenLawrence - 25 December 2012 11:48 AM

This is what happens when you decide not to celebrate Christmas.

Merry Christmas, Stephen

Signature

I am, therefore I think.

 Profile

 Posted: 25 December 2012 01:02 PM [ Ignore ]   [ # 1885 ]
Sr. Member
Total Posts:  6851
Joined  2006-12-20
kkwan - 25 December 2012 01:01 PM
StephenLawrence - 25 December 2012 11:48 AM

This is what happens when you decide not to celebrate Christmas.

Merry Christmas, Stephen

Merry Christmas kkwan.

 Profile

 Posted: 25 December 2012 01:18 PM [ Ignore ]   [ # 1886 ]
Sr. Member
Total Posts:  2012
Joined  2007-10-28
StephenLawrence - 25 December 2012 12:59 PM

That must mean the amount in the selected envelope whatever it is. so A isn’t a fixed amount.

Whatever it is, A must be a constant because if A is not a constant, the TEP collapses to a non-TEP.

I’ve said what A is. And I’ve provided cogent arguments as to why it is a mind trick. In my example A is not \$20 even if \$20 is in your envelope!

I’ve said this also must be the case in the TEP because switching needs to work as a general strategy. I’ve said that however the pair is selected the position will be the same.

Now it’s up to you to come up with a way of selecting the pair that would change this.

really Kkwan at this point (much earlier actually) you should put up or shut up.

You have not provided any cogent arguments about what A is or why it is a mind trick.

Signature

I am, therefore I think.

 Profile

 Posted: 25 December 2012 01:25 PM [ Ignore ]   [ # 1887 ]
Sr. Member
Total Posts:  6851
Joined  2006-12-20
kkwan - 25 December 2012 01:18 PM
StephenLawrence - 25 December 2012 12:59 PM

That must mean the amount in the selected envelope whatever it is. so A isn’t a fixed amount.

Whatever it is, A must be a constant because if A is not a constant, the TEP collapses to a non-TEP.

Give a concrete example Kkwan ffs.

Say what the pair you select from is and how the money got into the envelopes.

Stephen

 Profile

 Posted: 25 December 2012 01:29 PM [ Ignore ]   [ # 1888 ]
Sr. Member
Total Posts:  6851
Joined  2006-12-20
kkwan - 25 December 2012 01:18 PM

You have not provided any cogent arguments about what A is or why it is a mind trick.

In my example it looks like if you have \$20 in your envelope that is A. But it isn’t. That is demonstrating the mind trick in my example. I’ve also explained that how ever the pair you are playing with is selected it will be the same and you can check that and see by coming up with concrete examples of your own.

What more could you reasonably want?

Stephen

[ Edited: 25 December 2012 01:34 PM by StephenLawrence ]
 Profile

 Posted: 25 December 2012 01:53 PM [ Ignore ]   [ # 1889 ]
Sr. Member
Total Posts:  2012
Joined  2007-10-28
StephenLawrence - 25 December 2012 01:00 PM

A does vary. The consequence have nothing to do with the truth of the matter.

No, A must be a constant to preserve the inherent symmetry of the TEP as either (X, 2X) or (X, 1/2X) and not beg the question.

That is the compelling rationale for A to be a constant and it is the truth of the matter.

Signature

I am, therefore I think.

 Profile

 Posted: 25 December 2012 02:03 PM [ Ignore ]   [ # 1890 ]
Sr. Member
Total Posts:  6851
Joined  2006-12-20
kkwan - 25 December 2012 01:53 PM
StephenLawrence - 25 December 2012 01:00 PM

A does vary. The consequence have nothing to do with the truth of the matter.

No, A must be a constant to preserve the inherent symmetry of the TEP as either (X, 2X) or (X, 1/2X) and not beg the question.

That is the compelling rationale for A to be a constant and it is the truth of the matter.

It’s no rational at all, it is just meaningless.

Set up a concrete TEP as I keep asking. Say what the two amounts in the two envelopes are and say how they got there. Then we can see if A is a constant.

I can already tell you it won’t be because you’ll need to take into account what would happen if you had various amounts in your envelope. The reason for that is you need to show that switching would work as a general strategy.

Stephen

 Profile

 126 of 132 « First Prev 124 125 126 127 128 Next Last »