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The two envelopes problem
 Posted: 25 December 2012 02:06 PM [ Ignore ]   [ # 1891 ]
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kkwan - 25 December 2012 11:18 AM
StephenLawrence - 25 December 2012 10:27 AM

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

Why not?

Because if it were true we should switch. But without looking switching makes no difference.

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 Posted: 25 December 2012 02:23 PM [ Ignore ]   [ # 1892 ]
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StephenLawrence - 25 December 2012 01:25 PM

Give a concrete example Kkwan ffs.

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

We will take either (\$20, \$40) or (\$20, \$10) as mutually exclusive events in the TEP.

Let A be a constant:

If A is \$20, then either (\$20, \$40) or (\$20, \$10) are in the envelopes which is consistent with the above, not beg the question and thereby preserve the inherent symmetry of the TEP which is of the form either (X, 2X) or (X, 1/2X).

Let A vary:

1. If A is \$40, then only (\$20, \$40) exist

2. If A is \$10, then only (\$20, \$10) exist

This is because the two events are mutually exclusive.

1 and 2 break the inherent symmetry of the TEP, begs the question and thus collapses the TEP to a non-TEP which is a travesty of the TEP.

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 Posted: 25 December 2012 02:42 PM [ Ignore ]   [ # 1893 ]
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StephenLawrence - 25 December 2012 01:29 PM

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?

That “it looks like, but it isn’t” is not convincing at all as a demonstration of the so-called mind trick.

Either something is or it is not. There is no mind trick.

So, if A is denoted as the amount in the selected envelope and you say “it looks like A, but it isn’t A”, it is clearly absurd and preposterous.

A is A and that is all there is to it.

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 Posted: 25 December 2012 02:51 PM [ Ignore ]   [ # 1894 ]
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StephenLawrence - 25 December 2012 02:06 PM

Because if it were true we should switch. But without looking switching makes no difference.

Not so. Step 2 being true or false does not influence the strategy to switch.

Even if you look, you are clueless wrt to switch or not to switch.

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 Posted: 25 December 2012 02:53 PM [ Ignore ]   [ # 1895 ]
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kkwan - 25 December 2012 02:42 PM
StephenLawrence - 25 December 2012 01:29 PM

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?

That “it looks like, but it isn’t” is not convincing at all as a demonstration of the so-called mind trick.

That’s what it means to say it’s a mind trick, so it is completely convincing.

A is A and that is all there is to it.

We know that is not all there is to it because that ignores that A might be \$10, \$20 or \$40 in my example.

Even with \$20 in our envelope A is not \$20.

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 Posted: 25 December 2012 02:55 PM [ Ignore ]   [ # 1896 ]
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kkwan - 25 December 2012 02:23 PM
StephenLawrence - 25 December 2012 01:25 PM

Give a concrete example Kkwan ffs.

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

We will take either (\$20, \$40) or (\$20, \$10) as mutually exclusive events in the TEP.

Which pair is in the envelopes and how did the pair get there?

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 Posted: 25 December 2012 02:58 PM [ Ignore ]   [ # 1897 ]
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kkwan - 25 December 2012 02:51 PM
StephenLawrence - 25 December 2012 02:06 PM

Because if it were true we should switch. But without looking switching makes no difference.

Not so. Step 2 being true or false does not influence the strategy to switch.

Of course it does because the famous equation follows from it.

Even if you look, you are clueless wrt to switch or not to switch.

Rubbish if you look you know what to do. \$10 switch \$20 switch \$40 don’t switch.

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 Posted: 25 December 2012 04:18 PM [ Ignore ]   [ # 1898 ]
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kkwan - 25 December 2012 10:52 AM

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.

Step 4! It claims that the value of the money you can expect to gain by switching is dependent upon the probability and it is 4/5A. The probability and the values of what’s inside those envelopes cannot be made into a mathematical statement.

kkwan - 25 December 2012 10:52 AM

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.

Although they are still irrelevant regardless of how you look at it, my analogy would not be heads(A,2A) and tails(A,1/2). The TEP gives the quantity of any first pick as the value, A, because it is indeterminate and then relates how it is to the other possibility. It is simpler and identical to label head = A (arbitrary) and tail = 2A. Then the first pick is either A or 2A (rather than just A). Once that is done, to switch is just to negate your first throw, indeterminately.

Also, if you pick A to name the first quantity, although to switch is certain to be either twice as much or half as much, it only counts as one alternative, not two because the second possibility is absolutely determined by the first.

If A = \$1, you will certainly get either \$0.50 or \$2.00 by switching if you knew what A was. But it is exclusively one or the other, not both possibilities added together. In other words, if A happens to be \$1.00, you will not get \$0.50 half the time and \$2.00 half the time! It would be \$0.50 half the time OR/exclusively \$2.00 half the time depending on which one is true.

[ Edited: 25 December 2012 04:20 PM by Scott Mayers ]
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 Posted: 25 December 2012 09:25 PM [ Ignore ]   [ # 1899 ]
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StephenLawrence - 25 December 2012 02:53 PM

That’s what it means to say it’s a mind trick, so it is completely convincing.

Conversely, if it is not a mind trick, it is completely unconvincing?

We know that is not all there is to it because that ignores that A might be \$10, \$20 or \$40 in my example.

In your example, A is not a constant and therefore A could be \$10, \$20 or \$40.

However, in the TEP, A is a constant as \$20 and therefore cannot be \$10 or \$40.

This is the crucial part of the TEP which I find rather disturbing and unsatisfactory.

Why does the TEP as it is, prevent us from selecting either \$10 or \$40?

Because, if we can do that, it breaks the inherent symmetry, begs the question and collapses the TEP to a non-TEP. Those are the barriers, but are they absolute?

Now, what if the TEP is the TEP only half the time and is not the TEP the other half of the time?  Why does it flip-flop like that? Information storage is the key as by doing so all possible information wrt selection can be stored:

In electronics, a flip-flop or latch is a circuit that has two stable states and can be used to store state information. Flip Flop is a bistable multivibrator. The circuit can be made to change state by signals applied to one or more control inputs and will have one or two outputs. It is the basic storage element in sequential logic. Flip-flops and latches are a fundamental building block of digital electronics systems used in computers, communications, and many other types of systems.

This means that when the TEP is the TEP half the time, A is a constant \$20 and it must be the only selection.

OTOH, in the other half of the time when the TEP is not the TEP, the inherent symmetry can be broken, the two events remain mutually exclusive but A is not a constant and as such:

1. We can select \$40 from (\$20, \$40) 1/4 of the time as (\$20, \$10) does not exist.

2. We can select \$10 from (\$20, \$10) 1/4 of the time as (\$20, \$40) does not exist.

So, like your non-TEP example, by considering all the time, switching is not beneficial but without the two mutually inclusive events which is the sticking point in exporting your solution to the TEP.

This is a hybrid resolution of the TEP. The switching argument is valid for half the time but it is not valid the other half of the time. It is a doable resolution with electronics and computers.

What do you think?

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 Posted: 25 December 2012 10:47 PM [ Ignore ]   [ # 1900 ]
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kkwan - 25 December 2012 09:25 PM

In your example, A is not a constant and therefore A could be \$10, \$20 or \$40.

Right and we need to be clear why. It’s because if we played over and over we would get different amounts in our envelope.

However, in the TEP, A is a constant as \$20 and therefore cannot be \$10 or \$40.

No, in the TEP it’s the same. If we played over and over we would get different amounts too. If we consider all the possible situations we could be in they cancel out.

Because, if we can do that, it breaks the inherent symmetry, begs the question and collapses the TEP to a non-TEP. Those are the barriers, but are they absolute?

Barriers in your mind I think.

But you could check with a concrete example. What you need to do is say what amounts are in the two envelopes and how they got there.

If you won’t there is no point in continuing.

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 Posted: 26 December 2012 05:02 AM [ Ignore ]   [ # 1901 ]
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StephenLawrence - 25 December 2012 02:55 PM

[
Which pair is in the envelopes and how did the pair get there?

Either pair could be in the two envelopes but we don’t know which pair.

Some entity put either of the two pairs in the two envelopes randomly..

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 Posted: 26 December 2012 05:18 AM [ Ignore ]   [ # 1902 ]
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kkwan - 25 December 2012 03:42 AM

That either (A, 2A) or (A, 1/2A) are in the two envelopes is the complete decription of the TEP which you deny.

Yes, I deny that. You can describe the two possibilities of TEP as (remember, <X,Y> means X is your first pick, Y your second):

1. <X,2X> and <2X,X>
or as:
2. <X,2X> and <Y, Y/2>, with the addition that Y = 2X (or X = Y/2).

Taking X = Y is wrong, it introduces 3 amounts.

kkwan - 25 December 2012 03:42 AM

Counting amounts from two mutually exclusive events is clearly wrong.

I assume your argument is that I must consider the mutually exclusive events jointly?

kkwan - 25 December 2012 03:42 AM

I don’t deny that with either (A, 2A) or (A, 1/2A0 in the two envelopes, if A is the selected amount, then A is either the smaller or the larger amount.

Yes. And the smaller amount is a different amount than the bigger amount, but you do as if they are the same.

kkwan - 25 December 2012 03:42 AM

It is fundamentally flawed and has no relevance to the TEP at all.

A mathematical exercise is fundamentally flawed? I ask you to calculate the chances of gaining by switching in the non-TEP situation. Your refusing to do it shows that you see your error, but you just cannot admit it. C’mon kwann, it is an easy task. Just show me how your solution for the non-TEP situation differs from TEP. It is not necessary to show me the differences between the situations, I am fully aware of them. You should give a clear argument, instead of throwing words as “fundamentally flawed”, “misleading” and whatever.

kkwan - 25 December 2012 03:42 AM

I have explained in post 1852. Please refer to it.

No kkwan, you only showed that the situations differ. You did not show why my solution is wrong. You cannot, because it is not wrong. Show me, mathematically, that it is wrong.

kkwan - 25 December 2012 03:42 AM

Of course, with total ignorance, the amount A in the selected envelope should be the reference.

No. TEP informs you that you have two amounts, one twice as big as the other. Or, what is equivalent, one amount is half of the other.
If you take the smallest amount as reference, then you have (X,2X), if you take the biggest as reference then you have (X,X/2). But the references differ: X in (X,2X) is not the same as X in (X,X/2). However, in both mutually exclusive descriptions (they are not different situations), you can pick the biggest or the smallest amount first:

1. <X,2X> or <2X,X>:—> your possible gain and loss are X
2. <X,X/2> or <X/2,X>:—> your possible gain and loss are X/2

You mixup description and reality, maybe this is what Scott is at.

kkwan - 25 December 2012 03:42 AM

You cannot assume that (10, 20) are in the two envelopes as it could be (10, 5) because you don’t know which event is actual. That is begging the question. It is that simple, but you don’t seem to get it.

It seems you don’t get you even don’t know you have 10.

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 Posted: 26 December 2012 05:45 AM [ Ignore ]   [ # 1903 ]
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Scott Mayers - 25 December 2012 04:18 PM

Step 4! It claims that the value of the money you can expect to gain by switching is dependent upon the probability and it is 4/5A. The probability and the values of what’s inside those envelopes cannot be made into a mathematical statement.

This is step 4:

4. If A is the smaller amount the other envelope contains 2A.

How is step 4 relevant to what you wrote?

Although they are still irrelevant regardless of how you look at it, my analogy would not be heads(A,2A) and tails(A,1/2). The TEP gives the quantity of any first pick as the value, A, because it is indeterminate and then relates how it is to the other possibility. It is simpler and identical to label head = A (arbitrary) and tail = 2A. Then the first pick is either A or 2A (rather than just A). Once that is done, to switch is just to negate your first throw, indeterminately.

It is simpler but it is incomplete and misleading if you consider (A, 2A) to be in the two envelopes and you beg the question by ignoring (A, 1/2A) as you don’t know whether (A, 2A) or (A, 1/2A) are in the two envelopes.

Also, if you pick A to name the first quantity, although to switch is certain to be either twice as much or half as much, it only counts as one alternative, not two because the second possibility is absolutely determined by the first.

What do you mean by “the second possibility is absolutely determined by the first”?

If A = \$1, you will certainly get either \$0.50 or \$2.00 by switching if you knew what A was. But it is exclusively one or the other, not both possibilities added together. In other words, if A happens to be \$1.00, you will not get \$0.50 half the time and \$2.00 half the time! It would be \$0.50 half the time OR/exclusively \$2.00 half the time depending on which one is true.

If A = \$1, A is known constant. Then either (\$1, \$2) or (\$1, \$0.5) are in the two envelopes. By switching, gain/loss is \$1/\$0.5.

With many repetitions of the TEP, half the time (\$1, \$2) and half the time (\$1, \$0.5) are in the two envelopes.

From step 7:

7. So the expected value of the money in the other envelope is

The expected value is 1/2 (2 x \$1) + 1/2 (\$1 divided by 2) = \$1.25

\$1.25 is more than A which is \$1. Therefore, it is beneficial to switch.

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 Posted: 26 December 2012 06:00 AM [ Ignore ]   [ # 1904 ]
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StephenLawrence - 25 December 2012 02:58 PM

Of course it does because the famous equation follows from it.

Step 2 does not influence the strategy to switch, but it is necessary to be true for the switching argument to be valid and my interpretation is it is true in your non-TEP example.

Rubbish if you look you know what to do. \$10 switch \$20 switch \$40 don’t switch.

You don’t know what are the amounts in the two groups. As such, whatever amount is in your selected envelope gives you no clue wrt whether to switch or not to switch.

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 Posted: 26 December 2012 06:16 AM [ Ignore ]   [ # 1905 ]
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StephenLawrence - 25 December 2012 10:47 PM

Right and we need to be clear why. It’s because if we played over and over we would get different amounts in our envelope.

It is because the setup of your non-TEP example requires A not to be a constant.

No, in the TEP it’s the same. If we played over and over we would get different amounts too. If we consider all the possible situations we could be in they cancel out.

The TEP is not synonymous to your non-TEP example.

Barriers in your mind I think.

Not so. They are tangible barriers which will invalidate any “solution” if they transgress them.

But you could check with a concrete example. What you need to do is say what amounts are in the two envelopes and how they got there.

I have done that with either (\$29, \$40) or (\$20, \$10) in the two envelopes which were chosen at random by some entity.

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