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The two envelopes problem
 Posted: 15 December 2012 02:23 PM [ Ignore ]   [ # 1816 ]
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StephenLawrence - 15 December 2012 11:54 AM

Yep.

Do you think this is relevant and applicable to the TEP?

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 Posted: 15 December 2012 02:51 PM [ Ignore ]   [ # 1817 ]
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StephenLawrence - 15 December 2012 12:01 PM

All I know is if you played the TEP you could have any one of infinite numbers in your envelope and which ever one it was that would be A.

The TEP is about money which is finite and discreet.

A must mean the amount in your envelope whatever that is.

Yes, but A must not vary for it to be designated as a reference amount.

If the pair you are playing with is picked from finite possibilites we can see that switching makes no difference as a general strategy, just like in my example.

As the TEP is about money, the possibilities are finite as money is finite and discreet.

We know it is true that switching makes no difference in the context of your non-TEP example whereby both (X, 2X) and (X, 1/2X) can occur at the same time, but is it also true in the context of the TEP whereby either (X, 2X) or (X, 1/2X) can occur but not both at the same time?

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 Posted: 15 December 2012 03:16 PM [ Ignore ]   [ # 1818 ]
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kkwan - 15 December 2012 02:23 PM
StephenLawrence - 15 December 2012 11:54 AM

Yep.

Do you think this is relevant and applicable to the TEP?

Yes. In my example we have \$20 in our envelope but we can’t get specific because we need switching to work as a general strategy to conclude we should switch. This is even if we look! Even if we look A is not fixed.

Once we check out all the possible combinations we see that they cancel out.

Because of this 2. is false in my non TEP example

This will be the case in any set up in which the amounts in the two envelopes are selected from finite options.

What happens playing out all the possibilities is adding up all the amounts you have when you gain, always comes to half the total of the amounts you have when you lose and you gain half the time so this always cancels out.

Just as in my example, the problem is the same in the TEP. We need switching to work as a general strategy to conclude we should switch and again because of this 2. is false and A is not fixed even if we look.

The TEP is different in that the two amounts in the envelopes are selected from infinite possibilites.

But still we see that in these infinite games the amounts we have when we gain always add up to half the amounts we have when we lose and so switching makes no difference.

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 Posted: 15 December 2012 03:19 PM [ Ignore ]   [ # 1819 ]
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kkwan - 15 December 2012 02:51 PM

As the TEP is about money, the possibilities are finite as money is finite and discreet.

Oh, well we don’t need to bother with infinity then.

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 Posted: 15 December 2012 03:22 PM [ Ignore ]   [ # 1820 ]
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kkwan - 15 December 2012 02:51 PM

Yes, but A must not vary for it to be designated as a reference amount.

Well, it’s not possible to fix A, which I’ve shown with my example. If the two envelopes are picked from any finite possibilites the situation will be just the same.

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 Posted: 16 December 2012 05:39 AM [ Ignore ]   [ # 1821 ]
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kkwan - 15 December 2012 02:13 PM
GdB - 15 December 2012 08:07 AM

OK, let’s try again, step by step. Say for every step if you agree or not, and why:

1. In TEP there are two unknown amounts: not one, not three but two.
2. From these two unknown amounts, you know that one amount is twice the other.
3. There are two possibilities: you pick the smallest amount first, or you pick the biggest amount first.
4. These two possibilities can be described as (X,2X) if you pick the smallest amount first, or as (X, X/2) if you pick the biggest amount first.
5. That means: in (X,2X) X stands for the smallest amount and in (X,X/2) X stands for the biggest amount.
6. In TEP there are exactly two unknown amounts: not one, not three but two. (Maybe you changed your mind, therefore I repeat it here)
7. As there are no more than two amounts, this means that the X in (X,2X) and the X in (X,X/2) are not the same: otherwise we have three amounts, and we know there are only two.
8. The two possibilities can also be described as (X,2X) and (2X,X).
9. This description is conform with the fact that there are only two amounts.
10. As the absolute value of X - 2X and of 2X - X is the same, namely X, one can conclude that possible loss and possible gain are equal.

1. There are two and only two amounts in two and only two mutually exclusive possible events: either (X, 2X) or (X, 1/2X) for any finite value of X.

Your two events are just ‘you pick the smallest amount first’ and ‘you pick the biggest amount first’. These are the ‘mutually exclusive possible events’.

kkwan - 15 December 2012 02:13 PM

2. Thus, there are two and only two amounts in each mutually exclusive possible event and not three or any other number

The interesting fact here is that you do not let me treat them ‘mutually exclusive’. If I say that for (X,2X) the possible gain and loss are the same, and in (X,X/2) too, you deny that fact. So then you do not treat them as ‘mutually exclusive’. See your next reaction:

kkwan - 15 December 2012 02:13 PM

3. Mutually exclusive events cannot occur at the same time. This means either (X, 2X) or (X, 1/2X)
occur but not both at the same time.

kkwan - 15 December 2012 02:13 PM

5. For any finite value of X, (X, 2X) and (X, 1/2X) are distinct and not equivalent events for the same value of X.

Let’s take an example. You do not know what is in the envelopes, but I know: I put the money in: I put \$6 and \$12 in the envelopes. Again: you do not know this! You pick one envelope: there are two ‘mutually exclusive events’. Either you picked the \$6, and by switching you gain \$6; or you picked \$12, and you loose \$6 by switching. You do not know what X is, but I at least know it is one of the two: \$6 or \$12. So X is \$6 or \$12: if X is \$6, then the other amount is \$12. If X is \$12, the other amount is \$6. Do you deny this?
Now you have no idea what is in the envelopes. An excessive strategy would be to evaluate all possible combinations of amounts. But for every pair you start with, you will see the result is the same. For (\$5,\$10) your possible loss/gain is \$5, for (\$1234,\$2468) it is \$1234 etc. So for every ‘mutually exclusive event’, the possible loss/gain is the same. If you consider one single pair, say (\$6,\$12), then there are no (\$6,\$3) or (\$12,\$24). You are not considering one pair, where the TEP clearly defines there are only two amounts.

kkwan - 15 December 2012 02:13 PM

6. (X, 2X) is equivalent to (2X, X). So if you assume that as the one and only one possibility, you are begging the question and contravening the basic nature of the TEP.

I of course mean the first variable meaning the envelope you picked first.

It is telling that you do not answered my next question:
TEP is defined as follows:

Let us say you are given two indistinguishable envelopes, each of which contains a positive sum of money. One envelope contains twice as much as the other. You may pick one envelope and keep whatever amount it contains. You pick one envelope at random but before you open it you are offered the possibility to take the other envelope instead. Should you switch?

This is not the same as:

You are given one envelope. Now you are offered a second envelope, which can contain half or twice the amount in your envelope. Should you switch?

kkwan: explain the difference between the two. Make for both the correct calculation for the possible gain or loss. Show where your calculations differ.

Again: if you say you are interested in TEP only, I know that you see the trouble: that your calculation is only valid for the second situation, but not for the TEP.

Also you should give your reason why you do not want to bet with me, according to my proposal. You are just avoiding. Don’t you want to earn \$1000? You are so sure about yourself! C’mon, kkwan, according to you you cannot loose! I am sure I am right, so c’mon, let’s bet! Or are you not sure enough?

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 Posted: 16 December 2012 10:14 AM [ Ignore ]   [ # 1822 ]
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StephenLawrence - 15 December 2012 03:16 PM

Yes. In my example we have \$20 in our envelope but we can’t get specific because we need switching to work as a general strategy to conclude we should switch. This is even if we look! Even if we look A is not fixed.

Once we check out all the possible combinations we see that they cancel out.

That is applicable only in the context of your non-TEP example, but inappropriate and inapplicable in the context of the TEP.

Because of this 2. is false in my non TEP example

I disagree, as it does not follow that if possible gain is equal to possible loss, step 2 is false.

OTOH with A denoted as the amount in the selected envelope and as we can compare A to the other amount in any one group, the probability is 1/2 that A is either the smaller or the larger amount because there are four possible values of A of which two are the smaller and two are the larger amount.

Therefore, step 2 is true in the context of your non-TEP example.

This will be the case in any set up in which the amounts in the two envelopes are selected from finite options.

What happens playing out all the possibilities is adding up all the amounts you have when you gain, always comes to half the total of the amounts you have when you lose and you gain half the time so this always cancels out.

Just as in my example, the problem is the same in the TEP. We need switching to work as a general strategy to conclude we should switch and again because of this 2. is false and A is not fixed even if we look.

The TEP is different in that the two amounts in the envelopes are selected from infinite possibilites.

But still we see that in these infinite games the amounts we have when we gain always add up to half the amounts we have when we lose and so switching makes no difference.

Not quite so.

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

In the TEP, there are two and only two mutually exclusive events: either (X, 2X) or (X, 1/2X) for any finite value of X. This means that in the TEP, only one event, either (X, 2X) or (X, 1/2X) can occur at one time

OTOH, in your non-TEP example with two mutually inclusive events, both events group 1 (X, 2X) and group 2 (X, 1/2X) do occur at one time.

In your non-TEP example, it is possible to select either X, 2X or 1/2X from either group 1 or group 2, but it is not valid to do that in the context of the TEP because if either 2X or 1/2X can be selected, it implies that we assume/know which is the actual event which is unjustified.

Hence, in the context of the TEP with total ignorance if we select an envelope, we don’t know whether the amount in it is either the smaller or the larger amount and whether either (X, 2X) or (X, 1/2X) are in the two envelopes.

As such, we can only denote the amount in the selected envelope as A and as A is either the smaller or the larger amount, then either (A, 2A) or (A, 1/2A) are in the two envelopes without begging the question.

Thus, if you say we can select either 2A or 1/2A instead of A, that implies you assume you know either (A, 2A) or (A, 1/2A) are in the two envelopes which is begging the question and fallacious reasoning.

Because the TEP is fundamentally not synonymous to your non-TEP example, your finding wrt to switching in the context of your non-TEP example is inappropriate and inapplicable to the TEP per se.

So, in the context of the TEP, it is valid to only deduce the following:

1. If A is the smaller amount, by switching, gain is A.

2. If A is the larger amount, by switching, loss is 1/2A.

And with step 2 clearly true:

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

8. This is greater than A, so I gain on average by swapping.

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 Posted: 16 December 2012 11:05 AM [ Ignore ]   [ # 1823 ]
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GdB - 16 December 2012 05:39 AM

Your two events are just ‘you pick the smallest amount first’ and ‘you pick the biggest amount first’. These are the ‘mutually exclusive possible events’.

Not so. It is inherent in the TEP that for any finite value of X, there are two and only two distinct mutually exclusive possible events: (X, 2X) or (X, 1/2X).

The interesting fact here is that you do not let me treat them ‘mutually exclusive’. If I say that for (X,2X) the possible gain and loss are the same, and in (X,X/2) too, you deny that fact. So then you do not treat them as ‘mutually exclusive’.

Not so. Your possible gain/loss wrt either (X, 2X) or (X, 1/2X) is based on the unjustified assumption that they are the actual events which begs the question and is untenable.

Let’s take an example. You do not know what is in the envelopes, but I know: I put the money in: I put \$6 and \$12 in the envelopes. Again: you do not know this! You pick one envelope: there are two ‘mutually exclusive events’. Either you picked the \$6, and by switching you gain \$6; or you picked \$12, and you loose \$6 by switching. You do not know what X is, but I at least know it is one of the two: \$6 or \$12. So X is \$6 or \$12: if X is \$6, then the other amount is \$12. If X is \$12, the other amount is \$6. Do you deny this?
Now you have no idea what is in the envelopes. An excessive strategy would be to evaluate all possible combinations of amounts. But for every pair you start with, you will see the result is the same. For (\$5,\$10) your possible loss/gain is \$5, for (\$1234,\$2468) it is \$1234 etc. So for every ‘mutually exclusive event’, the possible loss/gain is the same. If you consider one single pair, say (\$6,\$12), then there are no (\$6,\$3) or (\$12,\$24). You are not considering one pair, where the TEP clearly defines there are only two amounts.

Neither you nor anybody put/know the amounts in the two envelopes.

Thus, if you say you put \$6 and \$12 or any two amounts into the envelopes, you are fixing the game and/or begging the question wrt the TEP.

In the TEP, inherently either (X, 2X) or (X, 1/2X) are in the two envelopes.

Also you should give your reason why you do not want to bet with me, according to my proposal. You are just avoiding. Don’t you want to earn \$1000? You are so sure about yourself! C’mon, kkwan, according to you you cannot loose! I am sure I am right, so c’mon, let’s bet! Or are you not sure enough?

Let’s play the TEP as it is with either (X, 2X) or (X, 1/2X) in the two envelopes whereby neither you nor anybody put/know the amounts in the two envelopes.

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 Posted: 16 December 2012 11:32 AM [ Ignore ]   [ # 1824 ]
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StephenLawrence - 15 December 2012 03:22 PM

Well, it’s not possible to fix A, which I’ve shown with my example. If the two envelopes are picked from any finite possibilites the situation will be just the same.

Not with your example, but in the context of the TEP, A is a constant if a value is assigned to it and thus A can be used as a reference amount.

In the TEP, if A is \$20, then it does not vary irrespective of whether it is the smaller or the larger amount as in (A, 2A) or (A, 1/2A).

In your example, A can be either \$20 or \$40 or \$10 depending on whether A is the smaller or the larger amount in each group. So, the value of A will vary depending on circumstances which means there is no reference amount per se.

This is one fundamental difference between the TEP and your example and there are non-trivial consequences because of this difference.

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 Posted: 16 December 2012 06:02 PM [ Ignore ]   [ # 1825 ]
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Kkwan,

1) 2. Is false if it wouldn’t work if applied as a general strategy.

2) 2. wouldn’t work if applied as a general strategy.

3) Therefore 2. is false.
=====================================================================
1) Is self evidence

We can check 2) and see that it is true.

And the conclusion deductively follows from the premises.

What tricks us (mainly) is it seems the amount in our envelope is fixed if we look. But it isn’t because of 1).

And that as they say is that.

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 Posted: 16 December 2012 11:23 PM [ Ignore ]   [ # 1826 ]
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kkwan - 16 December 2012 10:14 AM

OTOH with A denoted as the amount in the selected envelope and as we can compare A to the other amount in any one group, the probability is 1/2 that A is either the smaller or the larger amount because there are four possible values of A of which two are the smaller and two are the larger amount.

Therefore, step 2 is true in the context of your non-TEP example.

No. This is the other mind trick, you’ve reminded me there are two in this puzzle.

You will indeed have the smaller amount half the time but 2. is still false.

If you have \$10 the probability of having \$5 in the other envelope is 0 and the probability of having \$20 is 100%  (not 50/50)

As 2, must be applicable what ever you have in your envelope 2. is false.

When you have the smaller amount, on average you will have half of what you have when you have the larger amount.

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 Posted: 17 December 2012 04:32 AM [ Ignore ]   [ # 1827 ]
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kkwan - 16 December 2012 11:05 AM

Not so. It is inherent in the TEP that for any finite value of X, there are two and only two distinct mutually exclusive possible events: (X, 2X) or (X, 1/2X).

kkwan, stop this bullshit. The TEP is clearly formulated:

Let us say you are given two indistinguishable envelopes, each of which contains a positive sum of money. One envelope contains twice as much as the other. You may pick one envelope and keep whatever amount it contains. You pick one envelope at random but before you open it you are offered the possibility to take the other envelope instead. Should you switch?

Full stop. Everything else is your addition. So answer the above question in the above described situation.

TEP is not defined as:

You are given one envelope. Now you are offered a second envelope, which can contain half or twice the amount in your envelope. Should you switch?

But that is the way you redefine TEP. You still owe me to show how you treat TEP and the above situation differently. Show me the calculations of both, and explain the difference to us. You are not able to do this, kkwan. For you, both situations are the same. But in the second is missing that you originally had the choice between the two envelopes, and that is exactly the fact you do not use in your ‘derivation’.

kkwan - 16 December 2012 11:05 AM

Neither you nor anybody put/know the amounts in the two envelopes.

What a nonsense. How can be guaranteed that the amounts in the envelopes conform to ‘one amount is twice the other’, if there is not somebody, or some automaton programmed to put the amounts in the envelopes? For you there is no difference in how it is done. The only important thing is that you do not know what the factual amounts are.

kkwan - 16 December 2012 11:05 AM

Let’s play the TEP as it is with either (X, 2X) or (X, 1/2X) in the two envelopes whereby neither you nor anybody put/know the amounts in the two envelopes.

That is not the TEP as it is. Maybe I should point you to your first posting again, where you clearly state (well, copy from Wikipedia…):

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.
...

Bold by me: everything following is part of the reasoning, and not of the setup.

C’mon kkwan, I want to win \$1000, why don’t you want to bet on it? Why are you so sure about yourself, and still do not take the chance to earn an easy \$1000?

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 Posted: 17 December 2012 10:18 PM [ Ignore ]   [ # 1828 ]
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StephenLawrence - 16 December 2012 06:02 PM

Kkwan,

1) 2. Is false if it wouldn’t work if applied as a general strategy.

2) 2. wouldn’t work if applied as a general strategy.

3) Therefore 2. is false.
=====================================================================
1) Is self evidence

We can check 2) and see that it is true.

And the conclusion deductively follows from the premises.

What tricks us (mainly) is it seems the amount in our envelope is fixed if we look. But it isn’t because of 1).

And that as they say is that.

Step 2 was specifically proposed as one of the steps in the argument to switch in the context of the TEP and it is true only in that context and nowhere else.

It is not the general strategy to switch per se.

The general strategy to switch is the totality of all the steps 1 to 8 and not just step 2. You have conflated the totality with one step. Obviously, it is crucial that all the steps be true to make the totality viable.

As such, your conclusion that step 2 is false because the general strategy of switching does not work in the context of your non-TEP example does not necessarily show step 2 to be false whether in the context of your non-TEP example or in the context the TEP.

Nevertheless, I have shown step 2 is true even in the context of your non-TEP example, but the designed set-up in your non-TEP example (with mutually inclusive events) is such that it will inevitably lead to the conclusion that switching is not beneficial, notwithstanding that step 2 is true.

So, your non-TEP example is a fixed game with no degree of freedom wrt selection and as such, it is not synonymous to the TEP which is not such a fixed game.

Hence, the unworkability of switching in the context of your non-TEP example has no bearing on the the workability/unworkability of switching wrt TEP per se.

It is like comparing apples with oranges.

====================================================================

1) is not self-evident, 2) is just part of 1) and 3) is superfluous as you asserted step 2 is false in 1) (without showing why it is false).

So, this is not that.

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 Posted: 17 December 2012 11:39 PM [ Ignore ]   [ # 1829 ]
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StephenLawrence - 16 December 2012 11:23 PM

No. This is the other mind trick, you’ve reminded me there are two in this puzzle.

You will indeed have the smaller amount half the time but 2. is still false.

If you have \$10 the probability of having \$5 in the other envelope is 0 and the probability of having \$20 is 100%  (not 50/50)

As 2, must be applicable what ever you have in your envelope 2. is false.

When you have the smaller amount, on average you will have half of what you have when you have the larger amount.

If the probability of A as either the smaller or the larger amount is 1/2, how is step 2 false?

Step 2:

2. The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2.

If you have \$10, it must be from group 2 (\$20, \$10) and obviously the probability of having \$5 in the other envelope is O as \$5 is not possible because it does not exist in either group 1 (\$20, \$40) or group 2 (\$20, \$10) whereas the probability of having \$20 in the other envelope is 1/2, not 1, because we must be in group 2 to have \$20 in the other envelope, upon selecting the envelope with \$10.

The amount in the selected envelope is either \$10, \$20 or \$40:

1. If it is \$10, the probability of it being the smaller amount is 1/2 as it must be from group 2.

2. If it is \$20, the probability if it being the smaller amount is 1/2 as it must be from group 1 and the probability of it being the larger amount is also 1/2 as it must be from group 2.

3. If it is \$40, the probability of it being the larger amount is 1/2 as it must be from group 1.

So, whatever you have in the selected envelope, the probability of it being either the smaller or the larger amount is 1/2.

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 Posted: 18 December 2012 12:19 AM [ Ignore ]   [ # 1830 ]
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kkwan - 17 December 2012 11:39 PM

1. If it is \$10, the probability of it being the smaller amount is 1/2 as it must be from group 2.

No, it is 100% because there is 0% probability of \$10 being the larger amount

3. If it is \$40, the probability of it being the larger amount is 1/2 as it must be from group 1.

No, it is 100% because there is 0% probability of \$40 being the smaller amount.

So 2. is false in my example. Also we can move to the TEP from here.

Stephen

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