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The two envelopes problem
Posted: 13 May 2012 02:50 AM   [ Ignore ]   [ # 886 ]
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GdB - 09 May 2012 10:24 PM

My small worry is that there seem to be some readers who still think that kkwan makes a valid argument, see e.g. here.

I think we start out with, we know the equation is wrong because we know it wouldn’t work in practice. We know this from imagining two players in opposite positions trying to utilise it, or from simply imagining switching twice.

But from there what I think is happening is people are trying to show what is wrong with it, apart from the above, and failing.

They assume they are succeeding because they know they are right that there is something wrong with Kkwan’s equation.

Stephen

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Posted: 13 May 2012 05:02 AM   [ Ignore ]   [ # 887 ]
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Gary T - 04 May 2012 06:12 PM

  A ceases to be a variable and becomes a constant at the point where a first envelope is selected.

Why?

From that point onward (and therefore during both step 6, where the components of the probability equation are specified, and step 7, where the equation is formalized) A cannot have both values simultaneously, i.e.: A cannot simultaneously have both the larger and the smaller values, which are the two values assigned during steps 4 and 5.

Of course not, just like it couldn’t before it was selected. So what?

Stephen

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Posted: 13 May 2012 07:13 AM   [ Ignore ]   [ # 888 ]
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kkwan - 12 May 2012 07:10 PM

They are not pejoratives. If you only imagine ($10, $20) are in the envelopes, whereas it is also possible that ($10, $5) are in the envelopes and as such, as you have omitted ($10, $5), it is a one-sided perspective, incomplete and misleading.

Listen kkwan: I say: “Imagine there are $10 and $20 in the two envelopes.” And then you say I must imagine $10 and $5 to it? Why should I? If I say to you that you must imagine a red piece of paper and a green one, then why does logic say I should also image a blue one?

kkwan - 12 May 2012 07:10 PM

If you cannot accept that in the context of the TEP, either ($10, $20) or ($10, $5) are possible events, then what sort of logic are you proposing?

The context of TEP says there are two amounts, not three. Your third amount is part of your wrong argumentation.

Here another one: imagine I have two envelopes, with $10 and $20 in it. Without knowing, I draw the envelope with $20 first. Why did you not mention just one time, that according to your formula, there is a chance of 0.5 that the other envelope contains $40? (A is the amount in the first envelope, so A = $20, so there is a chance of 0.5 the other envelope contains $40.)

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Posted: 13 May 2012 07:15 AM   [ Ignore ]   [ # 889 ]
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StephenLawrence - 13 May 2012 02:50 AM

I think we start out with, we know the equation is wrong because we know it wouldn’t work in practice. We know this from imagining two players in opposite positions trying to utilise it, or from simply imagining switching twice.

But from there what I think is happening is people are trying to show what is wrong with it, apart from the above, and failing.

They assume they are succeeding because they know they are right that there is something wrong with Kkwan’s equation.

Try the same as I am doing with kkwan: imagine I have two envelopes with $10 and $20 in it. Is there then a possibility that there is $5 (or maybe $40) in one of the envelopes? If you think so, did you imagine what I said?

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Posted: 13 May 2012 05:59 PM   [ Ignore ]   [ # 890 ]
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GdB - 13 May 2012 07:13 AM

Listen kkwan: I say: “Imagine there are $10 and $20 in the two envelopes.” And then you say I must imagine $10 and $5 to it? Why should I? If I say to you that you must imagine a red piece of paper and a green one, then why does logic say I should also image a blue one?

In the context of the TEP, in the argument to switch, with A denoted as the amount in the selected envelope and as A is either the smaller or the larger amount, the total amount in the envelopes is either (A, 2A) or (A, 1/2A).

Thus, if you imagine ($10, $20) are in the envelopes, because ($10, $5) is also possible, you should also imagine it.

That is not the same as imagining red, green and blue paper.

The context of TEP says there are two amounts, not three. Your third amount is part of your wrong argumentation.

In each mutually exclusive event, there are only two amounts.

Here another one: imagine I have two envelopes, with $10 and $20 in it. Without knowing, I draw the envelope with $20 first. Why did you not mention just one time, that according to your formula, there is a chance of 0.5 that the other envelope contains $40? (A is the amount in the first envelope, so A = $20, so there is a chance of 0.5 the other envelope contains $40.)

As mentioned above, the two envelopes could contain either ($10, $20) or ($10, $5).

If you select an envelope, it could contain either the smaller or the larger amount.

If it contains the smaller amount, the other envelope contains $20 and if it contains the larger amount, the other envelope contains $5.

In the context of the above, $40 is irrelevant.

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Posted: 13 May 2012 10:03 PM   [ Ignore ]   [ # 891 ]
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kkwan - 13 May 2012 05:59 PM

In the context of the TEP, in the argument to switch, with A denoted as the amount in the selected envelope and as A is either the smaller or the larger amount, the total amount in the envelopes is either (A, 2A) or (A, 1/2A).

This is the TEP:

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.

That you denote the smaller and bigger amount with A is already wrong. It is part of your wrong argument. You are right that there is more than one representation of TEP. But you should pick one consistently. In the example of $10 and $20, the total amount is $30. So A + 2A must be $30. But A + (1/2)A must be $30 too if A + (1/2)A represent the TEP. But then A = $20, otherwise the amounts do not add up to $30. So the pair of amounts can be represented by ((1/2)A, A), but then A = $20. You are arguing with ‘variable constants’.

kkwan - 13 May 2012 05:59 PM

Thus, if you imagine ($10, $20) are in the envelopes, because ($10, $5) is also possible, you should also imagine it.

That means that if I say that you should imagine $10 and $20, which makes $30, I still have to imagine that there is $15 total in the envelopes. So I can only say that you are cheating, you do not stick to the example.

kkwan - 13 May 2012 05:59 PM

In each mutually exclusive event, there are only two amounts.

We have one event, kwann, just one: $10 and $20.

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Posted: 14 May 2012 12:00 AM   [ Ignore ]   [ # 892 ]
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GdB - 13 May 2012 10:03 PM

This is the TEP:

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.

That you denote the smaller and bigger amount with A is already wrong. It is part of your wrong argument. You are right that there is more than one representation of TEP. But you should pick one consistently. In the example of $10 and $20, the total amount is $30. So A + 2A must be $30. But A + (1/2)A must be $30 too if A + (1/2)A represent the TEP. But then A = $20, otherwise the amounts do not add up to $30. So the pair of amounts can be represented by ((1/2)A, A), but then A = $20. You are arguing with ‘variable constants’.

Where is it in the TEP which says only ($10+$20 = $30) but not ($10+$5 = $15) is the total amount?

A, denoted as the amount in the selected envelope, is an unknown constant. However, it can be a constant and be either the smaller or the larger amount with no contradiction.

OTOH in your description of the amounts ($10, $20) as (A, 2A) or ($20, $10) as (A, 1/2A) your A is a variable, not a constant.

From the wiki on variable (mathematics)

In mathematics, a variable is a value that may change within the scope of a given problem or set of operations. In contrast, a constant is a value that remains unchanged, though often unknown or undetermined.

However, A denoted as the amount in the selected envelope, is clearly a constant and cannot be a variable.

As such, your A is incoherent.

That means that if I say that you should imagine $10 and $20, which makes $30, I still have to imagine that there is $15 total in the envelopes. So I can only say that you are cheating, you do not stick to the example.

It is because your example is one-sided and incomplete, not cheating.

We have one event, kwann, just one: $10 and $20.

No, there are two mutually exclusive events, either ($10, $20) or ($10, $5).

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Posted: 14 May 2012 01:25 AM   [ Ignore ]   [ # 893 ]
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kkwan - 14 May 2012 12:00 AM

Where is it in the TEP which says only ($10+$20 = $30) but not ($10+$5 = $15) is the total amount?.

Nowhere. But I gave $10 and $20 as example that fits the conditions of TEP. But at the moment I say “imagine the 2 amounts, $10 and $20” you imagine additional $5 and $10 to it.

kkwan - 14 May 2012 12:00 AM

A, denoted as the amount in the selected envelope, is an unknown constant. However, it can be a constant and be either the smaller or the larger amount with no contradiction.

No, it isn’t a constant. If the envelopes contain $10 and $20, then your A is $10 when it is the lowest amount, and $20 when it is the biggest amount.

kkwan - 14 May 2012 12:00 AM

OTOH in your description of the amounts ($10, $20) as (A, 2A) or ($20, $10) as (A, 1/2A) your A is a variable, not a constant.

No. I show you when you use the the descriptions (A, 2A) and (A, (1/2)A) consistently, i.e. mutually exclusive, as description of the 2 amounts $10 and $20, then it turns out that that A is a variable, and not a constant. It is your handling of the case that shows you use A as a constant.

kkwan - 14 May 2012 12:00 AM

No, there are two mutually exclusive events, either ($10, $20) or ($10, $5).

If there are $10 and $20 in the envelopes, what amounts are then in the envelopes?

As a general remark for those left who view this thread (not many I suppose): if there is somebody left who thinks kkwan’s arguments are convincing, then please let me know in this thread. Otherwise I might let rest the case. It has grown to great absurdity.

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Posted: 14 May 2012 07:57 AM   [ Ignore ]   [ # 894 ]
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GdB - 14 May 2012 01:25 AM

Nowhere. But I gave $10 and $20 as example that fits the conditions of TEP. But at the moment I say “imagine the 2 amounts, $10 and $20” you imagine additional $5 and $10 to it.

As ($10, $5) also fit the conditions of the TEP, it is equally possible and you should not ignore it.

No, it isn’t a constant. If the envelopes contain $10 and $20, then your A is $10 when it is the lowest amount, and $20 when it is the biggest amount.

In the TEP, you don’t know whether the selected envelope contains the smaller or the larger amount and what is the total amount in the two envelopes.

Hence, the amount in the selected envelope is denoted as A and it is a constant because the amount cannot vary, but A could be either the smaller or the larger amount.

However, in your analysis whereby you only consider ($10, $20) are in the envelopes, A is a variable. It is incoherent because A denoted as the amount in the selected envelope, should be a constant.

OTOH, if either ($10, $20) or ($10, $5) are in the envelopes, A is $10, it is a constant and it can be either the smaller or the larger amount with no contradiction.

No. I show you when you use the the descriptions (A, 2A) and (A, (1/2)A) consistently, i.e. mutually exclusive, as description of the 2 amounts $10 and $20, then it turns out that that A is a variable, and not a constant. It is your handling of the case that shows you use A as a constant.

(A, 2A) and (A, 12/A) refer to two mutually exclusive events ($10, $20) and ($10, $5) and not exclusively to ($10, $20) in which A is a variable. As such, A is a constant as $10.

If there are $10 and $20 in the envelopes, what amounts are then in the envelopes?

You cannot assume there is only ($!0, $20) in the envelopes as ($10, $5) is equally possible and cannot be omitted.

As a general remark for those left who view this thread (not many I suppose): if there is somebody left who thinks kkwan’s arguments are convincing, then please let me know in this thread. Otherwise I might let rest the case. It has grown to great absurdity.

Your argument is one-sided, incomplete and misleading.

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Posted: 14 May 2012 08:24 AM   [ Ignore ]   [ # 895 ]
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kkwan - 14 May 2012 07:57 AM

As ($10, $5) also fit the conditions of the TEP, it is equally possible and you should not ignore it.

Yes, and $3 and $6 fit too, and $20 and $40 also. And? I said there are $10 and $20 in it. Full stop.

kkwan - 14 May 2012 07:57 AM

However, in your analysis whereby you only consider ($10, $20) are in the envelopes, A is a variable. It is incoherent because A denoted as the amount in the selected envelope, should be a constant.

Where in TEP it says that the amount in the first envelope is a constant? The problem of your ‘solution’ (not of the way TEP is formulated!) is that you say it is a constant. If you pick the envelope with $10 first, then A = $10. And if you pick the one with $20 first, then A = $20. Which other envelope can you pick from one envelope, one containing $10, the other $20?

The confusion you make is that your formula fits to another situation, but that situation is not TEP. The situation is the following:

You get an amount of $10. Then there are two identical and closed envelopes, one contains $5, and the other contains $20. Should you switch your amount, your $10, with one of the envelopes?

Then it fits, $10 is a constant, but the situation is not TEP anymore. Maybe you can show me the difference in calculation of the two different games, the real TEP, and the above, TEP-like situation? If you say you are not interested in this other situation, I know you are not able to deal with the difference between the two.

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Posted: 14 May 2012 11:34 AM   [ Ignore ]   [ # 896 ]
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GdB - 13 May 2012 07:15 AM

Try the same as I am doing with kkwan: imagine I have two envelopes with $10 and $20 in it. Is there then a possibility that there is $5 (or maybe $40) in one of the envelopes? If you think so, did you imagine what I said?

I don’t see what you are getting at.

Imagine you’ve picked an envelope and you know it’s a $10 or a $20 but you don’t know which. Imagine that you’ve picked a $20. Is there a possibility that there is a $10 in your envelope?

Yes.

What’s the difference with your example and my example?

Stephen

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Posted: 14 May 2012 06:11 PM   [ Ignore ]   [ # 897 ]
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GdB - 14 May 2012 08:24 AM

Yes, and $3 and $6 fit too, and $20 and $40 also. And? I said there are $10 and $20 in it. Full stop.

If you imagine the envelopes contain ($10, $20), it is possible they also contain ($10, $5) but not $3 or $6 etc.

Where in TEP it says that the amount in the first envelope is a constant? The problem of your ‘solution’ (not of the way TEP is formulated!) is that you say it is a constant. If you pick the envelope with $10 first, then A = $10. And if you pick the one with $20 first, then A = $20. Which other envelope can you pick from one envelope, one containing $10, the other $20?

If you select an envelope, the amount in that envelope is clearly an unknown constant.

The problem with your “solution” lies with you insisting only ($10, $20) are in the envelopes and omitting the possibility that ($10, $5) could be in the envelopes.

The confusion you make is that your formula fits to another situation, but that situation is not TEP. The situation is the following:

You get an amount of $10. Then there are two identical and closed envelopes, one contains $5, and the other contains $20. Should you switch your amount, your $10, with one of the envelopes?

Then it fits, $10 is a constant, but the situation is not TEP anymore. Maybe you can show me the difference in calculation of the two different games, the real TEP, and the above, TEP-like situation? If you say you are not interested in this other situation, I know you are not able to deal with the difference between the two.

You have misinterpreted the situations. If you have $10 in the selected envelope, then there is either ($10, $20) or ($10, $5) in the envelopes and neither mutually exclusive event contravenes the TEP as $10 could be either the smaller or the larger amount.

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Posted: 14 May 2012 10:03 PM   [ Ignore ]   [ # 898 ]
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kkwan - 14 May 2012 06:11 PM

If you imagine the envelopes contain ($10, $20), it is possible they also contain ($10, $5) but not $3 or $6 etc.

So I put, for your eyes, a $10 note in an envelope, and a $20 note in another identical envelope, and then you say it is possible that there are $5 and $10 in it???

kkwan - 14 May 2012 06:11 PM

You have misinterpreted the situations. If you have $10 in the selected envelope, then there is either ($10, $20) or ($10, $5) in the envelopes and neither mutually exclusive event contravenes the TEP as $10 could be either the smaller or the larger amount.

Sorry, I am not ‘contravening’ TEP or so. I have given another situation, and I want an exact derivation of the chances, in order that you can show how this TEP-like situation differs from TEP. Derive a formula, and from this formula, say what the best strategy would be.

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Posted: 14 May 2012 10:07 PM   [ Ignore ]   [ # 899 ]
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StephenLawrence - 14 May 2012 11:34 AM

Imagine you’ve picked an envelope and you know it’s a $10 or a $20 but you don’t know which. Imagine that you’ve picked a $20. Is there a possibility that there is a $10 in your envelope?

Yes.

And is there a possibility that there is $5 in it? Or $40?

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Posted: 14 May 2012 10:25 PM   [ Ignore ]   [ # 900 ]
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GdB - 14 May 2012 10:07 PM
StephenLawrence - 14 May 2012 11:34 AM

Imagine you’ve picked an envelope and you know it’s a $10 or a $20 but you don’t know which. Imagine that you’ve picked a $20. Is there a possibility that there is a $10 in your envelope?

Yes.

And is there a possibility that there is $5 in it? Or $40?

Yes.

If the answer to my question is yes the answer to your question is yes.

Stephen

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