OK I’ll start with an example similar to the TEP in which this wouldn’t work and apply it to the TEP at a later stage. I’ll need to ask you questions to get there.

So imagine the TEP played with two pairs of envelopes $10 and $20 and $20 and $40. You have this information but don’t know which pair is in play nor which amount you have in your envelope.

The amount in your envelope is $20.

What should you do and why in this example?

Based on the information that either ($10, $20) or (20, $40) are in the two envelopes and with $20 in the selected envelope:

1. If $20 is the smaller amount, then ($20, $40) are in the two envelopes.

2. If $20 is the larger amount, then ($20, $10) are in the two envelopes.

In situation 1, on switching, gain is $20

In situation 2, on switching, loss is $10.

So, one should switch because the potential gain ($20) is twice the potential loss ($10).

OK, that would be true if you knew you had $20 in your envelope but you don’t you might have $20 you might have $10 and you might have $40.

So I’ll ask you to try again. Keep playing along for a little while and I think you will see the solution I’m proposing.

You don’t react on what I write: you deny that you know that (X - Y) = -1 * (Y -X) for all possible finite values of X and Y. If not, prove that to me.

That is irrelevant in the context of the TEP with two and only two possibilities: (X, 2X) or (X, 1/2X) for any finite value of X and total ignorance.

I don’t want to know if you find it irrelevant, I want to know if you agree. I know you do not react because you know it shows that your argumentation is wrong.

kkwan - 30 November 2012 09:48 AM

BS. I claim I know nothing about the amounts, except that they are fixed from the moment they are in the closed envelopes.

4 follows from 1 and 3; 5 follows from 4; and 6 follows from 5. I have no premise about the ratio between the amounts in the envelopes at all.

They are fixed, but you don’t know what are their actual values and whether (X, 2X) or (X, 1/2X) are in the two envelopes because of total ignorance.

As said, I presuppose less than you. And still I can prove that your possible gain and loss are equal.

Let’s all agree to stay away from Three-card Monte games and “Lets Make a Deal.” Nothing good can come from either one.

kkwan - 30 November 2012 09:48 AM

GdB, because of total ignorance, all possible situations must be considered together. Your 4,5, and 6 refer the universe of one and only one possible situation which is not the case in the TEP where there are two and only two possible situations for any finite value of X.

This is nonsense. I say that for every set of values (X,Y) it is valid that (X - Y) = -1 * (Y -X). This includes then also your (X, 2X) or (X, 1/2X).

You don’t react on what I write: you deny that you know that (X - Y) = -1 * (Y -X) for all possible finite values of X and Y. If not, prove that to me.

That is irrelevant in the context of the TEP with two and only two possibilities: (X, 2X) or (X, 1/2X) for any finite value of X and total ignorance.

I don’t want to know if you find it irrelevant, I want to know if you agree. I know you do not react because you know it shows that your argumentation is wrong.

kkwan - 30 November 2012 09:48 AM

BS. I claim I know nothing about the amounts, except that they are fixed from the moment they are in the closed envelopes.

4 follows from 1 and 3; 5 follows from 4; and 6 follows from 5. I have no premise about the ratio between the amounts in the envelopes at all.

They are fixed, but you don’t know what are their actual values and whether (X, 2X) or (X, 1/2X) are in the two envelopes because of total ignorance.

As said, I presuppose less than you. And still I can prove that your possible gain and loss are equal.

Let’s all agree to stay away from Three-card Monte games and “Lets Make a Deal.” Nothing good can come from either one.

kkwan - 30 November 2012 09:48 AM

GdB, because of total ignorance, all possible situations must be considered together. Your 4,5, and 6 refer the universe of one and only one possible situation which is not the case in the TEP where there are two and only two possible situations for any finite value of X.

This is nonsense. I say that for every set of values (X,Y) it is valid that (X - Y) = -1 * (Y -X). This includes then also your (X, 2X) or (X, 1/2X).

Lois, it is difficult to figure out which is your input on this post.

I don’t want to know if you find it irrelevant, I want to know if you agree. I know you do not react because you know it shows that your argumentation is wrong.

What you are expressing is tautological.

As said, I presuppose less than you. And still I can prove that your possible gain and loss are equal.

Not so.

This is nonsense. I say that for every set of values (X,Y) it is valid that (X - Y) = -1 * (Y -X). This includes then also your (X, 2X) or (X, 1/2X).

That is based on the assumption that there is one and only one possibility instead of two and only two possibilities which is an incomplete and misleading interpretation of the TEP.

OK, that would be true if you knew you had $20 in your envelope but you don’t you might have $20 you might have $10 and you might have $40.

So I’ll ask you to try again. Keep playing along for a little while and I think you will see the solution I’m proposing.

Not so, although it appears it might be so.

Any two and only two possibilities in the TEP must take the form of either (X, 2X) or (X, 1/2X) whereby X is the smaller amount in (X, 2X) and the larger amount in (X, 1/2X). For this relation to hold, X must be present in both the two possibilities.

Thus, without knowing what is the amount in the selected envelope, we denote the amount 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 which is consistent with the above requirement.

In your example, only $20 satisfy the above requirement whereas both $10 and $40 do not because they are not in both the two possibilities and hence are neither the smaller nor the larger amount in both the two possibilities ($20, $40) and ($20, $10).

Thus, the possibilities of selecting either $10 or $40 do not arise because:

1. If $10 is selected, then either ($10, $20) or ($10, $5) are in the two envelopes.

2. If $40 is selected, then either ($40, $80) or ($40, $20) are in the two envelopes.

Both 1 and 2 contradict what you proposed: that either ($20, $40) or ($20, $10) are in the two envelopes i.e. they are not possible situations.

Also, both $10 and $40 do not satisfy the above requirement and they are neither the smaller nor the larger amounts in both the two possibilities.

So, in your example, $20 must be the selected amount even if we do not know as only $20 satisfy the above requirement of either (X, 2X) or (X, 1/2X) whereby X is $20.

I don’t want to know if you find it irrelevant, I want to know if you agree. I know you do not react because you know it shows that your argumentation is wrong.

What you are expressing is tautological.

Well it is not, but whatever, I see you refuse to answer. Is the following true, kkwan:

For every set of values (X,Y) it is valid that (X - Y) = -1 * (Y -X).

The answer is yes or no.

kkwan - 01 December 2012 07:33 AM

As said, I presuppose less than you. And still I can prove that your possible gain and loss are equal.

OK, that would be true if you knew you had $20 in your envelope but you don’t you might have $20 you might have $10 and you might have $40.

So I’ll ask you to try again. Keep playing along for a little while and I think you will see the solution I’m proposing.

Not so, although it appears it might be so.

Kkwan, remember we are dealing with a case similar to the TEP. What I said is we need to move to this case and once we have the answers to apply it to the TEP.

I think I can show you the answer that way but you do need to play along to do it. It shouldn’t take too long.

So back to my example, what should you do and why?

You don’t react on what I write: you deny that you know that (X - Y) = -1 * (Y -X) for all possible finite values of X and Y. If not, prove that to me.

That is irrelevant in the context of the TEP with two and only two possibilities: (X, 2X) or (X, 1/2X) for any finite value of X and total ignorance.

I don’t want to know if you find it irrelevant, I want to know if you agree. I know you do not react because you know it shows that your argumentation is wrong.

kkwan - 30 November 2012 09:48 AM

BS. I claim I know nothing about the amounts, except that they are fixed from the moment they are in the closed envelopes.

4 follows from 1 and 3; 5 follows from 4; and 6 follows from 5. I have no premise about the ratio between the amounts in the envelopes at all.

They are fixed, but you don’t know what are their actual values and whether (X, 2X) or (X, 1/2X) are in the two envelopes because of total ignorance.

As said, I presuppose less than you. And still I can prove that your possible gain and loss are equal.

Let’s all agree to stay away from Three-card Monte games and “Lets Make a Deal.” Nothing good can come from either one.

kkwan - 30 November 2012 09:48 AM

GdB, because of total ignorance, all possible situations must be considered together. Your 4,5, and 6 refer the universe of one and only one possible situation which is not the case in the TEP where there are two and only two possible situations for any finite value of X.

This is nonsense. I say that for every set of values (X,Y) it is valid that (X - Y) = -1 * (Y -X). This includes then also your (X, 2X) or (X, 1/2X).

Lois, it is difficult to figure out which is your input on this post.

You don’t react on what I write: you deny that you know that (X - Y) = -1 * (Y -X) for all possible finite values of X and Y. If not, prove that to me.

That is irrelevant in the context of the TEP with two and only two possibilities: (X, 2X) or (X, 1/2X) for any finite value of X and total ignorance.

I don’t want to know if you find it irrelevant, I want to know if you agree. I know you do not react because you know it shows that your argumentation is wrong.

kkwan - 30 November 2012 09:48 AM

BS. I claim I know nothing about the amounts, except that they are fixed from the moment they are in the closed envelopes.

4 follows from 1 and 3; 5 follows from 4; and 6 follows from 5. I have no premise about the ratio between the amounts in the envelopes at all.

They are fixed, but you don’t know what are their actual values and whether (X, 2X) or (X, 1/2X) are in the two envelopes because of total ignorance.

As said, I presuppose less than you. And still I can prove that your possible gain and loss are equal.

Let’s all agree to stay away from Three-card Monte games and “Lets Make a Deal.” Nothing good can come from either one.

kkwan - 30 November 2012 09:48 AM

GdB, because of total ignorance, all possible situations must be considered together. Your 4,5, and 6 refer the universe of one and only one possible situation which is not the case in the TEP where there are two and only two possible situations for any finite value of X.

This is nonsense. I say that for every set of values (X,Y) it is valid that (X - Y) = -1 * (Y -X). This includes then also your (X, 2X) or (X, 1/2X).

Lois, it is difficult to figure out which is your input on this post.

What is it I’ve said that you don’t understand?

I don’t know what you said. I suspect it is embedded in the post somewhere, but the quote attribution boxes suggest that you didn’t say anything. I sometimes make mistakes in putting in the reply quotes in the correct way, so I thought you must have.

Let’s all agree to stay away from Three-card Monte games and “Lets Make a Deal.” Nothing good can come from either one.

Thanks, I brought up Monty Hall very early in this thread, and thought it was mildly amusing (which I consider good). But I think she is right, if she is suggesting that you avoid a 3-envelope problem, as that is a different issue.

You don’t react on what I write: you deny that you know that (X - Y) = -1 * (Y -X) for all possible finite values of X and Y. If not, prove that to me.

That is irrelevant in the context of the TEP with two and only two possibilities: (X, 2X) or (X, 1/2X) for any finite value of X and total ignorance.

I don’t want to know if you find it irrelevant, I want to know if you agree. I know you do not react because you know it shows that your argumentation is wrong.

kkwan - 30 November 2012 09:48 AM

BS. I claim I know nothing about the amounts, except that they are fixed from the moment they are in the closed envelopes.

4 follows from 1 and 3; 5 follows from 4; and 6 follows from 5. I have no premise about the ratio between the amounts in the envelopes at all.

They are fixed, but you don’t know what are their actual values and whether (X, 2X) or (X, 1/2X) are in the two envelopes because of total ignorance.

As said, I presuppose less than you. And still I can prove that your possible gain and loss are equal.

Let’s all agree to stay away from Three-card Monte games and “Lets Make a Deal.” Nothing good can come from either one.

kkwan - 30 November 2012 09:48 AM

GdB, because of total ignorance, all possible situations must be considered together. Your 4,5, and 6 refer the universe of one and only one possible situation which is not the case in the TEP where there are two and only two possible situations for any finite value of X.

This is nonsense. I say that for every set of values (X,Y) it is valid that (X - Y) = -1 * (Y -X). This includes then also your (X, 2X) or (X, 1/2X).

Lois, it is difficult to figure out which is your input on this post.

What is it I’ve said that you don’t understand?

I don’t know what you said. I suspect it is embedded in the post somewhere, but the quote attribution boxes suggest that you didn’t say anything. I sometimes make mistakes in putting in the reply quotes in the correct way, so I thought you must have.

This is what I said: “Let’s all agree to stay away from Three-card Monte games and “Lets Make a Deal.” Nothing good can come from either one.”

It happens that sometimes hitting enter a few times after the quote (after

, which is supposed to be the end of the quote) separates my comment from the quote when it appears. Sometimes it winds up in the middle of the post. I admit I don’t understand this. Sometimes my response winds up in the middle of a quote and sometimes it winds up at the end where it belongs for no rhyme or reason. Sorry for the confusion.

I have no idea where it will wind up in this one. I hit enter three times after

Let’s all agree to stay away from Three-card Monte games and “Lets Make a Deal.” Nothing good can come from either one.

Yes, thanks. That’s what I wrote.

When I wrote my first response to TimB I copied twice the

, which appears at the end of a quote, but it wound up blank in my response both times. So it looks even more incomprehensible.

ONE MORE annoying thing!!!!!

GOOD GOD, I can’t believe it. I actually typed out the symbol and word between the [] and the [] that appears at the end of a quote and it STILL didn’t show up in my post! I’ve been using computers for more than 20 years, have used and understood many programs, and have been fairly proficient, but this is the most impossible system I have ever had to deal with.