GdB - 01 December 2012 10:15 AM

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.

-1 * (Y - X) when expended is (- Y + X) which is equivalent to (X - Y).

So, (X - Y) = (-Y + X) which is equivalent to (X - Y).

Thus, (X - Y) = (X - Y) which is tautological and trivial. 

StephenLawrence - 01 December 2012 10:33 AM

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?

Your example is a particular instance of the TEP and should be treated as such.

So, without knowing what is in the selected envelope, we can denote the amount in the selected envelope as A and as A is either the smaller or the larger amount, either (A, 2A) or (A, 1/2A) are in the two envelopes.

We then proceed to consider whether to switch or not to switch as follows:

1. If we do not switch, we get A.

2. If we do switch, gain/loss is A/1/2A.

So, we should switch because the potential gain is twice the potential loss with the probability of gain/loss as 1/2. Also, with switching, the long run average is 5/4A which is more than A.

The rationale was stated in my post 1745 which I hope you have carefully read and digested.

Furthermore, with total ignorance, we cannot assume we can select either $10 or $40 because that implies we know what amounts are actually in the two envelopes. That is begging the question.

However, selecting $20 is not begging the question as the implication is either ($20, $40) or ($20, $10) are the two envelopes and we do not assume anything because do not know which event is actual.

This is the rationale of considering both the two possibilities together and not separately as by doing that, we are begging the question which is philosophically unsound and/or unjustified.

For the sake of argument, if you are omniscient and/or the above philosophical consequences are ignored (which is unwise and which contravenes the fundamental nature of the TEP), then:

In your example, if we can select $10, then $20 must be in the other envelope and if we can select $40, $20 must be in the other envelope.

So, if either $20, $10 or $40 can be selected, there will be 3 possible situations:

1. $20 is selected, gain/loss on switching is $20/$10.

2. $10 is selected, gain on switching is $10.

3. $40 is selected, loss on switching is $20.

In 1 and 2, one should switch. In 3, one should not switch.

The problem is, we don’t know which situation we are in.

But, as the probability of 1 is 1/2 and the probability of either 2 or 3 is 1/4, it is still beneficial to switch because the total probability of 1 and 2 is 3/4.

So, is that your “solution” of the TEP? 

kkwan - 02 December 2012 12:15 AM

GdB - 01 December 2012 10:15 AM

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.

-1 * (Y - X) when expended is (- Y + X) which is equivalent to (X - Y).

So, (X - Y) = (-Y + X) which is equivalent to (X - Y).

Thus, (X - Y) = (X - Y) which is tautological and trivial. 

What does that do there? I said it was true, and yes, you are right it is trivial. A derivation was not even necessary. But at least you have shown one of the principles mathematical proofs: rewrite a statement according to logical and mathematical rules, and see that you get at an obvious truth, e.g. a tautology. Great stuff.

So your answer is ‘yes’: for every set of values (X,Y) it is valid that (X - Y) = -1 * (Y -X).
Now the simple followup is that if this is true for every set of values (X,Y) it is also true for every subset of all pairs of (X,Y).
Again, kkwan, do you agree with that or not? Answer!
Now the set of all TEP pairs is such a subset of all pairs. Agree?

This is fully independent of my knowledge what TEP pair I have at the moment, if I describe it as (X,2X) or (X,X/2) does not matter. Agree?
For any pair, the possible gain or loss are the same. So also for these two. Agree?

In formula language:

Proposition 1: (X - Y) = -1 * (Y -X) (you proved it to be true)
To serve you, I do not know which pair I have: (X,2X) or (X,X/2), so I must check two propositions:
Proposition 2a: Y = 2X
Proposition 2b: Y = X/2

Derivation with 2a:
(X - 2X) = -1 * (2X - X)
-X = -1 * X
-X = -X —> which is obviously true

Derivation with 2b:
(X - X/2) = -1 * (X/2 - X)
X/2 = -1 * (-X/2)
X/2 = X/2 —> which is obviously true

So even with my total ignorance of which of 2 pairs I have, the possible gain or loss are equal. So it is not possible to improve my chances by switching.

Now, if you do not agree, you should point in which step of the derivation my error lies. Where did I not apply a valid mathematical rule?
If you cannot do that, then the conclusion is true.

kkwan - 02 December 2012 03:55 AM

So, if either $20, $10 or $40 can be selected, there will be 3 possible situations:

1. $20 is selected, gain/loss on switching is $20/$10.

2. $10 is selected, gain on switching is $10.

3. $40 is selected, loss on switching is $20.

In 1 and 2, one should switch. In 3, one should not switch.

The problem is, we don’t know which situation we are in.

But, as the probability of 1 is 1/2 and the probability of either 2 or 3 is 1/4, it is still beneficial to switch because the total probability of 1 and 2 is 3/4.

 

Dunno quite what you are doing above but it’s wrong.

Simpler to write it out like this:

1. $20 is selected gain on switching is $20

2. $20 is selected loss on switching is $10

3. $10 is selected gain on switching is $10

4. $40 is selected loss on switching is $20

Please try again.

Stephen

kkwan - 08 January 2012 09:27 AM

[
It is relative as follows:

If one does not switch, there is no gain or loss at all, but one loses the opportunity to double the money.

OTOH, if one switches, there are two possible situations:...

(Oh, the humanity… Why am I getting directly involved in this?...)

If one does not switch, there are also two possible differences:  One loses the opportunity to double the money.  But one also avoids the risk of cutting the money in half.

TimB - 02 December 2012 09:38 PM

If one does not switch, there are also two possible differences:  One loses the opportunity to double the money.  But one also avoids the risk of cutting the money in half.

Wow, Tim, a very good contribution to this thread! You should be honoured for this (with ‘u’ in ‘honour’: I learned English at school, not American…).

TimB - 02 December 2012 09:38 PM

(Oh, the humanity… Why am I getting directly involved in this?...)

The gut reaction is this is so silly why get involved. But actually it’s an interesting puzzle and all the more so because it seems so simple that it couldn’t possibly be.

If one does not switch, there are also two possible differences:  One loses the opportunity to double the money.  But one also avoids the risk of cutting the money in half.

So you should switch.

Imagine you have $10 dollars and are given the chance to swap for $8 dollars or $1 million dollars with 50/50 chance.

What should you do? 

The principle is just the same with the double or half bet.

Stephen

StephenLawrence - 02 December 2012 11:46 PM

TimB - 02 December 2012 09:38 PM

(Oh, the humanity… Why am I getting directly involved in this?...)

The gut reaction is this is so silly why get involved. But actually it’s an interesting puzzle and all the more so because it seems so simple that it couldn’t possibly be.

TimB: If one does not switch, there are also two possible differences:  One loses the opportunity to double the money.  But one also avoids the risk of cutting the money in half.

So you should switch.Stephen

Why switch? If there is $1 million in one envelope, and $500, 000 in the other, you’re going to get $500,000, at least.  And it doesn’t seem to me that switching increases the odds of getting $1 million, if you only get to play the game once.

GdB - 02 December 2012 11:42 PM

Wow, Tim, a very good contribution to this thread! You should be honoured for this (with ‘u’ in ‘honour’: I learned English at school, not American…).

I will accept honor, where I can get it, even if it includes a superfluous “u’, that is a vestige of English Imperialism.

TimB - 03 December 2012 12:07 AM

StephenLawrence - 02 December 2012 11:46 PM

TimB - 02 December 2012 09:38 PM

(Oh, the humanity… Why am I getting directly involved in this?...)

The gut reaction is this is so silly why get involved. But actually it’s an interesting puzzle and all the more so because it seems so simple that it couldn’t possibly be.

TimB: If one does not switch, there are also two possible differences:  One loses the opportunity to double the money.  But one also avoids the risk of cutting the money in half.

So you should switch.Stephen

Why switch? If there is $1 million in one envelope, and $500, 000 in the other, you’re going to get $500,000, at least.  And it doesn’t seem to me that switching increases the odds of getting $1 million, if you only get to play the game once.

Now you are expressing the argument for not switching.

Most of us accept that this is correct, but the puzzle is to properly explain why. There is the other way of looking at it, double or half, as you did in your previous post and if that’s right we should switch.

Stephen

What we have heeyah… is a failyah… to communicate.

I laughed out loud, Tim.  First time I’ve recalled “Cool Hand Luke” in twenty years. 

Occam

GdB - 02 December 2012 04:37 AM

What does that do there? I said it was true, and yes, you are right it is trivial. A derivation was not even necessary. But at least you have shown one of the principles mathematical proofs: rewrite a statement according to logical and mathematical rules, and see that you get at an obvious truth, e.g. a tautology. Great stuff.

So your answer is ‘yes’: for every set of values (X,Y) it is valid that (X - Y) = -1 * (Y -X).
Now the simple followup is that if this is true for every set of values (X,Y) it is also true for every subset of all pairs of (X,Y).
Again, kkwan, do you agree with that or not? Answer!
Now the set of all TEP pairs is such a subset of all pairs. Agree?

Of course, (X - Y) = (X - Y) for any finite values of X and Y.

This is trivially true and tautological in that it is of the form X = X.

So, if you derive a “proof” to show X = X (which is obvious and trivial) it is ludicrous.

This is fully independent of my knowledge what TEP pair I have at the moment, if I describe it as (X,2X) or (X,X/2) does not matter. Agree?
For any pair, the possible gain or loss are the same. So also for these two. Agree?

No, it is not so simple as you think so.

You have not considered the following problem wrt the TEP.

The problem is, you don’t know whether either (X, 2X) or (X, 1/2X) are actually in the two envelopes.

As such, if you select an envelope and denote the amount in it as A, you could be in either (A, 2A) or (A, 1/2A):

1. If you are in (A, 2A) and you switch, the gain is A.

2. If you are in (A, 1/2A) and you switch, the loss is 1/2A.

You cannot assume you are in situation 1 or 2 because you don’t know.

Agreed?

StephenLawrence - 02 December 2012 07:20 AM

Dunno quite what you are doing above but it’s wrong.

Simpler to write it out like this:

1. $20 is selected gain on switching is $20

2. $20 is selected loss on switching is $10

3. $10 is selected gain on switching is $10

4. $40 is selected loss on switching is $20

This is equivalent to what I wrote.

If $20 is selected, gain/loss on switching is $20/$10. Since the the potential gain is twice the potential loss and the probability is 1/2, it is beneficial to switch.

If $10 is selected, one should switch as the gain is definite. The probability is 1/4.

If $40 is selected, one should not switch as the loss is definite. The probability is 1/4.

Hence, if either $20 or $10 is selected, one should switch and the probability of selecting those amounts is 1/2 + 1/4 = 3/4.

So, as the probability of definite loss is only 1/4, it is a calculated gamble worth taking to switch.

TimB - 02 December 2012 09:38 PM

(Oh, the humanity… Why am I getting directly involved in this?...)

If one does not switch, there are also two possible differences:  One loses the opportunity to double the money.  But one also avoids the risk of cutting the money in half.

If you do not switch:

1. You missed the opportunity of getting double the money.

2. You avoided the risk of getting half the money, but you still get half the money.

The probability of 1 or 2 is 1/2.

But, it is double or half, not double or nothing.

So, why not decide to switch or not to switch by flipping an unbiased coin?

Head you switch or tail you do not switch or vice versa. 

OTOH, you do not know whether you are in situation 1 or 2, but you will definitely get some money without any outlay, so why not switch for the possibility of getting double the money?

As it is, you don’t know whether you have got double or half the money either by switching or by not switching.

This is the tantalizing conundrum of the TEP.

To switch or not to switch?