The Exchange Paradox and its interpretations by Ni, Ti, etc.

[uumlau]

Reference "The Exchange Paradox" on wiki. Or just google it.

In a nutshell, the paradox is as follows:

There are two envelopes. You know that each contains an amount of money, and that one contains twice the amount of the other.

• You pick an envelope at random.
• You open the envelope and see an amount of money, A.
• If you got the envelope with more money, then the other envelope has A/2 in it.
• If you got the envelope with less money, then the other envelop has 2A in it.
• The expectation value of the amount of money in the other envelope must be
  
  
  because choosing the other envelope will either halve or double your money.
• 5/4A is > A, therefore you should choose the other envelope.
• Except the exact same argument could be made if you chose the other envelope first and there are only two choices, that neither envelope is to be preferred over the other.

We know that the choice of envelope is random, and "choosing the other envelope" should make no difference at all. So what is the problem with the reasoning above?

Here's a quick and easy answer

The concept of "expectation value" has no meaning in this scenario. The values of the envelopes are predetermined, and you choose one or the other. It isn't as if you choose an envelope and then the amount in the envelope is set to twice or half the amount in your envelope.

Here's a more complicated answer

The real way to calculate expectation value is to say that one envelope has X and the other has 2X, and the expectation value of either envelope is 1.5*X. When we choose an envelope and look at its contents, we don't learn X. Therefore, the conditional probability implied by the 5/4A expectation value cannot apply.

Here's an even more complicated answer

http://www.pitt.edu/~jdnorton/papers/Exchange_paradox.pdf

One can google for even more complicated answers/explanations. Or read the talk/argument sections of the wiki.

My question to the forum is this:
Which explanation (of any that you find from any source) satisfies you the best?

My purpose in asking is that I look at this problem, and my instant reaction is the first, simple explanation. As far as I'm concerned, I've answered the question fully and succinctly, and any explanation beyond that is either misunderstanding that simple explanation or expressing the same point with much more detail. I believe that is due to my instinctive Ni approach, and that Ti types would want a more methodical approach: Ni cares about "meaning", while Ti cares about "proving". That's my hypothesis at any rate.

I am interested in anyone's replies, no matter type, though I am especially interested in replies from those who theoretically use both Ni and Ti (INFJ, ISTP, etc.).

The issue isn't about any sort of "right answer". It's obvious to all parties that something is askew. The question is what degree of explanation of the paradox satisfies oneself.

 

[cascadeco]

I liked the first answer too. The other responses seemed in the end kind of irrelevant (in the sense that while I could see response 2 as valid, I found it kind of 'annoying' at the same time. Was drawn to what I saw as 'getting to the point', which was the first. Of course I didn't read the paper! Knew that would go far beyond response 2! hahah)

 

[JocktheMotie]

I arrived at the first one, first.

Although the second basically says the same thing and also makes sense. So.

 

[Totenkindly]

Didn't read the third one yet, but my thought before I read the answers is that the value of the two bags is already set before i know what they are, and the problem proof above tries to define A differently for each scenario. while A already is a fixed amount. I don't really care about what the 'expectation value' is.

 

[garbage]

My thought was that the answer doesn't match up with intuition and that I ought to approach it in another way. So, I basically 're-derived' that second answer there. I would have rigorously re-checked the logic to that first approach in order to reconcile the approaches, but I'm lazy. (On second glance, it looks like A means different things at different times; it suggests a failure in logic.)
I didn't really think of that first answer myself; otherwise, I would have gone with it.

--

As far as the utility of the explanations? I tend to use the following rule of thumb--the simpler for the application, the better. The first is the 'best' unless we need to dig deeper; we don't need more rigor and detail until we come to situations where the nuances need to be explored. As different perspectives, they're all perfectly sound ways of viewing the situation, and they're all 'right'--but not all are useful for all applications.

I would not give the Norton paper to my eleven-year-old niece to explain this phenomenon, for instance.

--

Nor would I introduce someone to typology by throwing Jung's Psychological Types at them. Nor would I explain my thoughts on 'the spirit of the Christmas season' to a two-year-old; he'll probably be content to think of that 'spirit' as a physical being, Santa. Nor would I want to give a detailed excuse beyond 'because I said so!' to that same two-year-old who questioned why I do not want him to open his presents early.

The point is that there are more complicated ways of explaining all of these things; things can be explained from many different reference points; different reference points are useful for different reasons and in different situations; let's not delve into details until we need to, but let's also recognize when we do need to do so.

 

[SilkRoad]

You lost me the second any kind of math entered into the...er...equation. Sad, I know.

 

[INTP]

The sort of math that says that the chances of getting more money if you switch is flawed(atleast for this problem)

 

[Poki]

I simply arrived at the fact that its not possible to know which has more. It doesnt take math or statistics. Simply...the other envelope could have 1/2x or 2x...to me the question is "do you feel lucky?". What is your probability of being lucky...how lucky have you been in the past. How lucky have you been recently "your lucky streak may end...but you may have one or 2 more in your streak". Honestly the analysis you posted was more like "WTF" to me. It takes to much thought...

Here is my analysis process...
1. Open one envelope
2. See how much is possible to loose gain. If there is $100 then the other is either $50 or $200.
3. Do I even care about losing half of whats in my hand
3. Am I happy with the amount I have in my hand
4. If its very little I take the chance
5. If its alot I walk away happy
6. Do not even process "what ifs"...make a choice and be happy with it

edit: oh forgot the math... 0<.5x<2x taken that x > 0. The second I open the first envelope I just guaranteed whether or not X is greater then 0 simply by the fact that the envelope actually contains money. Either way I come out ahead of not opening either one.

Solve for x: 0<.5x
answer: x>0

This is logic(apparently Ti)

Other people are funny with all there fancy complicated calculations/analysis.

 

[Z Buck McFate]

I find the first answer most satisfying, probably because I really don’t see any practical purpose in considering it further. I could easily see myself getting distracted and pulled toward considering the others, though, if I were feeling bored. Or if some possible purpose came to mind (wouldn't have to be practical).

(edit: I might get distracted by 'is there anything that could possibly be close to this, in which knowing this could change outcome?'- i.e. similar problems, but as long as the outcome couldn't possibly change as a result then it would feel like beating a dead horse.)

 

[Poki]

I find the first answer most satisfying, probably because I really don’t see any practical purpose in considering it further. I could easily see myself getting distracted and pulled toward considering the others, though, if I were feeling bored. Or if some possible purpose came to mind (wouldn't have to be practical).

(edit: I might get distracted by 'is there anything that could possibly be close to this, in which knowing this could change outcome?'- i.e. similar problems, but as long as the outcome couldn't possibly change as a result then it would feel like beating a dead horse.)

beat it...beat it again...again *evil beavis laugh*

 

[uumlau]

Part of the issue that the mathematicians have with the problem appears to be that the problem is incompletely defined. Personally, I don't believe it is, because my reasoning ends at "why the heck would you use an expectation value in this case?" But, if we switch our perspective a bit, and assume (because the problem assumes) that an expectation value is OK to use, then it is fair to say that the problem doesn't give enough information to evaluate an expectation value.

Why? Because in the real world, there are upper and lower bounds to the amount of money that might realistically be put in the envelope, and potentially a probability distribution of particular amount pairs. E.g., if you found $0.01 in your envelope, you're not going to find 1/2 cent in the other, it will be 2 cents. (Or, there is a real US-minted half cent in there, which is probably worth a great deal if you can find a coin collector to whom to sell it!) In general, if one assumes even a rudimentary guesswork "money spectrum", if one's envelope amount is on the low end, there is a good chance that the other envelope has more, and as one approaches the high end, switching envelopes gradually becomes more likely to "halve" the initial value you see.

Of course, it all depends on the distribution. If it's a flat distribution with a fixed maximum and minimum, e.g., $1 - $20, no fractional dollars, then you have even more constraints that affect whether you should switch: any odd amount (which will always be < $10) will always be the lower of the pair, and any amount > $10 will always be the higher of the pair. If it's an even number of dollars less than or equal to 10, then, quite interestingly, the expectation value of switching really is 5/4 times the amount you see! Why? Because in this case we have 10 equally likely pairs, each with a 10% chance to come up. Looking at the value in the envelope you select then limits the possibilities down to only two pairs, each 50% likely. So, of the 20 cases (10 pairs, 2 possible choices for each) we know that switching doubles our value in 5 cases, halves our value in 5 cases, and has an expectation value of 5/4 our value in 10 cases. Note that this breaks the symmetry that the paradox implies, that switching back would also have a high expectation value, because one HAS to choose an envelope and open it to get the information needed to calculate the expectation value, and that expectation value isn't ALWAYS 5/4 the original value, but sometimes half or twice.

So, the mathematicians end up saying that if you assume a distribution that is any amount of money (unbounded, not necessarily integers of dollars or pennies), then you have an "improper prior", in which case an expectation value cannot be calculated. Or, stated more intuitively, if you know "how" the envelopes were filled, then looking at an envelope gives you enough information to determine whether you want to choose the other, and breaks the symmetry of using the same argument to switch. Lacking that knowledge of "how", the expectation value is undefined.

TL;DR version: If one's reaction to the problem is "It is wrong to use expectation values, here," then one tends to find the problem trivial. If one's reaction to the problem is "Why can't one apply an expectation value, here?" then one has a much lengthier deliberation and discussion.

 

[garbage]

When there's a 'paradox' in applying these sorts of mathematical concepts, it often means one of the following:

1. The 'hammer' (e.g. expected value) that we use to nail down the problem can't actually apply--because of some minor technicality or assumption that we're not used to seeing when we approach problems using that hammer. Essentially, we're so focused on seeing nails, but we're actually looking at a screw. Or.. whatever, use an analogy more poignant than that.
2. Related, and sometimes as a result of the above, we translate a word problem into a mathematical structure improperly--our definition of the problem is flippin' wrong.
3. Our logical and/or mathematical structures are incomplete or inconsistent (something that we don't often accept). This is when we must build new, fancier tools, which often means approaching the problem in a new way, shifting our assumptions, etc.

I accepted your first answer as 'complete enough.' However, I did look into the paper, the Wikipedia article, and other resources out of curiosity--but not because I felt that the detail was necessary. However however, my curiosity was driven by the fact that my master's studies focused on value and decision theory, and I find trying to frame decisions to be fascinating and useful. Hell, half of my career has been built around framing decisions, so I've got a vested interest in and curiosity about what's out there.

Ah, Ne

If we want to tie this to type, I would imagine that stereotypical Ti types would find the lynchpin, the Jenga block that stabilizes the whole structure, that knocks the whole argument in the OP's line of reasoning down--and pull it. They would then dust off their hands, leave the scattered mess on the floor and say, "Welp, my work here is done," and then walk away.

 

[Rasofy]

It seemed obvious to me that it was a matter of luck, even after knowing the content of one envelope.
I liked the 2nd answer better: ''1.5x expectancy''
The fact that I'm a poker fan probably made a difference.

 

[Little_Sticks]

I didn't look at the answers, but since this is about probability, isn't expectation value only used when you want to analyze the amount you would have if you grabbed the envelope X amount of times and want to know the average amount you would end up per grab, for a given number of X trials?

umm wait,

There are two envelopes. You know that each contains an amount of money, and that one contains twice the amount of the other.

2A = 4 (A/2). This is a misrepresentation of the problem? When was one value supposed to be 4 times the other?

Wouldn't it be?
(1/2)*A + (1/2)*(2A) = 3/2*A = 1.5A - that would make sense that after two trials you would have 1.5*A on average for each grab.

I'm missing the point, right? Oh wait, so the paradox is if you say one is A and the other is A/2? Then one is still double the other, right? Then

A/2*1/2 + A*1/2 = 3/4A = .75A; but I still don't think it matters because A is your assumed reference point and if the other value is 1/2 of that it will still provide the right answer; if one value is double A, then that too will provide a different answer, but it's a difference reference point to A?

dsfjsdalgjsdf I am now Satan. Good day.

 

[Little_Sticks]

I don't know. Actually I don't get it. Because a coin flip is always 50-50. If the envelopes are randomized each trial, then there's no way to assume that the average value of the grabs will converge anywhere specific in between A and 2A.

I guess. I don't know. I'm going to look at your answers now.

 

[Coriolis]

I am most satisfied with the second explanation, and it is the one that automatically occurred to me when I read the scenario description. The first seems imprecise; the third overelaborate.

 

[Cellmold]

You lost me the second any kind of math entered into the...er...equation. Sad, I know.

Yeah same, I just read up on it and basically came to something similar to the first solution, but that depends upon whether I understood it properly.

I also agree with [MENTION=5578]bologna[/MENTION] about people's level of understanding being important to what explanation would work best; mine is limited and simple so a simple explanation is best.

So can I ask if I understood this correctly: The idea is to work out which envelope contains which amount of money and then be offered a switch, so then you need to work out whether you should switch based upon 'expectation' of what the other one might contain? I suppose there isn't really an answer, it's just a 50/50 chance, right?

Meh im not very intelligent so you might have to hand hold me through this one.

 

[Andy]

I like the second answer best, as it uses nice simple algebra rather than expecting me to know exactly how an expectation values is defined! I have more reasoning skills than knowledge, you know.

 

[uumlau]

Yeah same, I just read up on it and basically came to something similar to the first solution, but that depends upon whether I understood it properly.

I also agree with [MENTION=5578]bologna[/MENTION] about people's level of understanding being important to what explanation would work best; mine is limited and simple so a simple explanation is best.

So can I ask if I understood this correctly: The idea is to work out which envelope contains which amount of money and then be offered a switch, so then you need to work out whether you should switch based upon 'expectation' of what the other one might contain? I suppose there isn't really an answer, it's just a 50/50 chance, right?

Meh im not very intelligent so you might have to hand hold me through this one.

The basic idea is that if you were given the situation where the 2nd amount was determined after you see the first amount, then the concept of "expectation value" applies. In simple terms, the expectation value is the "average" of all possible results. The expectation value of a fair roll of a 6-sided die is "3.5", even though there is no 3.5 on the die, for example.

The expectation value is useful when that concept of average winnings actually means something. And it's more than just how likely you are to win. For instance, I could have a 1/3 chance to win and a 2/3 chance to lose, but if I only pay $3 to play, and get $12 if I win, then the average amount I'd win per play is 1/3(1x$12+3x(-3)) = $1. Don't worry too much about the math, it depends a lot on specific wording. My point is that even though you're more likely to lose, on average, you still win, because the winnings when you win, on average, outstrip what you lose when you don't win.

Expectation value isn't foolproof. It is possible (google the St. Petersburg paradox) to have an infinite expectation value, but you'd never want to play because you'd run out of money before you ever won anything. (An infinite expectation value is why betting tables have a maximum possible bet, since you could just double your bet each time and eventually win, assuming you could borrow and bet any amount of money.) It is also possible, as in this thread's paradox, to simply misapply the math.

In the case of this paradox, you don't know what you're taking the average of in the first place. It's deliberately trying to trick you with fake reasoning by pretending that you're averaging doubling vs halving - the 1/2(A/2 + 2A) = 5/4 A is the average of doubling vs halving the value A - except you don't really know what A is. (A isn't the value in the envelope you see, it's that other number, X, that you see in the 2nd case, and you aren't really "doubling" or "halving" X.)


But yes, I suspect that the math aspect makes it hard to use for typing, because if one isn't comfortable with the math, one will go for the least mathematical explanation, regardless of type.

 

[Cellmold]

The basic idea is that if you were given the situation where the 2nd amount was determined after you see the first amount, then the concept of "expectation value" applies. In simple terms, the expectation value is the "average" of all possible results. The expectation value of a fair roll of a 6-sided die is "3.5", even though there is no 3.5 on the die, for example.

The expectation value is useful when that concept of average winnings actually means something. And it's more than just how likely you are to win. For instance, I could have a 1/3 chance to win and a 2/3 chance to lose, but if I only pay $3 to play, and get $12 if I win, then the average amount I'd win per play is 1/3(1x$12+3x(-3)) = $1. Don't worry too much about the math, it depends a lot on specific wording. My point is that even though you're more likely to lose, on average, you still win, because the winnings when you win, on average, outstrip what you lose when you don't win.

Expectation value isn't foolproof. It is possible (google the St. Petersburg paradox) to have an infinite expectation value, but you'd never want to play because you'd run out of money before you ever won anything. (An infinite expectation value is why betting tables have a maximum possible bet, since you could just double your bet each time and eventually win, assuming you could borrow and bet any amount of money.) It is also possible, as in this thread's paradox, to simply misapply the math.

In the case of this paradox, you don't know what you're taking the average of in the first place. It's deliberately trying to trick you with fake reasoning by pretending that you're averaging doubling vs halving - the 1/2(A/2 + 2A) = 5/4 A is the average of doubling vs halving the value A - except you don't really know what A is. (A isn't the value in the envelope you see, it's that other number, X, that you see in the 2nd case, and you aren't really "doubling" or "halving" X.)


But yes, I suspect that the math aspect makes it hard to use for typing, because if one isn't comfortable with the math, one will go for the least mathematical explanation, regardless of type.

Yeah that does make a bit more sense, but as you said, the maths side of things is difficult for me and I cannot understand the maths at all. I suppose I should watch my brother play on a fruit machine and try to work it out from that /badjoke.

You've intrigued me now though, im probably going to spend hours looking over this topic and even trying to understand the maths behind it. I can't help it, if im ignorant of something I have no problem admitting it but I desire to understand it so as to broaden my understanding.

Ive never understood why people cant admit to their igorance on something. I suppose it has something to do with hubris, but to me it always causes more harm than good.

Anyhow thanks for the explanation and your patience.