Talk:Two envelopes problem/Arguments

Variant with no constrains

https://docs.google.com/file/d/0B7Z3B3DjCNt6VjJLT284TDVSZGs/edit?usp=sharing

Suppose you are allowed to open the envelope before you make the decision to swap.

1. Suppose the amount in the envelope you opened is $10
2. The other envelope may contain either $5 (meaning the one you chose is the higher value of the two with P=.5) or $20 (meaning the one you chose is the lower value of the two with P=.5).
3. So the expected value of the money in the other envelope is

$E(X)=\$5*1/2+\$20*1/2=\$12.5>\$10$

This is greater than the one you opened. So you gain on average by swapping. This applied to any amount of money you discover in the envelope and suggesting you should always swap no matter what you find. This implies that opening the envelope and finding the amount is not needed in deciding the action of swapping.

This is contradictory to the fact that the probability of choosing the envelope with more money is .5, and swapping the envelop will not increase that probability (1-.5=.5).

Solution:

The probability, P=.5 of you choosing the envelope with the higher (or lower) amount does not imply that the other envelope have 50% chance containing 1/2 of amount you see and 50% chance containing 2X of the amount. There’s only 2 possible cases in this game.

Case #1: The game organizer put a $5 and $10 in each envelope. You have 50% chance of getting the $10 and once you chose the envelope of $10. The amount that the other contains is $5 with 100% probability.

Case #2: The game organizer put a $20 and $10 in each envelope. You have 50% chance of getting the $20 and once you chose the envelope of $10. The amount that the other contains is $20 with 100% probability.

The probability of the game organizer doing Case #1 is P and Case #2 is Q. The expected value of the other envelope once you open the $10 envelope is:

$E(X|X_{0}=\$10)=\$5*P+\$20*Q$

However, from your perspective, P and Q is unknown. So the expected value of the other envelope cannot be determined.

The probability of you choosing the higher value in either case is 0.5. So the probability of winning this game (chooing the envelope with the higher amount) is

$.5*P+.5*Q=.5*1=.5$

So no swapping is needed.

Modification to the setup:

To satisfy the original conclusion of "always swapping", the following is needed:

You are given the following information before the game:

The game organizer flip a fair coin.

1. if heads, they will put $5 and $10 in each envelope
2. if tails, they will put $10 and $20 in each envelope

Once you open the envelope that contains $10, the expected value of the other envelope is indeed higher and hence you should swap.

The philosopher's solution

Richard I have been reading Schwitzgebel and Dever and I do not think what they say matches what you say in your paper.

S&D, in their problem statement, refer to X as the amount of money in envelope A, the originally chosen one.

In their resolution on page 5 they refer to the expected value of X, in other words the expected value of your original envelope. This means to me that they are saying that the proposed calculation is:

E(B) = 1/2 ( 2 * E(A) + 1/2 * E(A) )

They then go on to point out that E(A) is not the same thing in the two cases. Again they refer to what you would expect in envelope A.

The correct calculation should be E(B) = 2* E(A | B >A) Prob(B > A) + 1/2 E(B | B <A) Prob(B < A)

In a reply to Caramella you say (of S&D), 'They say as clearly as can be said in words, that they are calculating E(B) = E(B | B >A) Prob(B > A) + E(B | B <A) Prob(B < A) and they rewrite this sometimes as 2 E(A ) 1/2 + 1/2 E(A) 1/2, in other words you seem to be saying that:

E(B | B >A) Prob(B > A) + E(B | B <A) Prob(B < A)

is the same thing as:

2* E(A | B >A) Prob(B > A) + 1/2 E(B | B <A) Prob(B < A)

Should this be obvious?

The second version seems much clearer to me. Martin Hogbin (talk) 14:02, 13 November 2014 (UTC)

First of all, the general rule which S&D refer to is: E(B) = E(B | B > A) Prob(B > A) + E(B | B < A) Prob(B < A). They know this rule.

Everyone is agreed that the two probabilities here are both equal to 1/2.

Next, since B = 2A when B > A and B = A/2 when B < A, the right hand side can be further developed as E(2 A | B >A) Prob(B > A) + E(A / 2 | B < A) Prob(B < A). S&D know this too.

Next, by linearity of expectation value, and also substituting the probabilities (both equal to 1/2) we end up with the *true* statement E(B) = E(A | B > A) + E(A | B < A)/4. S&D know this too.

So what is "wrong" (according to them) is that E(A | B > A) and E(A | B < A) have both been replaced by A (without expectation and without conditioning), which are *two* errors, not one. They do refer to both errors and they do correct both of them. I have communicated with them by email about all this and they don't disagree with my representation of their argument.

They claim that they are the first to show "what went wrong", but Falk (2008) probably beat them getting this particular resolution of this particular interpretation into peer-reviewed print. As far as I know, the mathematician's resolution of the mathematician's interpretation has been known for ages. Among mathematicians it is so trivial and well understood that people do not get academic credit by publishing solutions. Unfortunately, the philosophers and the amateurs do not understand the mathematicians. Martin Gardner apparently didn't understand the mathematical literature on TEP.

The second purpose of Schwitzgebel and Dever is to go on and add some (in their opinion) novel facts about expectations and linearity, but (AFAIK) nobody has done anything with this contribution and I have difficulty figuring out what they mean. Richard Gill (talk) 15:57, 13 November 2014 (UTC)

I follow that but would it not be easier to start with:

E(B) = E(A | B > A) + E(A | B < A)/4

in order to explain the error. In the WP version, A is described as 'the amount in my selected envelope' so it seems that some formula envolving the sume in the first envelope is being proposed.

As you say above, there are then two mistakes that cause the wrong answer. Martin Hogbin (talk) 19:47, 13 November 2014 (UTC)

How can you *start* with E(B) = E(A | B > A) + E(A | B < A)/4 ? It is a strange formula and it needs to be derived. By the way we also have E(A) = E(A | B > A) + E(A | B < A)/4, since E(B) = E(A). But it remains just as mysterious.

I was hoping we might be able to find a way to do that between us.

From the point of view of making sense of the philosphers' solution that is a much better place to start.

In the article, step 7 says:

'So the expected value of the money in the other envelope is:'

${1 \over 2}(2A)+{1 \over 2}\left({A \over 2}\right)={5 \over 4}A$

Later on the first resolution says:

'A common way to resolve the paradox, both in popular literature and part of the academic literature, is to observe that A stands for different things at different places in the expected value calculation'.

We can now make ir clear exactly what the two things that A stands for are and how they are different. The first A should be E(A | B > A) and the second one E(A | B < A). The two different A's are both conditional expectations of the value in the original envelope (which is what A purports to be), but with different conditions.

Surely this could make a simple, understandable, and mathematically sound explantion of what the philosphers are talking about. I do agree with you that the originators of the problem probably did not intend to suggest that A was an expectation.

One way to justify starting with what I suggested would be just to cite S&D they pretty much give that formula in words on page 5. Alternatively we could ourselves give the formula in words. This is for the simple philosophers' explanation of what went wrong'. The mathematical details can follow, as we have now. Martin Hogbin (talk) 12:53, 14 November 2014 (UTC)

I agree. And S&D agree with me (personal communication). And you can even cite my paper, if you find my self-published unfinished report a "reliable source". That's not for me to judge, but I am happy to provide you with some good arguments based on wikipedia's own definition of this term.

I think you will find that Falk (2008) also agrees, and that Tsikogiannopoulos (2012) also agrees. Hopefully, iNic and Caramella1 and Gerhard Valentin will agree, too. This explanation based on this interpretation is the common core in a whole host of reliable sources. Richard Gill (talk) 12:32, 15 November 2014 (UTC)

Anyway, above I have shown how to derive E(B) = E(A | B > A) + E(A | B < A)/4, in a way which seems very close to the steps of the switching argument. So: you could start by deriving this correct formula, and then you can go back and look at the switching argument and compare formulas and compare the two arguments, and then you can say, (assuming that the switching argument is based on a computation of E(B)), ah ha, now we can see the *two* mistakes in the argument! That is exactly what we have done in this conversation. I would like to see it as one of the two first solutions in the article. However some people "switch off" when they see a formula so would like to have the whole thing replaced just be words. Well, it becomes quite a lot of words then, and the concepts being manipulated are not familiar to everyone, anyway.

Intuitively, one can say that given that B > A, the amount A tends to be small: to be precise, it is then one third of the total. Given that B < A, the amount A tends to be rather larger: to be precise, it is then 2/3 of the total. All of one third plus a quarter of two thirds equals one third plus one sixth, and that sum equals one half: so altogether, and given the total in the two envelopes, the expectation of B is half of the whole. Which we knew already. But anyway, maybe this calculation makes the numbers 1 and 1/4 less mysterious. Richard Gill (talk) 06:12, 14 November 2014 (UTC)

Proposed introductory wording for the logical/philosophical resolution section

One solution is to assume that the 'A' in step 7 is intended to be the average sum in envelope A [S&D say 'expected' but this terminology is not widely understood. Is there any objection to using 'average'?] and that we are intended to calculate the average value in envelope B. It is pointed out that the 'A' in the first part of the calculation is the average sum, given that envelope A contains less than envelope B, but the 'A', in the second part of the calculation is the average sum in A, given that envelope A contains less than envelope B. The same symbol is used with two different meanings in both parts of the same calculation but is assumed to have the same value in both cases.

A correct calculation of the average sum in ( E(B) ) would be

E(B) = E(B) = E(A | B > A) + E(A | B < A)/4,

where: E(A | B > A) is the average sum in A, given that envelope A contains less than envelope B

and: E(A|B<A) is the average sum in A, given that envelope A contains more than envelope B

This is pretty much straight out of S&D. Martin Hogbin (talk) 13:25, 15 November 2014 (UTC)

You use the word "calculation" for a formula. But the formula is not a calculation. You don't even know how to compute the terms in the formula! I would use the word "calculation" for the steps which led to the formula; what some people would also called its derivation. You could say "a correct calculation actually leads to ...".

Yes OK.

But I think that the *correct derivation* should be done step by step. Here you just report a final result of a failed attempt to calculate E(B) in a rather indirect way. That doesn't help anyone. You don't explain the steps which led to the "good" (but useless) result E(B) = E(A | B > A) + E(A | B < A)/4, and you don't discuss the result.

Why do I need to explain how I got E(B) = E(A | B > A) + E(A | B < A)/4 here? I have a reliable source for it. It is what S&D say in words, for example, '...the expected value of X in the "2X" part of the formula (where envelope A is the envelope with less) ...'. You can explain all the details later.Martin Hogbin (talk) 19:54, 15 November 2014 (UTC)

I don't see what is the point of replacing the word "expectation" by "average". The word "expectation" is already used in the argument for switching. S&D use it too. If you replace the word "expectation" by average, then what are you averaging over? Are you only averaging over the two possible configurations A > B and A < B or are you also averaging over possible amounts x, 2 x?

The point is that 95% of readers will not understand what is meant by 'expectation'. The natural language meaning of the term is along the lines of, 'I expect there will be £20 in the envelope (this time)'. Average is a well known and roughly understood term. We could try to use things like 'on average' or 'over the long term' maybe. Martin Hogbin (talk) 19:54, 15 November 2014 (UTC)

The same applies to "expectation" and of course the main problem is that the switching argument never explains what it takes as random and what it takes as fixed, and hence never actually says what it means by expectation at all. By replacing one undefined word by another, you do not make progress. Richard Gill (talk) 17:08, 15 November 2014 (UTC)

Yes, I was just about to ask you about that very point. I have assumed that it is over all possible amounts but I agree that it is another weakness in the philosophers' argument.

I am not trying to make progress in the sense that you might, I am trying to give the average non-mathematical reader some idea of what the phlosophers' resolution is. If you think it would be better to remove all mathematical notation I could try to do that and say it all in words. Have you read the current text; it means nothing at all! Martin Hogbin (talk) 19:54, 15 November 2014 (UTC)

Why say E(B) = E(B) ? You could say E(A) = E(B) = E(A | B > A) + E(A | B < A)/4 but now I think it is really incumbent on you to show that the complicated, correct, and completely useless formula E(A | B > A) + E(A | B < A)/4 can be quickly reduced to E(A). Answer: suppose the amounts in the two envelopes (fixed but unknown) are x and 2x. Then E(A | B > A) + E(A | B < A)/4 = x + 2x / 4 = 3 x/ 2, but also, directly, E(A) = 1/2 x + 1/2 2x = 3x/2. Richard Gill (talk) 17:15, 15 November 2014 (UTC)

No, I do not want to criticise this solution here, just give some idea what it is. I am trying to give our non-mathematical readers some idea what is meant when philosophers say 'A' stands for two different things. It is far from obvious that A is intended to be an expectation. Martin Hogbin (talk) 19:54, 15 November 2014 (UTC)

OK, I see your point. Indeed, since a bare "A", without any averaging, is on the right hand side, this seems to me to make the philosopher's interpretation quite ludicrous. Clearly if the writer has got a bare "A" on the right hand side of the equation he must have been calculating the expectation conditional on the value of A. However, this is an alien concept to many readers, including, it seems, most philosophers. Moreover I have noticed that many amateurs simply cannot grasp that one can mathematically talk about E(B | A = a) without *seeing* a and without *fixing* a. Richard Gill (talk) 07:40, 16 November 2014 (UTC)

There is a big difference between "average" and "expectation." Expectation is a meaningful concept even in a single case according to some interpretations of probability and is thus not connected to averages at all. This is the most central issue about this problem/paradox. Anyone that thinks that this problem is a purely mathematical problem hasn't understood the problem at all. iNic (talk) 10:58, 20 November 2014 (UTC)

Another idea

On reflection, perhaps it would be better to assume the expectation is based on just two envelopes containing x and y=2x. This seems to produce something much more like the philosophers solution. Now E(A | B > A) = x and E(A | B < A) = y = 2x. This is what you have done in your mathematical details section. Martin Hogbin (talk) 11:10, 16 November 2014 (UTC)

Of course. I have revised the article accordingly. The advantage of the philosopher's solution is that there is no averaging over possible values of x and no conditioning on (no subsetting over) possible values of A. It is technically much much more simple. On the other hand, it is simply not the case that this was what the original TEP was about. It always was a paradox in Bayesian (subjective) probability connected to many other famous paradoxes of improper prior distributions. Apparently at some point the history was forgotten (or it was too difficult to understand) and the philosophers came up with their own new "paradox", which is really boring ... till you get to the version by logician Smulyan. Then at last it starts to get amusing, at least. Richard Gill (talk) 17:29, 16 November 2014 (UTC)

More thoughts on classifying resolutions (and therefore problems)

Richard, I think you are absolutely right with your Anna Karenina point, there are many ways it can go wrong. I still think that the fundamental distinction should be between bounded and unbounded, even though you point out that mathematically speaking this is not an important difference.

The problem in most of its forms refers to money or the value of objects. I suspect that that is how it was intended by its original proposers and how it is understood by most (non-mathematician) readers.

For the bounded version it is easy describe the problem clearly and to give a simple resolution that is mathematically sound and can be put into words so that most people can understand it. I asked you to confirm that solution just to make sure that I had not missed anything but it is exactly as I thought. If the sum in your envelope should be more than half the upper bound it must be the larger sum so swapping will result in a certain loss, and a large one. No more is required to completely resolve the real-world version of the paradox.

Once you move to an unbounded sum, you need some mathematics. It makes no sense to say, 'Tell me in simple language what the problem is if there is no upper bound'; some understanding of mathematics and infinity is essential. In the powers of two version, which I still find the hardest to deal with, most people have no conception of what, no upper bound would mean in practice. With a lower bound only, not only would it be almost certain that the sums could not be fitted into the envelopes, but it would be almost certain that the sums could not even be described, even if the envelopes were the size of the known universe. This quickly takes us a very long way from two gentlemen comparing the values of their neckties. With no lower bound we almost certainly still could not describe the sums in the envelopes and (as you pointed out) expectations are meaningless. You will, to all intents or purposes, have nothing or an infinite amount, so swapping is pointless.

I am not suggesting that we do not discuss these purely mathematical versions just pointing out that, although P(B=2A | A = a) is not 1/2 for all a, applies to both bounded and unbounded versions there is still a big difference between the two cases. Martin Hogbin (talk) 13:43, 27 November 2014 (UTC)

Your pont of view is also common in the literature. Still I would like to point out that the Littlewood problem, which derived from Schrödinger and seems to be the original exchange paradox, was about an impoper prior, and improper priors have been at item of controversy since Laplace. So if you deal with the problem by treating the bounded case as if that is all there is, you are cutting off the problem at its roots. Richard Gill (talk) 07:32, 28 November 2014 (UTC)

Do you have any record of exactly what Schrödinger version of the problem was? I cannot find any clear reference to the problem in Littlewood either, although I do agree with one thing he says, 'Mathematics (by which I shall mean pure mathematics) has no grip on the real world ; if probability is to deal with the real world it must contain elements outside mathematics'. Kraitchik and Gardner both refer to real-world physical objects and money. The problem as stated in the article also refers to envelopes and money. It is contrived paradox based on a propsed line of reasoning that is mathematically ill-defined. In the real world the paradox is easily resolved, you should not always swap because the sum in your envelope might be above half the maximum sum allowed.

Once you try to get round this by having no upper bound you move unavoidably into the world of pure mathematics. I have never seen a mathematically well defined statement of this paradox (I hope I would know one if I saw it) and I suspect that one does not exist. So, once you get into the world of mathematics, the answer is 'What paradox?'. Martin Hogbin (talk) 19:47, 28 November 2014 (UTC)

See Littlewood, page 4: "Mathematics with minimal raw material" Example (4).

If you have never seen a mathematically well defined statement of this paradox then you haven't read my paper. Nor understood any of Boris Tsirelson's comments on the problem. Expectations are *not* meaningless without an upper bound. They are being used in this situation throughout science and needed in this situation througout science. And this is usually not problematic.

Dear Martin, whenever you use "real numbers" and theorems of calculus in the real world, you are also in the realm of pure mathematics and infinities. The calculus was invented because it is easier to compute integrals than sums. But in the real world everything is finite and discrete. Integrals and derivatives are pure mathematical fictions. You are short-sighted. Laplace wanted to solve practical problems. He introduced the principle of insufficient reason in order to solve practical problems. His solution breaks down in the two envelopes problem. So Laplace has a big practical problem even if you don't see any practical problem. If this particular situation shows that Laplace's principle has to be abandoned - or perhaps more accurately - sometimes needs to be abandoned - then we need to know how to distinguish problems where it works from problems where it doesn't work. That is a very very practical issue.

Every time a scientist uses a normal distribution and talks about its expectation value or standard deviation they are taking expectations of unbounded random variables. It seems you want to have 99.99% of modern scientific literature thrown in the rubbish bin. Richard Gill (talk) 12:18, 29 November 2014 (UTC)

If any amount a is a possible amount to be in Envelope A, then it should have positive probability. But moreover, if amount a is possible, then surely twice that amount is possible too? Even if much less likely. So there can be no definite bound to the amount which might be in Envelope A. (An unbounded distribution can still easily have a very small expectation value). In my opinion, the argument that the amount has to be bounded is facile. It fails to distinguish carefully between actual unknown amounts and our beliefs about them. Richard Gill (talk) 05:59, 30 November 2014 (UTC)

[edit conflict]I am not trying to make such strong claims as you seem to think. I have never said that expectations are meaningless without an upper bound.

One understanding of the argument for switching is based on the assumption that P(B=2A | A = a) is 1/2 for all a. The unbounded powers of two distribution achieves this assumption. Having an upper bound makes the resolution simple; having a lower bound does not resolve the paradox and makes the problem a bit more realistic. The problem with the distribution with only a lower bound is that the probability that either envelope contains a sum that could be put into it, or even represented in any way inside it, is zero. This is a far worse problem than unbounded expectations in normal distributions or infinities in calculus or Zeno's paradox. Unbounded numbers can be used in practical calculations in cases where they are almost certainly not goint to happen but in the TEP, right at the start of the problem, both envelopes contain a sum that is physically imposssible. The unbounded problem is a non-starter.

At least for non-mathematicians, and that is who most of our readers will be. For the mathematical reader I no not want to remove the mathematical details or dumb down the paradox in any way, but for the average reader, we should not start with an imposssible situation. My point is that to many people there is a natural division between bounded and unbounded, even if mathematically this is not so significant.

Regarding what I said about a mathematically well defined statement of this paradox I was talking about pure mathematics, which does not have money or envelopes, or, in my opinion and that of some others, probability. I do not think it is possible to create the paradox in pure mathematics although I have to admit that, if I saw one, I would probably not understand it. Martin Hogbin (talk) 12:19, 30 November 2014 (UTC)

In fact the mathematical issue here is that if the probability of twice as large an amount as a given amount rapidly declines, then expectation values are finite, and already in small samples, averages are close to expectation values. Moreover, truncating a distribution by some high cut-off hardly changes the mean value. So for practical purposes the bounded/unbounded distinction is *irrelevant*. The important thing is that the tail of the distribution falls off rather rapidly. This situation is often met in practice. But also we often meet in practice the opposite situation. So-called heavy tailed distributions. If you know that some value exceeds some large value, it becomes more and more likely that it exceeds it by more and more, as that value increases! Black swans. The financial markets. Hurricanes and global warming. So: unbounded distributions with infinite expectation values are actually very very appropriate is mathematical models for all kinds of significant real world phenomena! Of course, the height of an extreme storm-flood at the Netherlands North-Sea coast is bounded since it can't be larger than what it would be if all the water in the world were stacked in the North Sea. Yet from a prediction and mathematical modelling point of view a good model does *not* have an upper bound, but on the contrary, has a heavy tail ... Richard Gill (talk) 09:35, 30 November 2014 (UTC)

Yes but are these not different problems from the basic one? Martin Hogbin (talk) 12:19, 30 November 2014 (UTC)

All specific problems are different from one another. I am mentioning these other problems as counter-example to your argument, according to which all distributions with infinite support are banned from practical science and supposed only to be of interest within pure mathematics. Richard Gill (talk) 08:28, 7 December 2014 (UTC)

You are exaggerating my position somewhat. I am not trying to make any mathematical points or to ban the infinite from from practical science, I am trying to find a good way to explain the resolution of the TEP to the average reader. If you prefer, I am making an arbitrary distinction to aid readers' understanding. Many people do not accept or understand the infinite so the division into bounded and unbounded will be quite natural for them. No mathematics need be lost or corrupted. As an example, we are free split the proof of Pythagoras' theorem into two parts, one where the hypotenuse was over 7 cm long and the other where it was not, if we wish. That would, of course be completely pointless, but we could do it.

In the bounded TEP there is a very simple resolution that everyone can understand. I suspect that most readers will stop there and not worry about the unbounded case. Now I suspect that you would like to try and make them consider the unbounded case, because the resolution does not rely on the distribution being bounded. We cannot force them to do that, although we might try to encourage them. The first step, I suggest, would be simply to state immediately after that bounded resolution that, mathematically speaking, the resolution of the paradox does not rely on the boundedness and that there is a mathematically sound resolution that applies equally to the unbounded case. That statement alone might beenough for many readers but we could then go on to give the full resolution. Martin Hogbin (talk) 10:45, 16 December 2014 (UTC)

How about a very short general resolution first

If we start by saying:

For the proposed argument for switching to be valid, it is necssary that the probability that the unchosen envelope contains twice the sum in the chosen one must always be 1/2, regardless of what sum might be in the chosen envelope. It can be shown mathematically that, if this is the case, the expected sum in both envelopes must be infinite and switching is therefore pointless.

As an example, consider the case where the sum in each envelope must be a positive power of 2 (2, 4, 8, 16, 32 ...). If there is no upper limit on the sums that might be in the envelopes, then the expected sum in both envelopes is infinite. If there is an upper limit, for example, 512, then in the case that the chosen envelope contains 512, the other envelope must contain less, so the statement that the unchosen envelope always contains twice the sum in the chosen envelope is false and the argument for switching fails.

For all possible ways of filling the envelopes, it can be shown that if the probability that unchosen envelope contains twice the sum in the chosen one is always 1/2, regardless of what sum might be in the chosen envelope, the expected sum in both envelopes must be infinite. Martin Hogbin (talk) 09:59, 17 December 2014 (UTC)

Discussion with Evgeniy E

I don't see why the conclusion that the sum in any envelope has no mathematical expectation — in other words, that the expectation is infinite — is destructive for the argument. The argument does not rely on existence of this expectation. I would like to point out that the argument for switching must fail whatsoever the distribution of the sums is. Reasons that make people not switch are merely logical, not ones of probability: we know that there is no difference between the envelopes, whichever envelope is initially chosen. All difference that may exist may follow not from the mental act of initial choice, but from some properties of the envelopes themselves; the formulation of the problem gives no such special properties, whatsoever the distribution of sums is. So, the distribution of sums in these envelopes is not even important; we can safely assume that the sums may take any real values from zero onwards, probabilities non-assigned. No? - Evgeniy E. (talk) 12:31, 17 December 2014 (UTC)

The argument for switching that is presented to us explicitly calculates an expectation as a reason to swap. Martin Hogbin (talk) 14:10, 17 December 2014 (UTC

You mean, implicitly… Well, the words "the expected value of the money" are indeed present, but no probabilities are distributed across the ranges of possible sums, which makes think that these values are not taken in money units. The probability variable we have to deal with is discrete, not continiuous. We are given an amount of money in the initially chosen envelope, but the value of this amount is not significant and is not subject to probability laws, existence of this amount should only aid imagination. That is my reading. As you see, it depends on how you read it. - Evgeniy E. (talk) 14:35, 17 December 2014 (UTC)

It is hard to see how the words "the expected value of the money" can be taken to have any meaning other than an expectation value, given the proposed calculation.

You say, 'no probabilities are distributed across the ranges of possible sums' and that is correct. Howver we are lead to believe that the probaility of the other envelope being double the chosen one is always 1/2, but this cannot be , unless the expected sums are both infinite. That is the trick Martin Hogbin (talk) 15:23, 17 December 2014 (UTC)

I am under impression that this probability is not derived from any calculation; otherwise, prerequisites for such calculation would be given and the calculation itself expressed. This probability is only assigned on the stage of formulating the mathematical face of the problem. The words "value of the money" do not specify the unit of assessing the amount of money; assuming, that this unit is something like dollars, pounds, rubles or tenges, makes the initial amount float across different experiments, which is contrary to the spirit of the paradox. (I. e. why we need it, why we are interested). So, I am led to see that this unit is supposed to be the amount in the initially chosen envelope.

If I understand you correctly, you say that in the conditions of the paradox as you see them, mathematical expectations do not exist ('infinity' is not a numeric value), and therefore their utilisation in any kind of argument makes no sense. - Evgeniy E. (talk) 11:36, 18 December 2014 (UTC)

This is not my argument it is that of many mathematical sources. You can find links to them on the article page.

Articles in WP must be based on reliable sources not our own private research or opinion. Martin Hogbin (talk) 10:23, 19 December 2014 (UTC)

Articles indeed, but I understand that this page is rather a "safety valve to let steam go", as we say in Russian, a method for directing unavoidable arguments about the problem somewhere where they provide no danger, than a way to discuss the article itself, at least the wording on the main talk page made me think so. Sorry for inconvenience, if any is provided. - Evgeniy E. (talk) 13:40, 19 December 2014 (UTC)

The problem with the paradox is that it is not clear exactly what the argument for switching is meant to be. We as solvers of the paradox are therefore forced to try to fomulate exactly what the proposer of the argument for switching is trying to say. Much of the literature on the subject falls into two categories in this respect.

One category assumes that the proposer of the switching argument is asking us to compare the (unconditional) expectation in the chosen envelope with (unconditional) expectation in the other one. I, and some others here, do not much like that interpretation of the bogus argument for switching because it do not seem to be in accord with a natural interpretation of the problem or with its history. Nevertheless we have included this interpretation, and its associated resolution, in the article.

The other interpreatation, which I prefer and which is the interpretataion being addressed above, is to take it that the proposer is trying to calculate the conditional expectation in the unchosen envelope, given a specific (but unknown) sum in the chosen envelope. This interpretation seems to me to match the wording of most versions of the problem better.

This argument for switching only works if the sum in the unchosen envelope always has a probability of 1/2 of being twice the sum in the chosen envelope. If this is true, the (unconditional) expectation in both envelopes is infinite/undefined. So, yes, the argument for switching does actually work in this case but, in the long run at least, swapping is pointless. Martin Hogbin (talk) 11:47, 19 December 2014 (UTC)

I think, the main conclusion we have to learn from this discussion is that no interpretation of this problem is natural, any interpretation depends on the personal history of the interpreter. From my side, I confess that I do not understand how an argument for swapping may both work and not work at the same time, but that may have something to do with me, and I leave it to you. :-) - Evgeniy E. (talk) 13:40, 19 December 2014 (UTC)

Yes I agree. I have said before that this paradox is a self-inflicted injury. In fact it is doubly so. Firstly the paradox has to be created from an obviously non-paradoxical situation by proposing a bogus argument for swapping, then the resolver of the paradox has to decide exactly what the bogus argument is supposed to be before the paradox can be resolved. But, as Richard Gill points out, it was devised to tease people. Martin Hogbin (talk) 15:15, 19 December 2014 (UTC)

The argument for swapping does not both work and not work at the same time. It does not work, finito. The problem is to explain why it doesn't work. Martin should not have said "the argument for swapping only works if ...". He should have said "the argument for swapping only appears to work if ...". In that case, one simply has to think a little bit further. The argument derails, a little bit later. Richard Gill (talk) 08:53, 21 December 2014 (UTC)

YADD: an argument resumed

Answering this one. To Richard Gill.

The envelopes were labelled A and B before the choice was made, and they were labelled C and D after the choice was made, C being the envelope that was initially picked; I am sorry if I caused confusion, I was sure I had described the difference. Let me say that A is the envelope in the left hand, and B is one in the right hand. You say I need to prove that the envelope which enters the right-hand position (B) has, with equal probability, either twice as much or half as much money as the envelope which enters the left-hand position (A). However, that is what the word "indistinguishable" means to me: there is no sign that one envelope is likely to have more money than the other one, and the hand the envelope initially enters is not such sign either. Another thing is that I do not see why anything should depend on the exact amount of money ("values of a") in an envelope howsoever defined. Thank you for your feedback. - Evgeniy E. (talk) 15:35, 15 December 2014 (UTC)

There is a different problem with my argument that I have just noticed, though: mean(C) and mean(D) are defined in the units of the money amount in A, whose values in money units should differ in different experiments; one may set $x to be the same always, but this is not realistic. However, this is a pseudo-problem, because I may set as the random variables denoted C and D the ratios of the money amounts in envelopes C and D to the money amount in envelope A, rather than the money amounts in envelopes C and D themselves, and proceed with saying that I am interested to know mean(C)/mean(D) and mean(D)/mean(C) rather than mean(C/D) and mean(D/C). In any case, treating the problem as a linguistical one rather than a mathematical one and treating different definitions of the same objects as completely different things, that have nothing in common, that is to say do not bind each other to have the same mathematical properties, is an approach that is both interesting and rational… - Evgeniy E. (talk) 18:49, 15 December 2014 (UTC)

Now, the question is which step is wrong in the argument in the article. The answer is simple: the mistake is in the step 8, since "the swapped envelope" is not "the other envelope". Why? Because the operation of swapping works on actual, real envelopes (let me call them this way; A and B in my designation), while the relationship of otherness works on virtual envelopes (C and D in my designation), those existing in my head only and related to strategy setups in my mind. The previous step establishes that the ratio of the amount in the other envelope to the amount in the picked envelope is 1.25, while the ratio of the amount in the picked envelope to the amount in the picked envelope is 1. Yet, the operation of implementing a choice presumes finding out the amounts of money on the base of real envelopes, not on the base of strategic, imaginary ones; so, the conclusion in the previous step has nothing whatsoever to do with this operation, and this operation is not ordered to happen. Then, the argument halts. Cure? Be more careful in interpretation of mathematical results. The analysis in the step 7 did not describe how to choose the right outcome, it only described how the outcomes were distributed: the average of the ratio of the two outcomes was greater than one. - Evgeniy E. (talk) 21:18, 15 December 2014 (UTC)

I have read here references to some arguments involving the distribution of money amounts in given out envelopes; I admit I did not really understand the causing nature of these arguments, but I do not think any such considerations are relevant. The purpose of the player is not to gather as much money as possible, it is to be at a win as often as possible. So, the probability variable whose mean is evaluated must not be any amount of money, it must be a special "success" variable, which is dimensionless. How to define the latter variable? I had the way of making it equal to the ratio of the eventually chosen amount of money to the amount in the left-hand envelope (no matter what this amount is) given the strategy in a given experiment; if the player wins, that means that his strategy (take C or take D) results in the greater value of this variable than the other possible strategy. More specifically, this variable took, with equal probabilities, four values: 1, 2, 1, and 0,5, as it is both unknown which kind of relationship exists between the two envelopes and which envelope is initially chosen from the two. As to assigning the probability values: the two couples of events are pairwise independent; the probabilities of the former two events are assigned a half each per the meaning of the word "indistinguishable", while the probabilities of the latter two events are a more complicated issue. It might seem that any distribution is possible; however, the solution only works out if the two events are assigned probabilities a half each, otherwise one of the strategies turns out to be better, which makes no sense... Now, to find out which strategy is best, we calculate the average of this success variable over many experiments given the strategy and then choose the strategy that resulted in the better mean. Analysis showed that both strategies provided the player with the same mean. There is no reason to believe that assigning such probabilities inflicts any consequences on reasoning about the probability distribution of money amounts in all envelopes to be potentially given, because the matter to be considered is rationality of the choice, not the amount of money received. All reasoning is not holeless as of now, but it seems to me amusing at least. I am done with it by now, though. Thanks to all for listening, if anyone has. - Evgeniy E. (talk) 16:16, 16 December 2014 (UTC)

The problem with this puzzle is that it is a self-inflicted injury in that it is quite obvious to most people that you should not swap, and that is also mathematically correct. It is necessary to present a bogus argument for swapping in order to create a paradox at all. Without the bogus argument there is no paradox; everyone agrees that it is pointless to swap. Resolving the paradox does not require you to show that swapping is pointless, that is obvious. To resolve the paradox, you must show exactly where, in the (mathematically vague) line of reasoning the fault lies. Depending on exactly how you interpret the proposed logic for swapping, the fault will be different. Martin Hogbin (talk) 17:01, 16 December 2014 (UTC)

Well, for me the reason I visited that page at all was a question I asked to myself: "why, if the risk of some action is considerable, but the risk appears unlikely, I still may feel like opting for the action, provided I know that the trouble simply will either happen or not?" This is a question of my feeling, of my internal brain/mind-functioning, this is not a question of reason… I turned to the conclusion that, since the brain has to choose the same way of functioning in every dubious situation of similar nature (just because it has a "mechanical" nature and cannot act differently in the same circumstances), it had to evolve so as to choose such a way that leads to the best results over the course of many dubious decisions in total. That means that the brain must inherently "understand" the basics of the probability theory, that is the conclusions of this theory must be intuitive for it; whenever our intuition fails us, the reason must lie in being short of taking into account some side notions not pertaining to the probabilistic estimation of the expectation itself. Then, I was interested to think, not very successully, on the question, what kind of simple automatic "reasoning" must induce the brain into making the correct estimation that the strategies are equal; the simple ad absurdum argument is probably not enough; evidently, the brain ought to see that the strategies are not just equal, but the same from a certain point of view. Hence these ponderings. As to your next question, I'll try to find out how to make my reply more clear… - Evgeniy E. (talk) 17:45, 16 December 2014 (UTC)

I do not understand your argument, 'the mistake is in the step 8, since "the swapped envelope" is not "the other envelope"'. Martin Hogbin (talk) 17:04, 16 December 2014 (UTC)

In solving a problem with assistance of mathematics there are three stages: 1) from a natural inner representation of knowledge derive a mathematical structure that represents its shape, 2) find out substructures in the mathematical structure that describe the shape of the answer, 3) interpret the mathematical answer by assigning this shape to knowledge about our intuitive notions. Mistakes may happen in any of these stages, however forming a wrong mathematical structure on stage 2 is not yet a fault. To be able to prove that any mistake was made on stage 3, one needs to know how mental reflections of objects, i.e. definitions of things, are organised in our mind; since we don't know that, usually this problem is decided intuitively, we see that "this mathematical result is not about that!" and merely declare it. The right approach would be to come up with at least a half-formalism to describe reflections of objects in our mind for this case, and with a description how this half-formalism should work with the notions of this problem, to show why I think the mathematical result in question does not provide us with a reason for any action. However, so far I try yet again an intuitive description. The question is what constitutes a reason for selecting an envelope. Yep, I know that if I compare (by taking a ratio) any couple of choices (the initial one and the switched one) each time an experiment is made, then the average is greater than one; that follows merely from the nature of the choices. Naturally, this argument works in both directions. However, the values that were assigned to the corresponding variable, both times 1 for the first choice, and 2 and 0.5 for the second choice, were not derived from any rigid unit describing real envelopes; hence, they did not describe the chosen envelopes ("the swapped envelope"), but only the choices themselves ("the other envelope"). They were derived from the definition: "the defining feature of A is that I picked it; the defining feature of B is that it is the other envelope I did not pick". In these definitions, there is nothing to describe A and B as real envelopes with properties, communicated to such envelopes by the formulation of the problem. In order to make a decision, I need to compare a couple of envelopes, not a couple of choices; so, there is no ground provided for a decision yet. The idea of those who follow the original argument (provided we understand "A" in 5A/4 and in A as a kind of unit, the real values being 5/4 and 1) is apparently along the lines "since the definition so formulated is applied to a real-world object, the properties of the real-world object and the properties of the thing I defined belong to the same thing that behaves the same and shares all the same properties"; that is not a convincing idea, and the question "what constitutes a reason for an action" still holds.

By the way, it is interesting to note that in my approach of yesterday, only when the probability distribution of the initial choice is 0.5 to 0.5, the correct conclusion is reached; if it is 1 to 0, then my approach coincides with the approach described in the article; if it is 0 to 1, then it degenerates into another paradox in which it is disadvantageous to change the choice. All other distributions provide slighter kinds of advantage to the first or to the second strategy. Apparently, we must assign probabilities 0.5 and 0.5 to these events of initial choice, but I cannot explain why and I cannot interpret this behaviour. - Evgeniy E. (talk) 07:24, 17 December 2014 (UTC)

Evgeniy, you said "Now, the question is which step is wrong in the argument in the article. The answer is simple". The answer is not simple. Because of the Anna Karenina principle, there is not one particular way to try to repair the switching argument, hence there is not one particular way to say it goes wrong. Read my paper. http://www.math.leidenuniv.nl/~gill/tep.pdf There are several different interpretations of what the writer of the switching argument was trying to do, and hence several different interpretations of where his argument goes off the rails. Richard Gill (talk) 08:58, 21 December 2014 (UTC)