Talk:Two envelopes problem/Archive 1

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Archive This is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page.

Contents

Merge

Definitely a good idea to merge the two pages. I'm a bit new at it all, so if anyone wants to do it, that would be good. Can we make it so that searching for "two envelopes" will find the "envelope problem"?

John

I don't know if they should be merged necessarily, but it might be a good idea to define the difference more sharply and compare the two throughout. Personally I'm still at a loss for why the two envelopes problem is still open considering that the envelope paradox is solved. The difference seems to be that in the latter you're allowed to look into the envelope before making a decision, whereas the first is a more recent variant that asks about the analysis before one takes a peek. If that's correct then readers should be directed to the paradox before they take on the open problem. Personally though I'm confused if that's the only difference. I don't understand why looking would change anything. The amount of money in your envelope might be an arbitrary value, but if you reason about that variable then the reasoning must include every possible value that variable might have, and each assumption of a value is analogous to a hypothetical peek. Davilla 07:35, 8 February 2006 (UTC)
I'm having a long and heated discussion with the two authors of envelope paradox for the purpose of merging the two pages. It's not the case that there are two different problems, one called the "two envelopes problem" and another called the "envelope paradox." It's just two different names of the same thing. Actually, the problem/paradox have gotten many different names over the years. Accordingly, there should be only one page at wikipedia devoted to this problem/paradox. All the different names should be redirected to one and the same page. On this we agree, but we disagree on some key issues about what the merged article should contain and state. INic 03:31, 20 February 2006 (UTC)
Any status updates on the merge? It has been a while, and it really would be nice if the articles were merged.
... I went ahead and merged the pages by making the other page just redirect here -- that article is essentially the same as this one. If you have nitpicks to work out, work them out on one article -- keeping parallel articles for months because we can't settle minor disputes seems a bit absurd.
I agree! INic 00:12, 25 August 2006 (UTC)

Ranking of Problems

Why insist on ranking to problems as "hard", "harder", "hardest"? Any such ranking depends on how you think about them, and is implicitly POV, not to mention un-important. The parallel article has them marked simply as first, second, third. I ask this on the Talk page instead of just fixing it because it seems at least one contributor insists on the distinction. (comment from the history: reverted to "the hardest problem" as this is, in fact, the hardest to solve). Incidentally, the "hardest" problem looks trivial to me.

I agree that it's really not possible to rank the variants of the problem according to their difficulty in any absolute sense. However, when you read the papers published so far you will see that almost all authors think they are more difficult in this increasing order. This wikipedia article only reflects this fact. You are, of course, free to disagree with the majority of the authors in the references. I do. But to include your or mine personal opinions about this in the article would be a violation of NPOV. INic 11:15, 21 August 2006 (UTC)
Thanks for the clarification. Is it possible to cite which articles specifically list the third form as the hardest? I'd like to read a more detailed analysis written by someone who thinks the problem is not trivial, but didn't find it in a brief search through th refs -- Though there are many references listed, the article does nothing to indicate what information comes from which source.
I'll clarify "In fact, the hardest variant" to "A variant often considered hardest" in light of the discussion.
The hardest problem (which is the fourth variant) is hardest just because it doesn't contain any probability reasoning. It is thus immune to all explanations relying heavily on probability theory. I agree that the article could refer to the references to a greater extent. I will improve that shortly. INic 00:53, 25 August 2006 (UTC)
Thanks!
I would clarify by saying that it is immune from subtle errors in its own probabilistic reasoning, since it has none. I doubt that it is immune from explanations utilizing probabilistic reasoning (the A vs A/2 formulation sounds like an abstract example of Simpson's paradox...), although that doesn't mean someone has actually come up with one yet. Baccyak4H 17:49, 29 August 2006 (UTC)
POV is, however, exactly what they have in the parallel article. First of all they say that the problem is solved which it's not by any means. Secondly, they refuse to include the most basic and original version of the problem which here is presented first. Thirdly they have their very own invented "solution" to the hardest problem—nowhere to be found in the literature. INic 11:15, 21 August 2006 (UTC)
On reading the solution in the other article, I appreciate that this version does not include it -- it looks, at best, poorly stated -- but it would be nice if we could have some referenced discussion of the reasoning behind the third paradox.
The last (fourth) variant was invented by the logician Raymond Smullyan. As far as I know no one so far has managed to extend one of the proposed solutions for the other cases to this purely logical case. If anyone find that somewhere it would be great to have it included here as a proposed solution. INic 00:53, 25 August 2006 (UTC)
Thanks again. I'll keep an eye out, but it's probably not happening. As I read it, any extension of the proposed solutions wouldn't really work because Smullyan has fundamentally changed where the paradox is introduced -- it's not really the same problem anymore, so why would it have the same solution? (specifically: his version is just sweeping details under the rug and comparing things that aren't the same as if they were supposed to be ... but that's, of course, out of place 'original research,' so I won't bother to expand)
The point Smullyan wanted to make was to question if the paradox really has anything to do with probabilities at all. But if this is an entirely different problem someone has to show that. It's not obvious. I agree that it's very unlikely someone would manage to extend his/her other solutions to this case easily, but chances are that a good solution to this problem automatically solves the other variants. BTW, you are allowed to speak about your own ideas for a solution here at the talk page. INic 14:30, 27 August 2006 (UTC)
Cool -- so, if you don't mind, I'll spew some hopefully coherent ideas about solving the hardest problem:
The way I was thinking about the Smullyan variant, his first statement is saying that "if you hypothesize about what values you could get in the other envelope, compared to the value in the current envelope, you get two possible values -- one larger, one smaller. The distance to the larger hypothetical value, from the middle known value, is larger than the distance to the lower hypothetical value." His second statement is saying that, "if you consider the two actual values in the envelopes, (they have this relation to each other and) the distance between them is the same going either way." These two statements aren't contradictory at all. The only way you eek out a contradiction is by using loose enough language such that the two distances considered in the first statement are thought to refer to the same thing conceptually as the single distance in the second statement.
So, that's where I think the paradox comes in. ... did that make any sense? I hope so.
I don't quite understand your solution here. However, I just added a summary of the only solution to this variant I know of in the article. Please have a look and see if it's the same idea that you have. INic 03:16, 15 October 2006 (UTC)
It's a rather different way of stating it, but I think it's saying what I was grasping for (and saying it much better than I could!). (Or maybe it's saying something a bit different, but I like it better anyway.) So yeah. Thanks for that!

The introduction

I don't like the introduction:

The Two Envelopes Problem is a puzzle within the subjectivistic interpretation of probability theory. This is still an open problem among the subjectivists as no consensus has been reached yet.
Because the subjectivistic interpretation of probability is closer to the layman's conception of probability this paradox is understood by almost everybody. However, for a working statistician or probability theorist endorsing the more technical frequency interpretation of probability this puzzle isn't a problem, as the puzzle can't even be stated when imposing those more technical restrictions. Consequently, all published papers below with different ideas on a solution are written from the subjectivistic or Bayesian point of view.

I don't like it because it tells me the point of view I should have about the problem before expainig what the problem is. Comments about subjectivism should come later as a possible pont of view, not at the very beginning BEFORE everithing else!! --Pokipsy76 16:22, 24 October 2005 (UTC)

Good point! I moved the second paragraph to after the presentation of the puzzle and made it a comment. Is this good enough you think? If not, please feel free to improve it yourself. INic 01:14, 25 October 2005 (UTC)

A Harder Problem

The explanation given doesn't make sense. Let's say I open the envelope and there is a $100. If I switch I expect to gain $125. Now that I switched I do the following calculation: I have 1/2 * 50 + 1/2 * 200 = $125. In the other envelope I see $100. So why would I switch back? The fact that I openned the envelope and found a sum greater than 0 added additional information about the expected value in the game. There is no paradox in this case.

-- Martin

You are right, it should be made more clear in the article how the contradiction appears in this case. In essence it's the same contradiction as in the first case. The only difference is that here the explanation given for the first case doesn't work anymore. I'll try to improve the text. INic 13:28, 2 August 2006 (UTC)

Tad's original research

I have just done a total rewrite of this article. As far as I know this is the first complete solution to the puzzle. I have removed most of the pre-existing material in the interests of brevity. Perhaps the new solution should be abridged eg the examples of oscillating series and the treatment of the finite case in the postscript are not strictly necessary. Part of the problem of presentation is to make it intelligible to non-mathematicians yet to be rigorous at the same time. My solution attempts to cover both bases.

The previous write-up had the tollowing faults. The paradox statement was somewhat clumsy and open to variant interpretations. The article unnecessarily divided up the paradox into various versions. The 3rd paradox statement without probability makes no sense - on what basis could you possibly make a choice? Above all, the old page failed to provide the solution. I have merged "two envelopes problem" with "envelope paradox". I would prefer to call it "two envelopes paradox" but it is probably best to retain the old name so as not to confuse things.

Maybe it is worth keeping the winning strategy under the heading "Discussion"?

Tad Boniecki 15 April 2006 218.214.57.242 21:42, 14 April 2006 (UTC)

Tad, I agree personally with some of the conclusions you have. However, Wikipedia is not the place to publish original research. It's against the policy of this encyclopedia (as it is against the policy of any encyclopedia). If you want your ideas on wikipedia you have to publish your ideas in a peer reviewied journal first. Until that happens you are welcome to make additions to and change this article but based solely on ideas already published. Please see the references. INic 19:59, 17 April 2006 (UTC)
Tad, another option for you is to use Academic Publishing Wiki INic 09:00, 25 August 2006 (UTC)
Many thanks. I have just added my solution plus a discussion to the wikia Philosophy Journal. - Tad

The 'an even harder problem' does not make sense

I have a problem with the 'An even harder problem'. First of all, there are two major mistakes in the calculation of the expected gain:

1) the first 1/2 should be omitted

2) a factor (1 - q)^(n-1) is missing in the final expression. so it should be

(2^n q(1 - q)^n - 2^(n-1)q(1 - q)^(n-1))/R = 2^(n-1)q(1 - q)^(n-1)(1 - 2q)/R

which is still positive if q < 1/2

But most of all, it does not make sense to consider distributions with infinite mean. They don't have practical value and you can't make any conclusion based on calculations with these kind of distributions.

It is easy to prove that the expected gain of switching envelopes before opening them, should be zero for each 'real' (finite mean) distribution. The expected amount in the two envelopes is the same by symmetry which means that the expected gain of switching (difference of the two) is zero, if the expected amount in the envelopes is finite.

This means that for every 'real' distribution that the mean of the gain is zero. If there are amounts, with probability > 0, for which it is better to switch then there are other amounts, with probability > 0, for which it is better not to switch.

The only distribution for which the reasoning of the paradox holds, is the one for which the envelopes contain 0 with probability 1. But then, the gain of switching remains zero even if the amount is doubled in the other envelope.

To conclude, It is better to remove the section about 'An even harder problem'. Distributions with infinite mean really don't make any sense and it just confuses the reader.

Regards, Stefaan

I agree that the example presented here can be better. I will soon replace this with an example found in the literature. But I don't agree that the only admissible distributions are the ones with finite means. Cauchy distributions for example lacks both mean and variance but that doesn't make them 'unreal' or unuseful. INic 21:17, 30 July 2006 (UTC)


Recent Edits - lets' express the probelm clearly before going into Bayesiana -Do we need to START with Bayes?

I recently edited the intro so it is easily understood:

The two envelopes problem is the following:

  • You are shown two indistinguishable envelopes and asked to select one, which you do.
  • You are then told that each envelope contains money, and one contains twice as much as the other.
  • You are given the option to swap the envelope you have chosen for the other one.
  • What should you do?
Some writers analyse this as a puzzle or paradox within the subjectivistic interpretation of probability theory; more specifically within Bayesian decision theory. This is still an open problem among the subjectivists as no consensus has been reached yet.


However, this has now been removed so the definition is

a puzzle or paradox within the subjectivistic interpretation of probability theory; more specifically within Bayesian decision theory. This is still an open problem among the subjectivists as no consensus has been reached yet.

IMHO it's far better to explain the problem BEFORE going into how it might be analysed. (I've tried it on lots of people - they have had good analyses without a whiff of Bayesianism!

First of all the two envelopes problem is not the one you have written above ending with the question "What should you do?". Instead the very statement of the problem must include the reasoning leading to the contradiction. That is followed by the question where the flaw in the reasoning is. That is the problem. See also my answer in the section above with title "What's the problem?". INic 00:13, 31 August 2006 (UTC)
Secondly, it is actually considered good wikipedia practice to have a short introduction to set the subject of the article into the right context before going into the meat of the matter. Please see WP:LEAD. INic 00:13, 31 August 2006 (UTC)
Thirdly, wikipedia articles should only be based on the contents of already published papers in reputable magazines. That you have friends that have solved this problem in an entirely different way doesn't matter I'm afraid. All published suggested solutions so far are Bayesian in character, and that is a fact that should be reflected here. When your friends have published their solutions we will rephrase this article accordingly. INic 00:13, 31 August 2006 (UTC)

Thanks - that's helpful. We are I think using 'problem' in two different ways. To me it is the problem' as I posed it - ending "What should you do?". To you, the problem is in the solution.

The problem or puzzle is to solve the paradox or get rid of the contradiction. I haven't seen any disagreement here in the literature. The only disagreement here is about what to call it, not what it is. My personal opinion is irrelevant. INic 02:28, 1 September 2006 (UTC)

What you call the 'problem', I would call the 'paradox'. I hesitate to say which is 'correct', but I think it is useful to distinguish between the two issues.

I can't find any clear distinction in the literature between what you call 'problem' versus 'paradox.' These words are used interchangeably. INic 02:28, 1 September 2006 (UTC)

I feel that the 'problem' is easy to state. The 'paradox' (or dilemma or whatever) is fairly easy to recognise. And the 'solution' is difficult indeed - but Bayes only arises in the solution.

No, as is explained in the comment of the lead section Bayesianism enter the stage already in the set-up of the problem. The only exception to this is the last, hardest, variant of the puzzle which isn't connected to probability at all. INic 02:28, 1 September 2006 (UTC)

You say that "All published suggested solutions so far are Bayesian in character" I don;t know the literature so cannot argue with this. However the following is not Bayesian:

Start off as before:

  • You are offered an envelope which you open. It contains £10 say
  • You are then shown two other envelopes and told that ONE of them £5 and the other one contains £20.
  • Should you swap?

This has a clear non-Bayesian answer.

                                        Johnbibby 13:23, 31 August 2006 (UTC)
This is clearly not the same problem. This is seen by the fact that no contradiction or paradox can arise based on this situation. And that this problem merits an non-Bayesian analysis and solution is thus totally beside the point. INic 02:28, 1 September 2006 (UTC)

I agree I shd not quote my friends in Wikipedia! - but I think it's OK to refer to them in 'Talk'

                                  Johnbibby 13:23, 31 August 2006 (UTC)
Yes that is OK. INic 02:28, 1 September 2006 (UTC)

"solution" -> "discussion"

I suggest to replace the word "solution" in the paragraph titles by "discussion" because what we do is discussing the problem and it is not clear who is proposing the "pèroposed solutions" and who is suggesting that these arguments *are* actually solution.--Pokipsy76 13:32, 15 September 2006 (UTC)

I agree that "discussion" is more fitting, especially for the later paragraphs. However, for the first "proposed solution" there isn't much disagreement that this is the correct solution. Hence to replace that with "discussion" would be incorrect as no discussion exists here really. "Proposed solution" I think is appropriate for covering all cases, from almost total agreement to total disagreement. Another option would be to rename them "solution," "proposed solution" and "discussion" respectively. However, some might argue that this introduces an unnecessary element of POV to the article. INic 01:09, 16 September 2006 (UTC)

One stupid persons viewpoint

I was posed this paradox a while back (the "harder" version), and passed it on to a friend of mine. He then pointed me to this page. I'm no wiki expert (I only registered today) but feel that it is not readily comprehensible to non-specialist readers. Or someone as stupid as me.

My way to explain the paradox is to see that the values are chosen before you even see the envelopes. You having chosen the one with $512 in means that the value of the other envelope is already set. The chances of that envelope having a $1024 or $256 in it are not 50/50, they depend upon the generosity of the person filling the envelopes.

Choosing the envelope with $512 in it was the 50/50 part of it, not whether the other one is higher or lower. That is why the maths doesn't work, it is a failure to interpret the question correctly.

I think that this agrees with the proposed solution, but is there any chance that someone could attempt to make the page a little more accessable to the layperson? Situationist 00:48, 8 November 2006 (UTC)

Math is not psychological or material, it is conceptual. Generosity does not play any part in this, as evidenced by the following two concepts. 1) The envelope filler MUST put two amounts of money, one in each envelope. 2) One MUST ALWAYS be twice as great as the other. Your example, in which you use real dollar amounts, can be contradicted by the assumption the filler is wealthy beyond belief. Or look a it this way: Have him put in $10 and $20. If you pick $10, you would see his as "generous". But if you pick $20 the first time, you would see him as stingy. Things like this are all a matter of opinion anyway. Try using variables when examining the problem, and imagine a computer filling the envelopes as well. It may help you understand it better. MateoCorazon 01:42, 12 November 2006 (UTC)

Arguments page

I will soon clean up this talk page and remove everything that are discussions of the puzzle itself and not discussions of the article. Discussion of the puzzle itself can go on forever and has to be done elsewhere. Maybe someone can put up a link to a blog or discussion board dedicated to this puzzle at the top of this talk page? iNic 01:40, 17 November 2006 (UTC)

You might consider a solution like that chosen at talk:0.999..., with the following box near the top:
Info This is the talk page for discussing changes to the 0.999... article. Please place discussions on the underlying mathematical issues on the Arguments page. If you just have a question, try Wikipedia:Reference desk/Mathematics instead.
--Niels Ø 07:52, 17 November 2006 (UTC)

Yes, that would be a good solution here too! iNic 12:34, 17 November 2006 (UTC)

Proposed solution of the original problem

I suggest to add the following explanation:

"If the smaller amount is consistently denoted as X, the larger amount is 2X. Both envelopes contain either X or 2X with probability 1/2 each. Hence the expected value in both envelopes is 0.5 X + 0.5 · 2 X = 1.5 X and you don't gain on average by swapping."

In my opinion this is not self-evident or beside the point, but fundamental to understand the problem. --M.T. 20:07, 11 December 2006 (UTC)

As the situation is symmetric it's evident that it doesn't matter which envelope we pick. That is the very cause of the paradox as the reasoning in the text leads to a non-symmetric result, which obviously must be wrong. The problem is not to find another way to calculate that doesn't lead to contradictions (that is very easy), but to pinpoint the erroneous step in the presented reasoning leading to the contradiction. That includes to be able to say exactly why that step is not correct, and under what conditions it's not correct, so we can be absolutely sure we don't make this mistake in a more complicated situation where the fact that it's wrong isn't this obvious. Who in the literature claim that your calculation is fundamental to understand the problem? iNic 01:37, 12 December 2006 (UTC)
I agree that it's important and interesting to find the error in the presented reasoning with the result that the error lies in the calculation of the expected value. However, I see two reasons to mention the correct calculation: First, it's just interesting, at least for me, and presumably also for other readers. Second, and even more important, stating that the calculation of the expected value is wrong does not necessarily implicate that the consequence is wrong. In the literature at least Schwitzgebel and Dever who are cited in the article consider the correct calculation of the expected value noteworthy. They write: "The availability of such a non-paradoxical calculation is old news, of course; the novelty here is the identification of the crucial difference between the paradoxical and non-paradoxical calculation." --M.T. 17:06, 12 December 2006 (UTC)
Well, first of all Schwitzgebel and Dever clearly state what is the interesting problem (page 4). Our interest in the two envelope paradox [...] is not in providing a correct analysis of the decision problem — we take it that [...] this is trivial, and in the case of the two-envelope problem, it is well worked-over territory — but rather [...] in explaining what is wrong in the clearly absurd, but intuitively seductive, calculation that, in the paradox, envelope A is better than envelope B (or vice versa). Secondly, if the article should mention the correct calculation we have to be able to refer to some proof that this is the only correct calculation. All you have provided is a calculation that doesn't lead to paradoxes. Unfortunately, this is only a necessary condition and not a sufficient condition for a correct solution. Any calculation preserving symmetry will avoid the paradox. Thirdly, I doubt that there might be readers that doesn't realize the absurdity of the conclusion in the paradox. But if you think that should be stressed you can do that without mentioning the "correct" calculation. iNic 05:38, 15 December 2006 (UTC)
The calculation I have cited above is from a newer paper from Schwitzgebel and Dever which is called "The Two Envelope Paradox and Using Variables Within the Expectation Formula". This calculation of the expected value is also given in "Adom Giffin: The Error in the Two Envelope Paradox" and "Priest and Restall: Envelopes and Indifference". We don't have to proof it. However, if the article says that the presented calculation of the expectation value is wrong, this does not necessarily say that the consequence -- namely to swap -- is wrong. It could be possible that the correct calculation of the expectation value would be that the expected value of the money in the other envelope were 6/5 A, so one would gain on average by swapping. So we have to give a proof that the expected value of both envelopes is the same, so swapping will not lead to a gain on average. --M.T. 22:18, 25 December 2006 (UTC)
It's a little surprising that you view your calculation as the only correct one even after reading those papers. Priest and Restall consider three different calculations and concludes that all of them are correct, depending how the situation is interpreted; and without any interpretation at all the problem is underdetermined, i.e., not strictly solvable. Schwitzgebel and Dever defend their idea that each occurrence of a random variable in an expected value calculation should have the same mean, they do not have to be constants as in your calculation. Adom Giffin adds the assumption that the situation can be repeated an unlimited number of times. This assumption is a natural one to make for a physicist, but here it's actually not justified. You should in fact be glad if you got this generous offer only once. Hence, we can't really be sure how envelopes would be filled if the offer ever would be repeated. Giffin postulates that they'd be filled with the same amount every time. But fact is that we don't know that. iNic 05:33, 29 December 2006 (UTC)
If the correct calculation would lead to 6/5 A as the expected value for the other envelope, and the reasoning was correct otherwise, it leads to a contradiction. So your scenario isn't logically possible as you claim it is. Please feel free to add a comment about that in the article for those that doesn't see it directly. But please don't write that it's wrong to swap. It's no more wrong to swap than to stick. If you think this point should be stressed too for some readers please add a comment about that too. iNic 05:33, 29 December 2006 (UTC)
We are talking about the original problem, when the envelope is not opened. In this case all we know is that one envelope contains twice as much money as the other envelope, but we don't know which one, and the chance to get the envelope with the larger amount by picking one at random is 50 %, as well as the chance to get the envelope with the smaller amount (point 2 of the reasoning in the article). So the only correct calculation of the expectation value in this case is the one I have cited above.
OK, what is in that case wrong with the following: Denote by A what we have in our selected envelope and denote by Y the difference between the contents of the two envelopes. The other (not selected) envelope then either contains A-Y or A+Y. Both options are equally likely, i.e., has probability 1/2, so the expected value for the other envelope must be
{1 \over 2} (A-Y) + {1 \over 2} (A+Y) = A,
which shows once more that we don't gain anything by swapping envelopes. This is not the same calculation as yours, yet it seem to be equally valid. If your calculation is the only one that is correct, this must be false. iNic 03:59, 3 January 2007 (UTC)
This calculation is indeed wrong. It contains the same error as described in the article, namely that A once denotes the larger amount and once the smaller amount. --M.T. 19:21, 3 January 2007 (UTC)
Yes maybe, but do you think that Priest and Restall would have had the same analysis? Definitely not. Please note that the solution proposed by Schwitzgebel and Dever is only a suggestion; it's not at all at the level of encyclopedic 'truth.' Priest and Restall, for example, have a fundamentally different solution, which is seen more clearly, perhaps, when applied to this equation than the original one. As the article shouldn't violate the NPOV policy, it's a bad idea to promote some authors opinions as the "correct" ones. To tell the readers what the "correct" solution is without qualifying that with who is saying that, and even hide the fact that it's controversial, is to do exactly that. iNic 00:23, 8 January 2007 (UTC)
Priest and Restall describe three different mechanisms for putting money in the envelopes and how the expected value is calculated in those cases. They describe mechanism 1 as follows: "A number, n, is chosen in any way one likes. One of the two envelopes is chosen by the toss of a fair coin, and n is put in that; 2n is put in the other." This describes the situation presented in the article for the case that the envelopes are not opened. We have an amount n and an amount 2n, and we pick one with a probability of 50 %. For this case Priest and Restall calculate an expected value of 3/2 n. --M.T. 13:57, 11 January 2007 (UTC)
Priest and Restall postulate some different mechanisms for putting money in the envelopes. They do not suggest that one of these is the "correct" one as you do. On the contrary they say it's impossible to know what the correct mechanism is and hence the original problem is underdetermined. Please re-read their article. And when it comes to the formula above, 1/2(A-Y) + 1/2(A+Y) = A, do you think that Priest and Restall would claim that this formula is wrong because A has different mean in its two occurrences to the left? iNic 00:58, 15 January 2007 (UTC)
I do not suggest that one of the mechanisms is the "correct" one, but that the mechanism described in this article is equivalent to mechanism 1 described by Priest and Restall. And for each mechanism there is exactly one correct calculation of the expected value. Priest and Restall have intentionally formulated the problem in such a way that different mechanisms can be assumed. Especially, they describe a "mechanism 2" which leads to a formula for the expected value in the other envelope that is given in step 7 of this article: E = 1/2 · 2A + 1/2 · A/2. With this mechanism you really have a 50:50 chance to get either 2A or A/2 in the other envelope, while in your envelope you have the fixed amount A. And yes, I'm sure that Priest and Restall would claim that the formula you gave above is wrong for the problem formulated in this article because A has different mean in its two occurrences to the left, because they write: "In a computation of expectation, the designation of the variables must correctly refer to the elements of the state-space, and not slide around in the process." So the question rather should be: Who actually specifies another formula for the expected value for the case we are talking about? --M.T. 21:11, 15 January 2007 (UTC)
I do understand your frustration that we can't easily state the "correct" expectation in the article. I wish we could, too. The reason we can't do this is that what this problem/paradox really boils down to is a question about probability itself and how it should be interpreted. Different authors have different solutions depending upon what philosophical interpretation they adhere to. For example, Priest and Restall have a totally different solution than Schwitzgebel and Dever. Philosophically speaking they are from different planets. Your impression is that they basically say the same thing, but that is very far from correct I'm afraid. Trust me. And the NPOV policy of course forbids us to claim that one author or philosophy is more correct than any other. So the best thing to do here is to leave it up to the reader to decide what she thinks is the best solution to the problem. iNic 03:54, 21 January 2007 (UTC)
You ask "Who actually specifies another formula for the expected value for the case we are talking about?" Well, if you look further down in the article you find the idea that the correct calculation should depend on how we assign values to the envelopes, i.e., exactly how is n chosen when the envelopes contain n and 2n? One of those possibilities leads to the expected value 11/10x in the "other" envelope, where we have x in our chosen envelope. This holds whether we look at x or not, so it's still relevant for the case we talk about. iNic 03:54, 21 January 2007 (UTC)
If there are different opinions on which calculation is correct, the NPOV policy requires that we present both, so the reader is able to decide what (s)he thinks is the best solution to the problem. Schwitzgebel and Dever write in their first paper on page 6: "Thus, the second method of analyzing the two envelope problem (yielding the indifference result) meets both constraints and is correct." The case you have mentioned is a special case with a special distribution that is known to the player. But we are still talking about the original problem here. --M.T. 21:02, 25 January 2007 (UTC)
To repeat, what you call "the problem" isn't the problem here at all. The problem under discussion here is not what a 'correct' calculation would look like. Instead, what occupies all authors is where the presented argument (that leads to a contradiction) is wrong. This is a big difference. So not only would you have to start a new Wikipedia article for your problem, prior to that you would have to start an academic debate centered at that issue. THEN you can apply the NPOV policy to the potentially different views there. When it comes to the two envelopes problem, we already obey the NPOV policy. iNic 00:37, 30 January 2007 (UTC)
What I call "the problem" or "the original problem" is the problem that is described as "the puzzle" at the beginning of the article: "Let's say you are given two indistinguishable envelopes, each of which contains a positive sum of money. One envelope contains twice as much as the other. You may pick one envelope and keep whatever amount it contains. You pick one envelope at random but before you open it you're offered the possibility to take the other envelope instead." If you state that the only interesting question is to find the error in the following reasoning, this is your personal point of view. But obviously there is an interest on if and how the expected value can be calculated correctly, which is documented in several papers. Furthermore, the main question is often seen as "Should I swap the envelopes or should I not, and under which conditions (envelope opened, know distribution,...)?" So if you try to suppress this documented points of view, this is a violation of the WP:NPOV policy. --M.T. 10:36, 31 January 2007 (UTC)
I've clarified the introduction even more now so it shouldn't be any doubt what this paradox is about. But if you feel that it would be interesting to start a new wikipedia article about the decision problem per se, please do so. iNic 19:52, 31 January 2007 (UTC)
Priest and Restall describe three mechanisms by which money could be assigned to the envelopes. For the original problem (amendment: given in this article) only the first mechanism describes the problem correctly. Actually, already point 6 of the reasoning given in the article is wrong in this case ("6. Thus, the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2"). Priest and Restall write: "The paradox is standardly formulated by employing reasoning of FORM 2 in a context where a mechanism of FORM 1 is deployed, which of course gives counter-intuitive results." I think that this point is not made clear enough. --M.T. 13:55, 31 December 2006 (UTC)
The purpose of the Priest and Restall paper is to show the dangers when using the principle of indifference, and why it often leads to paradoxes. The use of the principle of indifference easily produces results that can't all be right at the same time. This is old news in itself (Bertrand's paradoxes). What is new is that they argue that the two envelopes problem is caused by the (mis)use of the principle of indifference. In other words; the two envelopes problem is just yet another example of a Bertrand paradox, according to Priest and Restall. iNic 03:59, 3 January 2007 (UTC)
Historically, the reactions to Bertrand's paradoxes was to abandon the principle of indifference from probability reasoning altogether. However, it survived to this day in elementary books on probability for children, as it is still considered the simplest approach to probability. Today, the only ones still seriously defending (a refined form of) the principle of indifference are the Bayesians. This is why the two envelopes problem is of interest to Bayesians first and foremost. iNic 03:59, 3 January 2007 (UTC)
If we claim in the article that there is one and only one correct calculation of expected values for the closed envelopes, we will get into trouble further down in the article. In the "even harder problem" for example, some Bayesians in fact claim that it's better to always switch envelopes, even if you don't look in the envelopes! They will claim that your expected value calculation above is correct only for prior distributions of a certain kind. So even if we can find some authors supporting your claim, it will still only be a POV. And considering the fact that this is a sidetrack, i.e., not what this problem is really about, I think we are inviting more trouble than we solve by adding this POV calculation. iNic 03:59, 3 January 2007 (UTC)

Use of 'we' and 'you'

Hi,

Congratulations on creating an informative article. One grips is that the Manual of Style says articles should where possible not address the reader directly using 'we' or 'you'. I think the tone of this article should reflect that using more passive statements and so on. Thanks Andeggs 22:19, 22 December 2006 (UTC)

Personally, I like this direct style and don't think it becomes too unencyclopedic. The disadvantage of using passive statements everywhere is that the text can seem too academic and abstract for many readers. In addition, the common way to express this paradox in the (academic) literature is in this direct way. I don't think wikipedia should try to give a more academic impression of a topic than the academic articles that it's the summary of. iNic 00:50, 23 December 2006 (UTC)
Actually having had a look at the article again and at the manuals of style I think its the use of 'you' that is more of a problem than the use of 'we'. As the main Manual of Style says: "Avoid second-person pronouns". I'm sure this is possible with a bit of rewording but I don't want to do it and then be reverted immediately. I think the point about the academic literature is a little bit weak because we are writing an encyclopedia but they are writing maths/philosophy books and articles - and these are necessarily different. Thanks Andeggs 12:03, 23 December 2006 (UTC)
Please go ahead and do the changes you find appropriate. English isn't my native language so at least I won't revert any language changes. Regarding the academic style I was thinking about another writer at the talk pages here requesting a summary of this article for non-specialists that he could understand. Even the guideline you refer to include skepticism about its own applicability: While opinions vary on how far this guideline should be taken in mathematics articles — an encyclopedic tone can make advanced mathematical topics more difficult to learn — authors should try to strike a balance between simply presenting facts and formulas, and relying too much on directing the reader or using clichés [...] iNic 12:37, 23 December 2006 (UTC)
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