Talk:Two envelopes problem/Archive 9

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.

I have made some corrections and expansions to the first resolution so that I hope it now makes sense to philosophers, to mathematicians, and to amateurs. There is no "own research" here, everything can be found in Schwitzgebel and Dever, Falk, and all others who assume that the writer is trying to calculate the unconditional expectation E(B) by splitting over the two equally likely cases A < B and A > B. The only randomness which needs to be involved is the fair coin toss (whether A contains the smaller of larger amount). The amounts in the two envelopes x > 0 and y = 2 x are fixed (known or unknown, makes no difference). Richard Gill (talk) 17:40, 16 November 2014 (UTC)

I added near the beginning of the mathematical stuff a simple mathematical resolution on the lines of the example that we have been discussing here in recent days. I performed some simple arithmetic in order to build this example. I also used Bayes' rule, following numerous publications in the mathematical literature on TEP. I will look for some suitable references, later. I hope it makes sense to philosophers, mathematicians, and amateurs. Richard Gill (talk) 10:05, 17 November 2014 (UTC)

I added the coin toss variation in the Logical resolution section. I think that it is an important variant of the problem which makes the switching argument correct. Caramella1 (talk) 21:06, 16 November 2014 (UTC)

I agree, and this also helps people who don't know anything about expectation value or probability to deepen their understanding. Richard Gill (talk) 07:28, 17 November 2014 (UTC)

I made some changes to improve the English and the logic. Richard Gill (talk) 17:36, 17 November 2014 (UTC)

Now that there are simple versions of both the philosophers' and the mathematicians' solutions, how about now pruning a lot of the accretions which the article got over the years? Seems to me that the two "collapsed" mathematical explanations can be removed altogether. If there is anything important in there it should be put back "in the open". After all, there is a big *mathematical* literature on TEP and part of our wikipedia article should be written with a similar level of mathematical sophistication as many other articles on more or less elementary mathematical topics. Richard Gill (talk) 07:33, 20 November 2014 (UTC)

Please remove. iNic (talk) 22:16, 20 November 2014 (UTC)

Does anyone have a copy of Laurence McGilvery, 'Speaking of Paradoxes . . .' or Are We?, Journal of Recreational Mathematics 19: 15-19? Richard Gill (talk) 11:42, 21 November 2014 (UTC)

Yes but only on paper somewhere. When I find it I can scan it and put in the box. iNic (talk) 12:24, 21 November 2014 (UTC

Yes please! But I think you are no longer sharing *my* TEP dropbox. I will re-invite you. Richard Gill (talk) 07:39, 22 November 2014 (UTC)

I am going to add a reference to the Tsikogiannopoulos paper (English translation: http://arxiv.org/pdf/1411.2823.pdf) in the main article with a short explanation of what is wrong with the switching argument according to him. He gives a reasoning without the need of prior beliefs or distributions. Anybody objects to that? Caramella1 (talk) 06:12, 15 November 2014 (UTC)

We already voted about this and everyone voted against. iNic (talk) 22:14, 20 November 2014 (UTC)

It's not about voting, according to WP rules. It's about consensus and the arguments each editor can bring to the subject. Your argument was that the source was not reliable or it didn’t exist at all. Now with the help of Richard it has been shown that the source does exist, is reliable and even Tsikogiannopoulos uploaded an English translation on arxiv, though he had not to, so everyone can read it. Caramella1 (talk) 06:36, 21 November 2014 (UTC)

So I think now the question is whether or not the reference is useful. As far as I can see Tsikogiannopoulos does not add anything new to the present "first resolution". He solves a number of new variant problems. He does not know a great deal of the literature. Nickerson and Falk (2006) already have a big paper with a large collection of variants, all systematically analysed. And as I said below, the new material which Caramella1 has introduced is, in my opinion, just a duplication of what we now already have in the first resolution, and anyway, that is where it would have belonged, if it is considered by editors here to be illuminating. Richard Gill (talk) 07:20, 21 November 2014 (UTC)

What is the "first resolution"? The formula you added in the Logical resolutions? Caramella1 (talk) 07:39, 21 November 2014 (UTC)

There is a section called "logical resolutions". It is the first resolution discussed in the article. The core of the first resolution is that, if we assume that the writer wanted to compute E(B) by use of the formula E(B) = E(B | B < A) P(B < A) + E(B | B > A) P(B > A) then he got the probabilities right but screwed up with the two conditional expectations. E(B) = E(B | B < A) = E(A / 2 | B < A) = E(A | B < A)/2 != A / 2 and similarly for the other. That's what went wrong. The rest (all kinds of calculations connecting various things together) is just pedagogical. The *solution* is to note that E(A | B < A) != A. Richard Gill (talk) 11:19, 21 November 2014 (UTC)

I see what you mean by "first resolution". Tsikogiannopoulos explanation is *not* a duplication of the "first resolution". It follows a different logical path. He talks about two different events (or games) that do not belong in the same formula and he gives a revised simple formula that corrects this error. Maybe this explanation belongs to the logical resolutions after all, because it is both logical and simple. Caramella1 (talk) 09:17, 22 November 2014 (UTC)

Since the first and the second mathematical resolutions assume prior distributions, I made a new section "Resolutions without prior probability distribution" and put Tsikogiannopoulos resolution in there. Caramella1 (talk) 07:31, 16 November 2014 (UTC)

This solution is already contained in the "logical solution". Tsikogiannopoulos is simply carrying out the computation which I already added there. The smaller of the two amounts is x, fixed, known or unknown doesn't matter. The only randomness is whether A = 2 x, B = x or A = x, B = 2x, both of which have probability half. A correct computation of E(B) by splitting over the two events goes as follows:

${\displaystyle E(B)=E(B\,|\,A<B)P(A<B)+E(B\,|\,A>B)P(A>B)=E(2A\,|\,A<B){1 \over 2}+E({1 \over 2}A\,|\,A>B){1 \over 2}=E(A\,|\,A<B)+{1 \over 4}E(A\,|\,A>B)}$

So it ends with something a bit weird and not particularly useful. But now we can continue with

${\displaystyle E(B)=...=E(A\,|\,A<B)+{1 \over 4}E(A\,|\,A>B)=x+{1 \over 4}2x={3 \over 2}x}$

so (of course) different routes to calculate the same thing all give the same answer. A direct calculation of E(B) and E(A) gives of course the same answer.

You can simply add a reference to Tsikogiannopoulos at that point. Richard Gill (talk) 07:41, 20 November 2014 (UTC)

We have "non-Bayesian solutions" based on the interpretation that the writer is computing E(B) or more precisely E(B | X = x) where "E" is *not* averaging over prior beliefs about the smaller amount x but only over the coin toss which decided whether x or 2x went into A. We have "Bayesian solutions" based on the interpretation that the writer is computing E(B | A = a) where "E" is also averaging over prior beliefs about the smaller amount x. Tsikogiannopoulos is following the first interpretation; as he says, he follows Schwitzgebel and Dever, and Falk (2008). The idea that the logical solutions are not mathematical and that Bayesian = mathematical is nonsense. Both kinds of solution can be presented verbally and both kinds can be expressed in mathematical language. Richard Gill (talk) 07:47, 20 November 2014 (UTC)

Moreover, Tsikogannopoulos is totally unaware of the Bayesian interpretations and the prehistory of the two envelopes problem so his paper does not serve the purpose of providing a comprehensive neutral review. Richard Gill (talk) 07:26, 21 November 2014 (UTC)

Richard, the computation you gave in the Logical resolutions section seems correct to me. Is it from the Eckhardt paper? If not, can you add a reference to it?

I think you should connect the first with the third formula because together they show that E(B) = 3x/2. The middle formula is useful of course, but not there because it does not help to show what is wrong with the switching argument. In the switching argument the calculation is based on the amount A in the player's envelope, and not on x and 2x. The easiest method of calculation is not important there; the same result should be shown by any method one could follow.

As for the reference to Tsikogiannopoulos next to the formulas you gave, it wouldn't be correct because he follows a different path to end up at the same result. But I could move the explanation I added to the Logical resolutions to show that there is also another line of reasoning to show the fault in the switching argument. Do you agree with that? Caramella1 (talk) 07:20, 21 November 2014 (UTC)

Caramella1, go ahead, try moving it and integrating it with what is there now. Of course feel free to edit what is there now, too. But I disagree that Tsikogiannopoulos gives another line of reasoning to show the fault. Everyone agrees (within the unconditional, x fixed, interpretation, what is the fault. Different people like to embellish this in different ways by making different supplementary calculations. Tying things together in different ways. Getting the "right answer" by all kinds of different routes. ie for pedagogical purposes. Richard Gill (talk) 07:26, 21 November 2014 (UTC)

I think that the first and the third formula you added should go to the "Resolutions without prior probability distribution" section I made. This is because this reasoning does not presuppose prior probability distribution and is quite technical to be considered as "Logical resolution". But it should be referenced to a published source, so please place a reference. The middle formula may belong to the logical resolutions as an alternative way to think of the problem without trying to confute the switching argument. Caramella1 (talk) 08:40, 21 November 2014 (UTC)

The section should not be called "logical resolution". It is the section devoted to the resolution which is favoured by the philosophy literature, but also turns up in the maths, statistics, and pedagogical literature, and in the amateur literature. The resolution consists (from the point of view of a mathematician) of remarking that E(A | B < A) != A. Write that out in words and then you can call it a "logical" resolution and then the philosophers and many amateurs will understand it. This resolution is also discussed in the mathematical literature and also people who are not scared of formulas will want to see the resolution expressed in mathematical language. As Tsikogiannopoulos says, this resolution was given also by Falk (2008) and Schtwiztgebel and Dever. He does nothing new. He merely expresses these solutions in his own favourite words and calculations. Richard Gill (talk) 11:27, 21 November 2014 (UTC)

Tsikogiannopoulos paper does not refer to Falk. He does refer to S&D but only for the X, 2X explanation which is very common to various papers. The new idea that he provides is a very simple formula that gives the expected return when the player does his/her calculation based on the amount contained in his/her envelope, known or not. It can be easily re-written to give the expected value. Your formula does the same but you follow a different path with conditional expectations and a more technical way that is difficult to be followed by non-mathematicians. Caramella1 (talk) 09:05, 22 November 2014 (UTC)

This is what happens when a non-mathematician (iNic) hides the mathematical interpretation of his favourite (logical) solution because he doesn't understand it and thinks nobody will want to read it. The philosophers and mathematicians have to both understand one another's points of view and languages, otherwise the article will always be a mess ... Richard Gill (talk) 07:59, 20 November 2014 (UTC)

Well this is totally false, Richard. No solution presented in the article is my favourite solution so I'm not promoting any solution here, as you other guys do. I think we should keep mathematical formalism to a minimum simply because most ideas can be explained accurately without any mathematical formalism. Some can't and there we need to introduce some math of course. But this problem/paradox is a problem within philosophy and not mathematics so there is no need for a lot of math either. iNic (talk) 11:21, 20 November 2014 (UTC)

You have now for example written that "Yet in the calculation which leads to the paradoxical result that Envelope B contains on average more than in Envelope A..." But the expectation here is not the same as "on average." So this shows clearly that you are not understanding this interpretation of the problem at all. My suggestion was that Gerhard should write the text in this section because he understands it. This article will always be a mess as long as editors not understanding a type of solution nevertheless insist to edit that section. iNic (talk) 11:48, 20 November 2014 (UTC)

My apologies if I misinterpreted your (?) motives for hiding all the mathematical explanations of the two solutions, which actually had the purpose of bringing them together, not pushing them apart.

Huh, the expectation is not the same as the average? Mathematical expectation (as opposed to moral expectation) and average are synonyms. Conditional expectation is average within a subset. "Mean" is another synonym. This suggest to me, iNic, that you do not like mathematical language and hence do not understand solutions presented by mathematically fluent writers :-) Richard Gill (talk) 12:52, 20 November 2014 (UTC)

No, expectation and average is not the same thing. Expectation can be used as a guide for action in single cases, which the two envelope situation exemplifies. Pascal—one of the founders of the subject and inventor of the concept of expectation itself—used the expectation to argue for why one ought to believe in a god, for example. It doesn't make any sense to talk about an average in this case. Hope you agree. This is the original use of expectation and the interpretation as an average came much later. Decision theory and Bayesian philosophy still use expectation in this sense. It is absolutely crucial to know this for understanding much of the literature for this problem, and I understand now why you have had a hard time to grasp many of the papers. Do you understand now why it's not appropriate that you edit these parts of the article? iNic (talk) 21:56, 20 November 2014 (UTC)

I do know the ancient history, iNic. At some point people began to distinguish mathematical expectation from moral expectation. Christiaan Huygens built his whole theory of probability around the notion of (mathematical) expectation. His book De ratiociniis in ludo aleae ("On Reasoning in Games of Chance", 1657) remained the standard textbook on probability for more than 100 years. The literature on TEP, especially the popular and the pedagogical literature, is about mathematical expectation, as you can see in the switching argument itself. Once you have a probability model you can imagine the average in many repetitions. Even if it is a probability model for the existence of God. Obviously, amateurs might still be living in the 18th century. Martin Hogbin wanted to replace "expectation" everywhere by "average". The article now contains links to mathematical expectation. Richard Gill (talk) 23:27, 20 November 2014 (UTC)

This has nothing to do with the distinction between "mathematical expectation" and "moral expectation." That's a different story. This has to do with the probability concept itself, which wasn't even invented at the time of Huygens. The word 'probability' isn't mentioned once in the "standard textbook on probability" as you call it. Laplace was the first one defining 'probability' and making it the central concept of the theory. But it was still very far from the modern concept you have in mind here, involving randomness and repeatable experiments. The old concept has survived to our day in the form of Bayesian probability and the use of expectation within decision theory. The rationale is that we want to have a theory that gives us advice even in once-in-a-lifetime situations where the idea of repeating something doesn't make sense. The existence of a god is a good example. Either a god exists or it doesn't—it's not a repeatable experiment. All we can do is to gather evidence for and against the hypothesis and use decision theory to guide our belief. It doesn't make sense to think about this situation as any kind of repeatable experiment. What should in that case be repeated? If forced to answer this we need to invent an idealised model along the lines 'out of infinitely many worlds one is selected at random...' But as William Feller has noted, "Little imagination is required to construct such a model, but it appears both uninteresting and meaningless." iNic (talk) 03:53, 21 November 2014 (UTC)

I should have said "probability concepts". We have frequentist and subjective, ontological vs epistemological notions of probability. Of course the use of mathematical expectation (as a guide to action) needs to be motivated, in either interpretation of probability, in a way consistent with the probability notion in force. However both viewpoints lead to the same thing. Richard Gill (talk) 07:13, 21 November 2014 (UTC)

Of course not. Different viewpoints does not lead to the same thing. In this case some viewpoints can be used while other viewpoints can't be used at all. How on earth do you think that that would "lead to the same thing"? The current article is what happens when a mathematician illiterate in philosophy start to edit an article within philosophy. iNic (talk) 10:21, 21 November 2014 (UTC)

I would say the reverse. And calling TEP "an article within philosophy" is totally ludicrous. Richard Gill (talk) 11:28, 21 November 2014 (UTC)

I agree, as an article in philosophy the current article is totally ludicrous. Let's put the problem back in the context of philosophy, as it should be. iNic (talk) 12:30, 21 November 2014 (UTC)

Now I agree that we should keep mathematical formalism to a minimum. So please let's get back on topic. Caramella1 has, IMHO, introduced a new subsection in the wrong place which moreover duplicates what I had introduced in the right place. However, what he and I have in common is that we are editors who *do* feel a need also to express in mathematical formulas what can be expressed in words. Because we think that this is a service to many readers, who feel the same need as we do.

BTW I have argued that the conditional interpretation is the "mainstream" interpretation by referring to the literature and the history. I like presenting different solutions belonging to different interpretations. The idea that there is one right solution (and one right interpretation) and all the rest are wrong is useless. TEP is not somebody's copyrighted trademark. It's an organically changing living cultural phenomenon. Richard Gill (talk) 12:52, 20 November 2014 (UTC)

iNic, insulting an editor who does not agree with your opinion on this subject, by calling them 'a mathematician illiterate in philosophy' is uncivil. The history of this problem is that it is a mathematical puzzle. The whole puzzle is based on a proposed calculation, thus some mathematics is an inherrent part of the paradox. Many of the philosophical resolutions are so vague as to be meaningless. The statement 'stands for two different things' is of very little value unless we agree what king of thin A is intended to be. In mathematics there is a clear system of notation that distinguishes different kinds of quantity and tells us exactly how we camn manipulate them. THis is what the mathematical solutions do. There are notations used in some branches of philosophy but I am not aware of any philosophical solutions to this puzzle that use any form of rigorous notation. If you know of any, please give us some references.

Regarding 'expectation' it is used here in the standard mathematical sense of the long run avereage. That was obviously the intention of the original question setters. Etherial meanings of the term are not in any way helpful for resolving the paradox, just as philosophers have never made any progress at all on the existence of God. All the most general ramblings of philosophers mean is, 'we do not really know what the question is, so we cannot answer it'. Actually, I would have no objection to a section putting that viewpoint but let it not masquerade as some kind of logical resolution.Martin Hogbin (talk) 15:42, 21 November 2014 (UTC)

Richard, what you have written is a great improvement but still well beyond the understanding of 95% of our readers. We need some simple words, with minimal mathematical notation, to show exactly how the two resolutions show exactly what goes wrong in the proposed argument for swapping. As it is directly about how to improve the article, I will restart our discussions on the subject on the talk page. Martin Hogbin (talk) 15:34, 21 November 2014 (UTC)

As far as I know, the Ali Baba variant was first introduced by Nalebuff in 1989 in his paper "The Other Person's Envelope is Always Greener". This variation is about two players (Ali and Baba) where one of the amounts is depended by the other one by the flip of a coin, but each player looks in his envelope before deciding if he will switch or not. This is more specific than the simpler coin variation where there is only one player and he doesn't look in his envelope. I think that it is better to name this section "Coin toss variation" and you can place the Ali Baba variant in there if you want. Caramella1 (talk) 07:40, 23 November 2014 (UTC)

Nalebuff also said that the arguments do not require one to look in the envelope. Ali Baba is what this variation is called in the literature. Who has called it the coin toss variation? We could also call the standard TEP a coin toss problem. There are two envelopes one with x and the other with 2x and we name them A and B by tossing a coin, then give A to the player. The expression "coin toss problem" does not distinguish these two families of problems. Richard Gill (talk) 08:01, 23 November 2014 (UTC)

Nickerson and Falk (2006) (2x2x2 = 8 variants of Nalebuff's problem, some of them equivalent to earlier versions) "in *all* the versions considered [the players] know that the amount in E2 is either half or double that of E1, depending on the toss of a fair coin". The point is that in half of Nickerson and Falk's variants, the players don't know if they have E1 or E2. Richard Gill (talk) 08:45, 23 November 2014 (UTC)

I propose we name the two main resolutions: "simple resolution" and "Bayesian resolution". I propose we name Nalebuff's variant the "Nalebuff asymmetric variant". The main part of the article is focussed on the situation where the player does not look in his envelope. I think this corresponds to the main line of the history of TEP. For instance, the two owners of those smart ties did not know how much their wives paid for them. They did not know how much money they had in their pocket, either. Symmetry is a big clue throughout the main family tree (ties, wallets, envelopes). Richard Gill (talk) 10:01, 23 November 2014 (UTC)

I am not sure whether this is the resolution of all philosophers but it is a 'rationalised' explanation by two philosophers that makes mathematical sense. I agree with Richard that this is probably not what the puzzle's originators intended.

Here my proposed resolution again, based mainly on what S&D say, with some changes and notes based on subsequent discussion. We might preface this by mentioning that a general problem with the proposed argument is that it is not at all clear exactly what kind of quantity 'A' is intended to represent.

One solution is to assume that the 'A' in step 7 is intended to be the expected value in envelope A...

[The link shows that 'expected value' is a long run average in this example, which is undoubtedly what S&D meant by the term].

...and that we are intended to produce a formula for the expected value in envelope B.

Step 7 shows states that the expected value in B = A + A/4,

It is pointed out that the 'A' in the first part of the formula is the expected value, given that envelope A contains less than envelope B, but the 'A', in the second part of the formula is the expected value in A, given that envelope A contains less than envelope B. The flaw in the argument is that 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 would be

Expected value in B = Expected value in A (given A is larger than B) + Expected value in A (given A is smaller than B)/4 [ref S&D]

We could follow this by a simple example in which only two fixed values are allowed, say £10 and £20. We might then add a more general case of £x and £2x.

What do you think? Martin Hogbin (talk) 16:41, 21 November 2014 (UTC)

I think this is fine. And I agree that this is S&D's (and many others') explanation. One can go on to say that if the amounts in the two envelopes are x and 2x then the expected value in A given it is less than B is x, and the expected amount in A given it is more than B is 2x. Hence we find x + 2 x / 4 = 1.5 x, which is the same as we would have got by a direct calculation 1/2 x + 1/2 2x. Richard Gill (talk) 07:49, 22 November 2014 (UTC)

I agree with the x, 2x explanation, but I think that the "simpler" £10, £20 explanation is confusing. The £10, £20 example is valid only if the player knows the content of both envelopes but doesn't know which envelope contains which amount. If the player doesn't know the content of either envelope then the £10, £20 example has no point and the x, 2x is appropriate. If the player knows the content of the envelope A (say A = £10) then the £10, £20 example is half the truth. Caramella1 (talk) 08:29, 22 November 2014 (UTC)

I was going to add the first part of my explanation to the article but I cannot find where S&D has been published. Has it been published, if so where? If not, do we have a better source saying the same thing? Martin Hogbin (talk) 11:01, 22 November 2014 (UTC)

I found a reference to it in the Tsikogiannopoulos paper: Schwitzgebel & Dever (2008). The Two Envelope Paradox and Using Variables Within the Expectation Formula. Sorites 20, 135-140. I searched it online and found a web page containing an html version of it. Caramella1 (talk) 11:47, 22 November 2014 (UTC)

To be a good source for WP S&D needs to be published in a repectable publication. Martin Hogbin (talk) 14:06, 22 November 2014 (UTC)

It is. In the Sorites journal. Caramella1 (talk) 14:33, 22 November 2014 (UTC)

It has already been cited four times. More than Ruma Falk (2008) (three citations). http://scholar.google.com/scholar?cites=8756422601188170373&as_sdt=2005&sciodt=0,5&hl=en One of four is Tsikogiannopoulos (arXiv eprint). When I have put my paper on arXiv it will be five times. Richard Gill (talk) 16:34, 22 November 2014 (UTC)

There are no good sources, in the sense of wikipedia. There are only primary sources (sources presenting new contributions). There are no recent and neutral comprehensive reviews (good secondary sources). There are no tertiary sources at all. The primary academic sources are very little cited. The primary popular sources do not give satisfactory resolutions and there are no popular comprehensive reviews. Richard Gill (talk) 07:03, 23 November 2014 (UTC)

Give an x, 2x example first and for those who don't like algebra show what the formulas say, for instance, when x = 10. You can give a numerical example without supposing that the player knows what the numbers are. (This seems a very difficult point for many people to grasp). Neither the (faulty) switching argument, nor ways it can be corrected, change the premises of the puzzle: the *player* doesn't know the amounts. The player is however allowed to imagine amounts, and we are too. Richard Gill (talk) 16:23, 22 November 2014 (UTC)

I have added it now. Martin Hogbin (talk) 18:03, 22 November 2014 (UTC)

I have done a lot of pruning of dead wood, and also moved some live wood which had been put in odd places to better locations. Richard Gill (talk) 07:00, 23 November 2014 (UTC)

But you accidentally deleted the Tsikogiannopoulos explanation. I restored it to the simple resolutions section. Caramella1 (talk) 07:29, 23 November 2014 (UTC)

I deliberately deleted it. It seemed to me superfluous. You have put it back, in the wrong place. In the basic TEP there is no looking in envelopes!!! Richard Gill (talk) 07:35, 23 November 2014 (UTC)

First of all, don't delete other editor's contributions before discussing your intention in the talk page! Secondly, the case where a player makes his calculations based on the amount A is everywhere in the simple resolution section. It doesn't matter if the player actually looks in his envelope or not. Caramella1 (talk) 07:50, 23 November 2014 (UTC)

Indeed it doesn't matter if the player looks in the envelope or not, for this resolution, but sometimes it does matter, and this issue always causes confusion. Lots of probability-theory-challenged writers seem to think that you can only calculate E(B | A = a) if you look in envelope A, but this is nonsense, as Nalebuff already said long ago. So we have to be clear throughout the first part of the article that we are discussing the problem where there is no looking in envelopes. Which does not prohibit anyone from imagining looking, and doing the calculations which they would do if they looked. Thanks to algebra, we can do those calculations! Richard Gill (talk) 08:09, 23 November 2014 (UTC)

I did discuss my intention to prune dead wood and several editors agreed. But I have now put it back where it seems to belong now. I am not sure that it is notable, and I am not sure that it qualifies as a reliable source by wikipedia policy. In fact it is a brand new original research contribution which has not been cited by anyone yet, it is a primary source which should not be used by us. But I am not too bothered by that issue. For me the important point is: is the contribution useful for our readers, yes or no? Is the reference useful? It's a nicely written paper with a systematic study of a lot of new variations, and an original way to do the calculations in the "common resolution", but it misses all of the "Bayesian" literature. So it could be useful for some readers but it gives a very unbalanced view of the whole literature. Richard Gill (talk) 07:58, 23 November 2014 (UTC)

Nickerson and Falk (2006) give a systematic study of 8 variants. Their paper is published in English in an international, academic, peer reviewed journal. It has been around for 8 years now and has been cited 14 times. This is the kind of paper we should be using as reliable source. Richard Gill (talk) 08:32, 23 November 2014 (UTC)

It's amusing that many papers nowadays cite Wikipedia as their source and take the wikipedia problem statement to be "the canonical problem". So wikipedia has come full circle and is itself a reliable source now. Richard Gill (talk)

It is not good for Wikipedia. Martin Hogbin (talk) 18:13, 24 November 2014 (UTC)

I suggest that we consider three cases, I think they cover all posibilities:

The possible sums are bounded

I realise that there is no mathematical basis for this distinction but there a clear logical/practical/common sense case for treating this case separately. Many sources make this distinction and for a prcatical point of view it is the only possibility. Many readers may not want to go further.

I am trying to get to the solution in simple steps, starting with non-standard variations, but which are mentioned in various sources.

1) Start with the case where there are two envelopes containing £10 and £20 and the player knows this.

The player chooses an envelope at random and, in this variation, is allowed to look in his envelope before deciding. Before he looks he concludes that probability of there being double the money in the other envelope is 1/2 and that the probability of there being half the amount is also 1/2

Then: he looks and sees £10. Swap

Or: he looks and sees £20. Don't swap.

This is obvious to everyone

2) Exactly as 1 but the player is not allowed to look inside before deciding.

The player now knows that the although the probability of there being double the money in the other envelope is 1/2 and that the probability of there being half the amount is also 1/2, that probability is not independent of the sum in his envelope. He therefore knows that the proposed calculation is wrong as the expected value in the other envelope depends on what he has in his envelope.

3) Extend the case to many envelopes in which the sum is bounded.

We could give Richard's example here.

Not my example. The example (a sequence of equally likely amounts going up by powers of 2) goes back (at least) to Falk and Konold (1992). Which is a publication which all editors ought to study carefully. At the time comprehensive, and still one of the most authoritative sources. Richard Gill (talk) 17:06, 23 November 2014 (UTC)

We could give that example though, with the sources that you suggest.

what about the explanation above? I am trying to make it very simple for people to understand by taking it in stages.

I am happy with the idea to go in four? or is it five? steps, but you had better get the explanations right! I don't understand your step 4, 5 explanations. Richard Gill (talk) 11:35, 24 November 2014 (UTC)

BTW the example with the powers of 2 is not just an example ... it's part of the *proof* that there is no proper prior distribution on the smaller amount x which makes P(A > B | A = a) = 1/2 for all a. The proof uses Bayes' theorem in the form posterior odds = prior odds x likelihood ratio. If the amount in A were 32, then there are two possibilities for the pair of amounts: (16, 32) and (32, 64). Under either possibility, the chance to see A = 32 is the same, half, so the Bayes factor (aka Likelihood ratio) is 1. Therefore posterior odds = prior odds. So if posterior odds = 1 : 1, prior odds = 1 : 1. So... if P(A > B | A = a) = 1/2 for a = 32 then the prior odds for (16, 32) and (32, 64) are equal. Now repeat the argument with a = 64 ... and then with a = 128 .... and also with a = 16 ... and then with a = 8 ... That's why this example is so important. The example is at the same time the easy proof that P(A > B | A = a) = 1/2 for all a cannot be true, if we start with a proper prior distribution on x. It's a proof by contradiction. Suppose there is such a proper prior. Then Bayes rule is applicable. Therefore ... Richard Gill (talk) 13:49, 24 November 2014 (UTC)

So when Chalmers says, 'In particular, the distribution g(x) = x^(-1.5), cut off below a lower bound L and normalized, allows the paradox to arise. The distribution has a finite integral, and even though for most n, p(B>A|A=n) < 0.5, it is still the case that for all relevant n, E(B|A=n) > n', this is this new paradox, that you refer to below, in which P(A > B | A = a) = 1/2 for all a is not true but E(B|A=n) > n is true, this latter case being resolved by infinite expectations. Martin Hogbin (talk) 22:34, 26 November 2014 (UTC)

Richard, I have had some more thoughts on this subject, which might also be relevant to your paper. Martin Hogbin (talk) 13:51, 27 November 2014 (UTC)

Unbounded and P(B=2A) is 1/2 for all A

Here we just say that it has been shown that in all such cases the expectation on both envelopes is infinite. [ref Chalmers]

I think you mean, P(B = 2A | A = a) = 1/2 for all a. This case is that of the possible values of x being powers of two, extended on both sides to infinity, all values equally likely. There is no such probability distribution, expectation values are not defined. You could just as well say it is 0 as infty. Chalmers is a rather poor reference. Falk and Konold already said this in a much more precise and correct way. Richard Gill (talk) 17:06, 23 November 2014 (UTC)

Unbounded and P(B=2A) is not 1/2 for all A

The explanation here is similar to the bounded case, where it is obvious that P(B=2A) is not 1/2 for all A is not true

You mean P(B=2A | A = a) is not 1/2 for all a? It's a theorem that this is impossible. The "powers of 2" example is the proof. (A proof by contradiction, so ordinary folk won't understand it, but still it is a fantastic and powerful way to prove theorems). So what do you want to say? Richard Gill (talk) 18:14, 23 November 2014 (UTC)

Or are you here talking about the new variants where we have E(B | A = a) > a for all a? See Chalmers' second (and properly published and more interesting) paper http://consc.net/papers/stpete.html There are proper probability distributions over x (the smaller of the two amounts) such that E(B | A = a) > a for all a. Obviously, they cannot satisfy P(B=2A | A = a) = 1/2 for all a, we already knew that is impossible. So we are actually talking now about a new paradox. We have left the old reasoning and instead we have correctly calculated E(B | A = a) for all possible a. And found that it always exceeds a. It's also possible at the same time that P(B > A | A = a) > 1/2 for all a, so if you knew what was in your envelope, you would have several good reasons to switch! Yet clearly, if you don't know, there is no good reason to switch. A short resolution which satisfies mathematicians is that if E(B | A = a) > a for all a, or if P(B > A | A = a) > 1/2 for all a, then E(A) = E(B) = infty. The expectation value is then not a good guide to once-off decisions. Even if you repeat the experiment a finite number of times your average gain won't look anything like the expected gain. We are now hard up against the Saint Petersburg paradox. See David Chalmer's later publication http://consc.net/papers/stpete.html Richard Gill (talk) 17:06, 23 November 2014 (UTC)

So is your third case the new paradox (with proper distributions), or the original paradox (with the powers of 2). The first two cases belong together. Your explanations for the last two cases are wrong, unless by the third case you mean the infinitely many equally likely powers of 2 improper distribution. It is all a bit more subtle. Interesting mathematics! (almost "rocket science"). And in the third case (with the new examples) we head into philosophy and economics. The Saint Petersburg paradox has been around for quite a while and isn't going to go away soon. (Though for a mathematician it might be considered to be adequately resolved long ago). But nor for philosophers. For the latest contribution see the rather interesting http://philpapers.org/archive/ERGTEO.pdf Richard Gill (talk) 17:12, 23 November 2014 (UTC)

PS I don't see the need for major revision to the article once one is past the simple resolution and the simplest discussion of the Bayesian resolution. Richard Gill (talk) 17:14, 23 November 2014 (UTC)

I hope you can see what I am trying to do, even if I am not doing it properly. I want to give a simple resolution of the basic problem (real world, real money) that most people will be able to understand. Then we can explain how the paradox is resolved in other (which you might find more interesting) cases. Martin Hogbin (talk) 18:18, 24 November 2014 (UTC)

I do, I agree! Keep at it. Richard Gill (talk) 12:53, 25 November 2014 (UTC)

You have three Cases and you split the first Case you split into three Steps. I think the three cases should not be separated. Your third step of the first case leads nauirally to the powers of two example which is your second case. In other words, the step from bounded to unbounded is a tiny step and you were nearly there. I don't understand your third case at all. I think it doesn't belong here. Richard Gill (talk) 07:33, 26 November 2014 (UTC)

can I just check with you. If the sums in the envelopes are bounded then it is not possible that E(B | A = a) > a for all a.

If the sums are bounded then it is not possible that E(B | A = a) > a for all a. Proof: choose a equal to the largest possible amount in either envelope. Richard Gill (talk) 08:30, 4 January 2015 (UTC)

If the sums are bounded then it is not possible that Prob(B > A | A = a) = 1/2 for all a. Proof: choose a larger than half the largest possible amount in either envelope. Richard Gill (talk) 08:33, 4 January 2015 (UTC)

Under "Bayesian Resolutions : Proposed Resolutions", the article states: "... Indeed, that conditional expected value is larger than what's in A. But because the unconditional expected amount in A is infinite, this does not provide a reason to switch, because it does not guarantee that on average you'll be better off after switching. One only has this mathematical guarantee in the situation that the unconditional expectation value of what's in A is finite. But then the reason for switching without looking in the envelope, E(B|A=a)>a for all a, simply cannot arise." As I understand this section, it means that unconditionally (i.e., without looking at A = a -- or equivalently, without imagining looking at A = a), we know that E(B) = E(A) = infinity, so there's no reason to switch envelopes. However, conditional on any particular A = a that we observe (or imagine observing), we find E(B|A=a) is strictly greater than A = a.

Isn't that still paradoxical? In other words, isn't it paradoxical that for all possible conditional states of the world (A = a) we prefer to switch envelopes, but unconditionally we are indifferent to switching? 217.44.123.189 (talk) 12:10, 12 February 2015 (UTC)

I think it like many things. Dividing by zero for example. 7>3 and 7/x > 3/x unless x=0. Should one prefer £ 7/0 to £ 3/0 ? Martin Hogbin (talk) 13:57, 12 February 2015 (UTC)

I agree that, by itself, preferring 7/X to 3/X for X>0, and being indifferent when X=0, isn't paradoxical. But if X is a random variable with E(X) = infinity, then there does seem to be a problem. For example, let X = 1/U, where U is uniform on the open interval (0,1). Then, conditional on X=x, we always prefer 7/X to 3/X. However, unconditionally, we're indifferent because E(7/X) = E(3/X) = infinity. It seems the problem comes from using expected values as a basis for comparison. 217.44.123.189 (talk) 15:58, 12 February 2015 (UTC)

I am sure that Richard Gill, if he is watching will have a reply to your comment.

I sort of agree that the problem comes from using expected values as the basis of comparison but what else can we do. We do not know what is in the unchosed envelope so we can only use an expected value

We don't have to use the expected value. Is the utility of money linear? Anyway, what about "regret"? The costs of winning / losing bets are not just the amount of money. There is also "face". The paradox shows that unquestioned use of expectation value of net dollar gain can be stupid. 62.163.25.81 (talk) 09:11, 22 February 2015 (UTC) Richard Gill (talk) 05:39, 26 February 2015 (UTC)

I think the problem comes from trying to use natural everyday logic in cases that have infinities in them; it is well known that you cannot do that. There are many cases im mathematics where that applies but the TEP contrives to get people to try to apply natural logic in such a case. For example, take the series of all positive integers and the series of all even positive integers. Every term in the second series is twice the corresponding term in the first but the average value in both series is the same. Mathemetics is full of such things and they cause no problems to mathematicians but ordinary everyday logic fails.

I know Richard does not agree with me but I think that we should consider the bound and unbound cases differently. In the bound case the solution is simple. The unbound case is physically imposssible in that we, almost certainly, would not be able to get the money into the envelopes or even represent the sum in any way because there is not enough ink in the universe to do so.

In my opinion the unbounded TEP is an attempt to present a division-by-zero trick as a real-world problem. Martin Hogbin (talk) 17:48, 12 February 2015 (UTC)

Your point about applying "everyday logic" to problems involving infinities is well taken. I think the last paragraph of "Bayesian Resolutions : Proposed Resolutions" addresses this issue to some extent (although I don't think the argument need be limited to "economists").

My original observation (above) is that the new paradox doesn't appear to be resolved mathematically in the section "Bayesian Resolutions : Proposed Resolutions". I think that's because a complete resolution requires disallowing the use of expected values to make comparisons in such cases. This point is actually made in the subsequent section -- "Bayesian Resolutions : Foundations of Mathematical Economics" -- but I think it should appear earlier; or perhaps the two sections could be merged.

By the way, I want to compliment you and Richard Gill for your extensive analyses of this problem. I now have a proper user name, and hope to become more active in Wikipedia. ParrotOx (talk) 11:30, 13 February 2015 (UTC)

Thanks. Richard is the expert. I try to make it understandable to the average reader, which he is not so good at.

We both dislike the problem because there is no paradox until you present a bogus argument for there being one. Even then the solver has to complete the bogus argument himself.

Why not make the changes you suggest yourself? Martin Hogbin (talk) 18:43, 13 February 2015 (UTC)

I'll give it a try (after doing some research). ParrotOx (talk) 12:52, 15 February 2015 (UTC)

Good luck ParrotOx! NB we are not allowed to present *own research*. Only to report reliable sources. [As a professional in this field, I do do my own research, but I mustn't push it on wikipedia]. 62.163.25.81 (talk) 09:13, 22 February 2015 (UTC) Richard Gill (talk) 05:39, 26 February 2015 (UTC)

Understood. By "research", I meant relevant reading. ParrotOx (talk) 11:00, 26 February 2015 (UTC)

I made the suggested changes, combining the "Proposed resolutions" and "Foundations of mathematical economics" sections, and clarifying the need to replace expected value as decision criterion. ParrotOx (talk) 06:12, 7 March 2015 (UTC)

I like it! Richard Gill (talk) 19:25, 7 March 2015 (UTC)

"The Two-envelope Paradox: Asymmetrical Cases", by Chunghyoung Lee, abstract at http://mind.oxfordjournals.org/content/early/2013/05/22/mind.fzt023.abstract is a 2013 publication in Mind (journal), published by Oxford Press, that adds to the literature. (And so in this Wikipedia article I revised the wording that was presenting Bliss (2012) as something like "perhaps the most recent publication".) Maybe there are more publications since 2012, also. --doncram 13:45, 26 April 2015 (UTC)

Thanks, interesting. Behind a paywall. I have a copy from my University Library. Richard Gill (talk) 06:15, 1 June 2015 (UTC)

The total amount in both envelopes is a constant c=3x, with x in one envelope and 2x in the other. If you select the envelope with x first you gain the amount x by swapping. If you select the envelope with 2x first you lose the amount x by swapping. So you gain on average G=1/2(x)+1/2(-x)=1/2(x-x)=0 by swapping. We have to add this resolution to the article. --TotalClearance (talk) 14:38, 9 April 2016 (UTC)