Talk:Two envelopes problem/Arguments/Archive 2

Two envelopes without probability: summary of solutions
There are quite a number of competing "solutions" to Smullyan's paradox by logicians/philosophers, generally using ideas from the theory of counterfactual reasoning, and all of them very technical. It is very difficult for a non-expert to give succinct summaries.

The situation: you pick one of two closed envelopes, one of which contains twice the amount of money in it as the other, and both amounts are greater than zero. Smullyan claimed that you can say both:


 * 1) By switching, I gain or lose the difference in the amounts in the two envelopes, thus whether I gain or lose, I gain or lose the same amount.
 * 2) By switching, I either gain an amount equal to what's in my envelope or I lose half of what's in it; these amounts are different.

But how can the amount I gain if I gain both be equal, and be not equal, to the amount I lose if I lose?

Actually, to my (scientist's) mind the paradox merely exposes the inadequacy of common language: write out a mathematical description of the situation and the paradox vanishes. Moreover I think that Smullyan's *problem* is not actually probability-free (though his *analysis* - the arguments by which he arrives at these two statements - is probability free). In the beginning of the problem we are told that we pick one of two closed envelopes. The symmetry of this situation and the arbitrariness of our choice is an ingredient which many would formalize using subjectivist probability. This ingredient tells us that we are perfectly indifferent to swapping our envelope (closed) for the other. Neither of Smullyan's two arguments bring this crypto-probability ingredient into play. Their apparent contradiction shows, to my mind, the meaninglessness of the comparison between what you would win if you would win with what you would lose if you would lose. A statement which is meaningless has a meaningless negation and no contradiction results by arriving at both. Is it worth while to pick apart the argument and say where exactly it goes wrong?

Well that is exactly what is done in the philosophy/logic literature on the problem. I recommend especially Chase (1992) and Yi (1999). They disagree (strongly!) on technical details but both agree that there is no logical argument leading from Smullyan's assumptions, without the probability ingredient, to his two conclusions. Yi "fills in" the chain of formal logical steps which must be behind Smullyan's derivations and shows that both are false. Chase adds the probability ingredient and uses it to patch one of the two derivations in order to get to one of the conclusions - the one which is compatible with our indifference to switching. Richard Gill (talk) 14:22, 28 June 2011 (UTC)

Schwitzgebel and Dever
I have been (trying to) read the Schtwitzgebel and Dever paper carefully. It contains nonsense. It is full of logical non-sequiturs and mathematical errors. It does not explain the two (closed) envelopes paradox.

The main point of the paper is that if

E(Y|Z=z)=a(z) E(X)+b(z) and E(X|Z=z)=E(X), both for all z, then E(Y)=E(A).E(X)+E(B)

Here, A=A(Z) and B=B(Z).

OK, it is easy to check that this is a true theorem. The authors give some "exchange"-type examples where this could be thought to be useful, in the sense that a faulty intuitive reasoning also gives that answer, and the conditions of the theorem are satisfied, so the answer is correct despite the wrong intuitive reasoning.

(Non-)Application to the two envelope problem: in the two envelope problem, we take
 * Y=amount in second envelope (we want to relate E(Y) to E(X))
 * X=amount in first envelope
 * Z=amount in first envelope (the condition)

So *if* the expected value of the amount in the first envelope, given the amount in the first envelope, were constant,* then* it would be the case that: if the expectation of the amount in the second envelope given the amount in the first envelope is 5/4 times the amount in the second, then the expectation of the amount in the first is 5/4 times the expectation of the amount in the second. But the very first "if" is not true, so we aren't obliged to draw the "obviously wrong" conclusion that E(Y)=5/4 E(X).

So first of all, they claim that they have shown exactly where the original deduction that you must switch goes wrong. However just because the assumptions of their "new" theorem are not satisfied in this case, does not mean that the assumptions of someone else's theorem are not satisfied! Their new theorem does not explain at all why the argument leading to "keep on switching indefinitely" fails. Their new theorem simply does not apply to TEP so doesn't explain anything about TEP. They didn't show that the conditions of their theorem are not only sufficient but also necessary. They don't know the difference between A implies B and A is equivalent to B. Amazing. But of course, they are philosophers, not logicians, nor mathematicians.

However there is a simple way to see why the original TEP argument is actually in some sense OK.

Let Y be the amount in the second envelope and X be the amount in the first envelope. By symmetry we know that E(Y)=E(X). The argument purports however to show that E(Y)=(5/4) E(X) and hence you must switch. But there is no contradiction between E(Y)=E(X) and E(Y)=(5/4) E(X) when X and Y are two nonnegative random variables. There are in fact *two* solutions: E(X)=E(Y)=0 and E(X)=E(Y)=infinity. So the conclusion of that logical argument is not "you must switch" but "you may switch". And of course, you may switch back and forth (without opening envelopes) as many times as you like. As a friend of mine said today: TEP (closed envelopes) just shows you that if E(X) is infinite, and I give you X, you'll always be dissappointed.

In words: the assumption used at step 6 does correctly imply that the expectation of the amount in the second envelope is (5/4) times the expectation of the amount in the first. Since we know that both amounts are larger than zero, it follows that both expectations must be infinite. So there is no paradox. You may switch. You may switch as often as you like. It doesn't actually make any difference. When you look in your envelope you'll get less than the expected value, so you're in for a dissapoinment anyway.

I will write to Schwitzgebel and Dever pointing out the logical mistakes and the mathematical errors in their paper. I hope they will be prepared to write a correction note. Richard Gill (talk) 15:14, 12 July 2011 (UTC)


 * They are trying to solve the philosophical problem in the closed envelope case, not any specific mathematical problem. If you don't understand the philosophical problem but see everything as mathematics this paper is very hard to read indeed. iNic (talk) 07:46, 13 July 2011 (UTC)

They are trying to solve a philosophical problem by doing mathematics. And they fail: the same criticism which they give to other earlier such solutions can be leveled at theirs.

The paper is hard to read because it is badly written and contains a lot of nonsense. Are you a philosopher? It has been cited three times by other authors in order to criticise their solution. I do understand what you seem to want to call a philosophical problem, though I would say that it is a problem of logic, or of semantics.

IN fact, most people should find TEP uninteresting! It is a conundrum about logic. It is only interesting if you are interested in logic and possibly also in semantics and mathematics. There is no vast popular literature on it. There is only a vast technical literature. And interestingly, there is almost no *secondary* or *tertiary* literature on TEP. It is almost all purely research articles, each one promoting the author's more or less original point of view, and each one criticising earlier "solutions". And so it goes on. The three papers which cite those two young US philosophy PhD's do so in order to criticize their solution, and to propose an alternative. According to wikipedia guidelines on reliable sources, the article on TEP should be very very brief and just reproduce the comments in a couple of standard (undergraduate) textbooks on TEP. EG David Cox's remarks in his book on inference. I have no idea if there is a standard philosophy undergraduate text which mentions TEP. iNic hasn't mentioned one. We should take a look at other encyclopedia articles on TEP. I think I will write one for StatProb, and then wikipedia editors can use it. Survey papers are admissable secondary sources for wikipedia provided they do not promote the author's own research. They are a primary source for the latter. Ordinary people won't be perplexed by TEP. They know by symmetry that switching is OK but a waste of time (if you don't open your envelope). They don't really understand probability calculations anyway, so they know there is something wrong with the argument, but don't care what. Regarding Smullyan's version, they also know the answer (it doesn't matter whether you switch or not) so they know which argument of Smullyan's is correct. As writer after writer has stated, the problem of both original TEP and of TEP without probability is using the same symbol (original) or the same words (Smullyan) to denote two different things. It's a stupid problem and has a simple resolution. Well, and if we are allowed to look in our envelope, then everything is different. But no longer very interesting for laypersons. It turns out to be a fact of life that there are "theoretical" situations (but I can simulate them on a computer for you, they are not that theoretical!) where according to conditional expectation value you should switch, whatever. OK, and this is just a fact about a random variable with infinite expectation value: if I give you X you'll always be disappointed compared to its expectation. But to get its expectation I would have to give you the average of millions and millions of copies of X. Eventually the average will be so large that you'll prefer the swap whatever the value of the X I gave you first. Then there are all kinds of nice calculations about whether or not you should switch given a prior distribution of X and there are cute things about what to do if you don't want to trust a prior ... then you should randomize. It's all rather technical, isn't it. Only interesting for specialists. By the way this is *not* theoretical since I can approximate a distribution with infinite expectation with one with very very very large finite expectation. I can create approximately the same paradox without actually using infinite values. Syverson does that. It's very technical and hardly interesting for ordinary folk. Richard Gill (talk) 16:30, 13 July 2011 (UTC)

Open letter to Schwitzgebel and Dever
I have written to Eric Schwitzgebel and Josh Dever, see (posted on my talk page). If they permit it, I will post their reply on my talk page, too. Richard Gill (talk) 18:44, 17 July 2011 (UTC)

Schwitzgebel and Dever's simple solution
The Anna_Karenina_principle: "Happy families are all alike; every unhappy family is unhappy in its own way." Here it's the same: a logically sound reasoning can be logically sound in only one way, the reasoning which leads to a "paradox" can be wrong in many."

SD not only published a paper in the philosophy literature, but also have a little web page with a simple solution to TEP,. Let me summarize their simple solution here. It's quite good. First let me repeat the original TEP paradox, I'll call it TEP-I:

TEP-I

 * 1) I denote by A the amount in my selected envelope.
 * 2) The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2.
 * 3) The other envelope may contain either 2A or A/2.
 * 4) If A is the smaller amount the other envelope contains 2A.
 * 5) If A is the larger amount the other envelope contains A/2.
 * 6) Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2.
 * 7) So the expected value of the money in the other envelope is (1/2) 2A + (1/2)(A/2) = 5A/4.
 * 8) This is greater than A, so I gain on average by swapping.
 * 9) After the switch, I can denote that content by B and reason in exactly the same manner as above.
 * 10) I will conclude that the most rational thing to do is to swap back again.
 * 11) To be rational, I will thus end up swapping envelopes indefinitely.
 * 12) As it seems more rational to open just any envelope than to swap indefinitely, we have a contradiction.

For a mathematician it helps to introduce some more notation. I'll refer to the envelopes as A and B, and the amounts in them as A and B. Let me introduce X to stand for the smaller of the two amounts and Y to stand for the larger. I think of all four as being random variables; but this includes the situation that we think of X and Y as being two fixed though unknown amounts of money. It is given that Y=2X>0 and that (A,B)=(X,Y) or (Y,X). The assumption that the envelopes are indistinguishable and closed at the outset translates into the probability theory as the assumption that the event {A=X} has probability 1/2, whatever the amount X; in other words, the random variable X and the event {A=X} are independent.

Schwitzgebel and Dever seem to assume that that steps 6 and 7 together are intended to form a computation of E(B).

In that case, the mathematical rule about expectation values which is being used in step 7 is
 * E(B)=P(A=X)E(B|A=X)+P(B=X)E(B|B=X).

This means that we can get the average value of B by averaging over the two complementary situations that A is the larger of A and B and that it is the smaller, and then weighing those two so-called conditional averages according to the probabilities of the two situations. Now the two situations have equal probability 1/2, as mentioned in step 6, and those probabilities are substituted, correctly, in step 7. However according to the this interpretation, the two conditional expectations are screwed up. A correct computation of E(B|A=X) is the following: conditional on A=X, B is identical to 2X, so we have to compute E(2X|A=X)=2 E(X|A=X). We are told that whether or not envelope A contains the smaller amount X is independent of the amounts X and 2X, so E(X|A=X)=E(X). Similarly we find E(B|B=X)=E(X|B=X)=E(X).

Thus the expected values of the amount of money in envelope B are 2E(X) and E(X) in the two situations that it contains the larger and the smaller amount. The overall average is (1/2)2E(X)+(1/2)E(X)=(3/2)E(X). Similarly this is the expected amount in envelope A.

Schwitzgebel and Dever's executive summary is "what has gone wrong is that the expected amount in the second envelope given it's the larger of the two is larger than the expected amount in the second envelope given it's the smaller of the two''. Indeed. It's rather obviously twice as large!

This is perfectly correct, and very intuitive. But it's not the only thing that goes wrong, in that case. They take no note at all of the fact that the writer finishes with a solution expressed in terms of A, not in terms of E(A). Their explanation of what went wrong is seriously incomplete.

However there is another way to interpret the intention of the writer of steps 6 and 7, and it is also very common in the literature.

Since the answers are expressed in terms of the amount in envelope A, it also seems reasonable to suppose that the writer intended to compute E(B|A). This conditional expectation can be computed just as the ordinary expectation, by averaging over two situations. The mathematical rule which is being used is then
 * E(B|A)=P(A=X|A)E(B|A=X,A)+P(B=X|A)E(B|B=X,A).

In step 7 the writer correctly sustitutes E(B|A=X,A)=E(2X|A=X,A)=E(2A|A=X,A)=2A and similarly E(B|B=X,A)=A. But he also takes P(A=X|A)=1/2 and P(B=X|A)=1/2, that is to say, the writer assumes that the probability that the first envelope is the smaller or the larger doesn't depend on how much is in it. But it obviously could do! For instance if the amount of money is bounded then sometimes one can tell for sure whether A contains the larger or smaller amount from knowing how much is in it.

Note that some of the literature focusses on the first explanation (the expected value of the amount in the other envelope is different when it is the larger, than when it is the smaller) while some of the literature focusses on the second explanation (the chance that the amount in the first envelope is the larger cannot be assumed to be independent of the amount in that envelope). The original writer was simply mixed up between ordinary expectations and conditional expectations. He is computing an expectation by taking the weighted average of the expectations in two different situations. Either he gets the expectations right but the weights wrong, or the weights right but the expectations wrong. Who is to say? Only one thing is clear: the writer is mixed up and he is using the same symbol to denote different things! Depending on what we imagine he is really trying to do, we can give different analyses of what are the things he is giving the same name, which are actually different. Is he confusing random variables and possible values they can take? Or conditional expectations and unconditional expectations? Who's to say?

Anyway, this suggests to me that original TEP does not deserve to be called a paradox (and certainly not an unresolved paradox): it is merely an example of a screwed-up calculation where the writer is not even clear what he is trying to calculate, hence there are different ways to correct his deriviation, depending on what you think he is trying to do. The mathematics being used appears to be elementary probability theory, but whatever the writer is intending to do, he is breaking the standard, elementary rules. Whether we call "explaining" the "paradox" an exercise in logic or in mathematics is unimportant. Whether it is performed by philosophers, mathematicians or logicians is irrelevant too. It seems to me that original TEP belongs to elementary probability theory.

The idea that there should be a unique solution to this paradox is a wrong idea. The paradox is not a paradox, it's a mess. There are a lot of ways to clean up a mess. Richard Gill (talk) 16:51, 18 July 2011 (UTC)


 * The job is not to clean up the mess, that is easy (and can be accomplished much faster than what you have done above). Instead, the job is to pinpoint the erroneous step in the presented reasoning, and to be able to say exactly why that step is not correct and under what conditions it's not correct. And of course when it's correct. We want to be absolutely sure we won't make this mistake again in a more complicated situation where the fact that it's wrong isn't this obvious. In other words, a correct solution should present explicit guidelines on how to avoid this trap once and for all, in any imaginable situation. This is the true two envelope problem, and it is still unsolved. S&D have at least understood this--what TEP is all about--while you together with some of the authors of TEP papers haven't. iNic (talk) 01:26, 19 July 2011 (UTC)


 * The job is to say which if any steps are wrong. There are several wrong steps. They have all been identified in the past. S&D claim no novelty in their results, only novelty in their focus. No mistake could have been made if the author had distinguished random variables from their outcomes, probabilities from conditional probabilities, expectations from conditional expectations. And if he'ld known the standard rules for computing expectations. So there is no danger of making the same mistake in more complex situations. Philosophers do not do actual calculations for hard real problems. Those are left to the professionals, who do their calculations professionally. Philosophy has a big problem as long as it ignores the fact that standard probability calculus and standard notation and standard concepts were introduced precisely in order to avoid such mistakes (Kolmogorov, 1933, solving Hilbert's so-manyth problem, to provide an axiomatic mathematical foundation for probability theory). TEP is not a paradox, it's a problem for the student. A finger-exercise for the beginner in probability theory! Anyway, I'm having a nice correspondence with S&D at the moment as well as with two other philosophers / logicians and with several mathematicians. I think some nice new results are coming out but obviously they are not for Wikipedia for the time being. Richard Gill (talk) 01:45, 19 July 2011 (UTC)


 * OK but I don't get it. If "TEP is not a paradox" and only "a problem for the student" and a "finger-exercise for the beginner in probability theory," why on earth are you, four philosophers and several mathematicians writing on some new results about TEP? You are bringing even more contradictions to the table. iNic (talk) 02:07, 19 July 2011 (UTC)


 * Because it's what I'm paid for, and because it's fun. You should think of TEP, or rather TEP-1, as a kind of joke. After all jokes are built on getting a surprise. Then new people who like that joke and are creative, create new jokes in similar spirit. The whole TEP franchise should be seen as a running gag. Richard Gill (talk) 14:49, 19 July 2011 (UTC)


 * I have never thought of TEP as a joke, but it never surprised me either. What have surprised me, however, are all the crazy and contradicting ideas people have come up with to try to explain what is surprising to them. They are often quite amusing. I look forward to read your forthcoming paper! iNic (talk) 23:59, 19 July 2011 (UTC)


 * And I look forward to hearing your comments and criticisms! I'll be posting something it on my home page soon, and I'll put a mention of it on the TEP talk page. Richard Gill (talk) 01:50, 20 July 2011 (UTC)

TEP-2
Just like a great movie, the success of TEP led to several sequels and to a prequel, so nowadays when we talk about TEP we have to make clear whether we mean the original movie TEP-I or the whole franchise.

The first sequel was based on the discovery that there exist distributions of X such that E(B|A)>A. Necessarily, these distributions have to have E(X) infinite. The resolution of TEP-2 is that you will always disappointed when you get X if you were expecting E(X). In practical terms, you don't expect the expected value at all.

Going back to TEP-I, obviously the expected value of what is in the second envelope is bigger when it's the larger of the two, than when it's the smaller of the two. But at the same time it's the same if all these expected values are infinity. One call this decision theory if one likes, but to me this is just more elementary probability.

Actually there is a kind of "return of TEP-1" side to this part of the story. Maybe the original writer of TEP was a Bayesian and was representing his ignorance about X using the improper prior distribution with density (proportional to) 1/x. Argument behind this: we know nothing at all about the smaller amount x except x>0. Therefore we also know nothing about 1 billion times x, or x divided by pi ... in general, we know nothing about y=cx, for any c positive. So the distribution of X which represents our uncertainty must be unaltered when we transform it to the distribution of cX. There is only one such distribution. Anyway, the remarkable feature about this distribution is that it is now the case that given A=a, B is equally likely equal to a/2 or 2a, so it is indeed the case that E(B|A=a)=5a/4, or E(B|A)=5A/4>A. The expectation value does tell us to exchange envelopes. But this is not a paradox: what is in the other envelope has expectation value infinity. Whether we finally get X or 2X, we'll be disappointed when we compare it to its expectation value.

Improper distributions are a bit tricky but all these things are approximately true if we restrict to the amount x to any range say from some tiny epsilon (positive) to some enormous M, and take the density proportional to 1/x on this range. Our knowledge about x is roughly the same as that about cx provided c is not extremely large or small. The expectation value of X is close to M but X itself is almost never as large as its expectation. For most a, given A=a, B is equally likely a/2 or 2a.

Seen this way the paradox is not really a paradox about infinity, since one can see the paradoxical things happening without actually going all the way to infinity.

Also this story shows that the writer of TEP-1 need not actually have been making a mistake in steps 6 and 7 - he was a subjectivist using the completely logical and conventional but improper prior 1/x to represent *complete* ignorance about x>0. The "wrong" step was step 8 - he jumped to the conclusion that whether or not you should exchange can be decided by looking at expectation values. But if your ignorance about x is complete, its expected value according to your beliefs is infinite. And in that case you don't expect the expectation value. Richard Gill (talk) 15:04, 19 July 2011 (UTC)

TEP-3
Next we start analysing the situation when we do look in envelope A before deciding whether to switch or stay. If there is a given probability distribution of X this just becomes an exercise in Bayesian probability calculations. Typically there is a threshhold value above which we do not switch. But all kinds of strange things can happen. If a probability distribution of X is not given we come to the randomized solution of Cover, where we compare A to a random "probe" of our own choosing. More probability. Richard Gill (talk) 17:12, 18 July 2011 (UTC)

TEP-Prequel
This is of course Smullyan's "TEP without probability". The short resolution is simply: the problem is using the same words to describe different things. But different resolutions are possible depending on what one thinks was the intention of the writer. One can try to embed the argument(s) into counterfactual reasoning. Or one also can point out that the key information that envelope A is chosen at random is not being used in Smullyan's arguments. So this is a problem in logic and this time an example of screwed up logic. There are lots of ways to clean up this particular mess. Richard Gill (talk) 17:12, 18 July 2011 (UTC)

Conclusions
The idea that TEP is not solved, that the "paradox" is still not resolved, is wrong. There are a number of variant problems and each problem has a number of solutions. The different solutions to the same problem do not contradict one another. Original TEP is an example of a chain of reasoning which leads to an obviously false conclusion, hence there must be something wrong with the reasoning. The problem is to identify what is wrong. But there are several things wrong, and there are even several levels at which mistakes are being made. As soon as you have identified any of these mistakes you have given "a" solution. Original TEP was solved long ago, and then on the ruins new paradoxes were invented. There is no reason why that process should ever come to an end, either. It does not mean that TEP is *controversial*. It means that it is fruitful, stimulating. Richard Gill (talk) 01:29, 19 July 2011 (UTC)


 * It would be very interesting if you could show (in a published paper) how all the different ideas presented so far are logically equivalent. But until we have such a result we must confess that the proposed solutions are very different in character indeed. iNic (talk) 11:24, 20 July 2011 (UTC)


 * The different ideas here are *not* logically equivalent. That is the whole point. We are given what appears a sequence of logical steps. But if it is logic (or mathematics) it is informal logic (or informal mathematics). Definitions are not given. Hence assumptions are only implicit. The writer does not say at each step which theorem of probability theory, or rule of logic, is being used. The context is missing. Hence one can give a different diagnosis of "what went wrong" by construing different intentions (background assumptions...) on the part of the writer. Different contexts. So indeed: the proposed solution are different in character, and they correspond to philosophers, practical statisticians, Bayesian fundamentalists or whoever (educationalists, cognitive scientists, economists, ...), looking for a context for the argument, each supposing it to be the context most familiar to themselves and their intended readers, and showing "what goes wrong" if one takes TEP as an argument intended to take plcae in "their" context. One should compare TEP to the Aliens movie franchise, where each successive movie had a different director who each brought a very personal touch to their take on the basic story line. Richard Gill (talk) 13:56, 20 July 2011 (UTC)


 * The problem statement does follow absolutely normal rigor for being a recreational puzzle. Choosing one of two given envelopes from a table is very much within standard ways of formulating puzzle statements. The context is there for sure: choose one of two envelopes containing money. Are you really demanding that every puzzle statement in every book containing recreational puzzles have to be written in a way where they at each step refer to a theorem in mathematics or logic? That is just crazy. Or are you demanding this just for this puzzle and not for anyone else? In that case it begs the question. How should we know, in advance, for which puzzles we need to use a totally formalized language to even be allowed to state the question? iNic (talk) 23:29, 21 July 2011 (UTC)


 * iNic, are you saying that you believe that the TEP is an open problem, that is to say that the paradox has not been fully resolved. If you want to support this assertion, you must come up with a precise definition of the problem that results in a genuinely paradoxical situation (this comment placed by Marting Hogbin?).


 * This is not an open problem in probability theory or logic but it is an open problem in the philosophical systems where the problem can be formulated; decision theory and bayesian interpretation of probability theory. An open problem is not necessarily a problem for which no solution has been proposed, it can also be a problem where several different solutions have been proposed and where there still are no consensus which one of the proposed solutions is the correct one. For example, before Newton the problem of gravity was an open problem, but not because of lack of different theories of gravitation. iNic (talk) 23:29, 21 July 2011 (UTC)


 * Decision theory, and subjectivist probability theory, are mathematical theories, not philosophical systems. People coming from different fields (mathematics, philosophy, logic, economics, foundations of probability, education...) see different issues because that's their jobs, to see the issues which matter to them. All the problems seen by people from different fields are all real problems. There is no such thing as "the correct solution" because all (at least, almost all) the solutions which have been offered to all the issues which can be detected are all simultaneously correct. Nothing much new has been contributed for a long time. Just regurgitation of old issues, translations from one language to another of the same thing. Richard Gill (talk) 17:01, 22 July 2011 (UTC)


 * If decision theory and subjective interpretation of probability are mathematical theories please tell me where I can find their respective (uncontroversial) axiomatizations. You think more like an art professor than a math professor. Can you mention some other recreational math puzzle where people from different fields are free to invent their own solutions, as one solution is as good as any other? In modern art one interpretation is indeed as good as any other; you're totally free to invent your own interpretation. No interpretation is wrong because no single interpretation is correct, so the approach is not self-contradictory. You seem to apply the same kind of logic here, right? Someone mentioned postmodernism as a joke at the talk page. But maybe it's not a joke in your case after all? Could this in fact be the first example of true postmodernistic behavior in math and logic? Showcasing that 'truth' itself is not objective but subjective, not only in art but in math and logic too? Can someone please rescue me from this nightmare? I think I have to invite Alan Sokal into the discussion chasing all the irrational demons away! iNic (talk) 22:50, 22 July 2011 (UTC)


 * iNic, I did not say that one solution was as good as any other. I did not say that "truth" is not objective in mathematics. You used the word "valid", I hope that the TEP page will only report valid solutions of TEP, and you yourself know there are several different ones. There is not one unique way to translate words about an imaginary situation into a mathematical model or into a logical framework. But anyway, neither your opinions nor my opinions are of any relevance. What do the reliable sources say? There are reliable sources from philosophy who treat TEP like a philosophy problem. There are reliable sources from probability theory who treat TEP like a mathematical puzzle. Plenty of them. Axiomatizations of decision theory, subjective probability: ever heard of de Fineti, Savage? Read some books on mathematical economics, consumer choice... ? Richard Gill (talk) 21:25, 23 July 2011 (UTC)


 * Namedropping is a nice trick when trying to avoid a question, but it doesn't work on me. There are many candidates for competing theories of bayesian probability/decision theory: Ramsey, de Finetti, Carnap, Savage, Jaynes, Jeffrey, Lindley to just name a few. They have all produced theories of bayesianism that differ a lot. Some of them addresses TEP explicitly and they take care of this problem in very different ways, which, of course, is not surprising. This is why TEP is still an open problem within bayesian decision theory / bayesian probability. When the foundations of a subject is still not settled the subject is commonly referred to as philosophy. This is the case here. I'm glad you're not a postmodernist and I'm glad that you say that your personal views on the matter is of no relevance here. If you can show that the different proposed solutions (so far) are just different ways of translating the problem into a logical or mathematical framework, that would be very interesting. However, this is not the view any of the authors of the sources have, and no paper so far is promoting this view. Very few even mention other authors and their proposed solutions. Those who do, do it only to criticize their ideas. If you could show that all TEP authors on the contrary should embrace each other in a big hug it would be very interesting new research indeed. Until then, please treat this as your own opinion, and your own opinion only. iNic (talk) 21:08, 26 July 2011 (UTC)

Dubious importance
This is most certainly a paradox. Initially confusing, due to flaw in the construction of the argument, but ultimately obvious. There is no point in switching. It's a fun game to think about. However the article states that it's an unsolved problem in mathematical economics. To confirm that, there's a citation from a philosophy journal (seriously - wtf?). I don't understand why is so much space being given to a minor mathematical puzzle. The Balance puzzle is far more interesting and difficult, but no one is going into the kind of interminable detail this article does. I appreciate that there are one or two confused post-docs writing papers about this, but isn't that true of anything (especially in philosophy)? I would like to see a short article about the simple mathematical puzzle, with a simple explanation of the logical error. Delete the references to the boring and irrelevant papers, delete the feather-light reference to Smullyan's book, include one reference to a pop-math book, and move on. Alternatively, if we are really determined to explore every angle, why don't we bring in the postmodernists - we could even include a reference to "Toward a transformative hermeneutics of quantum gravity" (which gets 20 times more citations than Chalmers's, very nice, paper) --Dilaudid (talk) 15:03, 21 July 2011 (UTC)
 * What is 'the problem' and 'the logical error' in your opinion? Martin Hogbin (talk) 15:57, 21 July 2011 (UTC)
 * I mentioned the economic problem in the article - I don't think is a problem. The "Two Envelopes Paradox" is that there is a contrived "switching argument" which suggests it is sensible to switch, but this is in clear disagreement with a much simpler and more obvious argument - symmetry. The switching argument is further contrived to say it is sensible to switch ad-infinitum, which is in conflict with basic reasoning (how can it be advantageous to switch twice??). The logical error in the "switching argument" is the sloppy definition of A - is A a single number (in which case we must assign probabilities which are not 1/2 to it being the larger or smaller number) or is A two different numbers (in which case step 7 is a total mess). These kind of issues are common in basic probability if you don't define notation properly. --Dilaudid (talk) 17:01, 21 July 2011 (UTC)
 * A represents the (unknown) sum that is in the envelope that you hold. Martin Hogbin (talk) 20:16, 21 July 2011 (UTC)
 * What do you mean by "represent" it? Is it a number? Dilaudid (talk) 21:33, 21 July 2011 (UTC)
 * Cutting to the chase, if it's a number, then the probabilities are functions (not 1/2). If it's a function, then what is the thing on the right hand side of the expectation? This is why the (painful) concept of random variables was created, to stop people making basic logical fallacies like this. I don't want to upset you guys (it will only drive you further into your positions), but come on, does anyone with any actual knowledge of probability think this is a real "problem"? It's not exactly Dark Matter :) Dilaudid (talk) 22:16, 21 July 2011 (UTC)
 * Exactly. No one in probability thinks this is a real problem. It's an example of the mess you get into when you don't distinguish between random variables and their possible values and their expectation values, between expectations and conditional expectations, between probabilities and conditional probabilities. Any attempt to translage the argument into correct, conventional, probability terms, exposes one or two or more errors - it depends on where you think the writer was trying to go. Precisely because of this, any attempt to resolve the paradox in purely philosophical terms will be wordy, convoluted, unappealing - ordinary language is not refined enough that we can easily make the distinctions which are so crucial, our ordinary thinking is clumsy when we deal with subtle probability issues. Because the original argument is not formalized there are different interpretations possible concerning what the writer was trying to do. Therefore it will always remain an open problem to state what is the actual error which the writer was making, since we don't know what the writer was thinking. On the other hand it is closed: the writer gets in a mess because he uses the same symbol (or words) to stand for different things, whatever line of thought he had in mind. It was precisely to avoid this kind of mess that modern probability calculus was invented. Richard Gill (talk) 16:53, 22 July 2011 (UTC)
 * I understand (I think) but the question that I am trying to answer is, 'Is there a simpler resolution of the TEP that we should add to the article'. I do not think the, undoubtedly correct, statements that you both make can easily be put into a simple and convincing argument for the general reader. What seems to happen is that someone tries to add the argument but ends up by simply saying, 'you should not swap if you have more money in your envelope', or 'the average amount you lose must be equal to the average amount you gain'.  Both may be true but they do not resolve the paradox, as doing this requires you to show the fault in the presented line of reasoning not show that there is an alternative one..


 * On the other hand, I am happy to try to work out a way to explain the problem you have described to the general reader. The thing that started me on this thread was that there seemed to be an assertion by some that there was a really simple resolution that we were missing. I do not think this is the case.  There is a quick answer for experts (the question dos not make sense, or at least it is not clear what you are trying to assert) but if this is made too simple it can easily become invalid.  Martin Hogbin (talk) 09:20, 25 July 2011 (UTC)

Puzzled
Richard, mentions, 'examples of faulty logical reasoning caused by using the same symbol (or verbal description) to stand for different things at the same time' and Dilaudid seems to be alluding to the same thing yet I do not see that resolution of the paradox anywhere in the article. Can someone tell me if that argument is considered a good resolution of the paradox and, if so, could they present the argument clearly here. Martin Hogbin (talk) 11:48, 22 July 2011 (UTC)


 * We are not here to judge which proposed solutions are 'good' or 'bad.' Our job is to give a short unbiased account of the more common solutions mentioned in published sources. This proposed solution should definitely be mentioned in the article. I think it is the simplest solution and it should be mentioned as the first proposed solution in the article. iNic (talk) 13:57, 22 July 2011 (UTC)


 * Falk: "The assertion in no. 6 (based on the notation of no. 1) can be paraphrased ‘whatever the amount in my envelope, the other envelope contains twice that amount with probability 1/2 and half that amount with probability 1/2’. The expected value of the money in the other envelope (no. 7) results from this statement and leads to the paradox. The fault is in the word whatever, which is equivalent to ‘for every A’. This is wrong because the other envelope contains twice my amount only if mine is the smaller amount; conversely, it contains half my amount only if mine is the larger amount. Hence each of the two terms in the formula in no. 7 applies to another value, yet both are denoted by A. In Rawling’s (1994) words, in doing so one commits the ‘cardinal sin of algebraic equivocation’ (p. 100)." Richard Gill (talk) 17:11, 22 July 2011 (UTC)


 * What is your opinion of this argument? Martin Hogbin (talk) 20:18, 22 July 2011 (UTC)


 * I do not see how Falk's argument does not apply in this situation, when you should swap, once.


 * You pick an envelope from a selection containing different amounts of money. A person who knows what is in your envelope then tosses a coin and puts either 1/2 or twice that amount of money in another envelope.  You have the option of swapping.  Would you? I would. Falk's argument seems to me to show that you should not swap and it is therefore wrong. Martin Hogbin (talk) 22:53, 22 July 2011 (UTC)


 * In standard TEP, when your envelope is the smaller of the two, it contains on average half what it on average contains when it is the larger. This is also Schwitzgebel and Dever's explanation (not of what is a wrong step in the argument, but why its conclusion is wrong). In your scenario, Martin, your envelope is not different by being smaller or larger.  Falk puts into words that in standard TEP, what's in either envelope is statistically different when it's the larger than when it's the smaller. The wrong step in the argument assumes that whether your envelope is the smaller or larger is independent of what's in it. In maths: the wrong step is to assume that the event {A<B} is independent of the random variable A. Falk's explanation: this step is wrong, because A is not independent of {A<B}. It seems that SD are the first philosophers who say this. Falk knows it. The mathematicians knew it long before (they know the symmetry of statistical independence; philosophers don't). Mathematicians writing for mathematicians can leave their conclusions in technical language and perhaps only semi-explicit: their intended readers understand it.  But the philosophers don't understand. And vica versa. Ordinary folk understand neither. Hence the wheel is reinvented so many times. Richard Gill (talk) 17:14, 23 July 2011 (UTC)


 * If you are pointing out that there is no way to arrange the money in the envelopes such that the probability that B contains half of A and the probability B contains twice A are both equal to 1/2 and the situation is symmetrical then I understand and agree. If this is not what you are saying then could you explain some more please. Martin Hogbin (talk) 18:03, 23 July 2011 (UTC)
 * I'm not saying that. Put x and 2x in two envelopes, then completely at random write "A" and "B" on the two envelopes. The situation *is* completely symmetric. And the chances A contains twice or half what's in B are both equal to 1/2. What these folks are saying is that it's asking fir trouble to denote both by "a" what's in A when it's larger and what's in "A" when it's the smaller. Call them "2x" and "x" instead and you don't screw up. Alternativy, learn about conditional probability and do it with maths. The original TEP argument looks like probability theory but isn't. Philosophers think it's purely a problem of[ logic, not of mathematics. Then what I just explained is the solution. But another solution is to do it properly within probability theory. But in the end it all comes down to the same thing. [User:Gill110951|Richard Gill]] (talk) 18:14, 23 July 2011 (UTC)
 * How does Falk's argument not apply to my case then? Martin Hogbin (talk) 18:26, 23 July 2011 (UTC)
 * In your case, what's in your envelope when it's the smaller of the two is the same as what's in it when it's the larger of the two. Your envelope is filled first, say with amount a, only after that is 2a or a/2 put in the other envelope. In TEP two envelopes are both filled with amounts x and 2x. After that, you choose one at random. If it's the smaller amount it's x. If it's the larger amount, it's 2x. This is what Falk is saying, and SD, and a load of other people, though I must say they can find incredibly ugly ways to say it. Richard Gill (talk) 20:40, 23 July 2011 (UTC)
 * I have just added a bit to my statement above. It now says, ' there is no way to arrange the money in the envelopes such that, for every possible value of A, the probability that B contains half of A and the probability B contains twice A are both equal to 1/2 and the situation is symmetrical. Does this make sense now? Martin Hogbin (talk) 19:13, 23 July 2011 (UTC)
 * That makes sense to me, though some purists would prefer that you added the word "conditional" in front of probability, though. However, it is possible to arrange things so that it is almost exactly true. This brings us to TEP-2, "Return of TEP", or "Great Expectations". If you arrange things so that what you just asked is almost exactly true, then for almost every possible value of A, the expectation of B is exactly or approximately 5A/4, so it seems one ought to switch. We seem to have a new paradox. However in that case the amount of money actually in envelope A is almost always far, far, less than its expectation value - expectation values are absolutely useless as a guide to rational choice. You might object that this corresponds to subjective beliefs about the amount in envelope A, or the smaller of the two amounts X, which are totally unreasonable. However, there are a load of reliable sources which do state that this is not only reasonable, it's even prescribed. If you really know absolutely nothing about X, you also, I think, would be prepared to say that you know absolutely nothing about A, and absolutely nothing about X converted from dollars to Euro's or Renbini or Yen. There's exactly one probability distribution which has this property: it's the probability distribution which makes log X uniformly distributed, so log X is a completely unknown real number somewhere from minus to plus infinity. The thing is that knowing nothing about a positive number means that however large it is, it could just as well be billions of time larger, and however small it is, it could just as well be billions of time smaller. Because of this your expectation value of what it is, is infinite. So you'll always be disappointed when you get to see the actual value x, which is a finite positive number. I was just now talking about subjectivist probability: probability as rational belief. If you want to think of TEP with frequentist probability, then I just want to say that it is easy to arrange that the properties you listed almost exactly hold true. I can write a program on my computer which will print out two amounts x and 2x to go in the two envelopes, and though the event {A<B} won't be exactly independent of the amount A, it will be as close to independent as you like. We can play the game a billion times and never ever meet an occasion when the conditional probabilities that B=a or a/2 aren't each exactly 1/2 (given A=a). So despite what's written in the literature, the paradox of TEP-2 isn't just a paradox of infinity. Unfortunately wikipedia will have to wait 10 years before we can correct that illusion. Richard Gill (talk) 20:53, 23 July 2011 (UTC)
 * But surely that is still a paradox of infinity. It is only because the expectation is infinite that you do not expect to actually get it.


 * Also, I think we agree that Falk does not put the argument very well, especially for the layman. Martin Hogbin (talk) 22:04, 23 July 2011 (UTC)


 * I agree that Falk makes her point rather poorly. I disagree that the paradox in the case of infinite expectation values is "just" a paradox of infinity. Consider the following completely finite case. Put 1 ct on the first field of a chess board, 2 on the next, 4 on the next.. Pick a field at random and put that amount in one envelope, twice in the other. Pick an envelope at random... The expectation value of what's in both envelopes is finite. Only once in 64 times is the amount in the other envelope *not* equally likely to be half or double that in the first, given the amount in the first. 63 out of 64 times the expected content of B is 5/4 what's in A. (1 in 128 times it's twice, once it's half). The expectation value of A is close to its maximum. Most times, A is far, far smaller than its expectation value. Also the average of many, many independent copies of A is mostly far smaller than the expectation value. So also in this completely finite case, the expectation value is pretty irrelevant. "The long run" doesn't kick in till long after you're dead. This is not an academic example. In finance, meteorology, astronomy, ... one meets this kind of behaviour.Richard Gill (talk) 18:29, 24 July 2011 (UTC)

Three resolutions
Richard, would you agree that three resolutions cover all reasonable formulations of the problem?
 * This covers all of TEP with closed envelopes except for the Smullyan "no probability" variant. That variant is a problem of the logic of counterfactuals, a topic on which philosophers disagree. But for each popular formal approach to counterfactual reasoning, there is a solution to the paradox. It's all rather technical, and hardly of popular interest. TEP when you look in A before deciding is interesting for probabilists and for decision theorists, but hardly controversial. Tom Cover's random probe solution is fun. Richard Gill (talk) 18:58, 24 July 2011 (UTC)

The situation not symmetrical
If the situation is not symmetrical then it can be advantageous to swap, but only once.


 * But the problem description makes it symmetrical! That's why we know there is no point in swapping, and hence that there must be one or more errors in the reasoning. Richard Gill (talk) 18:33, 24 July 2011 (UTC)


 * Yes but it still needs to be said. It is easy to come up with formulations where the envelopes are not symmetrical. Martin Hogbin (talk) 22:13, 24 July 2011 (UTC)


 * Agree. It needs to be said very clearly. Richard Gill (talk) 22:51, 24 July 2011 (UTC)

The expectation is infinite
Intuitive arguments involving infinity can easily fail, one way or another.


 * It's not a problem with infinity, it's a problem of a probability distribution which is so heavy-tailed to the right that its expectation value is in the far end of the distribution and totally unrepresentative of typical values or even of typical averages of less than astronomically many independent values. In the long run we're dead. The mean value is the infinitely long run average. Just not relevant to the real world. Richard Gill (talk) 18:39, 24 July 2011 (UTC)
 * What about your 'Second resolution' where you say, 'The average amount of money in both envelopes is infinite. Exchanging one for the other simply exchanges an average of infinity with an average of infinity'? Martin Hogbin (talk) 22:37, 24 July 2011 (UTC)
 * No contradiction. A distribution with infinite expectation value is merely an extreme example of a probability distribution which is so heavy-tailed to the right that its expectation value is in the far end of the distribution and totally unrepresentative of typical values or even of typical averages of less than astronomically many independent values. Richard Gill (talk) 22:59, 24 July 2011 (UTC)

The expectation is not independent of the sum in the original envelope
This covers cases where the possible sums in the envelopes are limited. It is also my first explanation in the 'Introduction section' (which I will now amend slightly).

This same argument can be presented differently in what some have called the 'simplest' argument. However, the x/2x style argument needs to be presented very carefully to be technically correct. The very simple style presented here previously actually misses the point. If presented properly, this argument is only accessible to experts. Martin Hogbin (talk) 09:16, 24 July 2011 (UTC)


 * Do you mean expectation or probability? The *probability* that the sum of money in Envelop A is the smaller or larger of the two is (in general) not independent of the amount itself. Equivalently:, the amount is (in general) not independent of whether it's the smaller or the larger. (As my chess board example shows, it can be more or less the case, but only in a situation where the expectation value says nothing about what you can realistically expect in a finite lifetime.) Richard Gill (talk) 18:47, 24 July 2011 (UTC)
 * Yes I did mean probability, the point being that if the probability of doubling your money is not independent of the sum in your original envelope you must take account of this when calculating the expectation. Martin Hogbin (talk) 22:37, 24 July 2011 (UTC)
 * Right. One could add to this: intuitively, the more money in A, the more likely it's the larger amount. And conversely: given A contains the larger amount, there's typically more in it than when it contains the smaller amount. Richard Gill (talk) 22:57, 24 July 2011 (UTC)
 * As you can see, I am making this up as I go along but I think what I actually meant was that if the expected gain ratio (expectation/original value) is not independent of the sum in the original envelope. I think this applies to every finite distribution, even your example and, in my opinion, it is sufficient to resolve the paradox.  If every step does not always hold good, the simple but compelling argument fails.   Of course, in your example, there are other reasons why you might not choose to swap but what I am trying to do is to find the simplest way, with the fewest different resolutions, to cover every reasonable formulation of the problem.  There will always be a continuing cat and mouse game between formulators and solvers for those interested. Martin Hogbin (talk) 08:56, 25 July 2011 (UTC)
 * The expected gain ratio depends on the amount in the original envelope if and only if the probability the original envelope contains the smaller quantity depends on the amount in it. That's because it's a two point distribution on a/2 and 2a hence the mean is a times (1-P(A<B|A=a))/2 +2P(A<B|A=a) which is the same as a times (3P(A<B|A=a)-1)/2. The expected gain ratio is therefore ( 3P(A<B|A=a) - 1 ) / 2. It depends on a if and only if P(A<B|A=a) depends on a. So if A is bounded this quantity has to depend on a. So yes, also my chessboard example has P(A<B|A=a) depends on a, as I said. It is constant and equal to 1/2 for all possible values of a except for the smallest and the largest.  It's almost constant. According to conditional expectation, switching is almost always justified. But we agreed that switching is a waste of time, by symmetry. The chessboard example is an example for which you will amost always be disappointed when comparing the mean value of A or B with their actual values. Richard Gill (talk) 14:56, 25 July 2011 (UTC)
 * I understand that. It is very similar to playing a martingale.  With a very large bankroll you will win nearly every time, with an infinite bank roll you will win every time but you bankroll never gets any larger.


 * My point was that even with your chessboard version, the first resolution works in that it is 'simply impossible that whatever the amount A=a in the first envelope, it is equally likely that the second contains a/2 or 2a', even though this is almost always true. To resolve the paradox you only need to show that any step is not completely correct.  The fact that you almost certainly will not get your expectation is another reason not to swap. Martin Hogbin (talk) 15:40, 25 July 2011 (UTC)


 * On the other hand, I see that your chessboard version could still be used as an argument to swap. Oddly enough there is very nearly a real game using the chessboard version, and this is 'Deal or No Deal'.  If a contestant is given the option to swap their box at the start of the game (which they do not usually get, in the UK at least) the expectation value in their new box will, most likely, be greater than the value in the box they hold.  In most cases, however, if they swap they will be disappointed Get less than the expectation). Martin Hogbin (talk)

Question
Am I right in saying that the abbreviated expectation calculation given in step 7 in the article is something of a fraud? It only applies if step 6 holds for every possible vale of A. If this is not the case then the expectation must be calculated properly by taking the average of all possible sums, weighted by their probabilities. Martin Hogbin (talk) 08:30, 26 July 2011 (UTC)
 * You are right, under one common interpretation of what the writer is tring to do. But it's not clear whether the writer is computing the conditional (given A) or the unconditional expectation of B. Both can be computed (in principle) by separating out the two cases, the two complementary events, $$AB,A)P(A>B|A)$$. Now, given A and given $$AB$$, we know B=A/2. So the writer could then procede to $$E(B| A)=2AP(AB|A)/2$$. If this was his intention, he procedes to substitute $$P(AB|A)=1/2$$. So either (within this reconstruction) he is mistakenly assuming that the event $$A0$$) through the improper distribution with probability density proportional to 1/x. In that case, it's step 8 which is problematic - his prior distribution has infinite expectation, and whatever value of A or B you would actually see, you will be disappointed when you compare them to their expected value. Expectation of B, not conditional on A : the writer is presumably using $$E(B)=E(B|AB)P(A>B)$$. He correctly substitutes $$P(AB)=1/2$$ but mysteriously replaces $$E(B|AB)$$ by A/2. How can two numbers equal two random variables? Or does he not distinguish between A and E(A)? It would be correct to write $$E(B|AB)=E(X|A>B)=E(X)$$. This would finally lead to the (true) statement $$E(B)=3E(X)/2=E(A)$$. This seems to be the interpretation of the young philosophers Schwitzgebel and Dever, and leads them to their diagnosis "what goes wrong is that A is smaller when $$B>A$$ than when $$BA$$ but it's equal to 2X when $$A>B$$. On average, two times smaller in the first case. Most people in probability think that the writer is trying to compute the conditional expectation of B given A, but philosophers who don't know probability theory do not make these subtle distinctions between conditional and unconditional expectations and actual values. (That's what makes S&D's paper so hard for me to read, and why it's so weird they try to explain what is going wrong, in part, by doing some new probability theory of their own). Still it is true that the event $$\{AA$$ than when $$BB$$ or $$B>$$A does not tell us anything at all about the value of A. Conversely, the value of A tells us nothing about whether $$A>B$$ or $$B>A$$. Who's to say what the writer was intending? What he knew, what he didn't know? Was he a complete amateur who got more or less everything mixed up (he got more or less mixed up, depending on which way you imagine he was trying to go), or was he a very sophisticated Bayesian using an improper prior? Some people think that improper distributions have no place at all in mathematics, others think they are a vital part of the Bayesian's toolbox. Richard Gill (talk) 11:57, 26 July 2011 (UTC)

TEP Synthesis
The following discussion creates a synthesis between the solution of Schwitzgebel and Dever, representing the philosophers, and the solution of myself, representing the probabilists. We need to be aware of the symmetry of (statistical) dependence and independence. In particular, random variable A is independent of event {A < B} if and only if event {A < B} is independent of random variable A.

S&D's diagnosis of what goes wrong was that the amount A in the first envelope is different, namely smaller, when it is the smaller of the two ("equivocation"). I strengthened this to the observation that on average, it's twice as small as what it is, on average, when it is the larger.

This observation proves that the amount A is statistically dependent on the event {A < B}.

My diagnosis was that step 6 assumes that the fact whether or not A is smaller, is independent of the actual amount A. In other words: step 6 assumes that the event {A < B} is is independent of A.

By symmetry of statistical independence, this is the same thing as independence of A on {A < B}.

There's one tiny speck of dirt in these arguments. If E(X) is infinite then twice infinity is infinity, we can't necessarily conclude that what's in the first envelope is smaller, on average, when it's the smaller of the two. And there's an extreme case where our uncertainty about X is so large that whether or not it's twice as large makes no difference. Knowing we have the larger of the two amounts doesn't change our information about it, at all. Consider the uniform distribution on all integer powers of 2: ..., 1/8, 1/4, 1/2, 1, 2, 4, 8, ... Yes I know this probability distribution doesn't exist in conventional probability theory, but arbitrarily good approximations to it do exist: you might like to think of it as a "real number" addition to the set of rational numbers.

Working with this distribution (or with close approximations, and then going to the limit), it *is* the case that the probability distribution of 2X is the same as that of X, both have infinite expectation values, E(B|A=a)=5a/4 for all a (integer powers of 2), and P(B=2a|A=a)=0.5=P(B=a/2|A=a) for all a. So at the boundary of reality, steps 6 and 7 of the TEP argument do go through. The reason now why switching only gives an illusory gain is that when expectation values are infinite, you will always be disappointed by what you get. They are no guide to decision making.

Richard Gill (talk) 06:44, 1 August 2011 (UTC)

My simple resolution - has this been covered in the literature?
First let me define what I am talking about - in the current article there is a "Switching argument" with 12 steps, and a 4 sentence "basic setup". I will try to put the "basic setup" into mathematical language:


 * S, L are positive real numbers, where $$2S = L$$. A, B are random variables, distributed as $$P(A,B = L,S) = 0.5$$ and $$P(A,B = S,L) = 0.5$$.

Note that this world is perfectly symmetrical - the labels A and B can be swapped without making any difference. Note also that I actually used A in the setup, whereas in the paradox it does not appear until step 1 - this should make no difference, but makes things easier to explain.

I think there are actually two sets of statements here (call them TEP-1a and TEP-1b). TEP-1a consists of both the set-up and the switching argument, and deduces steps 1-6 from the setup. TEP-1b uses steps 1-6 as definitions of a random variable B' and ignores the set-up. I assert the confusion generated by the paradox results from switching between the two variants. This is because the expectation of B is actually different under the two variants.

Under variant TEP-1a, the set up can be used to show statements 2, 3, 4, 5 are true (e.g. 4 states if A = S, then A,B = S,L, therefore B = L). Statement 6 it is a stretch, as A is taking two different values, one a quarter of the other, in different parts of the statement. This makes statement 7 nonsensical. Under variant TEP-1a, from the setup it is clear that $$E(B) = E(A)$$.

Under variant TEP-1b, we take statements 1-6 as a definition of the world, so in summary (based on steps 1 and 6) we can say:


 * A is a positive real number, define B as a random variable distributed as $$P(B = \frac{1}{2}A) = 0.5$$ and $$P(B = 2A) = 0.5$$.

Here the expectation of B, $$E(B) = \frac{5}{4}A$$, but the problem is not symmetrical. So under TEP-1b it makes sense to switch once, and only once.

I'm satisfied with this resolution up to a point. What I find tricky is that there is a difference in the way that A is defined between TEP-1a and TEP-1b, and I'm not sure exactly how to express it.

Note that the paradox is always explained in terms of the set up, because the set up is a far clearer way to explain the problem than items 1-6 of the switching argument. However the set up is abandoned before we attempt to calculate the expectation, this is because without abandoning the set-up, it is too simple to calculate the expectation in a non-paradoxical way. Dilaudid (talk) 20:10, 8 August 2011 (UTC)
 * Whatever this is it does not look like a simple resolution to me. I cannot follow it at all. Martin Hogbin (talk) 20:48, 8 August 2011 (UTC)
 * Hi Martin, apologies, this wasn't really aimed at anyone who didn't specialise in maths. Let Richard have a look. If it makes some sense to him, I will have a go at putting it into a common language. One problem here though, is that the language of probability (random variables, etc) was invented to deal with these concepts with a degree of rigor. Dilaudid (talk) 21:52, 8 August 2011 (UTC)


 * It occurred to me after I wrote the comment above that the problem may not be with your maths, but with my explanation. So to summarise what I'm saying above, the set-up already defines the amount in the chosen envelope as a random variable (an amount drawn at random from two amounts). Therefore A is a random variable. Therefore it cannot be used as a value in an expectation of B (the amount in the other envelope), therefore step 7 cannot be deduced. That is the simple explanation. Dilaudid (talk) 07:52, 9 August 2011 (UTC)
 * I think it can be used but only with great care. For example the probability that the unchosen envelope contains 2A must be independent of the value in the chosen envelope. For a finite distribution this is not the case so the expectation calculation is invalid, which is pretty much what I say in my explanation in the 'Introduction to resolutions' section. Martin Hogbin (talk) 08:35, 9 August 2011 (UTC)
 * I think you are wrong here. Here's a variant on two envelopes - C is the result of a dice roll. D is the number on the other side of the dice. So if C is 1, D is 6 - so D = 6C. If C is 2, D is 5, so D = 2.5 C. If C is 3, D is 4, so D ≈ 1.333C. Continuing this logic we can see that
 * $$E(D) = \frac{1}{6}(\frac{6}{1}+\frac{5}{2}+\frac{4}{3}+\frac{3}{4}+\frac{2}{5}+\frac{1}{6})C \approx 1.858 C$$
 * Once again we hit a problem because we are trying to find expectations using another dependent random variable. Even if we try to take an expectation of C, using C as the value we find E(C) = 1/6(C + C + C + C + C + C) = C. This is just as wrong, as C is not a specific number, it's the outcome of an event. Expectations are numbers, not random outcomes. Dilaudid (talk) 09:01, 9 August 2011 (UTC)
 * It is my understanding that random variable can be used in expectation calculations subject to certain conditions, one of which is the independence described above. Perhaps Richard could comment on this. Martin Hogbin (talk) 16:33, 9 August 2011 (UTC)
 * I did notice you said - "For example the probability that the unchosen envelope contains 2A must be independent of the value in the chosen envelope." This statement is false. The set up defines A as being chosen randomly from two amounts, a lesser and a greater. If the amount is the lesser it is certain that the amount in the other envelope is 2A (So $$Pr(B = 2A|A = S) = 1 \neq 0.5 = Pr(B=2A)$$, which is the definition of dependence). Or to address your point in a different way, you are trying to find the expectation of B, using a random variable A as the value - and you have defined A and B to be dependent, since $$Pr(B = L|A = S) = 1 \neq 0.5 = Pr(B=L)$$ - this is clear from the "basic setup", and also clear from steps 4 and 2. Dilaudid (talk) 19:17, 9 August 2011 (UTC)
 * Dilaudid, I think you are misunderstanding me. When I say "For example the probability that the unchosen envelope contains 2A must be independent of the value in the chosen envelope." I am referring to a condition that must be met before a random variable can be used in calculating an expectation value.  In the TEP, this condition is not met, thus the simple expectation calculation is invalid.  In my dice throw example on the 'Arguments' page the condition is met, thus the expectation can be simply calculated as 5A/4. Martin Hogbin (talk) 08:41, 11 August 2011 (UTC)
 * The dice roll. D=7-C (i.e, when C=c, then D=7-c). Given C=c, D=7-c with probability 1. Therefore E(D|C=c)=7-c. Both left and right hand sides of this equation depend on c. I can think of them both as functions of c. I can take these same functions of the random outcome of the die throw, C. Notation: Compute E(X|Y=y). The result is a function of y. Take this function of Y. By definition, we call the result E(X|Y). In words, the average value which X has given that Y has the value which it actually does have, whatever that might be. Which is random because Y is random. Theorem: E(E(X|Y))=E(X). I've found E(D|C)=7-C. I can take expectation values on left and right hand sides to find E(D)=7-E(C). (Note that I frequently make use of the linearity of expectation values. You can take linear combinations before or after averaging - you'll get the same result. Richard Gill (talk) 11:03, 10 August 2011 (UTC)
 * Hi Richard, not sure what your point is here - I appreciate that it's not true that $$E(D) \approx 1.858 E(C)$$ - that was my point, that the kind of sloppy use of random variables in expectations which appear in the paradox, can also be used to create other paradoxes, which are even more ridiculous than the two envelopes paradox. I prefer to show that E(D) = E(C) by symmetry - they are opposite sides of the same dice being rolled. Of course E(C) = 7 - E(C) since E(C) = 3.5, and d = 7-c since opposite sides of a dice add up to 7. Dilaudid (talk) 11:36, 10 August 2011 (UTC)
 * Hi, yes I know you know the right answer, and you know quick ways to get it. My point is that your funny (wrong) derivation looks superficially like an attempt to compute E(D|C) and my point is that you *can* approach the dice problem, correctly, in this way, and get the right answer too. You just have to know the basic rules of probability. And use a notation which is expressive enough to be able to make the distinctions which are crucial to getting it right. To begin with, distinguish random variables and possible values thereof. Understand what E(D|C) actually means. To continue, I showed how one can correctly derive E(D|C)=7-C by something looking like your funny calculation, but doing it right. From this follows E(D)=7-E(C). Of course, that could also have been got directly, just taking expectations on both sides of the identity C+D=7. We also have by symmetry E(C)=E(D) so finally we can find E(D)=E(C)=7/2. So clever tricks can give short cuts to computing things, but you have to know how to use them. Training and understanding is needed to use sophisticated professional tools. You need to know which direction to point a gun in, before you try to shoot someone with it. Richard Gill (talk) 10:57, 11 August 2011 (UTC)

The theorem
The theorem using A is a positive real number, defining B as a random variable isn't just only sloppy formulated, forcing A as well as B to be zero or infty, it's more of that: it's a real mess.
 * The theorem
 * $$P(B = \frac{1}{2} \frac{1}{2}A + \frac{1}{2}2A) = \frac{5}{4}A$$     –  in this form  –  is nonsensical.

That theorem does not even correspond to real life's thinking nor to real life's saying.

Even in language of real life you never say just "B is either twice of of A, or B is half of A"  only, fullstop. - Simultaneously you are always (explicitly or implicitly) implying:
 * B is either twice of A (but only in case that A is the smaller amount, i.e. if A = B/2) –  or B is the half of A (but only in case that A is the bigger amount, i.e. if A = 2B).

These genuine implications are, from the outset, part of the statement.

"B is either twice of A or half of A" without the necessary implied restrictions is nonsensical.

Without distinguishing, no correct result. A nonsensical result, based on cheeky and sloppy constraints, is no paradox, imho it's insane. I saw Falk mentioning the "sin". Is this aspect emphasized with sufficient clarity and shown within the article? Thank you. Gerhardvalentin (talk) 13:03, 9 August 2011 (UTC)
 * Hi Gerhard, I think there might be an error in your formula - did you mean:
 * $$E(B) = \frac{1}{2}(\frac{1}{2}A + 2A) = \frac{5}{4}A$$
 * Dilaudid (talk) 13:26, 9 August 2011 (UTC)
 * Yes thank you Dilaudid, of course you're right, my fault, am correcting my mistake now. Gerhardvalentin (talk) 15:19, 9 August 2011 (UTC)
 * Let me correct Gerhard's theorem. I use X, Y, A, B to denote the two amounts in order of size, and the two amounts in Envelopes A and B. I suppose that X>0 and Y=2X were fixed in advance, and then a fair coin was tossed to determine whether (A,B)=(X,Y) or (A,B)=(Y,X). The coin was tossed independently of the determination of X. These preliminary remarks completely fix the joint probability distribution of (A,B) in terms of that of X. So in principle we can sit down now and compute E(B|A=a). Since B must either equal 2a or a/2 if A=a, corresponding to the complementary events B>A and A>B, we can write down the following: E(B|A=a)=2a P(B>A|A=a)+a/2 P(A>B|A=a). So *if* P(B>A|A=a)=P(A>B|A=a)=1/2 we find that E(B|A=a)=5a/4. This might in principle be true for some values a of A, or all, or none. Suppose it is true for all values of a. Apparently the writer of the argument believes that this is the case. If I average over all possible values a of A according to the probability distribution of A probability distribution, whatever it is, I come to the conclusion E(B)=5E(A)/4. I am using here the theorem that one can compute an unconditional expectation (i.c., that of B) in two steps: first by averaging while keeping some condition fixed, then by averaging with respect to that condition. Now by symmetry (remember the fair coin toss?) we know in advance that E(A)=E(B) and we also know that E(A)+E(B)=E(A+B)=E(X+Y)=E(3X)=3E(X). So we know for sure that E(A)=E(B)=3E(X)/2. But it now appears that we also have E(B)=5E(A)/4 and this appears to be a contradiction. But it isn't! There is one solution (given that we already know E(B)>0), and it is E(A)=E(B)=E(X)=infinity. Now if you read my paper you will discover that actually, within standard probability theory, it is not possible to find a probability distribution of X such that P(B>A|A=a)=P(A>B|A=a)=1/2 simultaneously for all possible a. However it is possible to get arbitrarily close to having this true, and it corresponds to more or less believing that log(X) is uniformly distributed over all the real numbers. In fact, if we take total lack of prior information about X>0 to mean that our prior information about cX is also nil, whatever c>0 might be, we "have" to use this non-existent uniform distribution of log(X) to express our prior knowledge about x. It doesn't exist, too bad, but distributions very very close do exist, e.g. take the uniform distribution between plus and minus one googleplex. For such a distribution it is almost the case that E(B|A=a)=5a/4 for all possible values a of A. And for such a distribution, E(X) is not only mega-astronomically large, it is actually in the very very far right hand edge of the distribution of possible values of X. If you read my paper you will find out that in this case, the probability that X is smaller than, say, a billionth of its expectation value, is practically speaking equal to one. So: the author of the original TEP argument might well have been entirely correctly using independence of A and the coin toss to reflect the fact that he is so totally ignorant of the value of X. The argument only breaks down at the point where the computation of conditional expectation is supposed to imply that you will do better by switching envelopes. Because your *sublime* ignorance about how large or small X might be means that you will actually always be highly, highly disappointed when you open either envelope and compare the amount in their to the amount which follows from computing 3/2 times the mathematical exectation of X. The mathematical expectation value is irrelevant to decision making. Expectations are supposed to approximate long run averages. But in this case the long run doesn't even kick in till far past the death of the universe, let alone yourself, so you do *not* expect the expectation value at all. Richard Gill (talk) 10:41, 10 August 2011 (UTC)

I have written a paper entitled Anna Karenina and The Two Envelopes Problem,. It expounds my explanation (the Anna Karenina principle) of why there is so much argument about TEP and contains some modest new results. Comments are welcome. It is not finished yet; when it is I'll submit it to a suitable journal. Next I must read and comment on Dilaudid's contributions here! Richard Gill (talk) 15:08, 9 August 2011 (UTC)


 * Gerhard and Dilaudid, how would you deal with this problem? You are given an envelope containing an unknown but finite sum of money.  A person who knows the sum in your envelope then places either twice or half the sum in a second envelope according to the toss of a coin and you are given the option to swap.


 * You call A a random variable representing the sum in your envelope and calculate the expectation in the other envelope as 5A/4. On that basis you decide to swap (once only).  The difference here is that the probability of doubling your money is independent of the sum in your envelope or, to put it another way, the probability of doubling your money is 0.5 for every sum that might be in your original envelope.


 * How does your argument not apply to this case? Martin Hogbin (talk) 16:50, 9 August 2011 (UTC)
 * Hi Martin, this isn't my argument so I won't defend it. As I said above, if you construct a random variable B in this manner, it makes sense to swap once (I call this situation TEP-1b). But this is very different to the "basic setup" which is supposed to set up the Two Envelopes Paradox in the current article, notably there is no symmetry in TEP-1b. Gerhard - I think your point is quite difficult to follow and I think Martin has a point. Could you re-read and rewrite what you originally wrote? Dilaudid (talk) 19:38, 9 August 2011 (UTC)
 * Martin's argument for this case is correct, at least, assuming that utility of money is linear in the amount. Personally I would be pretty indifferent between getting a trillion dollars and a billion dollars. In fact I am not sure if I would want either. But I'd be very happy with ten thousand. I'ld also be pretty indifferent between losing or gaining one or two dollars. This is the reason why it is *not* irrational to go to casino's or buy lottery tickets. Utility of money isn't linear. Richard Gill (talk) 10:53, 10 August 2011 (UTC)
 * Richard I took the liberty of correcting your url. Richard & Martin - having read "The Puzzle" and your comments on my Dubious Importance section it seems pretty clear you know almost everything I have worked out. Enjoy the 6 faces paradox - I hope that is new to you :) Dilaudid (talk) 19:38, 9 August 2011 (UTC)


 * Martin, the point is the "sin" of using "A" in the TEP theorem for all values of A, whatever A may be:
 * For "A being 2B" (and vice versa B = 1/2 x A) as well as for "A being 1/2 x B" (and vice versa B = 2A).
 * Imho with the effect that the result of that theorem "B = 5/4 x A" will only be valid if A=B and vice versa B=A (i.e. if both are zero or both infty). And I once more entered my question here because  –  as already expressed in my statement on the article talk page  –  I feel that this aspect isn't emphasized enough in the article. (And as to your question above: there in the article talk page I said what I would suggest as a possible reply).


 * Although Richard 17:11, 22 July 2011  in the article talk page did cite there Falk:
 * "The assertion in no. 6 (based on the notation of no. 1) can be paraphrased ‘whatever the amount in my envelope, the other envelope contains twice that amount with probability 1/2 and half that amount with probability 1/2’. The expected value of the money in the other envelope (no. 7) results from this statement and leads to the paradox. The fault is in the word whatever, which is equivalent to ‘for every A’. This is wrong because the other envelope contains twice my amount only if mine is the smaller amount; conversely, it contains half my amount only if mine is the larger amount. Hence each of the two terms in the formula in no. 7 applies to another value, yet both are denoted by A. In Rawling’s (1994) words, in doing so one commits the ‘cardinal sin of algebraic equivocation’ (p. 100)."


 * Imho this aspect ("cardinal sin") is the key to the so called "paradox" and imho this should be emphasized in the article. Gerhardvalentin (talk) 21:17, 9 August 2011 (UTC)


 * That is the point of a random variable, it can take different values. In my calculation I commit the same sin of using the random variable A to represent the sum in the original envelope whatever value it may have and regardless of whether it contains 2B or B/2.  However my calculation is fine, because the probability that A is the larger sum is independent of the value of A.  This is not so for the finite TEP. Martin Hogbin (talk) 22:02, 9 August 2011 (UTC)


 * Indeed the point of a random variable is that it takes different possible values, there is a probability distribution over the values it can take. In the first lessons in Probability 101 we spend our time, over and over, emphasizing this distinction, and cadjoling the students to use a notation which makes that distinction explicit. A stands for the random variable amount of money in Envelope A, a stands for any possible value it might have, E(A) stands for the weighted average of the possible values a, weighted according to the probabilities P(A=a). And I haven't even started yet on the lesson explaining E(B|A=a) and E(B|A). Standard TEP is built up as follows. First of all someone fixes the two amounts of money which are to go in the two envelopes. We are going to treat them as random, either because we are thinking of many repetitions in which each time possibly varying amounts are used, or because we are using probability to represent our prior guesses as to what those amounts of money might be. (For instance, a priori you might be 50% certain that the smaller of the two amounts is less than $100). So let me denote by X and by Y the two amounts of money in order of size, so we are told that Y=2X>0. Independently of the value x which X actually has in any specific case, a fair coin is tossed to determine which amount goes in Envelope A. Let C denote the coin toss, it is another random variable, taking the values h and t with equal probability 1/2. And it is statistically independent of X. Now we can define random variables A and B by A=X,B=Y if C=h; A=Y,B=X if C=t. This means that the joint probability distribution of A and B is completely determined in terms of that of X and C. And now we can do things like compute the expected value of B given that A=a for any value of a which can occur (any value of X and any value of Y=2X), in terms of the probability distribution of X. Notation: E(B|A=a) stands for the expectation value of B, given a particular value a taken by A. One weighs the possible values of B in the situation that A=a, which are clearly restricted (at least) to a/2 and 2a, according to the conditional probabilities, given that A=a. This is all completely elementary probability calculus, and the problem is that the original author of TEP is not using a notation which distinguishes random variables from possible values thereof (and possibly he also doesn't distinguish either from expectation values); nor does he appear to be aware of the distinction between unconditional and conditional probabilities. In other words, the argument is a typical total screw-up by a fresh student! Yes, the students of today find it very difficult, since their hand-writing doesn't distinguish capital from small letters! (Let alone the problems they have because they don't know the Greek alphabet). After a long long time we manage to get our mathematics students aware of these important distinctions. With biologists or medical doctors or physicists or philosophers or lawyers the message never comes across. Hopeless. I think no editor should work on the TEP page without first taking a refresher course in elementary probability calculus. It was invented so that scientists could move on from the stupid paradoxes which plagued informal probability theory in the 19th century. It seems that the philosophers still have not moved on. Richard Gill (talk) 10:13, 10 August 2011 (UTC)
 * And then all was very quiet ... Richard Gill (talk) 11:35, 19 August 2011 (UTC)

Two neckties as two envelopes: the math
Let me express the two necktie paradox in a similar mathematical notation to a popular notation for the two envelopes. Remember that the two rich men who have met in a bar don't know how much each of their neckties, presents from their wives, cost. We do know that they each believe the other to be as rich as themselves and I suppose they also imagine their wives are similar, too. Denote by A and B the prices which were paid by their wives for the two neckties. Let the probability measure P express the subjective beliefs of the first man about these two prices. I could use Q for the other. For instance, P(B<100) is the first rich man's personal probability that the other guy's tie cost less than $100. Now we are told that each person's beliefs about the prices of both ties are symmetric on exchange of the two ties. For instance, P(B<100)=P(A<100), the first guy is equally sure about his tie costing less than $100 as he is that the other guy's tie cost less than $100. The other guy has equal certainty too, but maybe his beliefs are a bit different. So we also have Q(B<100)=Q(A<100), but there's no reason why Q-probabilities and P-probabilities should be the same.

They are considering the following wager: we'll ask our wives what actually were the prices of our neckties, and the one who has the cheapest will get both. So the guy with the more expensive tie ends with nothing.

By the way, it is possible that the two ties turn out to be worth the same amount, but if so, the game is called off. So more precisely, I'll take P to represent the first guy's beliefs about the game conditional on the event that A and B turn out to be different. So from now on, P(A=B)=0.

First Guy reasons as follows. If I accept the wager, and if A < B, then I end up with A+B > 2A, but if A>B then I end up with 0. By symmetry of P these two possibilities are equally likely. I have probability 1/2 of losing A, and probability 1/2 of gaining more than A. Hence I'll take the wager.

Of course, Second Guy reasons exactly the same, using his prior beliefs Q (conditioned on the prices of the two ties being different).

Neither First Guy nor Second Guy are good at doing probability. They both want to compute the expected net gain, if they accept the wager. It seems that they want to do it by splitting up the expectation value according to the two complementary possibilities, A < B and B < A respectively. This is how that should be done, from First Guy's point of view; thus the symbol E for expectation value should actually have a subscript P attached, to indicate that expectations are being taken according to First Guy's subjective probabilities P.

P-Expectation net gain = E(-A|B < A)P(B < A)+E(B|A < B)P(A < B) = ( E(B|A < B)-E(A|B < A) )/2 = 0 At the first step (first equality sign) I used the standard rule for computing an expectation value by averaging over conditional expectations in two complementary situations, knowing that in situation B < A, First Guy ends up with A less than what he started with, while in the situation A < B, First Guy ends up with B more than what he started with. Next I substitute the subjective probabilities P(A < B)=P(B < A)=1/2, by symmetry of First Guy's beliefs about the prices of the two neckties, also when given that they are different from one another. Finally I use symmetry again of First Guy's prior beliefs to show that his expectation value of the price of the other guy's tie when it is larger than his is equal to his expectation value of the price of his own tie when it is larger than the other.

Of course we knew in advance that First Guy should find the wager uninteresting. Arguing directly he could have said that with probability half I'll end up with A+B and with probability half I'll end up with 0. By the symmetry of my beliefs about A and B together, I'm completely neutral as to who has the more valuable tie, even when you tell me the price of both of them added up together. So the expectation value of A+B given B < A is equal to the unconditional expectation; that of A and B are the same; so we quickly find ''E(A+B| B < A)P(B < A) + 0. P(A < B) = E(A)''.

The more important thing is to show what went wrong with First Guy's faulty reasoning. Well, by comparison with the correct logical steps which parallel his own obviously wrong steps, we see that he is confusing expectation values, conditional expectation values, and the values themselves. He wants to compare the loss and the gain parts and show that the gain part is larger. He could correctly write the gain part E(B|A < B) as E(A+(B-A)|A < B) = E(A|A  E(A|A < B). In fact the difference is precisely, by using symmetry to exchange A and B, and then combining the difference of two conditional expectations given the same condition as the conditional expectation of the difference, of course, E(B-A|A < B).

Conclusion. Necktie paradox is indeed essentially the same as Two Envelope Paradox, it relies on shortening the probability calculations so that the reader doesn't notice that actually the writer is using conditional expectations and that the expected value of one of two amounts when it is the smaller of the two, is smaller than when it is the larger of the two.

Kraitchik's original "explanation" is hardly an explanation, he just does a careful calculation, in the situation that P makes A and B independent and uniformly distributed over 1$, 2$, ..., 100$, of the actual expected gain by First Guy and shows it is zero by symmetry. Personally I prefer my general proof by symmetry of P that P-expected net gain equals zero. But others might prefer to work through a small example just to see how the symmetry kicks in.

My analysis of "what went wrong" is that Kraitchik's rich man does not realise that the expected value of either tie when it is the least costly of the two is less than when it is the more costly of the two. I don't see how he got to be so rich. Probably just married an heiress.

By the way, both guys could just as well have given a similar, wrong, argument, why they shouldn't take on the wager!

Richard Gill (talk) 16:21, 11 August 2011 (UTC)

PS Just as in Two Envelopes, the argument pro-wager would be correct up to some point if A and {A < B} are independent: knowing A gives no information as to whether it is larger or smaller. There are probability distributions P which have this property, corresponding to such extreme level of ignorance about the amounts A and B that E(A) and E(B) are infinite. TEP provides concrete examples of distributions of (A,B), symmetric under exchange, with this independence. Hence we do need the Two Guys to be sensible enough that they are both certain of some upper bound to the prices of the two ties. Indeed Kraitchik in his discussion starts off saying that there is a bound to the values. Each can mention some amount of money (not necessarily the same) such that they are 100% certain that the prices are less than that amount. Richard Gill (talk) 06:00, 12 August 2011 (UTC)

What goes wrong
Thinking about the neckties leads me to the following succinct verbal explanation of what goes wrong in the TEP reasoning.

Considering the two scenarios that your envelope contains the smaller or the larger amount does not change what is actually in it; but intuitively, it should change your beliefs about how much is in it. In particular, it should change your *expectation value* of how much is in it.

This can very easily be made completely rigorous. For it is trivially true that E(A-B|A-B > 0) > 0. So it follows (provided the second, "smaller", term is  finite) that  E(A|A > B) > E(B|A > B). By symmetry, E(B|A > B) = E(A|B > A). Which gives me what I want: E(A|A > B) > E(A|B > A).

Because the conditional expectations in the two situations differ from one another, they must differ from the unconditional expectation, and the corresponding conditional probability distributions P(A=a|A > B) etc. must be different from one another and from the unconditional distribution.

Moreover, because of the symmetry of statistical (in)dependence, P(A > B|A=a) can't be independent of a. Knowing A changes your chances that it's the smaller or larger of the two amounts.

Summary:

1. Our beliefs about A are necessarily different when we know if it is larger than B, or smaller, or know nothing.

2. Our beliefs as to whether A is larger or smaller than B must be different when we know what A is than when we don't. More precisely, they will depend on how much it is and not always be the same.

The complementary nature of 1. and 2. is an expression of the symmetry of statistical dependence and independence. One could say, it's just "the other side of the coin"!

The only exception to these rules might occur if E(A) = E(B) = infinity. But then expectation values are no guide to decision anyway.

It turns out that Two Envelopes is made easier to understand by going back to the Two Neckties! Going from Two Neckties to Two Envelopes, irrelevant details have been added to further disguise what is going on. The irrelevant details being a specific, somewhat involved, story leading to a situation with two positive, unequal random numbers, whose probability distribution is symmetric under exchange. Richard Gill (talk) 13:17, 12 August 2011 (UTC)


 * Are things clearer when expressed the other way round. In other words we say that, for the finite case, there must always be some times (values of A) when it is not true that the unchosen envelope is equally likely to contain half or double the sum in the chosen one. Martin Hogbin (talk) 20:14, 12 August 2011 (UTC)


 * That's exactly what my little theorem proves, Martin (that "there must be some values..."). Of course, a priori it seems intuitively right. And in the *bounded* case it is easy to check (that's the case that you call "finite"). My little theorem says it *has* to be true in general - just as long as expectation value is finite. And if expectation is infinite, it's useless as a guide to action. Richard Gill (talk) 09:30, 17 August 2011 (UTC)


 * In what you call the finite case, but I'ld call it the bounded case, and for TEP, it is easier to go the other way round. For Necktie problem you still have to make another step to argue that the expectation value of A when it is smaller than B is definitely less than the expectation value of A when it is larger than B. On the other hand, the argument I gave here deals with TEP and 2Neckties simultaneously, it also deals with TEP-2, and it has a weaker condition (finite expectation values rather than all values bounded). In fact exactly the condition which is necessary to invalidate the conclusion of steps 6+7. So from a mathematical point of view it is the final, definitive, comprehensive... solution to TEP-1 and TEP-2 and 2Neckties simultaneously (when supplemented with the observation that in the infinite expectation case, expectation values are no guide to decisions). That's nice! But for a beginner and/or layperson it is probably enough to settle the matter in the bounded case.  I think the analysis also allows us to say that some of the philosophers' solutions are also "right on". The actual amount in your envelope doesn't change when you are told (or you imagine) it is the larger or the smaller of the two. But your beliefs about it are impacted in the bounded case, and more generally, in the finite expectation case! Hence conversely, your guess as to which is smaller or larger must also (at least sometimes) be impacted on being informed (or imagining that) the amount in Envelope A is some specific quantity. What the philosophers call "equivocation". We have to realise that the argument whether to switch or not, is not based on the actual amount in our envelope (or the other), but on our *beliefs* about this amount. I think it all ties together now. (For me, at least). Richard Gill (talk) 19:12, 13 August 2011 (UTC)

The problem is logically flawed
The solution of the paradox is that it is incorrect to assume that, with just two amounts of money available, you have a fixed amount of money in your hand and still the possibility of getting a larger and smaller amount by swapping (this isn't changed at all by opening the first envelope).

As an example, assume that there are two envelopes on the table, one with $50 and one with $100. If you choose the $50 envelope first, you gain $50 by swapping, and if you choose the $100 envelope first, you lose $50 by swapping, so on average you neither gain or lose anything. <BR><BR> The formula resulting in a gain of 5/4 of the original amount applies only if there are three amounts of money available (with the ratios 4/2/1) and you have the middle amount in the first place. So if you have $50 and there are $100 and $25 on the table you gain (0.5*$100 +0.5*$25 -$50) =$12.50.<BR> If on the other hand you have initially $25, you gain (0.5*$100 +0.5*$50 -$25) =$50 ,<BR> whereas if you have the $100 initially, you lose (0.5*$50 +0.5*$25 -$100) = -$62.50 ,<BR> so on average you wouldn't gain or lose anything either with this 'three envelope' situation.<BR><BR> Thomas

my comment
Suppose that all we know is one envelope contains more money and the other less money. Then there is no paradox, right? This double/half version is just a special case of that, so there should be no paradox there either.

Mathematical Fallacy
I just became familiar with the two envelopes problem, while I was searching for another. The solution seem simple and clear to me: Step 8, which states 1.25A > A is mathematically wrong, because A is not a real number. From the description of the problem, expected value of the contents of an envelope does not exist, hence a priori A is undefined. 1.25 times an undefined object is also undefined and is not necessarily greater then the object itself.

As I see, one category of propsed solution is to let a prior distribution, which makes A calculatable, hence removes the paradox. Another category is the observation of A. Suppose we found out that A = $100. Expected value of the second envelope conditional on A = $100, is indeed $125 and you should switch. But again there is no paradox, because you cannot switch back, first envelope is already open. Unless, you regard the interesting fact that you will always take the 'other' envelope as a paradox. I submit that this just intresting, not a paradox. —Preceding unsigned comment added by 193.255.135.1 (talk) 13:40, 13 April 2010‎ (UTC)

Possible Solutions
The switching argument is a form of Ignoratio elenchi in that it does not address the envelope problem as described. Instead, it describes a problem where we are given a fixed amount of money and have the possibility of either walking away with the given amount or of gambling and obtaining either twice or half the initial amount. There are thus three possible outcomes of this game:


 * 1) We leave with the amount in the first envelope (X)
 * 2) We gamble and obtain twice this amount (2X)
 * 3) We gamble and obtain half this amount (X/2)

The probability analysis as given is only valid if all three outcomes can be obtained over repeated trials. Since there are only two envelopes with two (presumably fixed) amounts, it is not possible to have three separate outcomes.

Instead, a more appropriate analysis would label the smaller amount as X and the larger amount as 2X. The expected value of each envelope is thus:


 * $${1 \over 2} X + {1 \over 2} {2X} = {3 \over 2}X$$

and neither envelope should be preferred over the other. This analysis only permits two possible outcomes over repeated trials: X and 2X, as we would expect from the problem description.

Put another way: if someone were to hand you $100 and then give you the choice of walking away with that money or of flipping a coin and obtaining $200 or $50, you would be well-advised to take the gamble as it has an expected return of $125 compared to the $100 you would obtain without gambling. If you were to repeat this experiment multiple times, each time beginning with $100 and sometimes obtaining $50 after the coin toss and sometimes $200, the probabilities would work out as given in the original analysis. However, in the envelope game this would never happen. If you have $100 in your envelope the other envelope will either always contain $200 or always contain $50 and your expected value will either always be $150 or always be $75. In this case, whether you switch envelopes or not, the expected value (and the value that you will obtain on average over multiple trials) will be constant and unaffected by your decision.

Mathematical Analysis
If we denote the possible outcomes for the first envelope as set $$\mathcal{A}$$ and the possible outcomes for the second envelope as set $$\mathcal{B}$$, and use A and B to represent variables sampled from these sets, the proper analysis then calculates:


 * $$E[ A | \mathcal{A}=\{X,2X\} ] = {1 \over 2} X + {1 \over 2} {2X} = {3 \over 2}X$$


 * $$E[ B | \mathcal{B}=\{X,2X\} ] = {1 \over 2} X + {1 \over 2} {2X} = {3 \over 2}X$$

The envelopes thus have equal expectations and neither is to be preferred over the other.

The paradoxical analysis assumes that the value in the first envelope is X (or 2X) and calculates:


 * $$E[ A | \mathcal{A}=\{X\} ] = X$$


 * $$E[ B | \mathcal{A}=\{X\}, \mathcal{B}=\{X/2, 2X\} ] = {1 \over 2} 2X + {1 \over 2} {X \over 2} = {5 \over 4}X$$

As noted, this analysis involves three possible envelope values: $$\{X,\ X/2,\ 2X \}$$, only two of which would actually be obtained in repeated trials of the experiment. As the laws of probability only describe expected outcomes over a large number of trials (see Law of large numbers), the probabilities obtained are not applicable to the experiment as described.

This analysis can be repeated with the assumption that $$\mathcal{A}=\{2X\}$$, but this will require that $$\mathcal{B}=\{X, 4X\}$$ in order to provide a positive expectation for the swapping argument. This line of reasoning allows us to demonstrate the error in the infinite swapping argument. If we assume that the first envelope has amount X in it, the presumed possible values for the second envelope are $$\mathcal{B}=\{X/2, 2X\}$$. After the swap we have two possible cases:


 * 1) $$B=X/2$$, in which case $$A=X$$
 * 2) $$B=2X$$, in which case $$A=X$$

Because we began our analysis with the assumption that $$A=X$$, it should be impossible to find any other value or expectation for A. Doing so gives rise to the paradox, and is only possible by assuming that the two cases instead permit the possibilities:


 * 1) $$B=X/2$$ and $$\mathcal{A}=\{X/4, X\}$$
 * 2) $$B=2X$$ and $$\mathcal{A}=\{X, 4X\}$$

Such possibilities would result in the infinite swapping argument. However, we have now extended our analysis to include 5 possible amounts in the envelopes: $$\{X/4,\ X/2,\ X,\ 2X,\ 4X\}$$ and it should be clear that there is a problem in our reasoning. Put simply: the two envelopes only allow for two possible outcomes to be present in our problem. If we assume the amounts to be fixed (that is, they do not change between trials of the experiment or after we have swapped envelopes) an analysis involving more than two possible outcomes is not applicable to the envelope problem as described as these outcomes could not all be obtained in repeated trials of the experiment.

Understanding the nuances of this problem can be difficult, and there does not seem to be any solution which is accepted by all scientists. The confusion and arguments surrounding this problem are not similar at all to those of the Monty Hall Problem (although some writers seem to see a similarity). The infinite swapping argument is a falsidical paradox in that it seems to make sense to the casual observer but upon careful examination is easily proved to be flawed. The infinite swapping argument assumes an ever-changing and internally conflicting domain for the set of possible outcomes of each envelope, and it mis-applies the laws of probability which are intended to provide long-term averages over multiple samples from a constant probability space.

Informal solution
A careful analysis shows that the probability assumptions used in the argument for switching are inconsistent with the laws of (subjective) probability. No rational person could ever have the beliefs which are assumed in the outset of the problem statement. To be specific, it could never be the case that the other envelope is equally likely to be double or half the first envelope, whatever the amount in the first envelope. This silly assumption led to a silly conclusion (namely, keep on switching for ever).

Suppose one of the envelopes could contain the amount a. Then apparently, a/2 and 2a are also possible. From this, also a/4 and 4a must also be possible amounts of money, and so on. It turns out that these amounts must not only be considered possible, but by use of the laws of probability it turns out that all these amounts of money must be exactly equally likely.

This certainly leads to a contradiction with a common sense approach to real world problems. The amount of money in the world is bounded so there definitely is an upper limit to the money which could be in the envelopes. Also, we don't allow for arbitrarily small amounts of money in the real world.

However, one need not invoke pragmatic principles to defuse the paradox. Also in mathematics, it is not possible to have uniform probability distributions over infinite, discrete sets. You just can't divide total probability 1 into an infinite number of pieces which are both positive and equal. So also within the abstract world of mathematics, the paradox is resolved by saying that the 2a or a/2 equally likely assumption, whatever a, can never be true. No assignment of probabilities to all possibilities can have this feature.

Formal solution
We now give the same argument in a more rigorous mathematical form.

In the Bayesian paradigm, a person who is reasoning with uncertainty in a self-consistent way is supposed to expresses his or her uncertainty about the possible pair of amounts of money in the two envelopes according to a probability distribution over all pairs. Everything is determined by the probability distribution of the smaller amount of money, since this fixes everything else (the other envelope has twice that amount, and the first envelope given to our subjective Bayesian is equally likely to be either). Given this probability distribution, we can compute the conditional probabilities that the other envelope contains 2x or x/2, given that the first envelope contains x. Suppose that these two conditional probabilities are both 1/2, for any value of x which could occur. It follows that if a random envelope contains an amount of money between 1/2 and 1, the other is equally likely to contain the doubled or halved amount, which is between 1 and 2, or between 1/4 and 1/2 respectively.

Now the smaller amount of money has to be between 1 and 2, or between 2 and 4, or between 4 and 8, ... or between 1/2 and 1, or between 1/4 and 1/2, or between 1/8 and 1/4 ... where in each case let's include the lower end-point of the interval and exclude the upper end-point. To say it a different way, we express the smaller amount of money in binary and look at the position of the leading binary digit; we now may as well round off the rest of the binary fraction to zeros. Binary 101.111 is rounded down to binary 100 or decimal 4 (a number at least as large as 4, and strictly smaller than 8). Binary 0.00101 is rounded down to binary 0.001 or decimal fraction 1/8 (a number at least as large as decimal fraction 1/8 and strictly smaller than 1/4).

After this rounding down, the only amounts of money possible in either (and both) envelopes are ..., 1/8, 1/4, 1/2, 1, 2, 4, 8, ... and all amounts are possible.

Now suppose that the probability that the smaller envelope (rounded down) has an amount $$x=2^n$$ with probability $$p_n$$, where n is any whole number (negative or positive). The probability an arbitrary envelope has x is $$(p_n+p_{n-1})/2$$, the probability that it is the smaller of the two is $$p_n/2$$. The conditional probability that it is the smaller amount can only be 1/2 if $$p_n/(p_n+p_{n-1})=1/2$$, and simplifying this equation tells us $$p_{n-1}=p_n$$, for all n=...-2,-1,0,1,2,...

Thus the amount of money in the smaller envelope (rounded down as explained above) is equally likely to be any of an infinite sequence, but there is no probability distribution with this property: we can't divide total probability 1 into an infinite number of both equal and positive probabilities.

The resolution of the paradox is therefore simple: though the assumptions appear reasonable, they are actually inconsistent with one another. No rational person who describes his uncertainty about the world in self-consistent probabilistic terms could ever believe, given that the amount of money in one of the envelopes is x, that the other envelope is equally likely to contain 2x or x/2, whatever x might be.

In the language of Bayesian statistics the uniform distribution over all (rounded down) amounts of money is an improper prior; improper priors are well known to lead (on occasion) to improprietry; in particular to self-contradiction.

Point of view from probability
From the point of view of mathematics, in particular of probability theory, Smulyan's paradox illustrates why not only possibilities but also probabilities must be taken into account when deciding on action in the face of uncertainty. The two points of view in his paradox correspond to focussing on the amount in the smaller envelope, A and the amount in the first envelope, Y. A single probability model, such as described before, incorporates both points of view in one self-consistent mathematical model. Moreover it incorporates prior beliefs about the amounts of money (and if desired also the utilities thereof) into the decision making process.

Great expectations (Broome example)
Suppose you and I play the TEP game. I fill the two envelopes by secretly tossing a biased coin with success chance 1/3 as long as it takes to get the first success. I'll put two to the power of number of failures (0,1,2, ...) before first success Euros in one envelope, twice that in the other, and let you pick an envelope at random. Suppose I have a capital of 1 million Euro's and I want to offer this game to lots of people. How much should I charge people to play the game? And how much should I charge for the opportunity to look in your envelope and decide whether or not to switch on the basis of what you see? These prices should be fixed so that on the one hand lots of people are attracted to play the game (like lots of people buy lottery tickets) while I make a steady profit (like the guys who run lotteries). Should I take out insurance on being forced to pay out more money than I have? The first problem is the basic problem of every actuary: how to fix insurance premiums. The second problem is a problem of re-insurance. And in both cases with fat tails (aka heavy tails).

Exercise: compute the first 20 probabilities (i.e. that the smaller of the two amounts is 1, 2, ..., 2^19 Euros). Note that 2^19 is a bit more than half a million, and the chance that the smaller envelope contains this amount is 15 in a hundred thousand. Small, but ... there is a 3 in ten thousand chance that the smaller amount would be even more, ie, more than 1 million Euro's! I'm going to have to take out a big insurance on having to pay out more than my capital... I am not sure I could get this insured ... in fact we run into the same problem again ... tiny chances of massive losses. It looks to me that I cannot offer the game at all, without truncating it quite heavily.

Now investigate the feasibility of offering this game at a casino in some truncated form. For instance, let's investigate truncating it at 2^15 Euro's (smaller amount), by which I mean that if I need more than 16 coin tosses I start again, tossing this coin, and waiting for the first success. And if that fails within 16 tosses, I do it again....

The maximum prize is now 2^16 Euro, ie 66 thousand Euro. The expected amount of money in the envelope with the smaller of the two amounts is 100 Euros, in the other is 200, so I could charge say 300 Euros for playing the game once and make an expected profit of 150 Euro's per game. In view of tax and overheads and bonus and shareholders etc that's not unreasonable. But I think the game could look attractive to players. The ratio of maximum prize to entrance fee is about 200 and the chance of getting that maximum prize is much better than in a usual lottery. And it can be improved by say paying another 100 Euro's to be allowed a peek in your envelope and deciding whether or not to switch. This could be a cool game in a high class casino. You can imagine other ways of paying for information. For instance, pay 10 Euro's in order to hear whether or not your envelope contains more than 100 Euro. Decide to stay or switch ... or pay more for more information...

OK so now we have a truncated version which is more or less realistic. Now change the truncation level say two steps up. Or change it two steps down. You'll find that two steps in either direction produces dramatic changes in the charges which the bank should ask to play the game, and the attractiveness of the game to gamblers. Indeed, in whether or not we can come up with a game which is interesting for the casino because it is interesting to players while generating a good income to the casino. For all these truncated games it's the case that unless the player's envelope contains the largest possible amount, it's to his advantage - in terms of expected value - to switch to the other envelope. And the advantage is by a fixed ratio 11/10 (except when the amount in the envelope is the smallest possible, 1 Euro, when the ratio is 2). Not as big as 5/4 but still clearly bigger than 1.

Morals: the untruncated game cannot be played. The results of analysis of the truncated game - for practical application - depend sensitively on where it is truncated. (And probability theory tells us that these features are not special to the particular example, but are generic). Compare with Saint Petersburg paradox. Same story. As many writers on TEP have observed. Richard Gill (talk) 13:50, 6 November 2011 (UTC)

Locate the error in steps 1 to 12
There are two envelopes. One of them, you don't know which one, contains twice the amount of the other envelope, what means that one of those two envelopes, you don’t know which one, is to contain half the amount of the other envelope.

This scenario necessarily requires that, for every single game, one of those two envelopes must first have been allocated with some unknown arbitrary amount X, with the aim that this amount X has also to determine subsequently the content of the dependent other envelope, that equally likely is to be allocated thereafter with either exactly 2X or alternatively with exactly X/2 then, so on average with 5/4 X, i.e. with either exactly duplicate or alternatively exactly half that determining amount X. Scale of proportion of those two possible dependent alternative amounts in the dependent envelope is 4:1 resp. 1:4.


 * So any determining envelope will contain an arbitrary determining amount of say: "X", while
 * any dependent envelope in consequence will contain on average 1,25 X (either exactly 2X or alternatively exactly X/2.

In the Original TEP, no one will ever know which one of the two envelopes contains the yet unknown but fixed "determining amount X" and which one of the two envelopes is the "dependent envelope" that is to contain the dependent amount of 5X/4 or 1,25 X on average, i.e. either exactly 2X or alternatively exactly X/2. For other variants e.g. the misleading paper of Barry Nalebuff Winter 1989 "Puzzles: The Other Person's Envelope is Always Greener" (where it is exceptionally well known who is holding the determining amount X and who is holding the dependent amount of on average 1,25 X) see below. Now you choose one of those two indistinguishable envelopes and call it "envelope A", and you call the unchosen one "envelope B".

You don’t know the amount of envelope A, nor whether it is the determining amount X or one of the two unknown alternative dependent amounts of 2X or X/2 in that game. You just know that – independently of "determining amount" or "dependent amount" – in any case envelope A will either contain half the amount of the unchosen envelope B, or with equal probability will contain double the amount of envelope B, and vice-versa. That’s all you know in the standard TEP.

Possible scenarios:
 * Envelope A contains the determining amount X:
 * then envelope B either contains the dependent amount of 2X resp. 2A and consequently envelope A contains B/2
 * or envelope B alternatively contains the dependent amount of X/2 resp. A/2 and consequently envelope A contains 2B.


 * Envelope B contains the determining amount X:
 * then envelope A either contains the dependent amount of 2X resp. 2B
 * or envelope A contains the dependent amount of X/2 resp. B/2

As to the consequences of swapping from envelope A to envelope B, you have to allow for two quite different scenarios, that means you have to consider both resp. all four different scenarios.


 * Either envelope A had been filled FIRST with any arbitrary determining amount X, then you can expect the contents of envelope B to be (2X + X/2) / 2 = 5/4 X. In that case you can also write (2A + A/2) / 2 = 5/4 A. But you never will know – in the given game – which envelope contains the determining amount X.
 * So this theorem will never be applicable in the ordinary standard TEP. It is applicable only in case that you should know for sure that envelope A contains the determining amount X.


 * Or envelope B had been filled FIRST with any arbitrary determining amount X, that means that your envelope A is equally likely to hold 2X or X/2, on average 5/4 X, whereas envelope B contains the determining amount of "X", say (2X /2 + 1/2X * 2 ) / 2 = "1X",  resp. in that case (2A/2 + (A/2 x 2)) / 2,5 = 0,8 "A". But, as said, you never will know – in the given game – which envelope contains the determining amount X and which envelope contains one of the two possible dependent alternative amounts.
 * So neither this theorem will ever be applicable in the ordinary standard TEP.

As – in the ordinary original standard TEP – you can never know which envelope will contain the determining resp. the dependent amount, all you can say is that swapping from A to B will give you (2X + X/2 + 2X/2 + 2X/2) / 4,5 = 1X  or (2A + A/2 + 2A/2 + 2A/2) /  4,5 = 1A

Result of switching to envelope B: either 5/4 A or 4/5 A (in case that you know for sure which envelope contains the determining amount of X) or 1A respectively (millionfold verified):  Only if envelope A contains the determining amount X, then envelope B can be expected to be (2A + A/2) / 2 = 5/4 A (as the article – without distinguishing - incorrectly claims for all types of envelopes) Only if envelope A contains the dependent amount, then envelope B can be expected to be (2A/2 + (A/2 x 2)) / 2,5  = 4/5 A If type of envelope A (determining or dependent) is unknown, then envelope B must be expected to be ((2A + A/2 2A/2 + (A/2 x 2)) / 4,5 = 1A

Other variants
Other versions (Barry Nalebuff's misinterpretation e.g.):

Player Ali gets one envelope containing some hidden determining amount X. Then a hidden coin is flipped and, depending on the outcome (head or tails), either double X or half X comes into the second envelope that is given to player Baba. In this version it is clear that Ali's envelope contains X and Baba’s envelope on average will contain 5/4 X.

Nalebuff misinterpreted this fact.


 * I do not think Nalebuff's version is meant to be an interpretation of the standard TEP but a different problem, where Ali should switch once. Martin Hogbin (talk) 22:40, 29 November 2011 (UTC)
 * Yes you are right, Martin, in this special "Ali-Baba-Variant"" it is clear that Ali, holding the determining amount of X, should switch to the dependent amount of 5X/4. But I wanted to say that Nalebuff didn't even realize that fact, but tried to show that switching from 5X/4 to X could be an option indeed. And I criticized that he did not even realize what stuff he wrote. Gerhardvalentin (talk) 23:28, 29 November 2011 (UTC)


 * You have fallen into the trap of explaining why you should not swap (which can be done in one word) rather than showing exactly where the flaw in the proposed line of reasoning is. Before you can do this even, you must first say what you think the proposed line of reasoning to be. Martin Hogbin (talk) 18:42, 29 November 2011 (UTC)


 * Dear Martin, you had better listened to what I say above: Experience enables you to be suspicious. Sometimes you hear things of which you know just from the outset, that they impossibly can be true, as in the TEP. If you see and know that things impossibly can be true, then you have to refute right away any single line, from the preamble until the end. Line 6 e.g. says "Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2" - but if you guess that "A" in step 6 designates the actual contents of your envelope "A" then you misconceived it and got it all wrong. "A" in step 6 means the "actual content of your envelope A" only in max. 50 % of cases, but never in the other half of cases. And if you hear of "5/4 A in every case", then your own experience tells you that this can be "possible" only in max. one special half of cases, but never in the other half of cases, and never ever always, then you HAVE to question the whole rest of those cheeky claims. You don't want to do that? Join doing it. Regards, Gerhardvalentin (talk) 21:34, 29 November 2011 (UTC)
 * I was not saying that you were wrong just that you have not explained it well. Martin Hogbin (talk) 22:40, 29 November 2011 (UTC)
 * Great, thank you. Kind regards, Gerhardvalentin (talk) 23:28, 29 November 2011 (UTC)

The heart of the matter
Let A and B denote random variables whose joint probability distribution encapsulates our uncertainty as to the actual amounts a and b in the two envelopes. I do not need assume here that A is half or twice B. I just assume that A and B are always different and that their distribution is symmetric under exchange. The following facts can therefore be used for two envelopes (all symmetric versions), two neckties, two-sided cards; with or without subjective probability, with or without finite expectations. The derivation is elementary. The results are not surprising. The point is that they are general results. Many solutions take a particular prior distribution by way of example and show that certain of these facts are true. That is a bit unsatisfactory because it doesn't prove that the results always have to be true, hence leaves a doubt in the mind of the reader. For example, this is why Martin Gardner felt that neither Kraithchik's problem nor TEP were properly solved at the time when he wrote about them. He had only seen particular examples but this does not prove that what we see in those examples always has to be true.

Theorem


 * (1) Under symmetry, E(A)=E(B).
 * (2) Under symmetry, if E(A) is finite, then it is impossible that E(B|A=a) > a for all a.
 * (3) Under symmetry, it is impossible that P(A < B|A=a)=1/2 for all a.

Proof

(1) is obvious (symmetry!)

(2) proof by contradiction with (1). If E(B|A) > A then E(B)>E(A) or both are infinite or undefined.

(3) proof by symmetry of "stochastic independence" between r.v. A and event { A < B }. Because if P(A < B|A=a)=1/2 for all a, then the event { A < B } is independent of the random variable A. Now replace A and B by A' = g(A), B' =g(B) where g is a strictly increasing function from the real line into a bounded interval of the real line (for instance, the arc tangent function). All the assumptions we made about A and B also hold for the transformed versions, but now we can be certain that expectation values are finite. From now on, I drop the "prime" and just write A and B for these transformed versions. Consider the trivial inequality E(A-B|A-B > 0) > 0. By finite expectation values, this can be rewritten as E(A|A > B) > E(B|A > B) = E(A|B > A) where the last equality uses symmetry. This inequality shows that A is statistically dependent on the event { A > B }, hence the event { A > B } is statistically dependent on the random variable A. Transforming back to the original variables this remains true.

Corollary (an exercise for connoisseurs/students of probability theory). Let g be a strictly increasing function and let A' = g(A), B' =g(B). Then the theorem also applies to the pair A' and B' . Extend to not necessarily strictly increasing g by approximating by strict and going to the limit (strict inequalities need no longer be strict in the limit). We find


 * (4) The probability distributions of A|A < B, of A, and of A|A > B are strictly stochastically ordered (from small to large).

These facts take care of the main variants of the two envelopes problem as well as all its predecessors two neckties, two-sided cards. The only way to escape the facts is to assume improper distributions. But they are ... improper. In fact, they are: ludicrous, according to Schrödinger, Littlewood, Falk, and just about everyone.

Restricting attention to TEP: if the writer of the paradoxical argument is trying to calculate E(B|A=a) in order to prove E(B|A=a) > a for all a, then he seems to be using in the course of his calculation that P(A < B|A=a)=1/2 for all a, but by result 3 this can not be not true. If he is trying to calculate E(B) in order to prove E(B) > E(A), then he seems to be using in the course of his calculation that E(A|A > B)=E(A)=E(A|A < B), and furthermore he is confusing E(A) with A. But in any case, the only way that we could have E(A|A > B)=E(A)=E(A|A < B) would be when all three expectation values are infinite. But in that case, optimizing expectation values cannot be used to optimize decisions, since one will always be disappointed with what one gets relative to the expectation value anyway. Alternative approach in this situation: the utility of having arbitrarily large amounts of money does not increase indefinitely. The player's utility for any amount of money x, say, is some function g(x) which increases as x increases but is bounded. We should replace A and B by g(A) and g(B). But these two transformed variables satisfy all the assumptions of the theorem too. In particular the expected utility of what's in a given envelope is strictly smaller when that envelope contains less than the other, than when that envelope contains more.

This material is also posted on my university home page at. Richard Gill (talk) 14:06, 1 December 2011 (UTC)
 * That seems to cover all the angles nicely. Martin Hogbin (talk) 17:47, 4 December 2011 (UTC)


 * You say in the introduction that it doesn't matter if the distribution has infinite expectation or not, but in your theorem you explicitly exclude that case. So it does indeed matter. This is the first lie. Then in the end you say that the only way to escape your solution is to assume improper distributions. But hey, do you really think that all distributions having infinite (or undefined) expectations are improper distributions? I know you know better than that. This is the second lie. Also, you have forgotten to make all your assumptions explicit, like stating that all values can be mapped onto the (infinite) real line. This assumption is not true in practice why your theorem (2) breaks down in reality. And you don't take account here for the fact that you believe that utility is bounded above at some arbitrary level (as long as it is finite). And where is your opinion represented that you say that we have to truncate the support for some proper distributions to escape the paradox? Here nothing is bounded and nothing is truncated. I guess you are in your math Nirvana now that has nothing to do with reality, right? In that case this is about as interesting as theology. iNic (talk) 08:45, 6 December 2011 (UTC)


 * Only statement (2) makes an assumption of finite expectations. A and B are random variables, the terminology implies they are real-valued. Every single TEP variant is about monetary values or utility, always supposed to be real valued. My longer paper shows how the mathematical facts mentioned here deal with every knoen TEP variant, including the other exchange paradoxes (neckties, cards). Statement (4) can be used to deal with the infinite expectation variants of TEP. The utility of money is a strictly monotone, bounded, function of the actual amount. Statement (4) implies that it is impossible that E(B|A=a) > a for all a, where A and B are the utilites of the amounts of money in the two envelopes, not the amounts themselves. So, iNic, this little bit of mathematics is useful for anyone who studies TEP and understands elementary probability theory. But for those who want to study TEP but can't appreciate elementary probability theory, it is difficult to say anything useful at all, I'm sorry about that. Richard Gill (talk) 17:01, 9 December 2011 (UTC)


 * (1) is just a statement of symmetry and not a result. The interesting question is if you claim that (1) is true even after looking in one envelope. Most authors say that symmetry is then broken. I don't see the novelty in the results (2) and (3). They are known and proved by other authors since long ago for the symmetric case (not opening any envelope). And once again, are they valid even after looking in one envelope? If so, why? And how do you apply these probability calculations to the non-probabilistic variant? All this will be interesting to read in your forthcoming paper. So the usefulness of these results to anyone studying TEP depends 100% on how you manage to actually show that usefulness. I also note that you now say that utility is a "strictly monotone, bounded, function of the actual amount." However, just a few days ago (20 November) you happily claimed that the utility of a cheque of money "might even start decreasing again and eventually fall close to zero, since all cheques written for astronomically large amounts of money are worth about nothing." Did you learn something new about elementary utility theory since November 20? I really hope so, otherwise you are just contradicting yourself (once again). iNic (talk) 09:44, 10 December 2011 (UTC)

Why steps 9 to 12 of the TEP "switching argument" is complete nonsense, while step 7 can only be correct if unknown presuppositions were given.
Steps 1 to 8 are correct only under the never to be given strict precondition that the distribution is KNOWN to be extremely asymmetric, in every game with some "fixed invariable amount" as an asset base contained in envelope A, serving as the reference base for determining the contents of the dependent envelope B that was filled thereafter.

then in 1/2 of cases envelope B has later been filled either with the "dependent amount" of 2A and, in the other 1/2 of cases, envelope B has later been filled with the other "dependent amount" of 1/2 A. Only in this one very special and never to be given fictive case the relation  A : B  will be  1 : 5/4  resp. 4/5 : 1. And this fact, in turn, makes steps 9 to 12 nonsensical: For in step 9 within this game, quite contrary to the original assumption, suddenly envelope B is explicitly assumed now to contain a fixed unchangeable amount serving as the asset base for this game, whereas envelope A suddenly is said to having been filled later and to dependently contain, with equal probability, one of the two alternative amounts that are newly dependent on the sudden "fixed invariable reference base" contained in envelope B: The newly "dependent" envelope A contradictorily is said now to contain either 2B or 1/2 B.
 * If envelope A is explicitly KNOWN to contain some fixed INVARIABLE AMOUNT in every game, as an invariant asset base for that game,

As the starting precondition of  "envelope A is KNOWN to be the unchangeable asset base ..."  can nor will ever be given, all one ever really must assume is a completely unknown and symmetric distribution with the relation "A : B"  and  "B : A"  to be  "1 1/8 : 1 1/8",  say 1 : 1. That should clearly be shown in the article.

The flaw in the TEP consists in an indistinct description of the situation, in changing reference assets and changing mappings of baseline and target value (without expressly admitting this conversion) and using this as a basis of a mathematical theorem that can be valid just under very special never to be given assumptions in just ONE direction. The flaw in the TEP especially is the absurd claim that a fixed asymmetric distribution is presupposed to be "KNOWN", with some "fixed" amount in envelope A as the only reference base, and any "dependent" amounts in later having been filled in envelope B, but then in step 9 abruptly changing that ungiven assumption contradictorily to its exact opposite. While, just from the outset, one indeed could as well unsecured presuppose that the amount in envelope B "could" strictly be fixed within every game, with any dependent amounts later having been filled in envelope A, where one had to expect that swapping from envelope A to envelope B on average will give 4/5 A only.

It should be shown that – as effectively nothing is known about the actual distribution – for this reason one forever has to act from the assumption of an unknown symmetric distribution with the result of "1 : 1". Gerhardvalentin (talk) 17:27, 6 December 2011 (UTC)


 * But once you open your selected envelope A the amount of money in that envelope becomes fixed, right? Let's say you find 512 monetary units in your envelope. This is a fixed number. Then for sure the other envelope must contain either 1024 or 256 monetary units, right? The other envelope only has these two possible contents and as we have no reason to believe in one of the cases more than the other, the probability for each possible amount must be 1/2, right? We use here the same mundane principle we use when we determine that the probability of the sides of a fair die is 1/6 each, for example. If that principle can't be used in this case we have to know why. We also need to know what new principle for determining P(B contains 1024) and P(B contains 256) we should use instead, and of course what probabilities that principle leads to. And last but not least we need to have a clear, explicit and exact rule for when this new principle should be used instead of the old one. Almost all authors miss this last step completely. iNic (talk) 09:47, 11 December 2011 (UTC)

Proposal:

Problem
The basic setup: 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 are offered the possibility to take the other envelope instead.

The switching argument: Now suppose you reason correctly as follows:
 * 1) I denote by A the amount in my selected envelope with unknown content that can either be "1x" or "2x".
 * 2) The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2.
 * 3) The other envelope may contain either 2A or A/2.
 * 4) If A is the smaller amount of 1x then 1x never can be "halved", but the other envelope contains 2A, i.e. 2x.   (In contrast to the claim it is never given that "any A" could be halved.)
 * 5) If A is the larger amount of 2x then 2x never can be "doubled", but the other envelope contains A/2, i.e. 1x.   (In contrast to the claim it is never given that "any A" could be doubled.)
 * 6) Thus both envelopes contain 1x with probability 1/2 and 2x with probability 1/2.
 * 7) So the expected value of money in envelope A, as well as in envelope B is   $$E(A) =E(B) = {1 \over 2} (2x) + {1 \over 2} x = {3 \over 2}x$$
 * 8) As both envelopes are expected to hold the identical amount of 3/2 x each, there is no incentive to swapping, and neither to swapping back.

The puzzle: The puzzle is solved if you present a correct theorem.

Regards, Gerhardvalentin (talk) 16:35, 9 December 2011 (UTC)


 * Sure, this is a correct argument, after your notation s corrected (distnguish A from E(A); write x instead of X). But most people think that the writer was trying to compute E(B|A=a), ie what you would expect to see in the other envelope if you peeped int he first and happened to see an amount a in there. The writer wants to show it is larger than a whatever a might be, hence you don't  have to peep. Gerhard, you still haven't grasped the Anna Karenina principle and you are still being careless with notation. And you are trying to solve a much less interesting interpretation of the problem. Richard Gill (talk) 17:13, 9 December 2011 (UTC)


 * Gerhard, this is very similar to one of the solutions presented on my page. The annoying thing about the TEP is that resolving the paradox requires you to first make clear exactly what you think the proposed line of reasoning is supposed to be.  In other words you have to decide exactly what sort of stupidity the proposer is guilty of.  As Richard has pointed out, they may mean something slightly different for your suggestion by their proposed argument and you need to also show why this is wrong.  In fact you need to show why everything the proposed line of reasoning could mean has a flaw in it and ideally show where that flaw is.   I have spent some time discussing this with Richard and would be happy to explain what I mean in more detail to you if you like. Martin Hogbin (talk) 18:13, 9 December 2011 (UTC)


 * Thank you Martin for your efforts, perhaps you could tell me there on my TEP-page. Regards, Gerhardvalentin (talk) 18:48, 9 December 2011 (UTC)

As the situation is symmetric you don't have to do a single calculation to show that there is no point in switching. This is not the problem at all. This is the setting of the problem, not its solution. iNic (talk) 19:09, 9 December 2011 (UTC)


 * The flaw:  The evidently wrong theorem is implying that "any A" forever can be doubled, and that "any A" forever can be halved, lapidary and shamelessly using the generalizing diction "2A" and "A/2", leading to 5/4 A. While this is valid only in one very special constellation like in Barry Nalebuff's "Baba"-version. Whereas, quite contrary, there could exist exclusively pairs consisting of each two related envelopes, each pair containing a total amount of precisely 3, e.g. always the amount of "1" in one of the two envelopes of each pair, and precisely the amount of "2" in the other envelope of each so called pair of two envelopes. If you select an envelope amounting to "2" it will never be possible to "double" this amount, a.s.o., a.s.o.  Gerhardvalentin (talk) 21:47, 9 December 2011 (UTC)

Please tell me which number that can't be doubled, and which number that can't be halved. That would be interesting to know. And besides, does the TEP argument really talk about doubling or taking halves "forever"? I don't think so. The argument only mention the double amount once (2A), and half of the amount once (A/2). So only one step in both directions are needed for the argument. That is quite far from an infinity of steps. iNic (talk) 00:37, 10 December 2011 (UTC)


 * Yes iNic, in effect the TEP in steps 1 to 8 is talking about on average doubling A and halving A in every game, yes, in any game. And in step 9 it even says that, on average, also any "B" can be doubled and halved in any game. Yes, in every game, without exception. The TEP simply is presenting "obvious nonsense" in saying so. Have a look to the above example: Two envelopes, the two amounts in any pair of envelopes are 1 and 2 in the above example. No knowledge about any determining amount nor about any dependent amount.
 * You choose one envelope and then call it envelope A, the unchosen envelope is called B.


 * A     B   (all supposable two permutations and total in this symmetric example:)
 * 2     1
 * 1     2


 * 3     3


 * If A=2 in the above example then, by swapping to envelope B,  A never can be doubled to "amount 4", but A only can be halved to "amount 1". But if A=1 then A never can be halved to "amount 1/2", but A only can be doubled to "amount 2". That "any" A can be doubled in 1/2 of cases and "any" A can be halved in the other 1/2 of cases simply is a false syllogism in every symmetric example.


 * If you know that the envelopes can only contain 1 or 2 of some amount of currency, then for sure you should stick if you get 2 and switch if you get 1. The whole point is that you don't know how much the envelopes contain. The symmetry is due to total ignorance what the contents might be. So your example is a bad model of TEP. iNic (talk) 00:14, 11 December 2011 (UTC)
 * I did not say peep in envelope A. The question is: What scale is A:B, and reality shows that, with symmetry in choosing your envelope A, this scale of A:B will be exactly 1:1. Gerhardvalentin (talk) 17:43, 11 December 2011 (UTC)


 * No, the proportion between A and B is by definition 1:2. As we don't know which is the biggest one it's true that we pick the larger one with probability 1/2 and the smaller one with probability 1/2. But this doesn't imply that the proportion between A and B is 1:1, that there is no difference between A and B. If someone picks you or your sister at random, it is equally likely that you are picked as your sister is picked. Sure. But this doesn't imply that you and your sister have the same sex. iNic (talk) 10:34, 12 December 2011 (UTC)


 * The "Baba"-example (that's what steps 1 to 8 of the TEP are based on, without saying so) shows quite another pattern: One envelope is filled with any arbitrary determining amount first (say "2" e.g.),  and this envelope is called "envelope A".
 * Only afterwards, after the determining amount had been fixed in this extremely asymmetric version, a second envelope called "B" is filled then randomly with the dependent amounts of 2A in 1/2 of cases and with A/2 in 1/2 of cases:
 * A     B
 * 2     4
 * 2     1


 * 4     5

But then, in step 9, the TEP abruptly is changing determining amount and dependent amount, saying now that the contents of envelope B is the determining amount and that envelope A now is holding one of the two dependent amounts:
 * A     B
 * 4     2
 * 1     2


 * 5     4


 * Nothing at all is changed abruptly in step 9. As the situation is totally symmetric w r t A and B it obviously doesn't matter which one you start with. This means that everything we say in terms of A can be said in terms of B as well. Step 9 only states this obvious fact. iNic (talk) 00:14, 11 December 2011 (UTC)
 * Reality shows that if you intentionally choose envelopes containing the determining amount then all "B's" will contain 5/4 A. And if you intentionally choose envelopes containing the dependent amount, then all "B's" will contain 4/5 A. Just do a reality test. Gerhardvalentin Gerhardvalentin (talk) 17:43, 11 December 2011 (UTC)


 * There is no "determining amount" and no "dependent amount." Please drop that notion from all your arguments. In fact, it doesn't matter the slightest bit how the amounts were chosen that goes into the envelopes. iNic (talk) 10:34, 12 December 2011 (UTC)


 * iNic, that's the error in the TEP. Vague language and unclear wording as a base of a mathematical theorem, based on bloody fallacies.  Gerhardvalentin (talk) 11:54, 10 December 2011 (UTC)


 * I think the language is very clear in the TEP argument. But I don't see any theorem there. iNic (talk) 00:14, 11 December 2011 (UTC)
 * I meant the theorem "E(B) = (2A + A/2) /2 = 5/4A". Gerhardvalentin (talk) 17:43, 11 December 2011 (UTC)


 * This is not a theorem. It is just a simple application of the formula for expected value. iNic (talk) 10:34, 12 December 2011 (UTC)


 * One aspect is: If the TEP would admit to be just "based on the extremely asymmetric AliBaba variant", then steps 1 to only 8 are fully correct. No doubt:
 * Swapping from Ali to Baba gives you 5/4 Ali on average. But the TEP is not just based on AliBaba, but it misleadingly claims to be "valid in general", even in a symmetric world, and that's the real "sin", besides the nonsensical claim of step 9 in abruptly changing determining and dependent amounts.


 * Step 9 is very far from nonsensical. Due to the symmetry between A and B step 9 is very reasonable. iNic (talk)


 * I see that the error lies in steps 1 to 12. The incorrect and ambiguous formulation of steps 1 to 6 do allow the wrong "general" conclusion of step 7 to be drawn, that forever is reserved just for the extremely asymmetric AliBaba-version only.


 * So it will be necessary to show  "how"  steps 1 to 6 are to be correctly formulated and interpreted correctly so that the incorrect general conclusion of step 7,  that only can be valid within extreme asymmetry, firmly based on the determining amount to be typically contained in envelope A, can be avoided to be mistaken to be valid for the symmetric world also. Subliminally, steps 1 to 6 always, without explicitly saying so, do imply "determining amount" and "dependent amount", although this is never given nor correct within a symmetric world. It is the misleading wording of steps 1 to 6 that, instead of explicitly excluding any "asymmetry with determining and dependent", even are expressively and typically implying it. And as said, without ever admitting that fact. That's the problem.  Gerhardvalentin (talk) 16:12, 10 December 2011 (UTC)


 * It is not true that the only way to arrange the two envelopes is to have one "determining" amount and the other one twice or half of that. You can for instance have a probability distribution on the real numbers that determines the result in both envelopes simultaneously. For example, lets say that x is a random number from such a distribution. With probability 1 we know that x must be strictly between two whole numbers, n < x < n+1. Put 2^n monetary units in one of the envelopes and 2^(n+1) in the other. iNic (talk) 00:14, 11 December 2011 (UTC)
 * Regrettably, this argument fails the test of reality, proven millionfold. The scale of proportion may be 1:2 or X:Y or X+Y or X-Y, or a mix of these scales. If there is symmetry, the total of all A's and all B's will be 1:1, no way out. Gerhardvalentin (talk) 17:16, 11 December 2011 (UTC)


 * What? "Fails the test of reality"? When did your opinions become part of reality? Of course there is symmetry, that's the whole point. In fact we have more than one symmetry to take care of here and that is what creates the paradox. If taking all symmetries into account we run into a contradiction. So the question is which symmetry is the most important one, and which can (or even should?) be ignored. iNic (talk) 10:34, 12 December 2011 (UTC)


 * Exactly, iNIc. We are just told that the two envelopes are filled with two amounts of money, one is twice the other. We are not told how. Now randomness comes in: we, the player, pick one of those two envelopes at random. Call the amounts in the two envelopes A and B; call the smaller amount X and the larger amount Y. So what we know is that Y=2X>0 and that (A,B) = (X,Y) or (Y,X) with equal probability 1/2, independently of the amounts X and Y. So if we want to use probability theory to analyse this situation, the only "free" ingredient is the probability distribution of X, the smaller of the two amounts. Everything else follows from this. Choose x. Double it, call that y. Pick one of the two at random, call that a. Call the other b. It doesn't matter how the two amounts are actually fixed, we can always think of the smaller amount as being fixed first. Because any procedure to fix both amounts fixes the smaller, and the other is then twice the smaller.


 * Well I don't agree with you that we are completely free to assume that X is picked from a probability distribution, i.e., that X is random. The most natural interpretation is that the donor donates one of two amounts that are picked deliberately by the donor. The reason is of course totally unknown for us but that is not the same as being random. If I don't know who your wife is it's not correct to assume that you picked your wife at random. On the contrary, I'm quite sure she wasn't picked at random. Same with donations. We could make a poll and ask donors big and small if they ever let a random process decide how much of their own money they should donate. Not likely they would say that they ever did that. So all justice we have to talk about probability distributions in this case is if we are bayesians (or if we are dealing with a version of TEP where the random mechanism for populating the envelopes are explicitly given). So let's say we are bayesians. iNic (talk) 12:30, 13 December 2011 (UTC)


 * Theorem (Gill's unified solution): if X has a proper probability distribution then it is impossible that P(A < B | A=a ) does not depend on a. In particular, it cannot be 1/2 for all possible values of a. Hence it is impossible that E(B | A=a) = 5a/4 for all possible values of a.


 * No, there are proper probability distributions for both these cases. The Broome distribution is a proper probability distribution satisfying the first one, and if we drop the unnessessary R assumption we can find nice, finite and proper probability distributions for the second case as well. iNic (talk) 12:30, 13 December 2011 (UTC)


 * Broome's distribution does not have P(A < B | A=a ) independent of a.


 * What do you mean? For the Broome distribution the expected value of B is larger than A given that A = a for every value of a. You mean that the case a = 1 where the gain is 100% differ from the rest of the cases when the gain is only 10%? Is that the crime? If so, why? iNic (talk) 20:22, 18 December 2011 (UTC)


 * iNic asks the question which symmetry is the important one, and which can be (or even should be?) ignored? The answer is easy. We are informed that one symmetry is true, for sure. The other symmetry is then invalid! That's all there is to say about it. Or ... you want to go outside of probabity theory. Well, then you are completely alone, and you are in Terra Incognita. The only thing everyone agrees on is that the only logical calculus of uncertainty is the standard calculus of traditional probability theory. Richard Gill (talk) 22:14, 12 December 2011 (UTC)


 * As soon as you see that both symmetries can be valid at the same time you realize that all proposed solutions so far are not correct. iNic (talk) 12:30, 13 December 2011 (UTC)


 * Both symmetries cannot be valid at the same time, if A and B are supposed to be numbers and we are working in the framework of conventional probability theory. I am looking forward for an interesting example by iNic where the "values" we are talking about are not numbers, but can be totally ordered. The set of possible values is going to have to be of cardinality greater than the real numbers. And because the values are not numbers the notion of expectation value does not exist. It is hard for me to imagine an appealing TEP paradox in this set-up. Richard Gill (talk) 09:21, 17 December 2011 (UTC)


 * Before I can reveal this version you have to answer the fish soup situation. The lesson from that example is important for the next step. iNic (talk) 20:22, 18 December 2011 (UTC)


 * Sure, as soon as you realize that everyone goes in the wrong direction you have to start going in the right direction, even if it feels like Terra Incognita in the beginning. However, to me this solid ground since long. iNic (talk) 10:36, 14 December 2011 (UTC)