Talk:Two envelopes problem/Arguments/Archive 4

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What is the paradox?

I have just thought of something that might shed some light on the disagreement on this problem.

What exactly do we think is required for the problem to be considered paradoxical? In the article itself the last steps in the argument are:

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.

I think we all agree that for the player to swap endlessly is an absurd situation and therefore we have a clear paradox. I also suspect that everyone agrees that this paradox has been resolved for all cases.

No, this paradox has not been resolved. iNic (talk) 13:07, 25 January 2012 (UTC)

INic can you give in a new section below, or on the arguments page, an example where you think there is still an argument endless exchange. Please give a full and clear description of the exact setup and of the line of reasoning that you believe leads to a paradoxical conclusion. State when and if any person looks in their envelope. Martin Hogbin (talk) 20:10, 25 January 2012 (UTC)

Please read the Wikipedia article, it should give you the information you are looking for. iNic (talk) 17:21, 26 January 2012 (UTC)

The WP article, quite obviously. does not tell me which version you have in mind. Martin Hogbin (talk) 09:25, 31 January 2012 (UTC)

Aha in my mind? Well, in my mind there are only one paradox, only one version. All the different "versions" in the literature are due to the fact that all proposed solutions are flawed. A bad solution doesn't solve the problem if the problem is tweaked just a little bit, or slightly restated. Hence the series of never ending new "variants" of the problem. But the correct solution solves them "all" in one blow. iNic (talk) 09:36, 31 January 2012 (UTC)

At the end of his paper, Nalebuff talks of what he seems to consider a paradoxical situation that is more like the two neckties paradox. There is no suggestion that the player would swap endlessly but he does seem to consider it paradoxical that both players should find it to their advantage to swap in some cases. This seems to me to be a weaker paradox and I (and Nalebuff it seems) wonder if this paradox has been fully resolved in all cases. Martin Hogbin (talk) 11:38, 25 January 2012 (UTC)

To swap back and forth is in itself not paradoxical. Many processes go back and forth endlessly without being a paradox. Think of a pendulum for example. The paradox here is that we have rational decision theory urging us, as it seems, to do a series of rational steps that when thinking of it as a whole doesn't seem to be rational at all. This is the paradox here. How can a theory of rational behavior immediately lead to irrational behavior? The "twin construct" that is used in the neckties version is a little different. There the paradox is that something must be wrong if both persons think that they gain value by swapping their neckties. It seems to create value out of nothing. But if both gentlemen happen to think that the other tie is prettier, then they do in fact both gain if they swap. So in this case one need to be more accurate when presenting the story so that a paradox is really created. I guess this is why the common formulation of the paradox today is in terms of envelopes and money and not neckties. However, when looking in one of the the envelopes the twin construct is still useful to derive the paradox. iNic (talk) 13:07, 25 January 2012 (UTC)

In the necktie problem the gentlemen are not interested at all in which necktie is more pretty but which one cost their wives more money to buy for them. They apparently both think for the same reasons that on average they will individually get richer, in expectation value, by switching. Since they can easily imagine themselves in the position of the other person, they both think that together they will get richer on switching, yet they know that together the total cash price of their neckties did not change. So there was a paradox.

To swap back and forth imagining that each time you swap you get richer (on average) is a paradox, since you know that on switching twice you definitely have not got richer.

I think that the paradoxical nature of the different versions is pretty clear and moreover I think they have all been adequately resolved more than 20 years ago. Since then there has been a lot of repetition, a lot of re-inventing wheels, and a lot of noise. But of course ... that is just my personal opinion. On the other hand I think it is a pretty objective fact that there have not been any new "solutions" to the version (studied by mathematicians and economists) in which we imagine the writer of the TEP argument being concerned with conditional expectations since 1989, and that there have not been any new "solutions" to the version as imagined by the philosophers (and to a lesser extent psycholgoists and educationalists) who, not knowing about conditional expectation, thought that the writer was calculating an unconditional expectation value, since about 1995. There has been one claim, by Schwitzgebel and Dever, that they are the first to explain what went wrong in the philosophers' version, but I don't see any novelty in it (note: nor has their proposal been enthusiastically adopted by authoritative writers from philosophy, it has not even been discussed. Just a few papers give the citation in order to cite all known published papers in the field.)

There have been two "new" paradoxes: Smullyan's word game without probability, and Cover's trick with a randomized way to increase your expectation values when you do look in your envelope (which was already known in 1992). Smullyan's paradox has been studied in a small number of very academic papers from philosophy. Writers who are using different theories of counterfactual reasoning naturally find different solutions. But they all do find solutions.

There has been a lot of "noise" in the economics literature about infinite expectation values, in which people are endlessly arguing whether infinite expectations makes sense in real world practise. And this discussion is a repeat of the Saint Petersburg Paradox endless discussion. On the other hand this discussion says nothing about the question of where is the *logical* mistake in the sequence of apparently *logical* steps leading to a clearly nonsense conclusion. So though in the wikipedia article one can discuss this question a little, it seems to me of tangential interest. The essential paradox is resolved by demonstrating the illogical steps of the reasoning. Those steps have been well understood for a long time now, even if they happen to be different according to different interpretations of what the writer is doing. Richard Gill (talk) 14:58, 25 January 2012 (UTC)

You may find the paradoxical nature of the different versions is pretty clear but it surely would be good practice, for the benefit of laymen and experts alike, to make absolutely clear in each case exactly what is considered paradoxical. It also would seem a good idea to decide on exactly what constitutes a 'resolution' of the paradox. In my opinion a proper resolution requires identification of the error in the proposed line of reasoning not merely showing that exchange is not beneficial. I am sure that some of the noise that you complain about comes from this confusion. Martin Hogbin (talk) 10:04, 26 January 2012 (UTC)

The word "paradox" means different things in different contexts. The Wikipedia article gives a good overview. iNic (talk) 17:21, 26 January 2012 (UTC)

Yes, I know what a paradox is and I can see the paradoxical line of reasoning in WP but there are other cases where the setup and proposed paradoxical line of reasoning might be different. It never hurts to make things crystal clear and thus avoid pointless arguments. To just show that you should not swap is pointless, we all knew that at the start by reason of symmetry. Martin Hogbin (talk) 18:09, 26 January 2012 (UTC)

Exactly! This was what I tried to tell you when you used this argument yourself. I'm glad you realize now how pointless it is. iNic (talk) 16:24, 27 January 2012 (UTC)

I would be happy to point out the error in your line of reasoning of only you would give it, with the exact setup. Martin Hogbin (talk) 19:20, 27 January 2012 (UTC)

Please start right there and replace your pointless argument with a valid one. That will be interesting! iNic (talk) 09:36, 31 January 2012 (UTC)

I think the article does do just that, now, Martin. It shows for each of three or more variants of the problem, or if you prefer, different ways to imagine what the problem is, where the logical error is in the reasoning that leads to a clearly nonsensical conclusion. Now the Broome version is one which, if you are a mathematician, is easily mathematicallly (=logically) resolved, but it still remains a weird phenomenon for ordinary people. I think that is the reason why many authors slap on top of the logical analysis the further, strictly speaking superfluous, claim that in any case, infinite expectation values are stupid / unreal / whatever. Nalebuff does this, Falk does this. Many others do. Like other weird constructions in mathematics (eg Banach-Tarski) it is weird. Banach-Tarski is called a paradox because it insults our imagination and intuition, but there is no actual logical contradiction. Just an impossibility for our little brains to visualise the thing.

So I guess we can say that the Two Envelope Problem provides, among its many variants, paradoxes of both kinds: things which are logically correct but hard or impossible to imagine hence hard to believe; and things which are logically incorrect and where it suffices to point out the logical error.

I think this is a new insight in the field of TEP Studies. Thanks! Richard Gill (talk) 19:12, 26 January 2012 (UTC)

Exactly what new insight did Martin give you now? I didn't get it. iNic (talk) 10:07, 27 January 2012 (UTC)

Richard, I do not disagree with anything that you have written but you seem to have missed my point. In the article the only claimed paradox that I can see is that, by the given argument, the player should swap endlessly. Nalebuff, however sees a different paradox for some versions, namely that both players (note that there is only one in the setup given) expect to gain by swapping. These are not the same thing. Martin Hogbin (talk) 19:53, 26 January 2012 (UTC)

Yes, in the article we are copying a particular, later, statement of the problem due to Ruma Falk. It's not the original. Earlier versions tended to have two people each in possession of an envelope (wallet, box, necktie ...) and both standing to gain, on average. We, outsiders, can follow both persons' arguments so apparently must support both of their conclusions, yet we know that the total does not increase - so how come can we advise both to switch? Richard Gill (talk) 14:01, 29 January 2012 (UTC)

The new insight I got is that a problem can be solved, since we can find the error in the reasoning, but there can remain a paradox, in the sense that intuition still does not agree with the result of logical deduction. Richard Gill (talk) 14:33, 28 January 2012 (UTC)

The Broome Two Envelope Problem can be resolved, we can see where the argument breaks down. But we retain a Broome Two Envelope Paradox, since it remains counter-intuitive that if we repeatedly create two envelopes containing two numbers according to the Broome recipe, and give them to Alice and Bob, then on all those occasions that Alice has a number a in her envelope, the average of the contents of Bob's envelopes is 11a/10 > a, while simultaneously, on all those occasions that Bob has an a number b in his enveloe, the average of the contents of Alice's envelopes is 11b/10 > b; for all possible values of a and b >1. (When a or b =1, replace 11/10 by 2).

Yet the distributions of the contents of both Alice's and Bob's envelopes is the same!. Each possible value 1, 2, 4, 8, ... occurs, relatively speaking, equally often. Richard Gill (talk) 14:49, 28 January 2012 (UTC)

It's funny because Albers (2003) has exactly the opposite opinion: "The two-envelope paradox is easily explained, but the corresponding problem is more difficult." (p 29). So the total disagreement among scholars abound. Not even a simple high level description of the situation is possible to agree upon. iNic (talk) 09:55, 29 January 2012 (UTC)

These authors are colleagues of mine in the Netherlands, I know them well. When they say two-envelope "paradox" they refer to the paradoxical (in the sense of leading to a counter-intuitive result) derivation of always switching. When they talk about two-envelope "problem" they mean the problem of figuring out what the owner of Envelope A should do when he or she actually looks in his envelope first, in a real world situation. To solve this problem one needs more input. For instance: a prior and a utility and the principle of maximizing expected utility. If you accept the principle and you take a prior and a utility function as given, the problem has become soluble, it's a well defined mathematical optimization problem. But how do you figure out a prior and a utility? Anyway, why should you accept the principle of maximizing expected utility? In this sense the TE problem is not solved. You would probably agree, iNic, that in this sense it is not solved. That's my opinion too. In fact I would say it cannot be solved in general, and it's a waste of time to look for a general solution. Albers et al also say that in some situations it can be better not to give an answer at all. I too agree with this. Often the most valuable result of a statistical consultancy is that the statistician can explain to the client that his problem is ill-posed, un-answerable, and that the client should re-consider what it is he wants to do. Richard Gill (talk) 13:55, 29 January 2012 (UTC)

PS Albers et al. are not referring to the Broome paradox when they make these remarks. Richard Gill (talk) 14:03, 29 January 2012 (UTC)

OK so you and your peers in the Netherlands agree that for the Broome version the paradox is not solved but the problem is solved, while for the rest of the variants it's the other way around? Very confusing indeed. iNic (talk) 10:50, 30 January 2012 (UTC)

No. The other versions are easier. The problem is solved, there is no paradox either.

As you said yourself, the word paradox has a number of different meanings. I would like to distinguish a problem to be solved (where does an apparently logical deduction go wrong), and a situation which it counter-intuitive. Nothing to be solved, though there is something to be learnt: intuition need not be correct.

But first of all, let's define the Broome problem and paradox precisely. Suppose two numbers are generated and written down according to Broome's recipe and put in two envelopes, which are labelled completely at random (independently of the contents) A and B. The recipe is: take a coin with probability 1/3 of heads. Toss it just as long till you see the first head. Call the number of tails before the first head t. It could be 0, or 1, or 2, .... The two numbers are 2 to the power t and 2 to the power t+1. Let A and B denote the random numbers inside envelopes A and B.

Now comes the problem: what is wrong with the following reasoning? The owner of Envelope A reasons (1) given what is in his envelope, say a, the expected contents of the other envelope are 2a if a=1 and otherwise 11a/10; in any case, it is strictly larger than a. He reasons that therefore, (2), without looking in his envelope, he will become better off by switching to the other envelope.

The conclusion is obviously false since the marginal probability distributions of A and B are identical. Exchanging one closed envelope to the other does not improve his situation at all - it leaves his situation identically the same to what it first was, since the bivariate probability distribution of the ordered pair A,B is the same as the bivariate probability distribution of the ordered pair B,A.

Our problem, as I said, is to explain where the apparently logical reasoning breaks down. The answer is that it breaks down at step (2). In step (2) we are silently using a theorem which requires a condition which is not satisfied in this case. The conditions of the theorem do not apply hence we cannot deduce that the conclusion of the theorem is true.

Still, we are left with the paradoxical (in the sense of counter intuitive) fact that in many, many repetitions, for any particular amount in Envelope A, the average of what's in Envelope B is larger, and vice versa.

Problem solved, paradox remains. The paradox that remains does not have a solution and does not need a solution. We have learnt that our intuition about averages can break down when infinities are involved. This has been known for several hundred years. This is not a paradox which can be solved, but it is a situation to which one can become accustomed ... in other words, one can develop a better intuition.

Von Neumann said: one does not come to understand new mathematics, one gets used to it. This is the sort of paradox which goes away by being aware of the possibility and getting used to it. It doesn't go away by logical deduction.

Unfortunately, many writers on TEP are not happy with this situation. They use overkill by trying to argue that the Broome situation could never turn up in the real world. In other words: sorry guys, we are stuck with the paradox, but don't worry your little heads about it, you won't ever have to deal with it in practice. Richard Gill (talk) 13:43, 31 January 2012 (UTC)

Richard I agree with what you say and I have a suggestion that might help in your mathematicians vs the rest argument. When talking about the finite, physically reasonable cases it is alright to use a real physical representation of the problem, in particular to use terminology like 'envelope, and 'money'. When referring to the unbounded versions I think it would be better to use strictly mathematical terminology, 'set', 'number' etc.

It is quite right to say that no one will never have to deal with the untruncated Broome game in practice. Infinite amounts of money do not exist, neither do envelopes large enough to contain them. On the other hand, we can deal with infinities mathematically.

The problem is caused by a desire to glamourise the more spectacular results of mathematics. In the formal section on the Banach-Tarski paradox [[1]] we see the problem expressed mathematically. The problem is that it does not look very exciting so we are tempted to talk of gold balls which can be doubled in size. Of course, real gold balls cannot be doubled but the mathematics is perfectly correct. The same problem exists in QM where writers attempt to explain QM results in terms of classical physics. This produces exciting paradoxes but is not particularly helpful in understanding the subject. Martin Hogbin (talk) 18:19, 31 January 2012 (UTC)

Richard, you say that "In step (2) we are silently using a theorem which requires a condition which is not satisfied in this case. The conditions of the theorem do not apply hence we cannot deduce that the conclusion of the theorem is true." What theorem is that? iNic (talk) 10:23, 1 February 2012 (UTC)

Theorem. Suppose E(X) is finite. If E(Y | X=x) > x for all x, then E(Y) > E(X). Richard Gill (talk) 20:58, 7 February 2012 (UTC)

The 'simple solutions' explained

The simple solutions, as exemplified by the solution given by Falk in her 2008 paper, are correct but only in particular circumstances which have not thus far been clearly identified. In her paper, Falk says:

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’.

So far this resolution is in accordance with Nalebuff's solution.

She continues:

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).

It is not quite clear to me exactly what this bit is saying. What exactly is, 'the smaller amount'?

Falk explains further by giving a revised version of the paradoxical line of reasoning:

1. I denote by A the amount in my selected envelope.

2′. Denote by S the smaller of the two amounts.

The probability that A = S is 1/2 and that A = 2S is also 1/2.

3′. The other envelope may contain either 2S or S.

4′. If A=S the other envelope contains 2S.

5′. If A=2S the other envelope contains S.

6′. Thus the other envelope contains 2S with probability 1/2 and S with probability 1/2.

7′. So the expected value of the money in the other envelope is 1/2 2S+1/2S=3/2S.

8′. This is equal to the expected value of A, which is 1/2 S+1/2 2S=3/2S, so I gain nothing on average by swapping, and I am indifferent between keeping and swapping.

My assertion is that this argument, and many similar ones, are perfectly correct iff 'the smaller amount', as represented by S above, is a constant. In other words it has a fixed value. This means essentially that the solver of the problem knows the possible sums in both envelopes. In other words, our envelope space contains only two elements.

Later Falk says:

Had step 2 been adjusted to say ‘the probability that A is the smaller amount, denoted As, is 1/2, and that it is the larger amount, denoted Al , is also 1/2’, then the same symbol A would not have been used for the two cases in steps 6 and 7, and the paradox would have been avoided.

What are As and Al here, random variables? If so the argument above does not really make sense. On the other hand if they are both constants everything is clear.

In the Broome version of the game, because of the way the sums in the envelopes are determined, it is not true that the sum in the unchosen envelope is equally likely to be half or twice the sum in the chosen one - unless, after the two sums in the envelopes have been determined, we are told what they are, reducing our envelope space to only two elements.

In my opinion this is the crucial fact missed by many of the authors of papers on the subject. You cannot ignore how the sums in the envelopes are determined in your calculations unless, after they have been determined, you are told both sums. What you can never, in my opinion do, is have variable constants.

Richard, I would be interested to hear your view on this observation. Prof Falk, if you are reading this, I am sorry for using your paper as an example of what I consider to be incomplete thinking, but I would be interested to hear from you also.Martin Hogbin (talk) 10:22, 8 February 2012 (UTC)

I agree. Except, that the analysis does not require the amounts to be known. We can do probability calculations conditional on the values of the two sums of money (the smaller and the larger), without knowing them. I am afraid I have had so many fruitless discussions with amateurs about the fact that one can usefully think about E(Y|X=x), the expectation value of one random variable given that another takes on a particular value, without that value actually being known or given. Some of those amateurs were professional physicists, who certainly could do mathematics better than me, in the sense of having a fantastic intuition for finding really smart ways to do extremely difficult calculations. Richard Gill (talk) 10:51, 9 February 2012 (UTC)

Some more thoughts on this. The word "variable" has a very precise meaning within rigorous (precise, formal, academic, professional, theoretical, abstract, ... ) mathematics. And then always accompanied by a quantifier "for all" or "there exists". Examples: for all x in some set, x satisfies some relation; or: there exists x in some set, such that .... Brackets indicate the precise scope of the quantifier. The word "constant" is not used, and a word like "fixed" or "known" also has no meaning. Careful mathematics is like careful computer programming. Mathematical variables are just like variables in computer code, they are merely place holders, and one is careful to show which places are supposed to be occupied by the same thing (scope; local versus global variables), and what this thing can be (type).

Within probability theory, but only within probability theory (and branches of mathematics built on probability theory, like statistics) we use the words "random variable" and "constant" with rigorous, precise, formal meanings. Within probability theory the notion of conditioning is useful. One can say things like "the probability of A, given B" but this does not mean that B is known to have occured. It simply means (if you like frequentist language) "imagine many, many repetitions of the probability experiment we are talking about. Restrict attention to all those occasions, and only those occasions, when B occurs (completely irrelevant whether anyone knows this to the case, or not). What is the relative frequency of A within this sub-family of repetitions?".

Unfortunately, ordinary folk think that variables must vary and constants must be fixed and possibly also known, and that we can only talk about the probability of A given B if B is known (known to someone, to have happened). But in probability theory, a constant is merely a random variable with a degenerate probability distribution. A random variable is a (deterministic) function from the set of possible atomic outcomes of some probability experiment to some set of possible values, typically real numbers. Conditioning on an event, does not mean that we force the event to happen. Working given X=x does not mean that we know or force X to take the value x. It merely means a conceptual restriction to all those events, and only those events, when the random variable X happens to take the value x. The symbol x denotes a mathematical variable. The context must show what it can be, and whether the name x is only being used locally or whether it is being used globally (or something in between).

Another word with precise technical meanings, various ones in fact, within mathematics, is "independent". Another recipe for confusion. Richard Gill (talk) 09:28, 10 February 2012 (UTC)

The philosopher's stone

[Moved from the main discussion page.]

I think you have missed a point though. You say, 'The solution above doesn't explain what's wrong if the player is allowed to open the first envelope before being offered the option to switch', but it also does not explain what is wrong if the first envelope is chosen from a non-uniform distribution, such as in the Broome game. We both know that the probability that the original envelope will hold the smaller sum is not, in that case 1/2. The 'common resolution' does work if there are only two possible sums in the envelopes (or if after they are chosen we are told what they are) in other cases it is nonsense. Martin Hogbin (talk) 19:16, 8 February 2012 (UTC)

The way I think about this (i.e., according to my own POV), the common resolution is not a resolution at all. But the second resolution is completely adequate. And the way I think about it, the Broome paradox is a completley new paradox, built on the second resolution. You thought you had flattened the bump in the carpet but then someone changes the rules a little, and a new bump came up behind your back. Broome does not suddenly make all the steps of the original TEP argument correct. We have just argued why they can't all be correct! Broome requires us first to rewrite the steps, with a correct computation, for his rather special circumstances, of the conditional expectation of what's in B given what's in A (which involves probabilities 3/5 and 2/5, not 1/2 and 1/2).

I think that Broome can justly be called a new variant. It is a different paradox, but clearly inspired by the original. The original is the original... (Nalebuff, Gardner). The original has been interpreted in two different ways in the literature: the second interpretation was the original interpretation, the first interpretation became dominant in the philosophy literature and to some extent the popular literature.

To make things more complicated still, Nalebuff already had a Broome-type example, in order to show that maybe the solution of our second interpretation isn't the whole story.

All in all, two envelopes problem is actually a rich family of related paradoxes; some are "just" logical paradoxes, easy to resolve by showing the incorrect logical step; others are brain-twisters, things which are hard to imagine but which can actually be true. These things which force us to expand our minds do have impact on thinking in applied fields, because the challenge to intuition needs to be addressed from the practical point of view. Richard Gill (talk) 20:07, 8 February 2012 (UTC)

I think you have missed my point. The common resolution makes sense in so far as it shows that the unconditional expectation in both envelopes is the same but it only does this if there are only two possible sums in the envelopes.

If the first envelope is chosen from, say, the set {2,4,6} the common resolution does not even calculate the unconditional expectation. Do you see what I mean? Martin Hogbin (talk) 22:24, 8 February 2012 (UTC)

I think I see what you mean, and I think in the common resolution the author is silently taking the two amounts of money in the two envelopes to be fixed. The writer does not distinguish random variables from constants. He writes A and X but he should write A and x. It takes years to teach students to distinguish between capital letters and small letters. Nowadays handwriting is so bad they don't even make any distinction in their ordinary writing.

There would be no TEP if the writer had carefully made the distinction, as many reliable sources have said!

So I have been bold again. I have added an expanded and mathematically respectable version of this common resolution as a sub-sub section to make all this explicit. Similarly I added more mathematical details of the second resolution as a sub-sub section. I hope in this way both laypersons can be satisfied, and mathematicians can be satisfied. The two resolutions are now written out as parallel stories which only differ (probabilistically speaking) according to whether we are conditioning on the pair of sums of money (e.g., on the value of the smaller amount), or we are conditioning on the amount in Envelope A.

For a probabilist this is the important distinction. The layperson does not really know this distinction. Instead, the layperson thinks of things being known and things being not known. Unfortunately, this puts the lay person on the wrong track.

The layperson who has a nodding acquaintance with elementary algebra and calculus thinks that there is an important distinction between variables and constants. True, the distinction is important, but here, another distinction is important. The distinction between random variables and possible values of random variables. Whether those possible values of random variables are thought of as constants or variables, and whether they are actually known or not known, is almost irrelevant.

I think here we are at the heart of the problems of the bifurcation of TEP into a philosopher's problem and a probabilists's problem. Many people have quite a lot of mathematical training but at the same time almost no training in probability. Such people can easily see TEP in the philosopher's way rather than the probabilist's way. An interesting example is given by Ruma Falk, who has taken both points of view in different papers over the years, with different co-authors. By training she is an educationalist and a psychologist, and not a statistician or a probabilist. Though she has written extensively on probability problems it seems to me that she writes more as a mathematician than as a probabilist, so occasionally making the wrong distinctions, and not using the optimal notation. Richard Gill (talk) 10:37, 9 February 2012 (UTC)

Richard, is the explanation of how the simple solutions arise and the precise problem that they do answer anywhere in the literature? It may seem obvious now but I have not seen this explanation in the literature that I have looked at. Martin Hogbin (talk) 16:33, 9 February 2012 (UTC)

Very good question. I am hoping that the paper by Thomas Bruss, which I have not yet read, does this. He is a well known fundamental mathematics oriented mathematical statistician, and statistical decision theorist. So I can't imagine that he has written something superficial or even wrong on TEP. I mailed him, and he has posted me an offprint of his 2006 paper. At the moment, I admit, this is an original synthesis by me (though it is completely elementary).

On the other hand, you could say that I am merely expressing in modern probability formalism, what a number of philosophers have said in lengthy articles full of long words and difficult philosophical concepts. I merely abstracted the mathematical essentials from their words. Made sense of their verbal stories in the only way possible which makes mathematical sense, hence presumably is indeed the intended meaning. These guys write papers which are bloody hard to read, but on the whole they do not actually write nonsense,Richard Gill (talk) 19:09, 9 February 2012 (UTC)

That is not quite what I was getting at. I have not seen this explanation offered anywhere until I mentioned it just above. Martin Hogbin (talk) 19:57, 9 February 2012 (UTC)

I think you have seen these explanations in my own draft paper! But writing for mathematicians you possibly did not catch what I was saying. By the way, Rawling is definitely thinking of the smaller amount as being fixed, he calls it s. And by his analysis the probabilities 1/2, 1/2, are correct, it is the substituted values which are wrong. So his analysis, with high probability, corresponds to what we now have.

He also uses words like "syncategorematic". I suspect that none of us actually understands his paper at all, but just grabs at a few fragments which seem recognisable. I suspect that many writers who cite him also hardly understand a word of what he says, except for the phrase "the cardinal sin of equivocation", which people love to quote. I suspect that none of them even saw the word "equivocation" before. (I think this particular paper really is a load of BS). Richard Gill (talk) 20:30, 9 February 2012 (UTC)

I think we must be talking about slightly different things. In your paper, you say, According to [the philosophers'] interpretation we are aiming at computation of E(B)... but that is just crazy. However we may propose to arrive at it a computation of E(B) is completely pointless. As I think you have pointed out elsewhere, it is a fixed number, there is absolutely no point in comparing this with random variable. I cannot believe that the philosophers are being this stupid. What I am saying that they do is that they are conditioning everything, in your notation, on X=x (or Y=y) right at the start of the problem. Martin Hogbin (talk) 09:47, 10 February 2012 (UTC)

Probably we are talking about different things. The philosophers are not mathematicians. My point of view is that the simple resolution does not assume the amounts in the envelopes to be known and fixed. It merely assumes that we are doing a probability calculation conditional (in the probabilistic sense) on X=x, where X denotes the smaller of the two sums, and x is a possible value thereof. The correct calculation shows that E(B|X=x)=3x/2. This statement is conventionally abbreviated to E(B|X)=3X/2. By averaging again, over all possible values x of X, one recovers E(B)=3E(X)/2. As a special case, X might be imagined to be constant. As an even more special case, it might be supposed to be known. Doesn't make any difference. Also, it doesn't make a difference whether probability is taken in subjective or frequentist sense, or as a mixture of the two. A frequentist might typically take X to be a constant random variable, equal to the fixed value x. Denoting the smaller amount by x, the name of a mathematical variable, does not imply that we think of x as being known or unknown, constant or varying. We simply to a calculation which applies to any situation, whatever x might be. Richard Gill (talk) 09:53, 10 February 2012 (UTC)

PS what amazes me is that the philosophers do completely fail to notice that there is another big thing wrong in their calculation. They are obviously computing an expectation (maybe a conditional expectation). But they end up on the right hand side of their equation with A, they seem to have forgotten that according to an earlier step A was apparently a random variable. So maybe quite a few of them are exactly that "stupid" as not to notice this. Richard Gill (talk) 09:56, 10 February 2012 (UTC)

We probably should be having this discussion somewhere else, on the arguments page for example.

I do not quite understand what you are saying. You say above, ' My point of view is that the simple resolution does not assume the amounts in the envelopes to be known and fixed', but you go onto say, 'It merely assumes that we are doing a probability calculation conditional (in the probabilistic sense) on X=x, where X denotes the smaller of the two sums, and x is a possible value thereof'. Is that not the same thing? If we condition on X=x then the smaller sum (and hence the larger sum) has a known fixed value. Martin Hogbin (talk) 10:14, 10 February 2012 (UTC)

One can meaningfully compute probabilities and expectations conditional on X=x, with or without actually knowing x. In fact that is the whole point of the second interpretation. Without looking in Envelope A we can imagine looking and seeing A=a. The symbol a is a mathematical variable. It stands for some number, some amount of money. Any amount of money. Similarly, doing probability calculations conditional on X=x carries no implication whatsoever that someone in this story actually knows x. That's the whole point of use of mathematical variables in doing algebra. You can say "let x be the solution" without knowing x and even without knowing if there is a solution or not, or many solutions or only one. Richard Gill (talk) 11:36, 10 February 2012 (UTC)

OK then I think I means something different from you. Let me explain exactly what i mean in two different ways then you can tell me how to express this more formally.

From a frequentist perspective we imagine the experiment to be repeated many times. The sum in one envelope is determined by some process (for example uniformly at random from a fixed sample space or by the Broome method) then double that sum is put in another envelope. Then a fair coin is tossed and according to the fall of the coin the player receives one of the envelopes.

If that process is repeated from the coin-toss stage only, that is to say with the same starting envelopes then, in my opinion, the philosophers' solutions are fine. From a Bayesian perspective this equates to the player being told the sums in the two envelopes, even though they may change from game to game.

This is not quite the same as saying that the sums in the envelopes are fixed, they are always fixed; nobody changes their value at any time. However, if they are fixed and known then Falk's resolution, for example, makes sense. Martin Hogbin (talk) 13:13, 10 February 2012 (UTC)

Yes, I agree! Falk's solution silently assumes we are thinking (in frequentist terms) of repetitions in which only the labeling (A and B) of the envelopes varies from repetition to repetition. For a probabilist this is the same as saying either that X and Y are taken to be constants (as opposed to random variables) from the outset (so maybe better to use lower case x, y); or that we are probabilistically conditioning on particular values X=x, Y=y. There's no mathematical difference, from here on. (Mathematically, the second possibility is more general - it includes the first).

The second resolution, thinking of A being fixed, you could say, makes more sense when X and Y are at the outset thought of as random (e.g., with probability distributions representing our prior uncertainty as to their actual values). But it seems that for many philosophers this was too sophisticated to contemplate. And possibly for many amateurs, too. That would explain the popularity of a resolution like Falk's. She can explain it much better to a class of young educational psychologists than the second. Richard Gill (talk) 10:50, 11 February 2012 (UTC)

I thought this might be an important point to make in the article but I am no so sure now. See below.

Why we will never find the philosopher's stone

My turn now to agree with you Richard. The literature (and consequently this article) is a mess. I think the reason is clear and returns us to one of the fundamental issues of statistics and probability, that without a well-defined question there can be no clear answer.

We start with an ill-posed setup. We then have a bogus proposal which we first have to interpret and then find the flaw in our own interpretation. Nalbuff (and maybe a few others) rounds up all this chaos and, giving reasonable interpretations of what it is supposed to mean, pretty well resolves it all.

Other writers, often philosophers, then come along and give alternative resolutions but invariably without saying exactly what paradox they are trying to resolve. Many produce long and mathematically dubious arguments to show something we knew from the start, that we should not swap. The aim should be not to persuade ourselves that endless swapping is pointless, that is obvious, the objective is to find the flaw in a proposed and plausible argument that we should. This, to my mind, requires that we should first clearly state the bogus argument and then show exactly where the error in that argument lies. Most of the literature completely fails to do this and works on the basis that everybody knows exactly what the paradox is. That is false; if it were true there would be no paradox.

So what to do with the article? It is a mess because Richard has made it reflect the literature on the subject. That is fine but it is of little value to the general reader. I would like to add a section for the general reader, at the start of the solutions section, based mainly on Nalebuff's paper on the subject and giving simple, plain language but mathematically sound resolutions for the paradox given at the start of the article. I would hope to get a consensus for this. Martin Hogbin (talk) 12:25, 11 February 2012 (UTC)

Personally I think the article is in quite good shape now. But then, I'm someone who (by now) already knows the literature and moreover has specialist interest in the technical niceties. So maybe it's a good encyclopedia article for an encyclopedia of statistics, but not for a general encyclopedia.

Maybe the comprehensive survey of the literature should be preceded by a short layman's article, restricting attention to highlights only, and concentrating on the original TEP (what we now call the second interpretation).

Why don't you just be bold and draft such an article, Martin? It could also go in Simple Wikipedia. Richard Gill (talk) 15:17, 11 February 2012 (UTC)

Sorry Richard, I meant it was incomprehensible for the layman. As an expert discussion I am sure it is fine. Martin Hogbin (talk) 00:49, 12 February 2012 (UTC)

No apologies needed. I agree with you! Be bold... Richard Gill (talk) 09:10, 12 February 2012 (UTC)

I like the new Intro !!!! Richard Gill (talk) 18:06, 13 February 2012 (UTC)

Thanks. I think it provides an explanation for the general reader. Martin Hogbin (talk) 23:02, 13 February 2012 (UTC)

This intro is bad in so many ways that I don't know where to begin. Best thing to do would be to delete it and improve the article instead. Make it readable and fun again, as it was several years ago. iNic (talk) 01:33, 14 February 2012 (UTC)

Perhaps you could start by saying what you think is factually incorrect about it. It is what the sources say. Martin Hogbin (talk) 09:40, 14 February 2012 (UTC)

It was placed in the same section as the "A common resolution," yet that solution wasn't even mentioned in your intro. Weird. The non probabilistic version wasn't mentioned either, of course. It's totally fine that you and Richard doesn't like or understand these ideas personally but it's a clear violation of NPOV to leave them out as you two constantly do (but Richard has improved in this respect lately). In addition you introduce a lot of Bayesian concepts in your intro without any explanation whatsoever what it is or why some authors think these concepts are needed and useful in this context. In particular they are not needed at all in the first "A common resolution" that appeared directly after your intro. They are needed in other common interpretations of the problem, but not all. This need to be explained, not just taken out of the blue. If we make the entire page more readable an intro of this kind is not needed. If written in this way it confuses more than what it helps. iNic (talk) 10:13, 14 February 2012 (UTC)

As I am sure Richard will agree, the 'common resolution' that we have now is not really a resolution of the stated paradox at all, it does not say where the error is in the proposed line of reasoning, it does not even use the same notation. It is just a reason why you should not switch, which is obvious.

My introduction is not a summary of all the resolutions but a statement of the most common resolutions from mathematicians. If you want a summary of the whole subject, the place for this is the lead. That could do with rewriting anyway. Martin Hogbin (talk) 17:04, 14 February 2012 (UTC)

So why don't you say that then? Replace the header "Introduction" with "My personal summary of all the resolutions I think are the most common from mathematicians." Then please sign your section so that all new readers will know who's view it is. iNic (talk) 10:13, 15 February 2012 (UTC)

This afternoon I was discussing TEP with two mathematicians with quite strong interests in philosophy. One of them said: the difference between mathematicians and philosophers is the following. All a mathematician needs is lots of pens and paper and a rubbish bin. All a philosopher needs is lots of pens and paper.

I plan to give TEP (wikipedia) a break for a while. It will always be hard to write a satisfactory article since the literature is so large, so much of it is very low quality, and there always will be people who look at is from a more mathematical point of view and people who look at it from a more philosophical point of view, and honestly, I think it may take another hundred years before these groups start to understand one another's point of view. There do not exist previously existing surveys of the whole literature which are both authoritative and neutral. This one on wikipedia is the first, ever! Let some new editors get involved. Richard Gill (talk) 21:08, 14 February 2012 (UTC)

Yes I agree. More editors need to be involved. Both with and without a rubbish bin. (At last I understand why it was such a big quarrel about the right to have a rubbish bin between Apple and MS a long time ago. Thank you!) iNic (talk) 10:13, 15 February 2012 (UTC)

PS we were talking about decision theory. The standard von Neumann-Morgenstern theory based on utility functions is simply no use for the real world. Suppose you were offered either 2 M Euro, or a 50/50 chance of getting either 0 or 5 M Euro. Unless you are Bill Gates, you would probably prefer the former. Suppose on the other hand you were offered either a lottery ticket which has a one in a million chance of winning two million Euros, or a lottery ticket which has a one in two million chance of winning five million Euros. Almost everyone would prefer the latter. Its a mathematical theorem that there does not exist a utility function (expressing the value of money) such that both of these preferences follow by supposing that people want to maximize expected utility. This is the Allais paradox. This is why "usual" Bayesian decision theory is a waste of time. Not only are the preferences I just mentioned the ones which most people actually do make in practice, I think we would mostly agree that they are entirely rational, too. Richard Gill (talk) 21:17, 14 February 2012 (UTC)

Yes Allais showed that the standard von Neumann-Morgenstern theory is flawed. But Allais didn't show exactly in what way it is flawed, how the theory can be fixed. The defenders of the theory has said that even if we have anomalies like this the vN-M theory is anyway the best decision theory we ever can get because any deviation from their axioms would lead to even larger irrational behavior. The interesting thing about TEP is that it shows the way to a real solution to this old problem, which Allais' problem never did. iNic (talk) 10:13, 15 February 2012 (UTC)

Well, according to the experts I talked to yesterday, there are well known ways to generalize vN-M so that Allais is no longer an anomaly. For instance John Quiggin's Rank-dependent expected utility. I think they correspond well to human psychology too. Of course the problem you get with generalizing a theory is that it fits everything and explains nothing. There are so many free parameters that you can always make it "explain" what you have seen in the past. The question is, is it useful for predicting things in the future? I don't know. Richard Gill (talk) 10:59, 15 February 2012 (UTC)

Yes true, but in these cases the resulting theory is purely descriptive. No trace of the rationality ingredient—that was so important for the early probabilists—are left. The vN-M theory is the last attempt to create a theory for the rational man. But I'm not an expert in this field so there might be later theories that I'm unaware of. I for sure need to read more in this area. iNic (talk) 14:38, 15 February 2012 (UTC)

As I understand it, Quiggins' approach has two components: utility is how one values money, but there is a different and new component which describes how one values risk. Nothing irrational in enjoying small chances of winning large amounts, and at the same time "playing safe" concerning large amounts. If only the bankers had behaved that way with our money! So one does not abandon rationality. On the contrary, the vN-M characterization of rationality simply did not do the job. I think Keynes already emphasized this. And vN-M were not dogmatic: they started with something simple. Before there was nothing. We now know their approach is inadequate. But still it forms a platform for further developments. Richard Gill (talk) 07:06, 17 February 2012 (UTC)

OK but for every new "refined" theory the concept of rationality becomes more and more diluted. It has now reached homeopathic levels of dilution. We can't distinguish a rational behavior from an irrational anymore using these theories. Everything goes. Any behavior can be "rationalized" afterwards as the theory has so many free parameters. I would say that not a single atom of the original ingredient of rationality is still left in this new homeopathic concept. Try to use Quiggins' theory during an evening at your local casino and you'll understand what I mean. iNic (talk) 14:13, 17 February 2012 (UTC) 

Sure, you don't have to tell me. Richard Gill (talk) 11:19, 18 February 2012 (UTC)

This is not a paradox.

Let us say you are given two indistinguishable envelopes, each of which contains a positive sum of money. One envelope contains twice as much as the other. You may pick one envelope and keep whatever amount it contains. You pick one envelope at random but before you open it you are offered the possibility to take the other envelope instead. You therefore swap envelopes.

The above statement is not a paradox. Nobody sees a reason to swap until a bogus line of reasoning is presented to them which appears to show a rationale for swapping. Consequently, just showing that you should not swap cannot be a resolution of a paradox. For a paradox to be resolved there must be one in the first place.

To resolve the TEP you must first state a paradoxical line of reasoning then show what is wrong with it. Anything else is pointless waffle. Martin Hogbin (talk) 09:40, 29 February 2012 (UTC)

Exactly, Martin. To resolve the TEP you must first state a paradoxical line of reasoning, and then show what is wrong with it. Literature in abundance.

If you take the two "colours" of red and green, then B can be red only under the condition that A is green, and B only can be green under the condition that A is red. Either, or, and no way out. Without any reference whatsoever to "values" or to "amounts".

In the switching argument of the TEP lines 1 to 6 are fully correct only for the asymmetric case that A clearly represents the determining amount, and B clearly just represents the derived, the dependent amount. That B entirely and fully depends on a predetermined A.

And lines 1 to 6 may also be "valid" for the case that there should be symmetry, but only as long as no one considers those lines to imply any asymmetry of any "determining" and "dependent". So these lines are very incomplete and too imprecisely worded to help prevent fallacies. For in the symmetric case, without any "determining" and "dependent" amounts,

the other envelope B can never contain 2A unless A = B/2, and

the other envelope B can never contain A/2 unless A = 2B. Just an excluding "Either - Or".

Meaning that "B = 2A" only can be possible if "A = B/2" (as it is with red and green), and "B = A/2" only can be possible if A = 2B. Just an excluding "Either - Or", and never any "as well as", like in the asymmetric case, adding both and dividing by 2.

So it is a complete illusion to conclude that B ever could be 5/4 A in the symmetric case.

To clearly see that, take my example of "one envelope contains the 200 fold amount of the other envelope." Then, supposed A determines B, switching from A to B will give you on average the 100.0025 fold amount of A, and switching from B to A will give you B/100.0025. But in the symmetric case, like "red and green", all you ever will know is that both, A and B likewise, can be expected to be (A+B)/2. And nothing else.

So at least line 7 is "the flaw", in misinterpreting lines 1 to 6 to mean the asymmetric case only of a strictly "determining A" and some strictly "dependent B". Literature in abundance. Gerhardvalentin (talk) 15:52, 8 March 2012 (UTC)

Please have a look to my attempt to distinguish "either/or" from "as well as". Gerhardvalentin (talk) 08:30, 30 March 2012 (UTC)

A possible fault

The 'flaw' I see is that by line 9 (switch, assume content to be B), people discard the previous lines 1-8. Line 9, following the same train of thought of lines 1-8, should give rise to the following:

9.1) Denote by B the amount in the currently selected envelope.
9.2) The probability that B is the smaller amount is 1/2, and that it is the larger amount is also 1/2.
9.3) If B is the smaller amount, then A is the larger amount: (A,B=A/2).
9.4) If B is the larger amount, then A is the smaller amount: (A,B=2A).
9.5) Thus the value in the previously selected envelope is A. This is smaller than the expected value of B (A*5/4).
10) Therefore, conclude that there is no further need to swap.

I think the paradox works that way: By discarding lines 1-8, one starts the entire argument from the beginning. By discarding lines 1-8, the important relationship between A and B is not established (that B is only either 2A or A/2). There may be other errors in the argument, but this is what I see as the central flaw of the argument. 112.205.100.28 (talk) 13:51, 24 July 2012 (UTC)

The stated problem ignores the value of the chosen envelope, which is lost on swapping.

You have selected one of two envelopes, it contains X or 2X each with a probability of ½. If you swap to the other envelope then you lose the contents of the envelope you are holding and gain the contents of the other one. So with a probability of ½ you lose X and gain 2X or with a probability of ½ you lose 2X and gain X. Your expected gain or loss is:

½(-X + 2X) + ½(-2X + X) = 0

In the long run, there is no gain or loss by swapping. --Norman Sheldon (talk) 08:31, 3 September 2012 (UTC)

I think this is the correct solution to the so-call paradox, because when you switch you expects gains not value of your envelope which is given to you at the first place. - Tom Cheung

Random Question

Wouldn't something along the lines of there being a cost (in terms of time) to switching also fix this problem? Regardless of how much money is in the envelope, wouldn't you eventually want to be doing something with your life than sitting there switching envelopes? Subtracting some amount, c (cost), with every switch should eventually make their utility from the problem negative. Jabberwockgee (talk) 19:59, 21 May 2012 (UTC)

Critique 1

I will switch if I selected the envelope with the smaller amount or stick if I selected the envelope with the larger amount. Knowing an amount only helps if it provides a bias between which one I selected (envelope with the smaller amount or the envelope with the larger amount).

Consider an adjustment to the basic set up that gives you even more information. You are told that there is a population of many pairs of envelopes such that one envelope contains twice as much as the other and your pair has been randomly drawn from this population. Together with the pair of indistinguishable envelopes you are given a spread sheet. The spread sheet lists every envelope pair in the population from which your pair has been randomly drawn, shows the number of each of these pairs in the population and (so you do not have to calculate it yourself) the percentage frequency this pair has within the whole population.

1. You have all the distribution data that you need for amounts in pairs of envelops, do you switch envelopes?

2. You open your selected envelope. It contains A; you are invited to switch. Now if the value A only appears in one place in the spread sheet, then you can determine if it is in the envelope with the smaller amount or the envelope with the larger amount. Now it is clear whether to switch or stick. Otherwise it appears twice, once as the smaller amount (in the pair A and 2A) and once as the larger amount (in the pair A/2 and A). Switching in the first gives a gain but switching in the second gives a loss. The percentage frequencies in the spread sheet will help you decide if you expect to gain or lose by switching. There is no guarantee that you made the right decision in this arbitrage.

3. You are told that one of the two envelopes contains 20 dollars, do you switch?

4. You are told that one envelope contains 20 dollars and the other envelope contains 40 dollars, do you switch?

Instead of the basic set up, you are given two indistinguishable envelopes. You are told that one envelope contains a US Government Bearer Bond to the value of 100,000,000 dollars and the other contains several sheets of used lavatory paper. Do you switch? Well the question you have to ask yourself is “Do I feel lucky?” Well, do ya, punk?

--Norman Sheldon (talk) 10:39, 12 September 2012 (UTC)

Critique 2

Specifically, in the switching argument, if I accept steps 1 to 8 as valid then step 9 is invalid. After the switch I agree that I can denote the contents by B, however I have arrived at this step by nominating the original envelope as containing A. The only values B can take that are compatible with this are 2A or A/2. Each has a probability of ½ so my expected value for the holding is 1.25A which is greater than A and I will not switch again.

However, I cannot accept steps 1 to 8 as valid. This is because the steps do not identify the correct entities to evaluate. The basic set up starts with two indistinguishable envelopes, i.e. an envelope pair drawn from a population of envelope pairs, yet all further analysis concerns itself with an individual envelope but not the pair as an entity. An envelope pair is made up of two envelopes whose contents are positive amounts such that the smaller amount is X and the larger amount is 2X. X is a variable whose value can be any of the smaller amounts available from an envelope pair in the population of envelope pairs. The probability distribution for X (i.e. the smaller amount) is given by P(X), hence the probability of the pair 2A and A is given by P(X=A) and the probability of the pair A and A/2 is given by P(X=A/2).

The switching argument appears to be an attempt to decide to switch when the selected envelope’s content is revealed and then the switch is offered. It goes off track from step 1.

Step 1 denotes the amount in my selected envelope as A. Remember that we have an envelope pair, the envelope holding the smaller amount contains X and the envelope holding the larger amount contains 2X. If we are solving the given question, then if A is in the envelope containing the smaller amount then A=X (and the other envelope contains 2X); if A is in the envelope containing the larger amount then A=2X (and the other envelope contains X). So let us see how the argument develops.

Step 2 says “The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2.” More correctly for the given question, this should say the probability that my selected envelope contains the smaller amount is 1/2 and that it contains the larger amount is also 1/2 i.e. the probability that A=X is 1/2 and the probability that A=2X is also 1/2.

Step 3 says “The other envelope may contain either 2A or A/2”. This is step of logic too far for the given question. This step is saying that we have an envelope pair containing A and 2A or an envelope pair containing A/2 and A i.e. we are considering two separate envelope pairs. This is a strong hint that argument is considering switching after revealing the content.

Steps 4 and 5 add nothing to the argument.

Step 6 is the point to stop and go through this.

I am given an envelope pair containing two envelopes etc. I assign a value A to the content of my randomly selected envelope. I have selected the envelope containing the smaller amount with a probability of 1/2 or the envelope containing the larger amount with a probability of 1/2. In the first case A=X and the other envelope contains 2X and the latter case A=2X and the other envelope contains X. There is no compelling reason to switch.

As before but I open the selected envelope, find that it contains A and then I am invited to switch. I consider that A could come from the envelope pair containing A and 2A or from an envelope pair containing A/2 and A. The first envelope pair occurs with probability P(X=A) and the second envelope pair occurs with probability P(X=A/2). The relative frequency that I have the envelope pair containing A and 2A is P(X=A) / (P(X=A) + P(X=A/2)) and the relative frequency that I have the envelope pair containing A/2 and A is P(X=A/2) / (P(X=A) + P(X=A/2)). By switching I will gain A with a probability of is P(X=A) / (P(X=A) + P(X=A/2)) or lose A/2 with a probability of is P(X=A/2) / (P(X=A) + P(X=A/2)). I will choose to switch if my expected gain exceeds my expected lose i.e. if

A * P(X=A) / (P(X=A) + P(X=A/2)) > A/2 * P(X=A/2) / (P(X=A) + P(X=A/2))

Since all the terms are positive, divide both sides by A and multiply both sides by (P(X=A) + P(X=A/2))

I will choose to switch if P(X=A) > 1/2 * P(X=A/2).

--Norman Sheldon (talk) 10:44, 12 September 2012 (UTC)

The not switching argument

The not switching argument: Now suppose you reason as follows:

1. I denote by B the amount in the envelope that I have not selected. 2. The probability that B is the smaller amount is 1/2, and that it is the larger amount is also 1/2. 3. The envelope I selected may contain either 2B or B/2. 4. If B is the smaller amount, then the selected envelope contains 2B. 5. If B is the larger amount, then the selected envelope contains B/2. 6. Thus the selected envelope contains 2B with probability 1/2 and B/2 with probability 1/2. 7. So the expected value of the money in the selected envelope is

   ½ (2B) + ½ (B/2) = 5B/4

8. This is greater than B, so I gain on average by not swapping.

Hence, if I know the content of the other envelope then I will not switch; and this is true whatever the value is of the other envelope.

We now have the contradiction that if we assign a value of A to the contents of the selected envelope then we will gain by switching but if we assign a value of B to the contents of the other envelope then we will lose by switching. So what is going on here?

We have two scenarios (call them Game A where we assign A to the contents of the selected envelope and Game B where we assign a value of B to the contents of the other envelope). In Game A we hold the value A and can either gain A or lose A/2 each with a probability of ½. So the odds being offered are 2 to 1 compared to an evens risk, this is very attractive to anyone. In Game B we hold either 2B or B/2 and can either lose B or gain B/2. So the odds being offered are 1 to 2 compared to an evens risk, this is unattractive even to an inveterate gambler. These two scenarios cannot co-exist, so when we say assign A to the amount in the selected envelope we are correct half the time (and ditto for assigning B to the other envelope).

Still not clear? Okay, suppose you know that the selected envelope contains A and the other envelope contains B, without using the words “if”, “whether”, “else” or “otherwise”, please answer do you want to switch?

To add a little to Donald Rumsfeld’s quote, there are also unknown knowns, i.e. we can assign known amounts A and B but we still do not know what they are. Until both envelopes are opened, we do not know (especially for a one-off situation) if we gain or lose by switching. Welcome to the quantum world.

--Norman Sheldon (talk) 18:27, 8 October 2012 (UTC)

The symbol A stands for different things???

One sentence in a proposed "resolution" of the paradox in the article reads as follows:

""'A common way to resolve the paradox, both in popular literature and in the academic literature in philosophy, is to observe that A stands for different things at different places in the expected value calculation, step 7 above."

This is complete nonsense. In all cases, A stands for the unknown amount of money in the first envelope selected before deciding whether to switch.

It is just like saying that a random variable that takes the values 1 and 2 with equal probabilities, both equal to 1/2, stands for "different things" depending on whether it equals 1 or equals 2.Daqu (talk) 20:02, 23 February 2013 (UTC)

Daqu, I tend to agree with you but the real problem is that the problem itself is not that well defined, it is a self-inflicted injury. In other words there is no puzzle initially, it is quite obvious that switching is not advantageous (certainly in the real world), the puzzle is created by proposing a bogus line of argument. The exact solution depends on exactly what the bogus line of argument is taken to be. Martin Hogbin (talk) 09:39, 8 March 2013 (UTC)

My solution @zazaalaza

Firstly I am offered the possibility to choose between switching and not switching. The problem in the line of reasoning above is that after line 8. the logic jumps directly to after the switch. After calculating that i gain more, why can't I calculate further? As I know that the opposite of the expected value is the expected loss, I want to calculate that too before switching. In other worlds, I want to calculate the expected value of the second possible action: I do not switch. Following this logic I can make the exact same calculation by denoting the other envelope with B and come to the realization that the expected loss equals 5/4. So if EV = 5/4 and EL=5/4 there is no reason to switch because even if i gain more, in the same time i lose more too.

Or in other words: expected value of switching = expected value of not switching — Preceding unsigned comment added by 5.15.29.71 (talk) 14:18, 13 March 2013 (UTC)

edit: it doesn't matter what the value of action A is untill it is equal to the value of action B — Preceding unsigned comment added by 5.15.29.71 (talk) 14:30, 13 March 2013 (UTC)

Flawed Logic

I don't understand why this paradox is such a big deal. I see this problem differently than how I understand the article. As I understand the problem, you are given a random Boolean value that you aren't allowed to check. You are then asked if you want to negate that Boolean value. Even if you can check only one of the envelopes, the amount of money contained within either envelope, whether measured relatively or absolutely, is not dependent upon the probability that it is the greater value. It is simply the amount of risk involved in switching. Another way of phrasing this problem would be to flip a coin and call heads or tails, then, without knowing the outcome, be given the ability to change your prediction. Without any knowledge of the outcome, you cannot make any sort of educated guess about switching; blindly changing whether you guessed favorably or not. If this logic is explained in the article, I haven't properly understood. I would subsequently like to be pointed to the section that states this and some articles be recommended to enhance my knowledge and understanding. Googol30 (talk) 01:42, 22 May 2013 (UTC)

I tend to agree with you. This paradox is essentially a self-inflicted injury. To most people it is obvious that swapping has no advantage and this does indeed turn out to be correct.

It is, however, possible to propose a line of argument in which there does appear to be an advantage in swapping. The problem is in finding the flaw in the given argument. The exact location and nature of the flaw in the proposed line of argument depends on exactly what the argument is. As expected, whenever the proposed argument for swapping is clearly defined, the flaw in it becomes apparent. Martin Hogbin (talk) 08:10, 22 May 2013 (UTC)

In general, I agree with both of you. The issue is less that the problem itself is difficult to grasp, and more that a "standard" way of examining the problem yields results that are clearly incorrect. Hence, we should be concerned about using those methods at least some of the time because we might unintentionally repeat the error in a way that would be less easy to catch. Dlepzelter (talk) 19:30, 31 October 2013 (UTC)

Odd Solution?

It would seem that with all of those multitudes of words spent on the topic, no-one came up with the obvious: you should always switch envelopes if the amount contained within your envelope is odd and never if it is possible for the amount to be odd, but you find an even quantity.70.90.204.42 (talk) 16:13, 1 May 2014 (UTC)

Unfortunately, as far as I know, no one has yet published such a solution. So wikipedia can't cite it. Richard Gill (talk) 16:14, 1 June 2014 (UTC)

That is correct only if we assume integer amounts. The general case is that the amounts can be continuous, so this distinction cannot be made. --Pan-wiki-tsik (talk) 18:21, 4 June 2014 (UTC)

Rationalization to the non-probabilistic approach

Labeling both envelopes (such as A and A/2 or X and 2X) is faulty because it presumes knowledge of both envelopes which, while adhering to pure logic, is not sound since logically the universe consists of knowledge of only 1 envelope. That said, knowing the contents of 1 envelope marks its contents as A (vs. 2A or A/2) rather than X or 2X (or A or A/2). This makes the first assertion true: either one gains A or loses A/2, in which case one should always choose the other envelope. It results in the same infinite loop as the probabilistic argument.

Of the 2 approaches, this is not a unique paradox. It is equivalent to the St. Petersburg paradox of infinite expected value. But I have to say, the expected value of a discrete sample space being the Euclidean mean of the values of that sample space is a somewhat ludicrous concept. One cannot *expect* a sum of money to be (2A+A/2)/2 = 1.25*A because it's not one of the options! I admit this flies against modern probability theory, but in the case of resolving a paradox it becomes reasonable.

The problem as it exists is faulty because allowing the player to make an infinite amount of decisions will result in exactly that; if the player is only allowed to make 1 further decision after his/her first (e.g. in the Monty Hall problem), s/he should opt for the other envelope regardless. — Preceding unsigned comment added by 173.168.18.15 (talk) 20:50, 8 June 2014 (UTC)

You are fully correct, IP 173.168.18.15, because the nonsensical term  (2A+A/2)/2=1.25*A  does a priory imply  "A≠A".  So a priori putting up with "2A"=2(A/2) and "A/2"=(2A)/2). Consequently you have to solve this nonsensical term as follows: "The expected value is (A+A)/2". Gerhardvalentin (talk) 07:33, 16 June 2014 (UTC)

Ali Baba Variant

The cited Nalebuff article lists multiple possible answers to the apparent paradox. The only one which implies either Ali or Baba should possibly want to trade is one in which there is a maximum apparent amount of money that might be in the envelope (Nalebuff, The Other Person's Envelope is Always Greener, 1989, pp. 175-177).209.6.139.197 (talk) 04:44, 17 October 2013 (UTC)David Lepzelter

Nalebuff shows the only scenario where the expected value of (2A+A/2)/2=1.25A can ever apply: If A is the "initial amount" contained in the first envelope (say A), and then the second envelope (say B) is equally likely to be filled – thereafter – with either 2A or A/2. Yes, if Ali got the "first" envelope, then he should switch and can expect to get 1,25*A. Whereas Baba, by switching must expect to get only 0.8*B.

But as, in general, "any" envelope is equally likely to having been picked first, the term (2A+A/2)/2 is nonsensical, because from the outset this implies  "A≠A". So a priori putting up with "2A"=2(A/2) and "A/2"=(2A)/2). Consequently you have to solve this nonsensical term as follows: "The expected value is (A+A)/2". Gerhardvalentin (talk) 08:22, 16 June 2014 (UTC)

What am I missing?

Surely any actual playable instance of this game has an underlying probability distribution of the amounts in the envelopes? And, because it has to add up to 1 over all amounts, it has to converge to 0 as the amounts get ever higher. So, given this probability distribution, and a value-in-the-first-envelope A, the probability that A is the smaller envelope cannot be exactly 1/2 over all possible values of A.

Consider:

• You play the game 1,000,000 times. You don't swap, just take the amount in the first envelope. You write down the highest amount you've seen so far.
• Now, on the 1,000,001th go, you're offered an amount higher than any you've previously seen. Intuitively, would you swap? — Preceding unsigned comment added by 217.150.97.26 (talk) 12:56, 15 October 2013 (UTC)

I think you've hit the nail on the head, the original argument to support switching assumes that 2X is equally as likely as (1/2)X, and that implies either an upper limit (and we know the maths don't work then, because with a limit of $100 you never swap $100) or constant probability over 0 to infinity with the implication that the expected winnings are infinite, or tend to infinity as we increase the value of the game limit towards infinity. Gomez2002 (talk) 12:39, 6 August 2014 (UTC)

Yes I agree. Either there is an upper limit, in which case it is easy to show that on average you do not gain by swapping, or there is not, in which case the expected value in both envelopes is infinite. I think this is the conclusion in the literature when you read it. Martin Hogbin (talk) 13:24, 6 August 2014 (UTC)

Randomized solutions

There's a basically equivalent but simpler formulation than the one in the main article. Select a strictly decreasing function, f(x), defined on the non-negative real numbers, which is 1 at 0 and approaches 0 as x approaches infinity; for instance, exp(-x), 1/(1 + x), 1/log(e + x), etc. If the originally chosen envelope contains an amount A, swap with probability f(A). For any two fixed amounts, A0 < A1, if the smaller amount is chosen, it will be more likely swapped than if the larger amount is chosen. The expected value is (A0 + A1)/2 + (f(A0) - f(A1))(A1 - A0)/2. If the amounts are A and 2A, the expected value is 3A/2 + (f(A) - f(2A))A/2— Preceding unsigned comment added by Thinkatron (talk • contribs) 06:46, 14 August 2014 (UTC)

Where is the problem?

One envelope is $10 another is $20. Pick first envelope... your expected value is now $10*.5+$20*.5=$15 great! Decide to switch to other envelope... your NEW expected value is STILL $20*.5+$10*.5=$15 ????

Where is the problem? Its so obviously devoid of a problem that I am shocked anyone would bother to write this artical... I don't care how important the person is who "discovered" the problem, that person was wrong, and their mistakes should not be allowed to be shown the light of day except in the context of ridicule and disproof. I would sugest this artical be removed from wikipedia. It was a complete waste of my time, I want it back! --THE TERMINATOR 24.13.179.59 (talk) 14:24, 16 August 2014 (UTC)

I actually bothered to read the "Example" section and THAT is where the problem is. The first thing I notice is how many dollar values are mentioned... lets list them: $10 envelope, $20 envelope, $40 envelope... BOOM that is 3 (yes THREE) envelopes! This violates the original premise of the "problem", namely that there are only TWO envelopes! The problem isn't with the "problem" the problem is with the math used to analyze the "problem"... If your "problem" has THREE envelopes then your in the WRONG wikipedia artical, this artical only has TWO envelopes. I don't know how much more painfully obvious I can make this! PLEASE PROVE ME WRONG! please... --THE TERMINATOR 24.13.179.59 (talk) 14:56, 16 August 2014 (UTC)

I partially agree. The expectation value of switch may be greater than the amount in the first envelope, but it's equal to the expectation value of the first envelope. This seems like a pretty obvious solution at least to the set-up where the first envelope isn't opened. I'm sure there's some mathematical nuance that I'm missing (seems somewhat more likely than having found a solution that no one else has thought of in the past 100 years), but perhaps that should be addressed in the article?jwanders^Talk 15:24, 26 August 2014 (UTC)

Flawed Analysis

Once you have selected your envelope with $20 in it, the contents of the other envelope is determined, there is no probability computation to be performed. If the other envelope contains $10, it's not advisable to switch. If the other envelope contains $40, it is advisable to switch, but there is no way to determine which of these is the case, so the best course is to keep the $20.

Yes, as you do not know whether you first have selected the envelope containing 1/3 of the total amount, or the envelope containing 2/3 of the total amount, you must expect to have got 1/2 of the total amount (3/3/2), so it is best to keep your envelope, as per switching you only can expect to get 1/2 of the total amount, again, see the literature. Gerhardvalentin (talk) 20:15, 15 October 2014 (UTC)

Solution idea by Tsikogiannopoulos

Another proposed explanation of the paradox is the following: Tsikogiannopoulos, Panagiotis (2014). "Variations on the Two Envelopes Problem". Hellenic Mathematical Society, Mathematical Review (77–78): 3–25. Suppose that both players see that the envelope of player A contains 100 euros and let's make a calculation using only numerical values: Once we know the amount of 100 euros, we conclude that the other envelope can contain either 50 or 200 euros. Now there are two possible events for the two fixed amounts that the game can be played with:

• Event 1: Amounts of 100 and 200 euros
• Event 2: Amounts of 50 and 100 euros

We assume that both players have no prior beliefs for the total amount of money they are going to share, so Events 1 and 2 must be considered to have equal probabilities to occur. In every variation of two fixed amounts where the players want to make their calculations based on the numerical value of the amount revealed to them, they will have to weigh the return derived from each event with the average fixed amount by which the game is played in this event. In the Event 1, player A will have a profit of 100 euros by exchanging his envelope while in the Event 2 will have a loss of 50 euros. The average fixed amount in the Event 1 is 150 euros while in the Event 2 is 75 euros.

The formula of expected return in case of exchange for player A that summarizes the above remarks is the following:

$E(A)={\frac {1}{2}}\cdot {\frac {+100}{150}}+{\frac {1}{2}}\cdot {\frac {-50}{75}}=0$

So, the decision of the exchange or not of his envelope is indifferent. Below is an explanation of why the players will have to weigh the return derived from each event with the average fixed amount:
In the Event 1, the player who will switch the amount of 100 euros with the amount of 200 euros will have a profit of 100 euros in a game that shares 300 euros in total. So we can say that this player played the game with a success factor of 100 euros / 300 euros = +1/3. Similarly, the other player played the game with a success factor of -1/3. In the Event 2, the player who will switch the amount of 50 euros with the amount of 100 euros will have a profit of 50 euros in a game that shares 150 euros in total. This player played the game with a success factor of +1/3 also and his opponent with a success factor of -1/3 also. In reality, when a player sees that his envelope contains 100 euros, he doesn’t know whether he is in the game of the Event 1 or in the game of the Event 2. If he is in the Event 1 and switches he will have a success factor of +1/3 and if he is in the Event 2 and switches he will have a success factor of -1/3. As we mentioned above, these two events have equal probability of 1/2 to occur, so the total success factor of player A considering both possible events is zero. This means that the decision of a player to switch or not switch his envelope is indifferent, even when he makes his calculations based on the amount of money that is revealed to him.

What does the new paper add to the body of knowledge on the subject?

The new paper shows a way to resolve the paradox if a player does a calculation with the amount of money he/her sees and not with some variables X, 2X, A, etc. It directly answers what is wrong with step 7 of the Problem as stated in the WP article. For example, if you see that your envelope contains $100 then you have A=100 and Step 7 would give you an expected value of 125 if you switch envelopes, which can't be right. The new paper shows a way to make your calculations based on the $100 you see and not end up with a paradox. I haven't found any other proposed resolution that only makes use of amounts of money and not some variables. Caramella1 (talk) 05:30, 13 October 2014 (UTC)

You say a value of £125 for the expected value cannot be right but what is the error in the calculation shown in the article and how does the paper show this. Martin Hogbin (talk) 09:15, 13 October 2014 (UTC)

The problem with the formula shown in step 7 is explained in the first paragraph of the Common resolution section. But in the Alternative interpratation the envelope of player A opens and the amount of money it contains is revealed. So the explanation provided in the Common resolution section is no longer valid to explain what is wrong if we replace variable A with the amount of money that is revealed to us. So if A=100 then the same formula would give us ${\frac {1}{2}}200+{\frac {1}{2}}50=125$

The paper explains this, by stating that each term of the above equation refers to a different event which shares a different amount of money. So each term should be divided by the mean amount of money of that particular event. I included the proposed formula in the Proposed resolutions to the alternative interpretation section. The only difference is that instead of computing the expected value, the author computes the expected return, that is how much money player A gains or loses if he switches envelopes. Caramella1 (talk) 12:54, 13 October 2014 (UTC)

The formula shows an expected value of £125 which suggests that you should swap. How does this find the flaw in the argument? Martin Hogbin (talk) 13:25, 13 October 2014 (UTC)

If by switching player A had an expected value of 5A/4 this would mean that the 5A/4 is the expected value of player B if he/she doesn't switch. If player B switches then he will have an expected value of the amount A, so player A wants to switch and player B doesn't. The positions of player A and B are completely symmetrical in this variation of the problem (there are other variations that they are not) so if the calculations were made for the player B envelope and the amount B it contains then we would have the opposite conclusion: Player B would want to switch and player A would not. If a player makes both calculations (for the amounts A and B) then he/she will have to switch and not switch at the same time. This is clearly paradoxical. That is the flaw of the step 7's formula. Caramella1 (talk) 05:13, 14 October 2014 (UTC)

The paper seems to be merely restating the problem. Of course it is paradoxical, that is why it is a puzzle. The problem is in finding the mathematical or logical flaw in the presented argument. Just saying that something must be wrong is not good enough, we all know that already. Martin Hogbin (talk) 07:52, 15 October 2014 (UTC)

I am not just saying that something must be wrong. I specifically explained why Step 7 of the switching argument is wrong for the variation it examines and I added a section of a paper that proposes an explanation in which there is no paradox and there is no apparent reason for player A to switch. Caramella1 (talk) 09:48, 15 October 2014 (UTC)

Does it actually resolve the problem, which is to find the flaw in a specific argument for swapping?

According to the author it does find a flaw in a specific argument for swapping. In Step 7 of the main argument the author states that each term of the equation refers to a different event which shares a different amount of money. So each term should be divided by the mean amount of money of that particular event. That way the expected return is 0, not 1/4 as follows from the step 7 of the Problem. Caramella1 (talk) 05:30, 13 October 2014 (UTC)

As I understand it A (the nature of which is very poorly defined) represents the sum in the envelope that you hold, which is £100. Martin Hogbin (talk) 09:14, 13 October 2014 (UTC)

Yes, correct. Caramella1 (talk) 12:26, 13 October 2014 (UTC)

So how it it a different event? Martin Hogbin (talk) 08:08, 15 October 2014 (UTC)

I said that each term of the equation refers to a diferent event. The first term of the equation is (1/2)(2A) and it refers to the event in which the two amounts are A and 2A. The second term of the equation is (1/2)(A/2) and it refers to the event in which the two amounts are A and A/2. In both of them player A has the amount A in his/her envelope, but in the first event the players will share an amount of 3A and in the second event they will share an amount of 3A/2. The paper states that these two terms should not belong to the same formula just as the are but they should be divided first by the specific event's mean amount to find player's A expected return in the case of swapping his/her envelope. Caramella1 (talk) 09:34, 15 October 2014 (UTC)

Carmella1 you are hitting the point, that's exactly what the literature says (Ruma Falk, Nalebuff and all others). The formula does not concern the problem of ONLY TWO (unknown) envelopes of A and 2A, nor does it concern the problem of ONLY TWO (unknown) envelopes of A and A/2.

Approaching the scenario of two unknown envelopes, as per the literature, must consider the total contents of either "1/3 and 2/3", or of "2/3 and 1/3" likewise. Period.

As per Nalebuff, the formula ${1 \over 2}(2A)+{1 \over 2}\left({A \over 2}\right)={5 \over 4}A$   (for quelconque A, be it – in the end – 1/3 or 2/3 of the total amount)  concerns quite another scenario –  it concerns strictly speaking the scenario of THREE envelopes, where the initinal envelope contains the initinal value of A, and "thereafter" two possible follow-up envelopes, one containing 2A and the other one containing A/2, whereupon one of both follow-up envelope is discarded randomly. Then B is either 2A of ANY "A" or A/2 of ANY "A". Only then the remaining follow-up envelope B will contain 1.25*A, as per the formula, and the initinal envelope A – in the end – contains 4/5*B. That's all. Gerhardvalentin (talk) 20:15, 15 October 2014 (UTC)

Exactly! The formula presented in the step 7 of the problem is valid only if the game is played by different rules, i.e. when the amount A is fixed and it is known that it will be given to player A and the amount of player B will be determined by the flip of a coin. If the coin lands heads then player B will have a (hidden) amount 2A in his/her envelope and if it lands tails will have the amount A/2. Then player A should switch and player B should keep his/her envelope. Notice that this case is no longer symmetrical for the two players. It is also covered in the Tsikogiannopoulos paper. Gerhardvalentin do you agree that the addition I made is useful for the readers of WP to understand the problem better? Caramella1 (talk) 06:40, 16 October 2014 (UTC)

Yes, but we should mention only the operative quintessence of the relevant literature (Falk, Nalebuff, Tsikogiannopoulos and many others), similar to your short explanation here, so that it can be recognised by grandma readers and editors, also. Gerhardvalentin (talk) 07:17, 16 October 2014 (UTC)

Proposed ideas of how Tsikogiannopoulos paper might be useful to the readers

Gerhardvalentin I didn't understand how would you like to change the proposed addition so that it will be more useful to the readers. Can you write your version of the addition in the main article and discuss it here? I asked user Martin Hogbin to do this also. Caramella1 (talk) 08:26, 16 October 2014 (UTC)

Quite easy: Let's say what the literature says: the problem of two unknown envelopes with either 1/3 and 2/3 of the total on one hand, or with 2/3 and 1/3 of the total amount on the other hand has to be solved by expecting to have first picked the average of 1/2 of the total amount, so switching makes no sense.

Whereas the "5/4-formula" never addresses that problem of two unknown envelopes with either 1/3 and 2/3, or with 2/3 and 1/3. Because if A is the larger amount of 2/3, then it cannot be doubled. And if A is the smaller amount of 1/3 then it cannot be halved. Doubling the larger amount breeds nonsense, as well as halving the smaller amount breeds nonsense. But as the formula applies to quelconque "A", be it – in the end (!) – 1/3 or be it 2/3 of the total amount, it is clear that consequently the "formula" addresses quite a different scenario, where A is known to be the fixed amount that in ANY case can be doubled resp. in ANY case can be halved, whereas B is known to be the dependent amount of either 2A or A/2, for quelconque A (!). Let's say what the qualified literature says. Gerhardvalentin (talk) 10:31, 16 October 2014 (UTC)

Your remarks are all correct. In the proposed variation of the problem I think that the Common resolution section gives a satisfying explanation which is similar to your remarks. The second variation of the fixed amount A which can be doubled or halved is not mentioned at all in the article I think. There are many references to it like the Priest, G. & Restall, G. (2007) “Envelopes and Indifference”. The explanation given in this paper is similar to that we both wrote, so perhaps we could add it also, but another section would be needed for that, named something like “One fixed amount variation”.

As for the subject, the Tsikogiannopoulos paper gives an explanation to the “Alternative interpretation” where one envelope opens. It proposes the idea that the two terms of the equation should be divided by the mean amount and it makes the calculations without variables (money and probabilities only), ideas that I have not seen before in the papers I have read. Do you agree to be placed in the alternative interpretation the way I worded it? If not, what is your proposal for changes in it? Caramella1 (talk) 11:38, 16 October 2014 (UTC)

The correct formula by merging fragments mentioned in the literature

Half of the total amount is essential, but "without variables" is not a sine qua non. Just put together the fragments that can be found in the literature, considering the two different scenarios.

And here, Vol. 9, No. 1 (2010) 1–8 World Scientific DOI: 10.1142/S0219477510000022) (Accepted 17 September 2009, Communicated by Natalia B. Janson), Derek Abbott, University of Adelaide, et al. say it frankly straightforwardly in "THE TWO-ENVELOPE PROBLEM REVISITED":

Imagine a player is presented with two envelopes, one containing x dollars and the other 2x dollars. The player randomly chooses an envelope and observes the amount inside. The question is ... ... The situation is entirely symmetrical and the average return is: R = 1/2 (x+2x) = (3/2)x. Thus the player’s expected return, R = 1.5x, no matter if the envelopes are switched or not.

They say that in the symmetrical scenario of unknown envelopes, all two envelopes can be expected to contain half of the total amount, period. So for this scenario of two unknown envelopes, the contents of each envelope has to be analysed a little bit closer.
Envelope "A" may contain the larger amount of A or the smaller amount of A/2 (=a), so on average the amount of (A+a)/2, and likewise envelope "B" may contain the larger amount of B or the smaller amountof B/2 (=b), so on average the amount of (B+b)/2.

Using now the contents of envelope "A" of (A+a)/2  (=half of the total amount) to get the expected value of envelope "B", and considering that A/2 (=a) can only be doubled, but cannot be halved and that A (=2a) can only be halved, but cannot be doubled, then the formula to get the expected amount of envelope "B" should read as follows:

${1 \over 2}(2a)+{1 \over 2}\left({A \over 2}\right)={3 \over 4}A$ resp. $={3 \over 2}a$   = again half of the total amount.

So the expected value of envelope "B" will be again half of the total amount:  $={1 \over 2}(A+a)={1 \over 2}(A+b)={1 \over 2}(B+a)={1 \over 2}(B+b)$

This standard scenario must clearly be distinguished from the scenario of any initial A and thus dependent B, being the only scenario where the 5/4-formula applies:

${1 \over 2}(2A)+{1 \over 2}\left({A \over 2}\right)={5 \over 4}A$     for any initial A, be it – in the end – 1/3 or 2/3 of the total amount, but that doesn't address the standard scenario.

The article should help the reader to see that a so-called "paradox" can arise through "incorrect mathematics", as the Royal Society London calls it. The article should clearly distinguish those two variants and say what the sources say. Gerhardvalentin (talk) 20:58, 16 October 2014 (UTC)

The important thing about this paradox that distinguishes it from, say, the Monty Hall problem is that it depends on a bogus line of argument being put up for switching. We all know that there is no advantage in switching, that is what our intuition tells us and what correct mathematics confirms (apart from the Ali Baba version which, as we both agree, is a different problem). The point is there is no inherent paradox. What most people think to be true turns out to be quite correct. There is only a paradox if we make one by putting up a bogus line of reasoning for swapping. I am not aware of any sources which say you should swap (apart, again, from the A B case). Do you see what I mean? Martin Hogbin (talk) 17:28, 22 October 2014 (UTC)

Martin, did you read the article? It does not distinguish the two known variants clearly enough (standard version vs. Ali Baba version) and consequently doesn't contrast them clearly enough. The article should show that contrast:

${1 \over 2}(2A)+{1 \over 2}\left({A \over 2}\right)$    versus

${1 \over 2}(2a)+{1 \over 2}\left({A \over 2}\right)$.    This contrast has to be shown. The article should clearly show "why" it is faulty to apply the 5/4-formula to the standard version, see Bruss(1996), e.g. Gerhardvalentin (talk) 18:28, 22 October 2014 (UTC)

Gerhard, I agree with you. The Ali Baba version is completely different from the standard version and should be completely separated from it. Treating it in the same way as the standard version will obviously lead to confusion.

Now what about my point. For the standard version there is no paradox unless a bogus line of arguing is put forward. It is obvious that you should not swap and that is waht correct mathematics shows. Correct mathematics though does not resolve the paradox, it merely confirms the obvious. Resolving the paradox requires finding exactly where the fault lies in a proposed line of reasoning. Martin Hogbin (talk) 19:07, 22 October 2014 (UTC)

Totally agreed. The flaw of trying to apply the 5/4-formula to the standard version must clearly be branded, sources in droves.

The 5/4-formula (only suitable for the Nalebuff version) doubles "any" ("initial!") amount, be it actually 1/3 or 2/3 of the total. This is the flaw regarding the standard scenario, where there is no "initial A", but only an unknown pair of "1/3:2/3" resp "2/3:1/3" amounts. Neither an "initial" nor a "dependent" amount. As Bruss(1996) clearly says: The fault is to double the contents of envelope "A" in any case, even if "A" actually holds the greater amount of 2/3, and to halve the contents of envelope "A", even if it actually contains the smaller amount of 1/3. Bruss says that the actual contents of envelope "A" can never be treated by using "a single variable". And the article should pay regard to this as clearly as possible. Some mathematicians and philosophers still hope for the loophole of A being either infty or zero, so the article should urgently care for clarity, and especially should stop commingling the two quite different scenarios with their quite different solutions. Gerhardvalentin (talk) 20:40, 22 October 2014 (UTC)

But you are only really restating the problem. We know the 5/4 answer is wrong but at exactly what step in the argument given in the article does the error occur? Martin Hogbin (talk) 21:48, 22 October 2014 (UTC)

??? Martin, didn't you read what I say to the 5/4-formula and its switching arguments, for one week now, since midst October? At first: in both variants,

• "envelope B" is equally likely to contain the double amount of "envelope A" (but for the standard variant only given that "envelope A" actually contains 1/3 of the total amount, for an actual amount 2/3 cannot be doubled), and
• "envelope B" is equally likely to contain the half amount of "envelope A" (but for the standard variant only given that "envelope A" actually contains 2/3 of the total amount, for an actual amount of 1/3 cannot be halved).

  The 5/4-formula, where ANY fixed initial A is KNOWN to be the initial value that previously can HAVE BEEN doubled or halved, and where "envelope B" is KNOWN to contain the dependent value, DOES NEVER address the standard version of two UNKNOWN envelopes. The cited "switching arguments" are indeed correct from item 1 to item 5 for both variants, i.e. for the standard version also.

  But, for the standard version ("1/3:2/3" resp. "2/3:1/3" without "a fixed initial A and dependent B"), item 6 is sloppy and false. Because "envelope B" contains the double amount of "envelope A" not for ANY A, but only IFF "envelope A" actually holds the small amount of 1/3, otherwise NOT. And likewise "envelope B" contains the half amount of "envelope A" not for ANY A, but only IFF "envelope A" actually holds the large amount of 2/3. Otherwise not. So items 6 to 8 do address only the Ali Baba variant of B=5/4A and A=4/5B. See Falk(until now), Bruss(1996), Bliss(2012) and many others.

In other words: in 50% envelope A contains 1/3 of the actual total amount.

If this is the case, then envelope B inevitably contains 2/3, difference +1/3 of the total amount. And in the other 50%, envelope A contains 2/3 of the total amount.

If that is the case, then envelope B inevitably contains 1/3, difference -1/3 of the total amount. So in the sum of all those two possible cases, all envelopes A contain on average 1/2 of the total amount available, and all envelopes B contain on average also 1/2 of the total amount available. So switching makes no sense.
But it is the fault of mathematicians to say that in every single case the probability that B=2A is 50% without regarding that this applies only for those cases where envelope A actually contains 1/3 of the total amount, otherwise not, and that in every single case the probability that B=A/2 is 50%, without regarding that this only applies for those cases where envelope A actually contains 2/3 of the total amount of both envelopes, otherwise not. Gerhardvalentin (talk) 17:09, 24 October 2014 (UTC)

The "switching arguments" are fully correct from item 1 to 3 for both variants, items 4 and 5 fully correct for the Ali-Baba version only, and with restrictions (because mistakable) correct for the standard variant.
Item 6 obviously being faulty for the standard variant, items 6 to 8 are correct only for the Ali Baba variant with ANY fixed initial A and ANY dependent B, resulting in the fact that that B will be on average 5/4A and A will be on average 4/5B. Once more: for the Ali Baba version, the dependent amount will on average be 5/4 of the fixed initial amount, and the fixed initial amount will on average be 4/5 of the dependent amount. So – in the Ali Baba veriant – you should swap from the initial amount A to the dependent amount B, but never swap back.
And it is clear that items 9 and following result by incorrectly trying to apply the 5/4-formula to the standard variant, also. Gerhardvalentin (talk) 00:12, 23 October 2014 (UTC)

Older discussion:

Again: In the scenario of two unknown envelopes ("1/3 + 2/3" resp. "2/3 + 1/3" of the total amount), it is true that in each and every case, and with equal probability envelope B contains the double amount resp. half the amount of envelope A. But it is careless to drop the subject that envelope B contains the double amount of envelope A only given that envelope A actually holds the small amount of 1/3, and likewise envelope B contains the half amount of envelope A only given that envelope A actually holds the large amount of 2/3. Otherwise not.

The 5/4-formula uses the variable "A" for both values of 1/3 (small) and 2/3 (large) at the same time, and Ruma Falk clearly calls this an inconsistency. So as per Falk the 5/4-formula is WRONG for the scenario of two unknown envelopes. But as per Nalebuff the 5/4-formula is VALID for the scenario of A being the initial value, whereas envelope B, with equal probability, received 2A resp. A/2, thereafter. The article should say what the sources say.

My proposal: the article should cite Ruma Falk and Nalebuff in the first line to get this clear and to name the flaw, and then of course the Tsikogiannopoulos paper might be useful to the readers. Gerhardvalentin (talk) 03:41, 17 October 2014 (UTC)

Why might the Tsikogiannopoulos paper be useful? Is it a notable contribution to the literature on this subject? What does it contribute that is not found in other prior references?Perswapish (talk) 04:04, 17 October 2014 (UTC)

The unclear article needs clarification. It is necessary to distinguish the two quite different possible scenarios: Nalebuff's "initial value A" (delivered to the player) and Nalebuff's "dependent value B", as an option to switch on, where the 5/4 formula is correct (A=B*4/5 and B=A*5/4), and the quite different scenario of two unknown envelopes where the player randomly picks one of them. And for the second scenario, every hint is useful that shows that it doesn't matter if the envelopes are switched or not. Yes, we should name those sources and mention their operative quintessence, that will help the article and will help the reader to comprehend what matters. Gerhardvalentin (talk) 04:56, 17 October 2014 (UTC)

Gerhardvalentin, given that you also aggree for the Tsikogiannopoulos addition I placed it back in the proposed explanations of the alternative interpratation. I also aggree for the changes you mention above about the scenario of initial and dependent values and I encourage you to start making these changes to the main article so that we can work on them. Caramella1 (talk) 07:50, 17 October 2014 (UTC)

In the first line it is necessary to distinguish the two conceivable but quite different scenarios:

• Two distinguishable envelopes A and B, where A contains some initial value, and thereafter B equally likely got either double or half of that value. As per Barry Nalebuff's asymmetric variant, often known as the Ali Baba problem: one envelope is filled first, called Envelope A, and given to Ali. Then a fair coin is tossed to decide whether Envelope B should contain half or twice that amount, and only then given to Baba. In this scenario of distinguishable envelopes the 5/4-formula is correct, on average B will contain 5/4*A and A will contain 4/5*B.
• The standard paradox: two indistinguishable envelopes, one of them is picked at random.

Nalebuff (1989), Christensen and Utts (1992), Falk and Konold (1992), Blachman, Christensen and Utts (1996), Nickerson and Falk (2006), pointed out that if the amounts of money in the two envelopes have any proper probability distribution representing the player's prior beliefs about the amounts of money in the two envelopes, then it is impossible that whatever the amount A=a in the first envelope might be, it would be equally likely, according to these prior beliefs, that the second contains a/2 or 2a. Doubling the larger amount as well as halving the smaller amount is impossible. Thus the 5/4-formula, which leads to always switching, is a non-sequitur, period. The article should show "how a two-envelope ‘paradox’ can arise through incorrect mathematics."(reference: Mark D. McDonnell et al., The Royal Society London). Any prime example to show that is welcome, and Tsikogiannopoulos may be cited as well. Gerhardvalentin (talk) 13:47, 17 October 2014 (UTC)

Your reasoning here is already clearly stated in the article in my opinion. First paragraph under "Proposed resolutions to the alternative interpretation." I personally don't see the advantage to include the Ali Baba problem here. That would confuse more than what it helps the casual reader, I'm afraid. There is no paradox connected to that scenario and it is already mentioned in the history section. Do you suggest that we should expand this section or rewrite it? If so, how? iNic (talk) 13:20, 18 October 2014 (UTC)

iNic, I can understand that you are content and satisfied with the article as is. But the obvious flaw in the 5/4-formula (that only depicts the special Barry Nalebuff variant of an initial A and a dependent B where qualsiasi A can be doubled resp. halved), but never addresses the standard paradox of a symmetrical scenario that can only be depicted by any 1:1-formula that provides for doubling only the actual small "1/3-amount" and for halving only the actual large "2/3-amount", is NOT clearly enough mentioned in the article. The article should clearly show the immortal flaw of mathematicians and philosophers to applying the 5/4-formula to the symmetric scenario of unknown envelopes also, where one of both is picked randomly. The article should say what the literature says: "... how a two-envelope 'paradox' can arise through incorrect mathematics" (reference: Mark D. McDonnell et al., The Royal Society London). So imo it is absolutely necessary to reworking the article. Gerhardvalentin (talk) 15:06, 18 October 2014 (UTC)

As an editor it is very important to be able to distinguish one's own conviction regarding the subject matter and one's role as Wikipedia editor who should always adopt a neutral point of view. So given that you can do that I'm very excited to see how you will improve the Common resolution section. iNic (talk) 15:13, 19 October 2014 (UTC)

???   iNic, the source was cited above, and for the case that you didn't see it, I just inserted the actual source here again. It's what the sources say, actually even The Royal society London. Gerhardvalentin (talk) 17:55, 19 October 2014 (UTC)

Sure, I saw your reference and i have of course read that paper. It is an interesting paper, among many other interesting papers in this subject. But what exactly are you trying to say now? Is your main objection that this paper isn't cited or mentioned enough in the article? Is this the reason the current article is confusing according to you? I don't get it. You need to be more specific regarding what you think is confusing and how you would like to remedy that. iNic (talk) 23:17, 19 October 2014 (UTC)

The article does not relativize enough the one-sided unilateral 5/4-formula that is known to addresses only the Ali-Baba variant of any initial A and any dependent B. The article does not show clearly enough that the 5/4-formula never addresses the standard paradox of two unknown envelopes. The standard variant of two completely unknown envelopes is much more intricate than the Ali-Baba variant. The article should stop to mangle those two diverse scenarios, but first and foremost the article should clearly distinguish those two quite diverse scenarios. Gerhardvalentin (talk) 09:07, 22 October 2014 (UTC)

I think the Common resolution section explains this idea quite clearly already. Can you please be more specific regarding how you would like to reword this section so that it becomes more clear in your view? iNic (talk) 11:41, 25 October 2014 (UTC)

No way clear, the laborious article needs a complete rewriting, see this talk page. Gerhardvalentin (talk) 13:58, 25 October 2014 (UTC)

Unless you explain explicitly how you want to rewrite the article no one will know exactly what you mean. Are there sections you want to delete completely? iNic (talk) 14:09, 25 October 2014 (UTC)

Gerhardvalentin, I think that it is almost impossible to convince INic about any change in the article, because he has set it the way he likes it and he doesn't want any change. I agree with you that the article is very confusing the way it is written and the propositions we both made are definetely in the way to clear things up for the readers. So I encourage you once again to stop trying to convince INic and work the changes with me and other users. I remind you one basic principle of the BRD cycle: "Bold editing is a fundamental principle of Wikipedia. No editor is more welcome to make a positive contribution than you are. When in doubt, edit!" Caramella1 (talk) 11:27, 19 October 2014 (UTC)

Simplest, verbal, resolution

The two envelopes problem is a *descendant* of the Necktie paradox. Mathematically, it is a *special case* of the Necktie paradox. The wikipedia page on the Necktie paradox has a completely adequate, brief, and verbal resolution. So maybe that is the answer to all our problems ... Richard Gill (talk) 19:01, 29 October 2014 (UTC)

Here it is, my translations from neckties to envelopes in parentheses: "In general, what goes wrong is that when the first man (the person holding envelope A) is imagining the scenario that his necktie is actually worth less than the other (envelope A contains the lower amount), his beliefs as to its value have to be revised, downwards, relatively to what they are a priori, without such additional information. Yet in the apparently logical reasoning leading him to take the wager (switch envelopes), he is behaving as if his necktie (the amount in his envelope A) is worth the same when it is worth less than the other, as when it is worth more than the other. Of course, the price his wife actually paid for it (the amount actually in the envelope) is fixed, and doesn't change if it is revealed which tie is worth more (which envelope has the larger amount). The point is that this price, whatever it was, is unknown to him. It is his beliefs about the price which could not be the same if he was given further information as to which tie was worth more. And it is on the basis of his prior beliefs about the prices that he has to make his decision whether or not to accept the wager." Richard Gill (talk) 19:05, 29 October 2014 (UTC)

The necktie paradox is easy to understand. The success of the two envelopes was to disguise the simple paradox by adding superfluous details - one envelope has twice the amount as the other - so that the reader would get lost in formula manipulation and calculations, instead of thinking. Richard Gill (talk) 19:11, 29 October 2014 (UTC)

Maybe; but if the point you "wrote a lot about in your paper" will not appear, I'll feel that the most deep and instructive aspect is lost. Boris Tsirelson (talk) 19:20, 29 October 2014 (UTC)

I agree that this point is the most *deep* point arising from this paradox, and it is the most instructive point for the professionals and their students, because all the rest is incredibly elementary. But I don't think that this deep point was the point which Littlewood, Schrödinger, Kraitchik, Gardner had in mind. Possibly the deep point that they had in mind the impossibility of putting a uniform distribution on a set of points with cardinality that of the natural numbers.

But anyway, I guess the article should go from simple to complex, and save the deep points for after the simple confusion is sorted out. After all, the paradox as stated is just a stupid mixup of conditional and unconditional expectations by a writer who vaguely knows some rules of probability calculus but cannot apply them consistenly because he doesn't take the trouble to distinguish between random variables and possible values of random variables. Every year I have a few first year students who only know how to write in capital letters. They can read small and capital letters, but they can't distinguish between them. They are lost from day 1 of the first course in probability. Richard Gill (talk) 05:36, 30 October 2014 (UTC)

Notice from the verbal resolution of the necktie paradox that the philosophers do have it right in saying that the problem is ultimately merely a problem of giving different things the same names. They have a long technical word for that. We have to distinguish between the actual amount a in envelope A (which when the envelope is in our hand is fixed, but unknown) and our beliefs about it, or if you prefer, our knowledge about it. If we are given some new information then our beliefs have to change accordingly. If we *imagine* being given some more information then we have to *imagine* how our beliefs would change. Whichever interpretation we give to step 7, the same *kind* of mistake is made. So maybe the article should start with the necktie resolution. Richard Gill (talk) 05:52, 30 October 2014 (UTC)

Common resolution

I think that the problem with the article is that it starts with a lousy solution. True, it is a common "solution" found in some part of the literature (philosophy) but I do not think it is a satisfactory solution, since it shows that the philosophers have just as many problems with probability calculus as almost anyone else. They do not clear up things, they only make things muddier.

The brief "Common resolution" which starts the article is too short and says some crazy things and does not match properly the mathematics which it is supposed to be interpreting. The final sentence is "Here X stands for the same thing in every term of the equation. We learn that 1.5X, is the average expected value in either of the envelopes. Being less than 2X, the value in the greater envelope, there is no reason to swap the envelopes."

First of all, why write "average expected"? Do you mean "average" or do you mean "expected"?

Secondly, 1.5X is smaller than 2X, but it is larger than 1 X, so why does this calculation tell us there is no reason to swap envelopes?

I propose we try to understand the literature which gives this "common resolution", and then actually report that common resolution in as favourable way as possible. See Section 2 of http://www.math.leidenuniv.nl/~gill/tep.pdf where I attempted to do exactly this. I denoted this kind of solution "the philosopher's solution" since it is the solution commonly seen in the academic philosophy literature. Here I tried to understand what the philosophers are trying to say, and convert it into simple mathematics, so that everyone can understand it. By the way, the philosophers don't know probability, so they did not interpret the problem in the same way as the original mathematicians who invented the problem! Philosophers don't know the difference between conditional and unconditional expectation value. The philosophers made things more confused, not less.

Personally, I think that the "common resolution" sucks. I would prefer that the article starts with the resolution which corresponds to what the original inventors of the problem had in mind, what I call "The probabilist’s choice". It is section 1.3 of my paper. Richard Gill (talk) 09:37, 29 October 2014 (UTC)

We have try try to keep our personal opinion about which proposals are good and which "sucks" aside as WP editors. In my opinion it's good to start with the Common resolution first not because most people or most editors think that the solutions in this category are the best but because it's the first idea most people come up with as a solution to the problem. So it's a good start to get people interested in continue to read the article. It also has the benefit of being the only category of solutions that tries to solve the problem as stated, that is the problem doesn't have to be rephrased in any way. A third benefit having this first is that in the process of showing why this idea can't be a complete solution of the problem we evolve the problem into the different Bayesians versions of the problem. After investigating all the different Bayesian versions we at the end state Smullyan's open question if the problem has anything to do with probabilities at all.

People from many different disciplines have shown interest in this problem, not only "mathematicians" and "philosophers." To squeeze economists, decision theorists, statisticians, Bayesians, and so on into one of these two categories seem a little old fashioned to me. It is also factually not correct that only "philosophers" have promoted solutions in the Common solutions category, and only "mathematicians" propose solutions in the "Alternative interpretation" category. There are both philosophers and mathematicians that support solutions in either of these categories. As well as others that are neither philosophers nor mathematicians. That said, of course I agree that the wording in the Common resolution section can be improved. Can we find a typical source and cite from that source maybe? iNic (talk) 12:18, 29 October 2014 (UTC)

It would be interesting to hear what other editors think. I'm not sure that this is "the first idea most people come up with" and I disagree that the mathematicians' version needs to rephrase the problem. The problem after all is to *analyse the presented argument* and to show what is wrong with it. No *rephrasing of the problem* is needed in order to do this, whether we take (what I call, for convenience) the mathematicians' or the philosophers' point of view. Those two points of views are merely two possible interpretations of what the person who wrote those lines of nonsense actually intended. The economists and decision theorists (and the statisticians, and the Bayesians) all take the mathematicians' interpretation, not the philosophers'. It seems to me that anyone with an "exact science" background will take the mathematicians' side. And the problem came from that side: Littlewood, Schrödinger, Kraitchik, Gardner, ... So in my opinion, the whole philosophy literature on the problem is an amusing but pretty irrelevant side-track. But I'm an arrogant professional mathematician, and that's just my opinion. Richard Gill (talk) 12:44, 29 October 2014 (UTC)

In other words, the "first idea most people come up with" is wrong, because most people don't have sufficient grasp of elementary probability in order to get anywhere with the problem at all. In other words, if you don't have some kind of understanding of what is an expectation value *and* of what is a conditional expectation value, you are completely lost, anyway. TEP started life as a mathematical puzzle. A puzzle invented by professional mathematicians in order to tease and amuse amateurs. Richard Gill (talk) 12:49, 29 October 2014 (UTC)

See also my remark on this matter here, page 73, footnote 1. Boris Tsirelson (talk) 14:27, 29 October 2014 (UTC)

Nice example, Boris! Richard Gill (talk) 14:53, 29 October 2014 (UTC)

You do indeed extend the original problem in what you call the probabilist's resolution below. Either by the aid of philosophy—Bayesian philosophy—where it is always permissible to introduce a prior probability distribution describing the subject's state of belief or knowledge, or by inventing a mechanism for distributing the amounts that can appear in the envelopes in the first place—if you think along frequentist lines. However, in the latter case it is not permissible to extend a problem like this according to frequentist philosophy. The original situation is about a single case and frequentist philosophy doesn't care about single cases. Single cases are undefined or even meaningless to talk about in their view. So you can't arrive at what you call the probabilist's resolutions without extending the original problem to either a repeatable experiment (which it wasn't originally) or to a problem strictly within a Bayesian philosophy of probability. So to describe this type of solutions as using purely mathematical ideas is misleading because they do not belong to the TEP solutions that are the least impregnated with philosophy. On the contrary, in this category of solutions even different philosophical views about infinity and how to solve the St Petersburg paradox pops up, making this category even more tainted with philosophy. The solutions of the unaltered problem ("Common resolution") as well as the the solutions of Smullyan's reformulation without probabilities tries to solve the problem at a more basic logical level. They too of course rely on philosophy, but they deal with more basic philosophical notions and doesn't have to dig into the problem of how to interpret the concept of probability nor do they have to ponder the true nature of infinity to solve the problem. I do however support your idea to rename the first category to something more interesting and less misleading. As you point out, we don't have any statistical basis for claiming that this solution is more common than any other. iNic (talk) 19:27, 5 November 2014 (UTC)

YADD (Yet another difficult description)

This puzzle is, first of all, a linguistical one rather than a mathematical one: it takes its strength from working with two kinds of terms, virtual and real envelopes, unduly confused into one kind of terms.

One has two envelopes, A and B. It is known that A has $x (the exact value is insignificant), and B has with equal probability either $2x or $½x. Let us call C the envelope that one chooses first, and D the other choice. The envelope C, with equal probability, is either A or B. Given the envelope C is the chosen one, what is the ratio of these mathematical expectations, mean(D)/mean(C)? For C, the mean value is composed of four equally probable events: for each identity of C, it may be either the case that B is $2x or $½x; i.e., we have ($x+$x+$2x+$½x)/4. For D, the mean value is composed of exactly the same events and is equal to mean(C): mean(C)=mean(D). That means that either strategy (to take C or to take D, retain the choice or switch the choice) is equally profitable. However, it is true that mean(C/D)=mean(D/C)=(2+½)/2. Yet, the ratio of the two amounts of money does not represent the gain in the action of switching the envelopes! Recall: the two envelopes were created as virtual ones, they only participate in those relationships that represent knowledge about the choice the picker made, they do not participate in relationships that define the real envelopes in the object domain. The non-chosen envelope is either half or twice the chosen envelope: yet, this information does not tell anything about the real envelopes. To proceed with this argument, we must assume that human notions are defined not by things in the outer world, but by the complex of inner mental relationships they participate in, which is a very plausible assumption in my view.

Now, we must find out the meaning of the operation of finding the mean. Both envelopes, C and D, are virtual: they represent the picker's strategy that is defined by his state of mind, namely by the answer which envelope is currently chosen in his mind, they do not represent anything real. The meaning of the mean is that, if one repeats an operation often enough, the average of its outcome will be close to the mean. What concerns the speaker is the amount of money taken each time: it is represented by mean(C) and mean(D) for either strategy. These means are equal; if the instances of the experiment are repeated, the average outcomes of each strategy will be close to each other. Now, if we repeat often enough the following experiment: pick an envelope and compare its contents with the contents of the envelope that was not picked, registering the ratio of the amounts, then, given the unequal character of the distribution of all possible amounts, the average value will be close to mean(C/D)=mean(D/C) and will not be close to one: whenever I win, the contribution of this instance of the experiment is further from one, whenever I lose, it is closer to one, but that makes no influence on whether I win or I lose, so the average is greater than one. This mean has nothing to do with the amount of money that the picker receives; that defies common sense, but that only means that the casual variation of our common sense is not adjusted enough for such artificial problems: it is more adjusted to evenly distributions of any value.

I hope this is interesting. - Evgeniy E. (talk) 09:23, 7 November 2014 (UTC)

You assume that if envelope A contains an amount a, then envelope B has, with equal probability, either 2a or 1/2 a. But this is not (necessarily) true. It needs to be proven. In fact, one can easily see that this cannot possibly be true for all possible values of a. That leads to contradictions. Richard Gill (talk) 11:02, 12 November 2014 (UTC)

Solution idea by Tsikogiannopoulos

Another proposed explanation of the paradox is the following: Tsikogiannopoulos, Panagiotis (2012). "Variations on the Two Envelopes Problem". Hellenic Mathematical Society, Mathematical Review (in Greek) (77–78): 3–25. An English translation is available on arXiv, http://arxiv.org/pdf/1411.2823.pdf Suppose that both players see that the envelope of player A contains 100 euros and let's make a calculation using only numerical values: Once we know the amount of 100 euros, we conclude that the other envelope can contain either 50 or 200 euros. Now there are two possible events for the two fixed amounts that the game can be played with:

• Event 1: Amounts of 100 and 200 euros
• Event 2: Amounts of 50 and 100 euros

We assume that both players have no prior beliefs for the total amount of money they are going to share, so Events 1 and 2 must be considered to have equal probabilities to occur. In every variation of two fixed amounts where the players want to make their calculations based on the numerical value of the amount revealed to them, they will have to weigh the return derived from each event with the average fixed amount by which the game is played in this event. In the Event 1, player A will have a profit of 100 euros by exchanging his envelope while in the Event 2 will have a loss of 50 euros. The average fixed amount in the Event 1 is 150 euros while in the Event 2 is 75 euros.

The formula of expected return in case of exchange for player A that summarizes the above remarks is the following:

$E(A)={\frac {1}{2}}\cdot {\frac {+100}{150}}+{\frac {1}{2}}\cdot {\frac {-50}{75}}=0$

So, the decision of the exchange or not of his envelope is indifferent. Below is an explanation of why the players will have to weigh the return derived from each event with the average fixed amount:
In the Event 1, the player who will switch the amount of 100 euros with the amount of 200 euros will have a profit of 100 euros in a game that shares 300 euros in total. So we can say that this player played the game with a success factor of 100 euros / 300 euros = +1/3. Similarly, the other player played the game with a success factor of -1/3. In the Event 2, the player who will switch the amount of 50 euros with the amount of 100 euros will have a profit of 50 euros in a game that shares 150 euros in total. This player played the game with a success factor of +1/3 also and his opponent with a success factor of -1/3 also. In reality, when a player sees that his envelope contains 100 euros, he doesn’t know whether he is in the game of the Event 1 or in the game of the Event 2. If he is in the Event 1 and switches he will have a success factor of +1/3 and if he is in the Event 2 and switches he will have a success factor of -1/3. As we mentioned above, these two events have equal probability of 1/2 to occur, so the total success factor of player A considering both possible events is zero. This means that the decision of a player to switch or not switch his envelope is indifferent, even when he makes his calculations based on the amount of money that is revealed to him.

What does the new paper add to the body of knowledge on the subject?

The new paper shows a way to resolve the paradox if a player does a calculation with the amount of money he/her sees and not with some variables X, 2X, A, etc. It directly answers what is wrong with step 7 of the Problem as stated in the WP article. For example, if you see that your envelope contains $100 then you have A=100 and Step 7 would give you an expected value of 125 if you switch envelopes, which can't be right. The new paper shows a way to make your calculations based on the $100 you see and not end up with a paradox. I haven't found any other proposed resolution that only makes use of amounts of money and not some variables. Caramella1 (talk) 05:30, 13 October 2014 (UTC)

You say a value of £125 for the expected value cannot be right but what is the error in the calculation shown in the article and how does the paper show this. Martin Hogbin (talk) 09:15, 13 October 2014 (UTC)

The problem with the formula shown in step 7 is explained in the first paragraph of the Common resolution section. But in the Alternative interpratation the envelope of player A opens and the amount of money it contains is revealed. So the explanation provided in the Common resolution section is no longer valid to explain what is wrong if we replace variable A with the amount of money that is revealed to us. So if A=100 then the same formula would give us ${\frac {1}{2}}200+{\frac {1}{2}}50=125$

The paper explains this, by stating that each term of the above equation refers to a different event which shares a different amount of money. So each term should be divided by the mean amount of money of that particular event. I included the proposed formula in the Proposed resolutions to the alternative interpretation section. The only difference is that instead of computing the expected value, the author computes the expected return, that is how much money player A gains or loses if he switches envelopes. Caramella1 (talk) 12:54, 13 October 2014 (UTC)

The formula shows an expected value of £125 which suggests that you should swap. How does this find the flaw in the argument? Martin Hogbin (talk) 13:25, 13 October 2014 (UTC)

If by switching player A had an expected value of 5A/4 this would mean that the 5A/4 is the expected value of player B if he/she doesn't switch. If player B switches then he will have an expected value of the amount A, so player A wants to switch and player B doesn't. The positions of player A and B are completely symmetrical in this variation of the problem (there are other variations that they are not) so if the calculations were made for the player B envelope and the amount B it contains then we would have the opposite conclusion: Player B would want to switch and player A would not. If a player makes both calculations (for the amounts A and B) then he/she will have to switch and not switch at the same time. This is clearly paradoxical. That is the flaw of the step 7's formula. Caramella1 (talk) 05:13, 14 October 2014 (UTC)

The paper seems to be merely restating the problem. Of course it is paradoxical, that is why it is a puzzle. The problem is in finding the mathematical or logical flaw in the presented argument. Just saying that something must be wrong is not good enough, we all know that already. Martin Hogbin (talk) 07:52, 15 October 2014 (UTC)

I am not just saying that something must be wrong. I specifically explained why Step 7 of the switching argument is wrong for the variation it examines and I added a section of a paper that proposes an explanation in which there is no paradox and there is no apparent reason for player A to switch. Caramella1 (talk) 09:48, 15 October 2014 (UTC)

Does it actually resolve the problem, which is to find the flaw in a specific argument for swapping?

According to the author it does find a flaw in a specific argument for swapping. In Step 7 of the main argument the author states that each term of the equation refers to a different event which shares a different amount of money. So each term should be divided by the mean amount of money of that particular event. That way the expected return is 0, not 1/4 as follows from the step 7 of the Problem. Caramella1 (talk) 05:30, 13 October 2014 (UTC)

As I understand it A (the nature of which is very poorly defined) represents the sum in the envelope that you hold, which is £100. Martin Hogbin (talk) 09:14, 13 October 2014 (UTC)

Yes, correct. Caramella1 (talk) 12:26, 13 October 2014 (UTC)

So how it it a different event? Martin Hogbin (talk) 08:08, 15 October 2014 (UTC)

I said that each term of the equation refers to a diferent event. The first term of the equation is (1/2)(2A) and it refers to the event in which the two amounts are A and 2A. The second term of the equation is (1/2)(A/2) and it refers to the event in which the two amounts are A and A/2. In both of them player A has the amount A in his/her envelope, but in the first event the players will share an amount of 3A and in the second event they will share an amount of 3A/2. The paper states that these two terms should not belong to the same formula just as the are but they should be divided first by the specific event's mean amount to find player's A expected return in the case of swapping his/her envelope. Caramella1 (talk) 09:34, 15 October 2014 (UTC)

Carmella1 you are hitting the point, that's exactly what the literature says (Ruma Falk, Nalebuff and all others). The formula does not concern the problem of ONLY TWO (unknown) envelopes of A and 2A, nor does it concern the problem of ONLY TWO (unknown) envelopes of A and A/2.

Approaching the scenario of two unknown envelopes, as per the literature, must consider the total contents of either "1/3 and 2/3", or of "2/3 and 1/3" likewise. Period.

As per Nalebuff, the formula     ${1 \over 2}(2A)+{1 \over 2}\left({A \over 2}\right)={5 \over 4}A$     concerns quite another scenario, it concerns strictly speaking the scenario of THREE envelopes, where the original envelope contains the original value of A, and "thereafter" two possible follow-up envelopes, one containing 2A and the other one containing A/2, whereupon one of both follow-up envelope is discarded randomly. Only then the remaining follow-up envelope B will contain 1.25*A, as per the formula, and the original envelope A contains 0.8*B. That's all. Gerhardvalentin (talk) 20:15, 15 October 2014 (UTC)

Exactly! The formula presented in the step 7 of the problem is valid only if the game is played by different rules, i.e. when the amount A is fixed and it is known that it will be given to player A and the amount of player B will be determined by the flip of a coin. If the coin lands heads then player B will have a (hidden) amount 2A in his/her envelope and if it lands tails will have the amount A/2. Then player A should switch and player B should keep his/her envelope. Notice that this case is no longer symmetrical for the two players. It is also covered in the Tsikogiannopoulos paper. Gerhardvalentin do you agree that the addition I made is useful for the readers of WP to understand the problem better? Caramella1 (talk) 06:40, 16 October 2014 (UTC)

Yes, but we should mention only the operative quintessence of the relevant literature (Falk, Nalebuff, Tsikogiannopoulos and many others), similar to your short explanation here, so that it can be recognised by grandma readers and editors, also. Gerhardvalentin (talk) 07:17, 16 October 2014 (UTC)

This "solution" corresponds to about half of the literature. There is another interpretation of the switching argument, and another resolution. Richard Gill (talk) 11:15, 12 November 2014 (UTC)

I have a pdf of the paper if anyone wants to read it (with the help of Google Translate). Richard Gill (talk) 11:12, 12 November 2014 (UTC)