Talk:Two envelopes problem/Arguments/Archive 5

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Richard's change to the problem in the lead

Richard you have changed the problem statement in the lead to:

Of two indistinguishable envelopes, each containing money, one contains twice as much as the other. The subject may pick one envelope and keep the money it contains. Having chosen an envelope at will, but before inspecting it, the subject gets the chance to take the other envelope instead.

Problem: what is the optimal rational strategy for maximising the amount of money to be gained?

As stated there is no problem, that is what I meant when we talked about the TEP being a self inflicted injury.

The optimum strategy for your problem is:

1) Keep your envelope.

2) Thank the envelope provider for the money.

I do not think anyone seriously challenges this Martin Hogbin (talk) 13:41, 8 November 2014 (UTC)

Indeed. The TEP paradox is not to solve the problem formulated in the lead but to show what is wrong with the argument for switching, which only came later. I did not touch the lead, I did not even look at it. I tackled the introduction, which came later. Weird. Whoever set the article up like that had absolutely no clue what TEP is about. I made necessary changes to the intro but I missed the lead. Richard Gill (talk) 14:11, 8 November 2014 (UTC)

I fixed the lead. Better now? Richard Gill (talk) 14:17, 8 November 2014 (UTC)

Sorry Richard, I noticed that you had none a lot of editing recently and assumed that you had made the change, which surprised me somewhat, I thought you had gone a bit crazy. I should have checked. Once again my apologies.

It is better now but I think it was better how it was before. I think we should sort out the body of the article first and then make the lead a summary of the article as it should be. Martin Hogbin (talk) 01:59, 9 November 2014 (UTC)

PS. I think I also saw this comment by you, 'The Introduction does not introduce the Problem. (did not! - I have changed it)'. Which made me think you had made the above change. Martin Hogbin (talk) 12:29, 9 November 2014 (UTC)

I think one does this work cyclically. Now it is time to go back to the body of the article and improve that. Then it will be time to go back to the lead. And so on ad infinitum. But it is important to establish that the article *starts* with some kind of canonical TEP.

Canonical TEP: There is no looking in envelopes. There is no information about possible amounts of money in the envelopes. The envelopes are filled with two different, positive amounts, one twice the other. You are allowed to pick an envelope completely at random. You do. You are *not* allowed to look in it. You *are* asked if you would prefer to have the other.

I agree. It would be good to describe that version as the canonical version if we have a source saying that. I would use a different word though 'formal/official/standard'. Martin Hogbin (talk) 12:22, 9 November 2014 (UTC)

The present article refers to a paper by Ruma Falk for the statement of the core / basic / standard TEP. You could read her paper and find out why she takes it as her starting point. Richard Gill (talk) 15:09, 9 November 2014 (UTC)

Richard, the description in the 2008 Falk paper is taken from Wikipedia! Do we have a source that does not refer back to WP? Martin Hogbin (talk) 19:35, 9 November 2014 (UTC)

Quite a few people (including Nalebuff) refer to Sandy Zabell, a Bayesian statistician, who has the Bayesian solution (of course) and says that he finds it hard to explain to non-mathematicians. Of course, Nalebuff did introduce his own new Ali-Baba game but he does refer to Zabell. Zabell thinks of both players looking in their envelopes and both desiring to switch. Nalebuff mentions that they do not even have to look in their envelopes to follow the (false) conditional probability reasoning and decide to switch, because they would wish to switch whatever they saw there. Falk (2008) says explicitly that she discusses Wikipedia's problem and Wikipedia's first solution. Nickerson and Falk (2006) studies a whole range of variations including our "canonical" TEP. They mention that the switching argument does not require one to look in the envelope, and they do know the Bayesian resolutions. Richard Gill (talk) 11:56, 14 November 2014 (UTC)

Just like Monty Hall problem, this is a briefly sketched "real world" situation. We have to make a decision. There is some built-in randomness in the problem - picking an envelope; there is also uncertainty or lack of knowledge or ignorance in the problem (what amounts were in the two envelopes to start with?). Informally, we use "probability" both for physical randomness and for uncertainty (we mix ontological and epistemological concepts of probability). As the solution to the necktie paradox makes clear, we have to distinguish between (a) the actual, unknown, amounts in the two envelopes; a and b; which are either equal to x and y = 2x or to y and x respectively, and which never change throughout the whole game, but which are not known; and (b) our knowledge about these amounts. Our knowledge about these amounts changes whenever we get some information. We can also *imagine* being given information (e.g. "there is $20 in Envelope A") and we can imagine how this changes our knowledge.

Whatever else is going on in the paradoxical (false) argument to switch envelopes, it is clear that (a) and (b) are being mixed up. *If* we want to "solve" the paradox by casting it in the language of the probability calculus (mathematics), which works just as well for ontological as for epistemological randomness as for a combo, the first thing we should do is to introduce mathematical notation which makes the distinction between (a) and (b) clear. Conventionally, this is done by using capital letters A, B, X, Y to stand for "random variables" and lower case letters a, b, x, y to stand for possible values which they can take.

The problem is a conjuring trick we should nat ask what was in the writer's mind when he gives the bogus line of argument for swapping but what for of misdirection he is trying to create in the mind of the audience. I still find it hard to believe that the presenter is trying to put in our minds a mangled calculation of E(B). We are specifically told about a (speculated) sum in the original envelope so I think it is much more likely that he is trying to confuse us with a muddled calculation of E(B|A=a). Are ther any sources, apart from you, that state that the given line of reasoning for swapping is intended to be a calculation of E(B)? Martin Hogbin (talk) 12:22, 9 November 2014 (UTC)

It wasn't my idea to suppose that the presenter is trying to put in our minds a mangled calculation of E(B). In my paper I call it the philosophers' solution because I came across it in Schwitzgebel and Dever (2007?) and a whole lot more articles in philosophy journals. Of course these people use pages and pages of words, and tend to mangle all their attempts at mathematics, in particular through inadequate notations. And they don't survey all the literature but just focus on a little corner of it. So they don't even realise that they are interpreting the problem differently from how mathematicians and statisticians had, earlier, interpreted it. But not all mathematicians take the same approach, either. For instance Ruma Falk, who works on the pedagogics of teaching elementary probability, also has a paper in which she uses the philosopher's interpretation (and another one where she takes the "mathematician's interpretation". Richard Gill (talk) 15:06, 9 November 2014 (UTC)

I think it is wrong to call the (presently) second solution a "Bayesian solution". It does not depend on using subjective probability. It works just as well for frequentist probability. It is nothing to do with Bayesianism. It does have to do with conditional probability and conditional expectation value, but that was already the case in the (presently) first presented solution.

I agree, in fact I think this should just be presented as the solution. The other one should be presented as popular/philosophical solution.

Right now the second solution is presented in a much too complicated way because the writer is thinking of a whole lot of variations at the same time (e.g. variations like Nalebuff's in which one does look in the envelope). The "Bayesian" solution of the canonical problem ought to be presented first, and after that, one can start moving on to variations of the problem and to new variations on the solution. Richard Gill (talk) 09:08, 9 November 2014 (UTC)

I agree again. One remaining problem is how to present a simple mathematically sound but non-mathematical description of that resolution. Martin Hogbin (talk) 12:22, 9 November 2014 (UTC)

An interesting quote from Gardner is, 'If we construct a payoff matrix as Kraitchik does in his book we see that the game is symmetrical and does not favor either player. Unfortunately, this does not tell us what is wrong with the reasoning of the two players. We have been unable to find a way to make this clear in any simple manner'.

There is also a hint that Gardner had the E(B|A=a) calculation in mind as he says, 'Does this paradox arise because each player wrongly assumes his chances of winning or losing are equal?'.

Martin Hogbin (talk) 13:19, 9 November 2014 (UTC)

It is easy to explain if you can use standard probability calculus notation and if the reader understands conditional expectation (and the relation between conditional and unconditional). The reader who doesn't have this background should not be reading deeper into this article. They will have to be contented with a verbal summary, as in the two neckties. It is a verbal summary which covers both of the first two interpretations. The same mixup is being made, either way. It is not wikipedia's job to explain basic elementary probability theory to its readers ... at least, not in an article which is not devoted explicitly to that topic. Richard Gill (talk) 15:14, 9 November 2014 (UTC)

Richard, I think a verbal summary which covers both interpretations is no summary at all but just confused verbage. I think we should try to explain the E(B|A=a) resolution in words that most readers can understand.

Assuming the possible sum in the envelopes is bounded seems a good place to start to me and is exactly the situation described by Gardner. He says, 'We may assume that each player has a random amount of money from zero to any specified amount, say $100'. Martin Hogbin (talk) 17:02, 9 November 2014 (UTC)

The article should say what the sources say. The formula ${1 \over 2}(2A)+{1 \over 2}\left({A \over 2}\right)={5 \over 4}$  (improper for the standard problem) is just only an incomplete fragment that perforce leads to false conclusions. As per the sources, this mistake is adding apples and oranges,

because the first term ${1 \over 2}(2A)$ is based on the claim that actually (A+B)=3A,

whereas the second term ${1 \over 2}\left({A \over 2}\right)$ is based on the claim that actually (A+B) = 3A/2,  being only one half of the first term. So this formula is addressing two quite differential "pairs of envelopes" with quite different total amounts of (A+B). Why not clearly say] what the sources say? Gerhardvalentin (talk) 16:30, 9 November 2014 (UTC)

Gerhard, suppose you open your envelope and see £100. Surely A is then 100. The other envelope must then contain either £50 or £200. What is wrong with the proposed calculation now? Martin Hogbin (talk) 17:02, 9 November 2014 (UTC)

Thank you. Pleae consider that, in the standard problem, A can either be the "small amount" of (A+B)/3, then "large B" is 2(A+B)/3.
Or A can be the "large amount" of 2(A+B)/3, then the "small B" will be (A+B)/3. In any "ONE" pair of envelopes, even if the value of A and the value of B both are unknown, their total is unknown but UNCHANGEABLE. The same applies to the actual total amount of (A+B) in case that A is known (no difference).
But the problematic formula (valid only for AliBaba) mixes two different "games" of TWO possible pairs of envelopes and treats both as if they were one single pair of envelopes. Term 1 addresses pair X and Term 2 addresses pair Y. Although any single pair of envelopes has got a given total amount of "A"+B, the formula uses two different total amounts of "A"+Blarge (of game X) and another of "A"+Bsmall (of game Y), i.e. two different values of B, resulting in a given total amount representing "3A" and, at the same time, in a given total amount representing "3A/2". Saying it is only ONE pair of envelopes. This is correct for the AliBaba variant, but weird for the standard version of the TEP. – Btw, the actual amount of A is unimportant. (I hope I made no typos.) See my "picture". Gerhardvalentin (talk) 18:41, 9 November 2014 (UTC)

In my example A is £100. It is not unknown it is fixed at £100. The other envelope may therefore contain £50 or £200. Do you agree?Martin Hogbin (talk) 19:21, 9 November 2014 (UTC)

Gerhard, you are wrong. The problem is not the amounts but the probabilities, if one interprets the aim of the argument as being to calculate E(B | A = a). Richard Gill (talk) 15:40, 10 November 2014 (UTC)

Richard, you are still wrong. See what I told on your talk page 3 years ago, in 2011. It is chastely on probability, and in this case we have to be very careful. As certified official comptroller of Swiss Compensation Fund I know what I just am talking about. We may not be perfunctory in this case. Consider a large amount of investigation. Yes, in 50% the other envelope contains twice, and in another 50% it contains half. But if you investigate just only a set of "1/2 of ALL cases"without constriction, then within this set, again in only 50% of this set the other envelope will contain twice, but within this set the other 50% will contain half, and the result of your investigation will be faulty. This is the flaw of the 5/4-formula.

Please be aware that it is not effectual to investigate in an undiscerning way just only "1/2 of ALL cases", but that it is absolutely necessary to investigate on the one hand explicitly only THAT subset of cases where the other envelope INDEED contains twice (50%, yes), and to investigate on the other hand explicitly only THAT subset of cases where the other envelope INDEED contains half, again 50%. Otherwise you stick with the 5/4-formula that is flawed for the standard version of the TEP. So please be aware of this subtle difference, of this subtle distinction. This is what we have to take into account.  Gerhardvalentin (talk) 03:02, 11 November 2014 (UTC)

Gerhard you are wrong. Read the literature. You are reporting OR (original research) by you, and it is flawed. It is perfectly possible that the 5/4 formula is correct for almost all values "a" of the amount in Envelope A. It is a theorem that the formula cannot be correct for *all* values. If the amount in envelope A is "a", then the 5/4 formula is correct (for the conditional expectation of B given A = a) if and only if it was a priori equally likely that the two amounts are (a/2, a) and (a, 2a). Richard Gill (talk) 16:16, 11 November 2014 (UTC)

Interdependency

I'm reminded to what I read above, Richard answered me:

And BTW you say: "if A actually is the larger amount (2/3 of the total amount), it cannot be doubled but can only be halved, and if A actually is the smaller amount (1/3), it cannot be halved but only be doubled". You're saying that if A is the larger amount, it's the larger amount. So what?

Two unknown envelopes, of only ONE pair of envelopes. If A is fixed at £100, then please consider that the total amount of both actual envelopes is fixed, also. Either the other envelope contains £50, then both envelopes contain £150, with A being 2(A+B)/3 in this case. This is valid for ONE pair of envelopes - as said with a total (A+B) of £150.

Or it is quite another pair of envelopes, where A also is fixed at £100, where the other envelope contains £200, then both envelopes contain £300, in this case A being only (A+B)/2. In the AliBaba version, these both scenarios are equally likely. But as to the standard problem, B can only be 2A iff A actually IS the smaller amount of (A+B)/3. Otherwise not.
And as to the standard problem, B can only be A/2 iff A actually IS the larger amount of 2(A+B)/3. Otherwise not.
This is mutually exclusive. Any "as well as" is only possible for two different pairs of ennvlopes.

Just to illustrate: Let's call the smaller of both amounts S, and the larger amount 2S, so the total amount T=3S. Suppose A=S (and B subsequently=2S)  :). Then the expected amount of envelope B is: ${1 \over 2}(2S)+{1 \over 2}\left({2S \over 2}\right)={3 \over 2}S$ = half of the total amount. So the expected value of envelope "B" will be half of the unchangeable total amount: $={1 \over 2}(3S)$. This looks rather simple, so I repeat my version of "2a=A) because it "looks more familiar" to the 5/4-formula:
${1 \over 2}(2a)+{1 \over 2}\left({A \over 2}\right)={3 \over 4}A$ 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.

Conclusion: A may not reach two different values at the same time within the same single formula (sin!) and, at the same time, the total amount of (A+B) may not reach two different values within the same single formula. A and B are interchangeable, but their total has to be constant, otherways weird, and only correct for AliBaba. Regards, Gerhardvalentin (talk) 20:57, 9 November 2014 (UTC)

Gerhard, you have not answered my simple question. If your envelope contains £100 the do you agree that the other envelope will contain either £50 or £200? Martin Hogbin (talk) 21:03, 9 November 2014 (UTC)

Martin, the answer is YES (but !!!). In a given pair of envelopes, if your envelope contains £100, then the other envelope in exactly 50% may contain £50, yes, BUT ONLY in those 50% where –at the same time – the actual total amount is £150. Repeat: only if the total amount actually is £150, only in THOSE 50%. Otherwise not.

And in a given pair of envelopes, if your envelope contains £100, then the other envelope in the other exactly 50% may contain £200, yes, BUT ONLY in those other 50% where the actual total amount is £300. Repeat: only if the total amount actually is £300, only in THOSE 50%. Otherwise not, please regard the (this) given inevitable inter-dependency, this strict coerciveness that applies to any single pair of envelopes. Any stubborn featherbrained "ignoring" of this evident interdependency can lead to post-hoc fallacies, just have a look to the literature.

A and B are only interchangeable.   In a given pair of envelopes, anything else is impossible. The total value of any actually given pair of envelopes is fixed and CANNOT be assimilated. It cannot be "adapted", this is utterly impossible, a complete impossibility, an oxymoron. In a given pair of envelopes, A and B are merely "interchangeable", fatto compiuto. Gerhardvalentin (talk) 21:24, 9 November 2014 (UTC)

Now you are getting at the heart of the problem. Gerhardvalentin, can you make a calculation of the expected return (profit or loss) without variables (a, A, S, etc.) and without prior beliefs, using the fact that one envelope contains £100? The same question goes to Richard Gill. Caramella1 (talk) 06:33, 10 November 2014 (UTC)

Caramellla1, thank you for your question. But there is no need to calculations, the expected return (profit or loss) clearly and evidently is zero, in the long run. Gerhardvalentin (talk) 11:57, 10 November 2014 (UTC)

Caramell1, of course I cannot make a calculation of expected return without specifying my expectations concerning what amounts could be in the two envelopes. But if you like I can give you the formula which you can use. You plug in your prior beliefs and then calculate. It's a very simple formula. If envelope A happens to contain £100 then we only need to know our prior beliefs that the pair of envelopes contain (£50, £100) vs (£100, £200). If our prior belief concerning those two possible cases is equal, then our posterior belief will be equal too. And the expected value of switching is £125. Richard Gill (talk) 15:54, 10 November 2014 (UTC)

Gerhardvalentin, can you prove what you say with a formula? Remember, no variables, no prior beliefs, only £100 in one envelope.

I'm just back, and yes, Caramella1, anybody can. With the correct formula, be it with positive amounts only, or be it with variables.

See the correct formula above (with variables "A and a" (please note that a=A/2 resp. A=2a – though Richad Gill controverts, but anyway, even M treats it that way). This "correct" formula says that the expected amount in envelope B is equal to the assumed amount in envelope A, being "half" of the total amount contained in both envelopes. Agreed? Though it isn't "necessary" that we know the positive amount contained in envelope A, let's agree that the amount in envelope A is known to be 100 (of any currency, be it £ or $ or €):
${1 \over 2}(2a)+{1 \over 2}\left({A \over 2}\right)={3 \over 4}A$ 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.

And with positive amounts only:

${1 \over 2}(2x100)+{1 \over 2}\left({2x100 \over 2}\right)={3 \over 4}2x100$ resp. $={3 \over 2}x100$ = again half of the total amount.

In other words, in exactly 50%, envelope A will actually hold the smaller amount (let's call it "S") of a=1/3 of the total amount contained in both envelopes (let's call the total amount "T"). ONLY IN THIS CASE, that envelope A actually holds T/3 (i.e. if envelope A actually holds the smaller amount of a=T/3 – and NOT in "ANY" 50%, i.e. NOT in the subset of those cases where envelope A actually holds A=2T/3 !), then envelope B will hold 2a (B=2T/3). And envelope B will hold A/2 not in ANY 50%, but only in that 50%, where envelope A actually holds the large amount of A=2T/3, and NOT in the subset of those cases where envelope A actually holds only a=T/3. – We have to pay regard to this immanent interdependency, the AliBaba 5/4-formula does not pay regard to this given immanent interdependence of one single pair of envelopes. Regards, Gerhardvalentin (talk) 23:07, 10 November 2014 (UTC)

Gerhardvalentin, if I understood you correctly, your formula suggests that the expected value of the other envelope is £150. So your envelope (suppose that you play the game) contains with certainty £100 and you expect that the other envelope contains £150. According to this reasoning you should switch because by switching you expect to have a profit of £50. Then how is the expected return zero? Caramella1 (talk) 05:02, 11 November 2014 (UTC)

Caramella1, it's because you just made evident that knowing the contents of the envelope picked, or not, makes no difference. You just made it evident that knowing the coincidental contents of the envelope picked can never be any indication on what the other envelope may actually contain. Gerhardvalentin (talk) 08:15, 11 November 2014 (UTC)

So, you say that it is impossible to include the £100 we know that is contained in one envelope to a formula of expected return and the result to be zero? Caramella1 (talk) 12:51, 11 November 2014 (UTC)

Richard, by "no prior beliefs" I meant that the player who sees £100 in his/her envelope has no reason to believe that the (£50,£100) scenario is more or less probable than the (£100,£200) so (s)he assigns equal probabilities 1/2 to both. But you covered that case and you say that the expected value of switching is £125? This is more than the £100 contained in the player's envelope so according to you (s)he should switch? So the "switching argument" presented in the main article which assumes no prior beliefs has no flaw? Please explain. Caramella1 (talk) 20:11, 10 November 2014 (UTC)

I explain, in my paper http://www.math.leidenuniv.nl/~gill/tep.pdf (and this is nothing new - there are many many reliable sources explaining why this goes wrong). Here is one try. Yes, *if* in advance the player thinks that (£50,£100) is equally likely to (£100,£200), then *if* they would find £100 in Envelope A they would switch. But now imagine some more amounts in Envelope A: going up by powers of 2: £200, £400, £800, ... and going down by powers of 2 £50, £25, £12.5, ... if the player thinks that ... (12.5, 25) and (25, 50) and (50, 100) and (100, 200) ... are all equally likely, then they would switch in any case, hence no need to look in the envelopes. But now the player apparently puts equal probability on (100, 200) x 2^n for all integers n (positive and negative). This is an improper prior distribution. Ordinary probability calculus breaks down.

For any proper prior distribution, the chance that Envelope A contains the smaller amount has to depend on the amount a in the envelope; it cannot be always equal to 50%. Richard Gill (talk) 15:22, 11 November 2014 (UTC)

The problem as stated in the main article doesn't mention about any prior distribution that the player had in mind before the envelope opens (if it ever opens for that matter). We could talk about this variant also, but let's focus on the simplest case first where the player thinks of NOTHING in advance. The moment the envelope opens, there are only 3 amounts that could be in the envelopes: £100 which is with certainty in the player's envelope and £50 or £200 which are supposedly in the other envelope. With no prior distribution and consequently no prior beliefs we agreed that the player could assign probability 1/2 to each outcome concerning the other envelope. You say that in this case the expected value of switching is £125 and this result indicates that the player should switch. Now, if the player forgets for a moment the £100 in his/her envelope and denotes by X the smaller amount and by 2X the larger one then the expected value becomes 3X/2 and this result indicates that the player has no interest in switching. Two different approaches that I think you consider both to be correct lead to two different results. How do you explain that? Caramella1 (talk) 17:54, 11 November 2014 (UTC)

The player does not look in the envelope and does not think in advance.

If you want to know how I explain things, read my paper http://www.math.leidenuniv.nl/~gill/tep.pdf

Of course the problem does not mention a prior distribution. That is the whole point. The argument for switching, however, does implicitly use a prior distribution. Implicitly, if we interpret the intention of the argumenter as being to compute E(B | A = a), he or she is assuming that a priori any particular amount is equally likely as the same amount multiplied by any power of 2 (any whole number, positive or negative). This leads into contradictions. Richard Gill (talk) 10:54, 12 November 2014 (UTC)

An example for you

Let us say the envelopes must contain more then £2 and less than £1000, and you know that. One envelope is picked randomly and the next one above is added as the second. You pick an envelope and open it to see £128. Should you swap? Martin Hogbin (talk) 21:29, 9 November 2014 (UTC)

Sorry, I don't get it. What means "the enelopes"? Single envelopes or pairs of envelopes with "1/3:2/3" resp "2/3:1/3"? One "single" envelope is picked out of how many envelopes? Out of a pile, or out of a crowd? What allocation? What scenario are you talking about, please help. In case of an accidental crowd of 995 envelopes (£ 2, 3, 4, 5 etc.) with equal allocation, it could be well to swap. Was this your question? But this is quite another issue, I suppose? Gerhardvalentin (talk) 22:12, 9 November 2014 (UTC)

I am saying that the two envelopes must contain more than £2 and less than £1000. I gave an example of the way in which the sums in the two envelopes may have been chosen.

If you prefer we could do it this way. We pick a sum between £2 and £500 and put it in an envelope. Then we put double that sum in another envelope. The player knows how the envelopes were set up but is given one of the two envelopes randomly.

He the opens his envelope and sees £100. He reasons that the other envelope is equally likely to contain £50 as £200, which means that on average he will get £125 by swapping. Is he correct in his reasoning? Should he swap? Martin Hogbin (talk) 22:38, 9 November 2014 (UTC)

Thanks. Disregarding the allocation of "£2 to £500" this is similar to the TEP. His reasoning to get on average £125 is nonsense, as you know, because in trillions he will neither win nor loose. But regarding the said (actual) allocation, and his £100 being closer to the lower end, he could conclude to actually be holding "the lower amount of both envelopes, also", so he actually should swap. But this is not the standard TEP. Btw, if this game is repeated millions of times, then on average he will neither win nor loose. Regards, Gerhardvalentin (talk) 23:21, 9 November 2014 (UTC)

In what way is my example not the standard TEP? Martin Hogbin (talk) 13:13, 10 November 2014 (UTC)

Disregarding the allocation of "£2 to £500", it is similar to the standard TEP that does not announce such "known allocation". Gerhardvalentin (talk) 13:21, 10 November 2014 (UTC)

Gerhard, the reasoning is not wrong. He reasons that on those occasions when his envelope contains £100, half of the time the other envelope would contain £50 and half of the time the other would contain £200. If he would switch every time, he would on average receive £125. All of this reasoning is perfectly correct, as long as the premise that £50 and £200 would occur equally often (among those occasions when Envelope A contained £100) is correct. And this is not unreasonable ... if the smaller envelope is equally likely to contain ... £25, £50, £100, £200 ... then, on those occasions when Envelope A actually contains £100, Envelope B is equally likely to contain £50 as £200. Richard Gill (talk) 15:46, 10 November 2014 (UTC)

Richard, thank you for your comments, but I know that you know better than what you just said. Since 12:24, 29 July 2011 I said on your talk page that it is necessary to detach us from the AliBaba variant, where this kind of reasoning is correct indeed, but wrong for the standard TEP. For the standard TEP, such kind of reasoning is utterly wrong indeed. The arguments (incorrect for the standard TEP) that lead to the 5/4-formula all is correct only for AliBaba. The formula ${1 \over 2}(2A)+{1 \over 2}\left({A \over 2}\right)={5 \over 4}A$ (for any A, regardless of A being – in the end – 1/3 or 2/3 of the total amount), addresses only AliBaba, but does not adress the standard TEP. Because "any A" means "any A", be A actually 1/3 of the total amount, or be A actually 2/3 of the total amount. For the standard TEP, this is wrong. This lapidary formula (not lapidary for AliBaba, but lapidary for the standard TEP) wrongly says that in "ANY 50%", B will be 2A, and it wrongly says that in "ANY 50%" B will be A/2. Please consider that, for the standard TEP, this is utterly WRONG.

The result E(B)=5/4A ignores that in one half of "ANY 50%", where it says that B will be 2A, this is not "given", as in exactly one half of those cases B cannot be 2A, as in this half of "ANY 50%", B actually is A/2. So you have to investigate the said predication a little bit closer:

only if atually indeed B=2A, meaning that B=2T(Total amounts)/3, then B=2A. This is valid only in this case. Otherwise not. And so on. In 1/4 of cases, the 5/4-formula assumes that if B=T/3, then A=T/3, and in 1/4 of cases, the 5/4-formula assumes that if B=2T/3, then A=2T/3. This is the flaw of the 5/4-frmula, regarding the standard TEP. As in the standard TEP, B=2A only if B=2T/3, and in the standard TEP, B=A/2 only if B=T/3. Not in "ANY CASE". Never ever. Comprising two quite different "1/4 of cases" into "1/2 of cases", and comprising two quite other different "1/4 of cases" into another "1/2 of cases" is correct for AliBaba only, but a flaw as to the standard TEP. Gerhardvalentin (talk) 17:18, 10 November 2014 (UTC)

Gerhard, I was trying to give an example where your argument applies but you should still switch. I have departed from the standard version in that the player looks in his envelope before deciding what to do, and I have considered the possible sums in the envelopes to have a maximum, as Gardner does. In all other respects, this is the standard TEP. Martin Hogbin (talk) 17:25, 10 November 2014 (UTC)

We have to detach from AliBaba. Looking in his envelope, or not, is irrelevant. In the standard TEP, there is no reason for him to swap envelopes. Gerhardvalentin (talk) 17:45, 10 November 2014 (UTC)

My example is not AliBaba. In that case the player picks an envelope then a coin is tossed to decide what to put into the other envelope. The player knows that he has the original envelope.

In my example there are two envelopes, one containing twice the sum that is in the other, but the player is given one randomly. That is the standard TEP. Martin Hogbin (talk) 18:18, 10 November 2014 (UTC)

Martin pardon, please help me: Is there any unclarity? Gerhardvalentin (talk) 18:35, 10 November 2014 (UTC)

In what? Martin Hogbin (talk) 18:50, 10 November 2014 (UTC)

Can I also suggest that we move to the arguments page. Martin Hogbin (talk) 18:51, 10 November 2014 (UTC)

In the standard TEP (you pick one of the two envelopes at random and you do *not* get to look what is in there) there is evidently no reason to switch. Yet there is an argument which seems to compel you to switch. The problem is to show what is wrong with the argument. There are different "solutions" depending on how you interpret what the designer of the argument had in mind. There is not a unique solution, because of the Anna Karenina principle. Richard Gill (talk) 20:24, 10 November 2014 (UTC)

Yes, I know that. I said just above, ' I have departed from the standard version in that the player looks in his envelope before deciding what to do'. Gerhard has replied that he does not consider looking in the envelope a significant issue in this discusssion, his words were, 'Looking in his envelope, or not, is irrelevant'. Do you agree that, apart from looking in the envelope, my setup refers to the standard TEP. Martin Hogbin (talk) 20:46, 10 November 2014 (UTC)

Gerhard, you are WRONG. Completely wrong. It seems you don't read the literature, and you don't understand Bayesian probability. In particular you didn't read my own paper. Richard Gill (talk) 09:34, 11 November 2014 (UTC)

Martin, again to the flawed expectation of B=5/4A, flawed for the standard TEP.
With probability of exactly 50% B will be 2A (only if A actually is T/3, i.e. 1/3 of the total amount – bear in mind that you have to account for your ignorance!), and with probability of exactly 50% B will be A/2 (only if A actually is 2T/3, the larger amount of both – bear in mind that you have to account for your ignorance!).

If A is T/3, then B=2T/3 (50%), and
if A is 2T/3, then B=T/3 (50%).
For A, this adds to T/2, and for B this adds to T/2, also.
So, for the standard TEP, you have to fix A (the contents of envelope A) to be T/2, and the expected value of B=A (and not 5A/4).

E(B)=A applies to whatever amount you see (or dont'see) in envelope A, be it 50 or 100 or 200. Just pay attention to the fact that you do not know whether A actually is the smaller amount or the larger one. You cannot lapidary say "with probability of 1/2, B will be 2A", knowing that this is valid only in case that A actually is T/3 indeed, and vice versa. In short: For the standard TEP, the 5/4-formula (A≠A) does not include your ignorance. Because you never can know whether actually A=T3 or A=2T/3, indeed. And in her 2009 paper Falk pays regard to the fact that it is the flaw of the 5/4-formula to ignore our ignorance. This has to be shown just in the beginning, to help understand the arising of the so called paradox. Regards, Gerhardvalentin (talk) 12:16, 13 November 2014 (UTC)

Gerhard, can you please improve the text in the article section now called Logical resolutions? iNic (talk) 13:20, 13 November 2014 (UTC)

Thanks, I will try to do so, but I'm away for some time now. Gerhardvalentin (talk) 13:55, 13 November 2014 (UTC)

Imagine that the player has no idea what are the amounts in the two envelopes. He has picked one completely at random (standard TEP). We call it "Envelope A". The player now *imagines* looking in Envelope A and seeing an amount there, let's for the sake of argument suppose that he *imagines* seeing 100 Euro there. He knows the other envelope contains 50 or 200 Euro. A priori, he believes it is equally likely that the two envelopes contained 50 and 100, as that they contained 100 and 200. So now, for him the other envelope is equally likely to contain 50 or 200. The expected quantity in the other envelope is 125 so he figures that he would switch. He gets the same conclusion, *whatever* amount he imagines being in Envelope A. So he doesn't have to look in the envelope: he knows that if he would look, he would decide to switch, whatever he saw there.

There is absolutely nothing wrong with this argument!!! But we know that switching (without looking in the envelope) is senseless. So we had better think about the assumptions which were used in the argument I have just given, a bit more carefully.

Is it reasonable to suppose that *whatever is in envelope A, the other is equally likely to contain half as to contain double that amount*?

There is no problem with the reasoning. There must therefore be a problem with the assumptions. Take the assumptions seriously and see where they lead. You will see that they lead to the nonsense assumption that the smaller amount of the two is equally likely equal to 50, to 100, to 200, to 400, ... and at the same time to 25, to 12.5, ... They lead to the assumption that the logarithm of the smaller of the two amounts of money is uniformly distributed between - infty and + infty. A so-called improper prior distribution. With such a prior distribution, expectation values make no sense. Probability calculus breaks down. The assumption was unreasonable and unrealistic, and leads to paradoxes.

Please, Gerhard, stop coming up with your own arguments, but instead, study the literature and *understand it*. Richard Gill (talk) 09:38, 11 November 2014 (UTC)

This discussion showcases perfectly the dichotomy found in the literature regarding how one ought to solve this problem. We as editors should not waste our time discussing the problem itself. If you want to continue to do that please do that at the Arguments page. Instead we should be happy that we have representatives of both views here as editors. I think that Gerhard can write the text for the more logical solutions of the problem, which include the "common resolution" and Smullyan's version, while Richard and Martin can write about the probabilistic/Bayesian approach. Which section is presented first in the article isn't that important. It's more important that the text is kept short with not too many mathematical formulas. The main ideas are easy to grasp and can be explained almost without any math at all. This page was much more fun to read some years ago I think. It was short and the main ideas were explained in a short and succinct way. I think it would be great if we could evolve the page to become less complex and easier to read again. iNic (talk) 12:54, 11 November 2014 (UTC)

I agree! Richard Gill (talk) 15:11, 11 November 2014 (UTC)