Talk:Two envelopes problem/Archive 6

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Falk (2008), Falk and Nickerson (2009), Nickerson and Falk (2005)

I have also been in email contact with Ruma Falk and have read the three papers on TEP of which she is author or co-author. They are all excellent. Note that they are written for different audiences. Nickerson and Falk (2005) is perhaps the first paper on TEP which seriously surveys all the solutions known to the authors at that time. It explicitly points out that the different solutions make different assumptions as to context and intention of the writer. It pays attention to the Bayesian context with and without improper priors. Explains that improper priors give infinite expectation values and hence unreliable advice as to decision or action. It's all there. All three are now in my dropbox, if anyone else is interested to read them. I think that Nickerson and Falk (2005) is obligatory reading for any wikipedia editor of TEP. It's the only paper with a complete and neutral survey of as many solutions as possible. Richard Gill (talk) 13:42, 4 November 2011 (UTC)

The error in the switching argument

Please help to check which sources say the following, and help to improve the article accordingly:

The basic setup does not accent in any way the fact that there is a huge difference  between having chosen the envelope "filled at first with any determining amount of money", i.e. containing the "original amount" of X  (let's call that amount "Xoriginal", and let us denote that amount with the value of "1") with probability of 2/4, or having chosen the envelope that was filled thereafter with the dependent amount, say "the second envelope", containing dependently either 2X ("Xdouble", with the denoted value of "2") with probability of 1/4, or with equal probability of 1/4, containing only X/2 ("Xhalf", with the denoted value of "0,5").

For only in the case that (with probability of 1/2) you should have chosen the envelope with the determining amount of "Xoriginal", then with probability of 1/2 the other envelope will contain dependently "Xdouble" (2X) or will contain with equal probability of 1/2 only  "Xhalf" (X/2). And note that only in that case that, by chance, in 1/2 of cases you should have chosen the envelope containing the determining "Xoriginal" amount, all the further conclusions of the TEP, including "5A/4") do apply, otherwise not.

But fully neglecting that draconic precondition, "the basic setup" is based on this last variant only. Without saying so. The basic setup and all further conclusions do fully neglect the scenarios of the two other possible constellations with quite other assigned prospects, having equal probability of 1/4 each: That you could have chosen the dependent "second envelope" of 2X ("Xdouble"), having quite other prospects assigned, or that you could have chosen the dependent "second envelope" of X/2, again having quite other assigned prospects, contradicting all further conclusions of the TEP.

The value of 5A/4 is only correct if the candidate, by chance, should have chosen the determining envelope "filled at first with any amount of money". Then the dependent envelope B (its "slave") can be expected to contain 5A/4.

But if the candidate should have chosen the dependent envelope filled thereafter only, remunerated either with considerable 2X or with X/2 only, then in that case the expected determining value of the other envelope can never be 5A/4 but will forever be 4A/5 only. That's a given fact:  E(B) = (2/2+1/2*2)/2,5 = 4/5 A.

And – without knowledge which one of those both envelopes had been chosen, the determining first one or the dependent second one – the expected value of the other envelope is exactly 1A:

E(B) = ((2*1+1/2)+(2/2)+(0,5*2))/4,5 = 1A.

Which sources do show that coerciveness? It could help to better articulate the inevitable facts. Who can / will help? Gerhardvalentin (talk) 18:56, 23 November 2011 (UTC)

The standard TEP story has two envelopes filled with two positive amounts of money, one twice the other. Then the player picks one of the two envelopes completely at random. How the envelopes are filled is not part of the story. Some authors, by way of example, discuss particular ways in which the host could have chosen the two amounts of money. They show that for their particular story, one or more of the steps of the argument is actually wrong. But such examples do not solve the paradox. For one thing, for many authors the important thing is not how the host filled the envelopes, but what the player believes about the amounts of money finally in the two envelopes. For another thing, showing that there is an inconsistency in the TEP argument in one particular case does not prove that the argument is incorrect in all cases. It might help you to discover what is the mistake in general, but it might not.

Anyway, the general mathematical principles behind all knowns solutions is given in Samet, Samet, and Schmeidler and later improved by Gill. Richard Gill (talk) 00:05, 24 November 2011 (UTC)

Thank you.  Am no prob. champ, nor do I know all the sources, but you helped me to see that the 5A/4  result can never be based on any symmetric distribution, because in that case of symmetric distribution the only valid result E(B) = ((2*1+1/2)+(2/2)+(0,5*2))/4,5 = 1A.

And you helped me to see that one thing is quite obvious:
The TEP story and any game is based on two quite different types of envelopes. Type "α" had been filled first with any certain determining amount "X" (let us say that this amount is "1" e.g., and only "thereafter" another envelope of type "β" has been filled either with dependent "2X" or with "X/2".

And the 5A/4 result is only valid in case you know for sure that the determining envelope A is of type "α", otherwise never: 1/2 * 2A + 1/2 * A/2 =5A/4
(but the "basic setup" of the TEP ignores that fact and doesn't even mention this "a priori given draconic restriction".)

Whereas in case you know for sure that envelope A is of dependent type "β" then E(B) = (2/2 + 1/2 * 2) / 2,5 = 4A/5 only.

And as long as you do not know for sure whether you hold a determining "α-envelope" or a dependent "β-envelope", as this is not "part of the TEP story", then only E(B) = ((2*1 + 1/2)+(2/2) + (0,5*2))/4,5 = 1A can be correct. That's it.

	Envelope A	amount  A	Envelope B	amount  B	result of switching  to envelope B
A of type α (determining envelope):	Xoriginal	1	Xdouble	2	2Aα of Xoriginal
A of type α (determining envelope):	Xoriginal	1	Xhalf	1/2	1/2Aα of Xoriginal	5/4 Aα of Xoriginal
A of type β (dependent envelope):	Xdouble	2	Xoriginal	1	1/2Aβ of Xdouble
A of type β (dependent envelope):	Xhalf	1/2	Xoriginal	1	2Aβ of Xhalf	4/5 Aβ
total  (type unknown resp. mixed)		4,5		4,5	1A (if type unknown / mixed)

Result of switching to envelope B:  5/4 A, 4/5 A and 1A respectively (millionfold verified):

Only if envelope A is of type "α" with determining amount, then E(B) = 1/2 * 2A + 1/2 * A/2 = 5/4 A (as the article – without destinguishing - incorrectly claims for all types of envelopes)
Only if envelope A is of type "β" with dependent amount, then E(B) = (2/2 + 1/2 * 2) / 2,5 = 4/5 A
If type of envelope A ("α" or "β") is unknown:    E(B) = ((2*1+1/2)+(2/2)+(0,5*2))/4,5 = 1A

Is that correct distinction mentioned anywhere in the sources?  Regards, Gerhardvalentin (talk) 01:52, 24 November 2011 (UTC)

This distinction is mentioned in many papers. You should read the sources if you want to work on this article. If you like you are welcome to join my dropbox folder which contains all the important papers and many less important.

There is a problem with your notation. You relate E(B) to A. But in many formulations A is also random. If you want to *do* probability calculus, please *learn* probability calculus. There are excellent free eBooks easily available. Many of the sources point out that the error in the TEP argument is the error of not distinguishing between random variables and possible values thereof. If you do not make this distinction yourself you cannot contribute to writing an article which correctly summarizes what is in the sources!

Secondly, your conclusions are not correct, because in the case that we use probability also to represent prior ignorance of, say, the smaller amount of money x in the two envelopes, then it is the case that if we a priori are so collosally ignorant concerning x that we use the improper prior distribution which makes log(x) uniformly distributed on the whole real line, then it is true that E(B|A=a)=5a/4 for all a. If that interpretation was the intention of the writer, then we must resolve the paradox in another way. Richard Gill (talk) 18:47, 24 November 2011 (UTC)

Yes, thank you Richard. I meant to ask whether – for a better understanding of the very  "nature of this paradox"  for grandma and grandson – any sources do clearly accent that "E(B)=5A/4" is fully valid,  but fully valid only if you know for sure that envelope A had FIRST already been filled with any arbitrary determining amount and only afterwards it was decided whether the dependent envelope B was filled with dependent half or dependent duplicate of the already fixed determining amount of envelope A. Because only then, although the already "prefixed" determinng amount in envelope A  will equally likely be double or half of the "dependent" amount in envelope B,  "5A/4"  for envelope B is perfectly correct then, but never otherwise. Because "E(B)=5A/4" does not tolerate any prefixed amount in envelope B with some "dependent" content in envelope A. This given restriction should be made obvious and may never be kept secret, otherwise creating an apparent paradox. This given axiomatic precondition not explicitly being "part of the story",  E=(B)=5A/4 should never unrestrained and nonrestrictive be laconically quoted. Do any sources clearly accent this basic fact, just to help grandma and grandson for a better understanding of the very nature of this just assumed "mystic paradox"? Gerhardvalentin (talk) 11:58, 25 November 2011 (UTC)

The present text of the article says explicitly: there are two envelopes, you pick one completely at random, and that is Envelope A, the other is B. Now, before you even look in A, you consider whether or not you'ld like to switch for B. There is no point at all thinking about the wrong scenario, except to underline the point that it is wrong.

Sure, there are plenty of sources which compare the true TEP scenario to the wrong scenario.

But another point: why do write E(B)=5A/4? Is A supposed to be random or fixed? Please think carefully about your notation! You have got to get this straight!

Richard, but it's the article that misleadingly states:

7. So the expected value of the money in the other envelope is
${1 \over 2}(2A)+{1 \over 2}\left({A \over 2}\right)={5 \over 4}A$

8. This is greater than A, so I gain on average by swapping.

It's the article that misleadingly says that you can expect B to be 5A/4 or 5/4 A. – It's the article that incorrectly says so. Gerhardvalentin (talk) 12:55, 27 November 2011 (UTC)

Yes, the reasoning would be easier to follow and easier to see where it went wrong if a good notation was being used. I think that the writer was trying to compute E(B|A=a). He gets the answer 5a/4 by supposing that, given A=a, the two possible values of B are 2a and a/2, and the two conditional probabilities thereof are 1/2, 1/2. And that's where he is wrong. Richard Gill (talk) 14:11, 29 November 2011 (UTC)

Next point: it is not only in the special (wrong) scenario you just described that E(B | A=a )=5a/4 can be true for some values a. It can be true in the proper TEP scenario for almost all values a. The statement E(B | A=a )=5a/4 is true if and only if, given that A=a, envelope B is equally likely to contain 2a or a/2.

For instance, the chess-board example: the host chooses one of the 64 squares of the chess board completely at random. The squares were pre-numbered 0 to 63. Given that he chose square r, he puts amounts of money 2 to the power r and 2 to the power r+1 into the two envelopes, shuffles them; you pick one and call it Envelope A. Unless there is 2 to the zero, or 2 to the 64, in your envelope, it is equally likely that Envelope B contains half or double the amount. E(B | A=a )=5a/4 for almost all values of a. With probability 63/64 therefore, E(B|A)=5A/4.

Now consider bigger and bigger chess-boards. As they get bigger and bigger, the probability that E(B|A) is not equal to 5A/4 goes to zero.

The precursors of two envelopes (Schrödinger and Littlewood's two-sided cards problems, Kraitchik two neckties) were thinking of this limiting situation.

Another example. The host tosses a biased coin with probability of heads equals 1/3 till the first time he gets heads. Call the number of tails he saw before he got heads r. He puts two to the r in one envelope and 2 to the r+1 in the other, shuffles them, and you choose one and call it Envelope A. Whatever amount a is in your envelope, it's the case that E(B|A=a)>a. Do the calculations yourself, please! You need to learn some elementary probability and the good way to learn it is by carefully working through some simple examples. Build up your probabilitistic intuition, practice using a good notation which distinguishes random variables and possible values of random variables. Install the statistical language R to your computer (www.R-project.org) and do some simple simulation experiments. Learn! Only after learning, can you teach. Wikipedia editors have to be good teachers, they need to be on top of their material or they'll only confuse their pupils. That would be like the blind leading the blind.

I already wrote some R scripts for playing with various scenarios. Let me know if anyone wants them. Richard Gill (talk) 11:57, 27 November 2011 (UTC)

Thank you, but my point is what the article says:

I showed above that it is fully right that you can expect the dependent envelope B on average to contain 5/4 of the determining amount in envelope A.

But regard this draconic restriction, that the TEP-article hides and reprobately keeps secret: This is only true if you know for sure that the "determining" envelope A, independently from the empty envelope B, has been filled with any random amount first, and only afterwards it was decided to fill the "dependent" envelope B with the dependent double amount of "determining amount in A", or with equal probability with dependent half that "determining amount of A".

(added): This will apply only in one HALF of cases, never in the second half. And to get efficient for any analysis you definitely have to KNOW FOR SURE that it applies in the special case, otherwise you may never use it in your theorem. If it is unknown it cannot be considered in any way.

As this restriction is nowhere mentioned in the TEP article, and you never can nor will know which envelope is the substantially "determining" one and which is the "dependent" one in the TEP,  A or B, this theorem leading to 5/4 A may never be laconically considered for the TEP. If you have chosen an envelope with "dependent amount", you forever have to expect to get only 4/5 A by swapping to B (twice if A is the dependent half amount of envelope B, half if A is the dependent double amount of envelope B). And as I said above - in lack of knowledge - you only can expect B to contain ((2*1+1/2)+(2/2)+(0,5*2))/4,5 = 1A.

5/4 A,  4/5 A  and  1A.  All of this is millionfold verified. If you put the cards on the table, then there is no paradox at all. No more need for infinity. The article should be accessible for grandma and grandson also. With open cards on the table, as per the sources (Ruma Falk, e.g.). Regards, Gerhardvalentin (talk) 13:22, 27 November 2011 (UTC)

What the article says

Gerhard, read the article! It says

There are two indistinguishable envelopes,

each of which contains a positive sum of money,

one envelope contains twice as much as the other.

You pick one envelope at random.

You denote by A the amount in your selected envelope.

In short: Envelope A is a randomly chosen envelope from an indistinguishable pair, Envelope B is the other.

The introduction to the article, where standard TEP is described (the problem which most of the sources, and in particular most of the popular sources, write about), doesn't say anything about how the envelopes are filled. It doesn't have to. You don't know how.

That's right, Richard. Not knowing how, and not having to know it. But alone your experience teaches you that there must have been a certain history, and "if" the window glass has been crushed from outside, makes other conditions for the insurance company as if it had been crushed from within the flat. There are quite other conditions for both variants.

And as  "5/4 A"  will forever be valid in only one special HALF of cases, and never in the second half, and only if you already know for sure, just from the outset, that your envelope "A" is the determining envelope resp. the determining amount, but never never ever otherwise, you have to admit that - not knowing this significant condition indeed to be given for sure - makes  "5/4 A FORFEVER and EVER !!! ? ! ? !"  a really ridiculous farce, makes "5/4 A"  an unforgivable error and a gross fallacy, please ask the insurance company to get their expertise. Gerhardvalentin (talk) 17:52, 29 November 2011 (UTC)

You don't know the two amounts, you don't even get to look in your own envelope. That's the whole point. Yes you convince yourself to switch. And then you could convince yourself to switch back. It's a paradox because the conclusion is obviously wrong. The problem is to show where the apparently logical reasoning goes astray. There are alternative solutions depending on what you imagine the writer was trying to do and what he was assuming. We don't know either. Over the historic development of the paradox, and as it spread to different cultures (from mathematical recreations to economists, statisticians, philosophers), people tended to have different ideas what the writer was up to.

That is why the article is hard to write: because there is not one solution, but many, and they are different. The problem is how to organise the different solutions. My own original research is an attempt to find synthesis. But wikipedia editors are not allowed to make new syntheses so I am not allowed to edit the article. But I offer my findings for discussion and I am likely getting them published in a good peer-reviewed journal soon. The results are still getting better and better, I think, thanks to discussions here and elsewhere.

Sure, some sources are confused, and very many sources are bloody confusing. One source of confusion seems to be with Nalebuff (1988, 1989) who introduced a new version of TEP in which we know that envelope A is filled with x>0, amount unknown, by the host, then the host tosses a coin to put x/2 or 2x in envelope B. The player is given Envelope A and asked if he would like to switch. Nalebuff then gives a correct reasoning with the host and owner, Ali, of Envelope A would like to switch, followed by the paradoxical (wrong) TEP-like reasoning why the owner, Baba, of Envelope B would also like to switch.

At the end of his problem description he also mentions the original or standard version - the version described in the wikipedia article, and from which he invented his new problem.

At the time Nalebuff wrote his article the story was being discussed by statisticians and mathematical economists (he heard it from colleagues). Gardner also heard about it at the same time, and put it into his 1989 book "Penrose tiles to trapdoor ciphers", p.147-148. Gardner had previously (1982) discussed the Kraitchik 1942, 1953 two envelopes version, converting it to two envelopes, in his 1982 book "Aha, Gotcha". Gardner (1989) discusses our basic, "proper" TEP, and most of the people who wrote in to Nalebuff, whose solutions Nalebuff discusses in his second paper, also discussed our basic, proper TEP. The TEP described in Wikipedia, the version described by Falk, who even quotes from Wikipedia, and the version most discussed in the literature. Please let's focus on the basic TEP. Not on a variant.

Of course the difference with the variant is interesting. Nalebuff's new Ali and Baba TEP problem is different. Your grandmother and grandchild will understand that Ali should want to switch and Baba should want not to switch. In the original TEP problem they understand that there is no point in anyone switching. But the problem is not to realise what the right answer is, it is to understand where the reasoning which leads to an absurd answer goes wrong. Richard Gill (talk) 07:32, 29 November 2011 (UTC)

Thank you Richard for your incredible patience. Yes, the real "shame" is that "the given actual content of envelope A" and "All possible contents of envelope A" are confusingly mixed together. Let me repeat again:

Only items 1–3 of the 12 claims of the article (1-12) do apply in every case.

But in one full halve of cases, forever when "A" inevitably is to be the dependent amount, all steps 4–12 are completely inconsistent with the given facts, no matter whether you know the actual content of envelope A, or not. Yes, everything seems to be consistent, but this consistency does not exist: step 4 does never apply in 1/4 of cases, and step 5 does never apply in the remaining 1/4 of cases.

Once more: Step 4 is completely incorrect in 1/4 of cases, whether you know the given actual content of envelope A, or not. And step 5 is incorrect in another 1/4 of cases, whether you know the given actual content of envelope A, or not. So steps 4–12 are false allegations like "The Emperor's new clothes", if you just have a look there. Why is it so hard to name those false allegations, that might apply in exactly 1/2 of cases only, but never in the remaining 50 % of cases, where either item 4 is completely incorrect in 1/4 of cases, or item 5 is completely incorrect in the other 1/4 of cases.

So you are required to add to step 4: "Incorrect in 1/4 of cases", and to add to step 5: "Incorrect in another 1/4 of cases". And, before step 6, to add "As a consequence, the rest is nonsense in 1/2 of cases".

Just have a look to the four equal likely scenarios. It should be possible to object items 4 and 5, and as consequence the rest of 6 to 12. Regards, Gerhardvalentin (talk) 14:39, 29 November 2011 (UTC)

You say "four equally likely cases". You are still assuming things we don't know. You are assuming first the host picks an amount, then he halves or double it, then randomly one of the two envelopes becomes envelope A. So you are doing probability with just two independent coin tosses in the whole story. But other people do the probability differently. Very many people suppose the host first picks a number x, then puts x and 2x in two envelopes. One of the envelopes is picked at random and called Envelope A. Some people "do the probability" only using the final random pick of an envelope. Many other people also use probability to describe their uncertainty as to what x might be.

I'm not saying your analysis is wrong. I'm just saying it's relative to particular assumptions about what is fixed and what is random, and what the randomness is. Your choice is not common on the literature! From a mathematical point of view, my unified solution is a good solution because it applies to every single interpretation I know. Unfortunately however it is mathematical so your grandmother or little grandson won't understand it. This is the problem with writing the article. It must start with easy solutions which are however restricted to a particular interpretation. And the particular interpretation must be made absolutely clear each time. Unfortunately many of the sources are not very explicit about which interpretation they are taking, so the editors of wikipedia have to do some "reconstruction".

The problem is, in fact, that there hardly exists an authoritative source which discusses all interpretations. Possibly the only one so far is the Nickerson and Falk paper. And my up-coming paper. Richard Gill (talk) 08:16, 30 November 2011 (UTC)

Yes Richard, you are correct again, thank you – I steadily just am trying ... and I didn't "see" those aspects before. But what you said shows clearly the direction, where to search. Thank you once more. Kind regards, Gerhardvalentin (talk) 12:40, 30 November 2011 (UTC)