Talk:Two envelopes problem/Archive 10

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What does the letter 'A' stand for?

In the article, sometimes 'A' is the amount in the selected envelope:"I denote by A the amount in my selected envelope.",
another time 'A' is the name of the selected envelope:"...given that envelope A contains less than envelope B."
If 'A' is a name, the expressions 2A or A/2 don't make sense. There is a considerable confusion concerning the usage of the letter 'A'. So both the sections "Problem" and "Simple resolutions" have to be revised. --TotalClearance (talk) 09:03, 9 April 2016 (UTC)

Yes, and the problem of A (envelope or envelope's amount) is that there is B also. Mathematicians use to disregard the obvious and easily perceived interdependency of the task presented, that is forced by the unchangeable total amount T of x + 2x (3x). No "average A" nor "any A" will ever be doubled – but only that very specific A that actually amounts to "x" (T/3), and in that specific case B consequently amounts to "2x", giving a difference of x.

On the other hand, no "average A" nor "any A" can ever be halved, but only that very specific A that actually amounts to 2x (2/3 T), difference between A and B again x.

So the confusing "expected value formula" (brazenly implying A=2A=A/2, hence ="zero" resp. =infinite, see Ruma Falk) evidently was just a smart joke to muck around with mathematicians and philosophers. Gerhardvalentin (talk) 09:42, 23 June 2016 (UTC)

This strikes be as the best resolution provided so far. The carrier of the money is confused with the contents of the carrier. All I can add is a starker example representation of the fallacy. Suppose the two choices are between a one dollar bill or a two dollar bill. With sighted people watching, a blind person is asked to give the argument presened in the actual article. For the process to work, sometimes there would have to be a conversion to a $0.50 bill or a $4.00 bill for the process to work. — Preceding unsigned comment added by PEBill (talk • contribs) 17:54, 10 July 2016 (UTC)

Yes, false switching argument. You are fully right in saying "if the two envelopes contain 1 and 2 (total=3), then the false argument says 0,5 and 2, resp. says 1 and 4".

The problem is that the "given switching argument" is tremendously defective and misleading for the described underlying symmetric task/variant of two indistinguishable, unknown envelopes with a fixed, unchangeable total amount T of 3x, where "envelope A" is no valid variable, but means x respectively 2x at the same time (of a total T of 3x).

The switching argument (expected value B=1,25A) only reflects the purposive asymmetric (!) one-way task/variant (Nalebuff: Ali vs. Baba), with some not yet decided total amount, where first of all only envelope A (that is "known" to be some predeterminated amount !) had already been fixed with any amount (only in that asymmetric task "A" is a valid variable) and is given to "Ali", and only thereafter the decision was made to equip the "known" to be the derived amount ! of envelope B equally likely with either double (2A, giving a total T of 3A) or half (A/2, giving a total T of 1.5*A). So solely in this latter task/variant with a not yet fixed total amount, on average envelope "B" will contain 1.25*A, and B is given to "Baba". Only in this purposive asymmetric one-way variant, where A is a variable, with B=1.25*A, you will gain A/4 (of "any A" !) and consequently get on average "5A/4" by switching from A to B, and you will lose B/5 and consequently get on average only "A" by switching from B to A. The diametrical difference of those two completely different tasks is thoroughly ignored by the present so-called "paradox". Most ignore that obvious and significant distinction.

So the trappy illusory switching argument does never apply to the presented symmetric basic setup of two indistinguishable, unknown envelopes with a fixed total amount. For the underlying symmetric variant of a fixed total (T=3x), the defective appraisal of the switching argument is fragmentary, incomplete, deficient and misleading.

As per Ruma Falk, the appraisal for the presented symmetric variant with a fixed total amount T=3x has to read instead:

3  The other envelope may contain either 2A (hence 2x) in case envelope A contains x, or A/2 (hence x) in case envelope A contains 2x. Total T in any case =3x.

4  If A is the smaller amount of x, then the other envelope contains 2A, hence 2x (difference =x)

5  If A is the larger amount of 2x, then the other envelope contains A/2, hence x (difference =x).

6  Thus both envelopes contain 2x with probability 1/2 and x with probability 1/2, hence both envelopes contain on average 3x/2 (T/2), so evidently no argument for switching.

For the presented basic setup of two unknown envelopes, the fragmented appraisal of the switching argument is incomplete and deficient, thus misleading.

In its fragmentary form, the switching argument addresses solely the asymmetric one-way variant of a not yet decided total amount (Nalebuff) of any envelope "A" (Ali) that is "known" to contain any already predeterminated amount, and any envelope "B" (Baba) that is "known" to contain the derived amount of 5A/4. Who will help to improve the awkward article. Sources en masse. Gerhardvalentin (talk) 14:06, 17 July 2016 (UTC)

How is this correct?

This problem is very similar to The Monty Hall problem. Only, there is only two envelopes. So you pick one. You have a 1/2 of getting the smaller amount of money. Then if you switch you would get the greater sum of money. However, there is also a 1/2 chance you got the greater sum of money. Then you would switch and get the smaller amount of money. The chance is equal!!!!!!!!!!!!!!!!!!!!!!!!! The article isn't bad, someone should just include this. Ghostana (talk) 21:06, 28 March 2022 (UTC)

Nvm. Ghostana (talk) 21:08, 28 March 2022 (UTC)

Nvm. I'm stupid. Ghostana (talk) 21:09, 28 March 2022 (UTC)

Nvm. I'm stupid. LOL Ghostana (talk) 21:09, 28 March 2022 (UTC)