Two envelopes problem: Difference between revisions

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{{Cquote|Imagine you are given two identical [[envelope]]s, each containing money. One contains twice as much as the other. You may pick one envelope and keep the money it contains. Having chosen an envelope at will, but before inspecting it, you are given the chance to switch envelopes. Should you switch?

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Since the situation is symmetric, it seems obvious that there is no point in switching envelopes. On the other hand, a simple calculation using expected values suggests the opposite conclusion, that it is always beneficial to swap envelopes, since the person stands to gain twice as much money if they switch, while the only risk is halving what they currently have.<ref name=":5" />

==Introduction==

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== History of the paradox ==

[[File:Greater Manchester Metrolink neckties.jpg|150px|thumb|Two [[necktie]]s.]]

The envelope paradox dates back at least to 1943, when Belgian mathematician [[Maurice Kraitchik]] proposed a puzzle in his book ''Recreational Mathematics'' concerning two men who meet and compare their fine neckties.<ref name=kraitchik>{{cite book |first=Maurice |last=Kraitchik |authorlink=Maurice Kraitchik |title="Mathematical Recreations" |publisher=George Allen & Unwin |___location=London |year=1943|url=https://archive.org/details/mathematicalrecr0000maur}}</ref><ref name=brown>{{Cite journal |last=Brown |first=Aaron C. |year=1995 |title=Neckties, Wallets, and Money for Nothing |journal=[[Journal of Recreational Mathematics]] |volume=27 |issue=2 |pages=116–122 }}</ref> Each of them knows what his own necktie is worth and agrees for the winner to give his necktie to the loser as consolation. Kraitchik also discusses a variant in which the two men compare the contents of their purses. He assumes that each purse is equally likely to contain 1 up to some large number ''x'' of pennies, the total number of pennies minted to date.<ref name=kraitchik/>

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== Other simple resolutions ==

A widely- discussed way to resolve the paradox, both in popular literature and part of the academic literature, especially in philosophy, is to assume that the 'A' in step 7 is intended to be the [[expected value]] in envelope A and that we intended to write down a formula for the expected value in envelope B.

Step 7 states that the expected value in B = 1/2(2A + A/2).

It is pointed out that the 'A' in the first part of the formula is the expected value, given that envelope A contains less than envelope B, but the 'A', in the second part of the formula is the expected value in A, given that envelope A contains more than envelope B. The flaw in the argument is that the same symbol is used with two different meanings in both parts of the same calculation but is assumed to have the same value in both cases. This line of argument is introduced by McGrew, Shier and Silverstein (1997).<ref>{{cite journal |last1=McGrew |first1=Timothy |last2=Shier |first2=David |last3=Silverstein |first3=Harry |title=The Two-Envelope Problem Resolved |journal=Analysis |date=1997 |volume=57 |issue=1 |pages=28–33 |doi=10.1093/analys/57.1.28 |url=https://academic.oup.com/analysis/article-abstract/57/1/28/139339|url-access=subscription }}</ref>

A correct calculation would be:

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which is equal to the expected sum in A.

In non-technical language, what goes wrong (see [[Necktie paradox]]) is that, in the scenario provided, the mathematics use relative values of A and B (that is, it assumes that one would gain more money if A is less than B than one would lose if the opposite were true). However, the two values of money are fixed (one envelope contains, say, $20 and the other $40). If the values of the envelopes are restated as ''x'' and 2''x'', it's much easier to see that, if A were greater, one would lose ''x'' by switching and, if B were greater, one would gain ''x'' by switching. One does not gain a greater amount of money by switching because the total ''T'' of A and B (3''x'') remains the same, and the difference ''x'' is fixed to ''T/3''.

Line 7 should have been worked out more carefully as follows:

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=== Nalebuff asymmetric variant ===

The mechanism by which the amounts of the two envelopes are determined is crucial for the decision of the player to switch her envelope.<ref name="Tsikogiannopoulos"/><ref>{{citation |last1=Priest |first1=Graham |last2=Restall |first2= Greg |year=2007 |title=Envelopes and Indifference |url= http://consequently.org/papers/envelopes.pdf |journal= Dialogues, Logics and Other Strange Things |publisher=College Publications |pages=135–140}}</ref> Suppose that the amounts in the two envelopes A and B were not determined by first fixing the contents of two envelopes E1 and E2, and then naming them A and B at random (for instance, by the toss of a fair coin<ref name=":0">{{Cite journal|last1=Nickerson|first1=Raymond S.|last2=Falk|first2=Ruma|date=2006-05-01|title=The exchange paradox: Probabilistic and cognitive analysis of a psychological conundrum|url=https://doi.org/10.1080/13576500500200049|journal=Thinking & Reasoning|volume=12|issue=2|pages=181–213|doi=10.1080/13576500500200049|s2cid=143472998|issn=1354-6783|url-access=subscription}}</ref>). Instead, we start right at the beginning by putting some amount in envelope A and then fill B in a way which depends both on chance (the toss of a coin) and on what we put in A. Suppose that first of all the amount ''a'' in envelope A is fixed in some way or other, and then the amount in Envelope B is fixed, dependent on what is already in A, according to the outcome of a fair coin. If the coin fell Heads then 2''a'' is put in Envelope B, if the coin fell Tails then ''a''/2 is put in Envelope B. If the player was aware of this mechanism, and knows that she holds Envelope A, but do not know the outcome of the coin toss, and do not know ''a'', then the switching argument is correct and she is recommended to switch envelopes. This version of the problem was introduced by Nalebuff (1988) and is often called the Ali-Baba problem. Notice that there is no need to look in envelope A in order to decide whether or not to switch.

Many more variants of the problem have been introduced. Nickerson and [[Ruma Falk|Falk]] systematically survey a total of 8.<ref name=":0" />

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==Conditional switching==

As an extension to the problem, consider the case where the player is allowed to look in envelope A before deciding whether to switch. In this "conditional switching" problem, it is often possible to generate a gain over the "never switching" strategy", depending on the probability distribution of the envelopes.<ref name="rspa">{{cite journal |last1=McDonnell |first1=M. D. |last2=Abott |first2=D. |title=Randomized switching in the two-envelope problem |journal=[[Proceedings of the Royal Society A]] |volume=465 |issue=2111 |pages=3309–3322 |year=2009 |doi=10.1098/rspa.2009.0312 |bibcode=2009RSPSA.465.3309M }}</ref>

== See also ==

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