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?

}}

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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===The puzzle===

The puzzle is to find the flaw in the line of reasoning in the switching argument. This includes determining exactly ''why'' and under ''what conditions'' that step is not correct, to be sure not to make this mistake in a situation where the misstep may not be so obvious. In short, the problem is to solve the paradox. The puzzle is ''not'' solved by finding another way to calculate the probabilities that does not lead to a contradiction.

== 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/>

The puzzle is also mentioned in a 1953 book on elementary mathematics and mathematical puzzles by the mathematician [[John Edensor Littlewood]], who credited it to the physicist [[Erwin Schrödinger]], where it concerns a pack of cards, each card has two numbers written on it, the player gets to see a random side of a random card, and the question is whether one should turn over the card. Littlewood's pack of cards is infinitely large and his paradox is a paradox of improper prior distributions.

[[Martin Gardner]] popularized Kraitchik's puzzle in his 1982 book ''Aha! Gotcha'', in the form of a wallet game:

{{blockquote|Two people, equally rich, meet to compare the contents of their wallets. Each is ignorant of the contents of the two wallets. The game is as follows: whoever has the least money receives the contents of the wallet of the other (in the case where the amounts are equal, nothing happens). One of the two men can reason: "I have the amount ''A'' in my wallet. That's the maximum that I could lose. If I win (probability 0.5), the amount that I'll have in my possession at the end of the game will be more than 2''A''. Therefore the game is favourable to me." The other man can reason in exactly the same way. In fact, by symmetry, the game is fair. Where is the mistake in the reasoning of each man?

| author = [[Martin Gardner]]

| source = ''Aha! Gotcha''

}}

Gardner confessed that though, like Kraitchik, he could give a sound analysis leading to the right answer (there is no point in switching), he could not clearly put his finger on what was wrong with the reasoning for switching, and Kraitchik did not give any help in this direction, either.

In 1988 and 1989, [[Barry Nalebuff]] presented two different two-envelope problems, each with one envelope containing twice what is in the other, and each with computation of the expectation value 5''A''/4. The first paper just presents the two problems. The second discusses many solutions to both of them. The second of his two problems is nowadays the more common, and is presented in this article. According to this version, the two envelopes are filled first, then one is chosen at random and called Envelope A. [[Martin Gardner]] independently mentioned this same version in his 1989 book ''Penrose Tiles to Trapdoor Ciphers and the Return of Dr Matrix''. Barry Nalebuff's asymmetric variant, often known as the Ali Baba problem, has one envelope 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.

Broome in 1995 called a probability distribution 'paradoxical' if for any given first-envelope amount ''x'', the expectation of the other envelope conditional on ''x'' is greater than ''x''. The literature contains dozens of commentaries on the problem, much of which observes that a distribution of finite values can have an infinite expected value.<ref>{{cite journal |last1=Syverson |first1=Paul |title=Opening Two Envelopes |journal=Acta Analytica |date=1 April 2010 |volume=25 |issue=4 |pages=479–498 |doi=10.1007/s12136-010-0096-7|s2cid=12344371 }}</ref>

==Multiplicity of proposed solutions==

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== Example resolution ==

Suppose that the total amount in both envelopes is a constant <math>c = 3x</math>, with <math>x</math> in one envelope and <math>2x</math> in the other. If you select the envelope with <math>x</math> first you gain the amount <math>x</math> by swapping. If you select the envelope with <math>2x</math> first you lose the amount <math>x</math> by swapping. So you gain on average <math>G = {1 \over 2} (x) + {1 \over 2} (-x) = {1 \over 2}(x - x) = 0</math> by swapping.

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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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This interpretation of the two envelopes problem appears in the first publications in which the paradox was introduced in its present-day form, Gardner (1989) and Nalebuff (1988).<ref>{{Cite journal|last1=Nalebuff|first1=Barry|date=Spring 1988|title=Puzzles: Cider in Your Ear, Continuing Dilemma, The Last Shall Be First, and More|journal = Journal of Economic Perspectives|volume = 2|issue=2|pages=149–156|doi = 10.1257/jep.2.2.149 |doi-access=free}} and Gardner, Martin (1989) '' Penrose Tiles to Trapdoor Ciphers: And the Return of Dr Matrix. ''</ref>)

It is common in the more mathematical literature on the problem. It also applies to the modification of the problem (which seems to have started with Nalebuff) in which the owner of envelope A does actually look in his envelope before deciding whether or not to switch; though Nalebuff does also emphasize that there is no need to have the owner of envelope A look in his envelope. If he imagines looking in it, and if for any amount which he can imagine being in there, he has an argument to switch, then he will decide to switch anyway. Finally, this interpretation was also the core of earlier versions of the two envelopes problem (Littlewood's, Schrödinger's, and Kraitchik's switching paradoxes); see [[Two envelopes problem#History of the paradox|the concludinghistory section, on history of TEP]].

This kind of interpretation is often called "Bayesian" because it assumes the writer is also incorporating a prior probability distribution of possible amounts of money in the two envelopes in the switching argument.

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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>

== History of the paradox ==

The envelope paradox dates back at least to 1953, when Belgian mathematician [[Maurice Kraitchik]] proposed a puzzle in his book ''Recreational Mathematics'' concerning two equally rich men who meet and compare their beautiful neckties, presents from their wives, wondering which tie actually cost more money. He also introduces 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. The men do not look in their purses but each reason that they should switch. He does not explain what is the error in their reasoning. It is not clear whether the puzzle already appeared in an earlier 1942 edition of his book. It is also mentioned in a 1953 book on elementary mathematics and mathematical puzzles by the mathematician [[John Edensor Littlewood]], who credited it to the physicist [[Erwin Schrödinger]], where it concerns a pack of cards, each card has two numbers written on it, the player gets to see a random side of a random card, and the question is whether one should turn over the card. Littlewood's pack of cards is infinitely large and his paradox is a paradox of improper prior distributions.

[[Martin Gardner]] popularized Kraitchik's puzzle in his 1982 book ''Aha! Gotcha'', in the form of a wallet game:

{{blockquote|Two people, equally rich, meet to compare the contents of their wallets. Each is ignorant of the contents of the two wallets. The game is as follows: whoever has the least money receives the contents of the wallet of the other (in the case where the amounts are equal, nothing happens). One of the two men can reason: "I have the amount ''A'' in my wallet. That's the maximum that I could lose. If I win (probability 0.5), the amount that I'll have in my possession at the end of the game will be more than 2''A''. Therefore the game is favourable to me." The other man can reason in exactly the same way. In fact, by symmetry, the game is fair. Where is the mistake in the reasoning of each man?

| author = [[Martin Gardner]]

| source = ''Aha! Gotcha''

}}

Gardner confessed that though, like Kraitchik, he could give a sound analysis leading to the right answer (there is no point in switching), he could not clearly put his finger on what was wrong with the reasoning for switching, and Kraitchik did not give any help in this direction, either.

In 1988 and 1989, [[Barry Nalebuff]] presented two different two-envelope problems, each with one envelope containing twice what is in the other, and each with computation of the expectation value 5''A''/4. The first paper just presents the two problems. The second discusses many solutions to both of them. The second of his two problems is nowadays the more common, and is presented in this article. According to this version, the two envelopes are filled first, then one is chosen at random and called Envelope A. [[Martin Gardner]] independently mentioned this same version in his 1989 book ''Penrose Tiles to Trapdoor Ciphers and the Return of Dr Matrix''. Barry Nalebuff's asymmetric variant, often known as the Ali Baba problem, has one envelope 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.

Broome in 1995 called a probability distribution 'paradoxical' if for any given first-envelope amount ''x'', the expectation of the other envelope conditional on ''x'' is greater than ''x''. The literature contains dozens of commentaries on the problem, much of which observes that a distribution of finite values can have an infinite expected value.<ref>{{cite journal |last1=Syverson |first1=Paul |title=Opening Two Envelopes |journal=Acta Analytica |date=1 April 2010 |volume=25 |issue=4 |pages=479–498 |doi=10.1007/s12136-010-0096-7|s2cid=12344371 }}</ref>

== See also ==

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