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{{Short description|Puzzle in logic and mathematics}}
[[File:Two envelopes.svg|thumb|The problem concerns two envelopes, each containing an unknown amount of money.]]
The '''two envelopes problem''', also known as the '''exchange paradox''', is a [[paradox]] in [[probability theory]]. It is of special interest in [[decision theory]]
The problem is typically introduced by formulating a [[Hypothesis|hypothetical]] challenge like the following example:
{{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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Now suppose the person reasons as follows:
{{ordered list
| type = decimal|Denote by ''A'' the amount in the player's selected envelope.|The probability that ''A'' is the smaller amount is 1/2, and that it is the larger amount is also 1/2.|The other envelope may contain either 2''A'' or ''A''/2.|If ''A'' is the smaller amount, then the other envelope contains 2''A''.|If ''A'' is the larger amount, then the other envelope contains ''A''/2.|Thus the other envelope contains 2''A'' with probability 1/2 and ''A''/2 with probability 1/2.|So the [[expected value]] of the money in the other envelope is
<math display="block">{1 \over 2} (2A) + {1 \over 2} \left({A \over 2}\right) = {5 \over 4}A</math>|This is greater than ''A'' so, on average, the person reasons that they stand to gain by swapping.|After the switch, denote that content by ''B'' and reason in exactly the same manner as above.|The person concludes that the most rational thing to do is to swap back again.|The person will thus end up swapping envelopes indefinitely.|As it is more rational to just open an envelope than to swap indefinitely, the player arrives at a contradiction.
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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.
== 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:▼
{{
| 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
==Multiplicity of proposed solutions==
There have been many solutions proposed, and commonly one writer proposes a solution to the problem as stated, after which another writer shows that altering the problem slightly revives the paradox. Such sequences of discussions have produced a family of closely related formulations of the problem, resulting in voluminous literature on the subject.<ref>A complete list of published and unpublished sources in chronological order can be found in the [[Talk:Two envelopes problem/Literature|talk page]].</ref>
No proposed solution is widely accepted as definitive
== 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
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).
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
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
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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The simple resolution above assumed that the person who invented the argument for switching was trying to calculate the expectation value of the amount in Envelope A, thinking of the two amounts in the envelopes as fixed (''x'' and 2''x''). The only uncertainty is which envelope has the smaller amount ''x''. However, many mathematicians and statisticians interpret the argument as an attempt to calculate the expected amount in Envelope B, given a real or hypothetical amount "A" in Envelope A. One does not need to look in the envelope to see how much is in there, in order to do the calculation. If the result of the calculation is an advice to switch envelopes, whatever amount might be in there, then it would appear that one should switch anyway, without looking. In this case, at Steps 6, 7 and 8 of the reasoning, "A" is any fixed possible value of the amount of money in the first envelope.
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)
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
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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Suppose for the sake of argument, we start by imagining an amount of 32 in Envelope A. In order that the reasoning in steps 6 and 7 is correct ''whatever'' amount happened to be in Envelope A, we apparently believe in advance that all the following ten amounts are all equally likely to be the smaller of the two amounts in the two envelopes: 1, 2, 4, 8, 16, 32, 64, 128, 256, 512 (equally likely powers of 2<ref name=":1" />). But going to even larger or even smaller amounts, the "equally likely" assumption starts to appear a bit unreasonable. Suppose we stop, just with these ten equally likely possibilities for the smaller amount in the two envelopes. In that case, the reasoning in steps 6 and 7 was entirely correct if envelope A happened to contain any of the amounts 2, 4, ... 512: switching envelopes would give an expected (average) gain of 25%. If envelope A happened to contain the amount 1, then the expected gain is actually 100%. But if it happened to contain the amount 1024, a massive loss of 50% (of a rather large amount) would have been incurred. That only happens once in twenty times, but it is exactly enough to balance the expected gains in the other 19 out of 20 times.
Alternatively, we do go on ad infinitum but now we are working with a quite ludicrous assumption, implying for instance, that it is infinitely more likely for the amount in envelope A to be smaller than 1, ''and'' infinitely more likely to be larger than 1024, than between those two values. This is a so-called [[
Many authors have also pointed out that if a maximum sum that can be put in the envelope with the smaller amount exists, then it is very easy to see that Step 6 breaks down, since if the player holds more than the maximum sum that can be put into the "smaller" envelope they must hold the envelope containing the larger sum, and are thus certain to lose by switching. This may not occur often, but when it does, the heavy loss the player incurs means that, on average, there is no advantage in switching. Some writers consider that this resolves all practical cases of the problem.<ref name=":2">{{Citation | first = Barry | last = Nalebuff |title = Puzzles: The Other Person's Envelope is Always Greener| journal = Journal of Economic Perspectives | volume = 3 | issue = 1 | pages = 171–181 | doi=10.1257/jep.3.1.171| year = 1989 | doi-access = free }}.</ref>
But the problem can also be resolved mathematically without assuming a maximum amount. Nalebuff,<ref name=":2" /> Christensen and Utts,<ref name=":3" /> Falk and Konold,<ref name=":1" /> Blachman, Christensen and Utts,<ref>{{cite journal |first1=NM |last1=Blachman |first2=R |last2=Christensen |first3=J |last3= Utts |year= 1996 | journal =The American Statistician |volume=50 |issue=1 |pages= 98–99 | doi = 10.1080/00031305.1996.10473551 |title=Letters to the Editor}}</ref> Nickerson and Falk,<ref name=":0" /> pointed out that if the amounts of money in the two envelopes have any proper probability distribution representing the player's prior beliefs about the amounts of money in the two envelopes, then it is impossible that whatever the amount ''A=a'' in the first envelope might be, it would be equally likely, according to these prior beliefs, that the second contains ''a''/2 or 2''a''. Thus step 6 of the argument, which leads to ''always switching'', is a non-sequitur, also when there is no maximum to the amounts in the envelopes.
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Under dominance reasoning, the fact that we strictly prefer ''A'' to ''B'' for all possible observed values ''a'' should imply that we strictly prefer ''A'' to ''B'' without observing ''a''; however, as already shown, that is not true because <math>E(B)=E(A)=\infty</math>. To salvage dominance reasoning while allowing <math>E(B)=E(A)=\infty</math>, one would have to replace expected value as the decision criterion, thereby employing a more sophisticated argument from mathematical economics.
For example, we could assume the decision maker is an [[expected utility]] maximizer with initial wealth ''W'' whose utility function, <math>u(w)</math>, is chosen to satisfy <math>E(u(W+B)|A=a)<u(W+a)</math> for at least some values of ''a'' (that is, holding onto <math>A=a</math> is strictly preferred to switching to ''B'' for some ''a''). Although this is not true for all utility functions, it would be true if <math>u(w)</math> had an upper bound, <math>\beta<\infty</math>, as ''w'' increased toward infinity (a common assumption in mathematical economics and decision theory).<ref>{{cite book|last1=DeGroot|first1=Morris H.|title=Optimal Statistical Decisions|date=1970|publisher=McGraw-Hill|pages=109}}</ref> [[Michael R. Powers]] provides necessary and sufficient conditions for the utility function to resolve the paradox, and notes that neither <math>u(w)<\beta</math> nor <math>E(u(W+A))=E(u(W+B))<\infty</math> is required.<ref>{{cite journal|last1=Powers|first1=Michael R.|title=Paradox-Proof Utility Functions for Heavy-Tailed Payoffs: Two Instructive Two-Envelope Problems|journal=Risks|date=2015|volume=3|issue=1|pages=26–34|doi=10.3390/risks3010026|
Some writers would prefer to argue that in a real-life situation, <math>u(W+A)</math> and <math>u(W+B)</math> are bounded simply because the amount of money in an envelope is bounded by the total amount of money in the world (''M''), implying <math>u(W+A) \leq u(W+M)</math> and <math>u(W+B) \leq u(W+M)</math>. From this perspective, the second paradox is resolved because the postulated probability distribution for ''X'' (with <math>E(X)=\infty</math>) cannot arise in a real-life situation. Similar arguments are often used to resolve the [[St. Petersburg paradox]].
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Byeong-Uk Yi, on the other hand, argues that comparing the amount you would gain if you would gain by switching with the amount you would lose if you would lose by switching is a meaningless exercise from the outset.<ref>{{cite journal |author=Byeong-Uk Yi |year=2009 |title=The Two-envelope Paradox With No Probability |url=http://philosophy.utoronto.ca/people/linked-documents-people/c%20two%20envelope%20with%20no%20probability.pdf |url-status=dead |archive-url=https://web.archive.org/web/20110929034017/http://philosophy.utoronto.ca/people/linked-documents-people/c%20two%20envelope%20with%20no%20probability.pdf |archive-date=2011-09-29 }}</ref> According to his analysis, all three implications (switch, indifferent, do not switch) are incorrect. He analyses Smullyan's arguments in detail, showing that intermediate steps are being taken, and pinpointing exactly where an incorrect inference is made according to his formalization of counterfactual inference. An important difference with Chase's analysis is that he does not take account of the part of the story where we are told that the envelope called envelope A is decided completely at random. Thus, Chase puts probability back into the problem description in order to conclude that arguments 1 and 3 are incorrect, argument 2 is correct, while Yi keeps "two envelope problem without probability" completely free of probability and comes to the conclusion that there are no reasons to prefer any action. This corresponds to the view of Albers et al., that without a probability ingredient, there is no way to argue that one action is better than another, anyway.
Bliss argues that the source of the paradox is that when one mistakenly believes in the possibility of a larger payoff that does not, in actuality, exist, one is mistaken by a larger margin than when one believes in the possibility of a smaller payoff that does not actually exist.<ref>{{cite
Albers, Kooi, and Schaafsma consider that without adding probability (or other) ingredients to the problem,<ref name=":4" /> Smullyan's arguments do not give any reason to swap or not to swap, in any case. Thus, there is no paradox. This dismissive attitude is common among writers from probability and economics: Smullyan's paradox arises precisely because he takes no account whatever of probability or utility.
==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
▲== History of the paradox ==
▲[[Martin Gardner]] popularized Kraitchik's puzzle in his 1982 book ''Aha! Gotcha'', in the form of a wallet game:
▲{{quote|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 the 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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