Two envelopes problem

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The two envelopes problem is a puzzle or paradox within the subjectivistic interpretation of probability theory; more specifically within Bayesian decision theory. This is still an open problem among the subjectivists as no consensus has been reached yet.

The puzzle: Let's say you are given two indistinguishable envelopes, each of which contains a positive sum of money. One envelope contains twice as much as the other. You may pick one envelope and keep whatever amount it contains. You pick one envelope at random but before you open it you're offered the possibility to take the other envelope instead.

Now, suppose you reason as follows:

  1. I denote by A the amount in my selected envelope
  2. The probability that A is the smaller amount is 1/2, and that it's the larger also 1/2
  3. The other envelope may contain either 2A or A/2
  4. If A is the smaller amount the other envelope contains 2A
  5. If A is the larger amount the other envelope contains A/2
  6. Thus, the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2
  7. So the expected value of the money in the other envelope is

    {1 \over 2} 2A + {1 \over 2} {A \over 2} = {5 \over 4}A

  8. This is greater than A, so I gain on average by swapping
  9. After the switch I can denote that content B and reason in exactly the same manner as above
  10. I will conclude that the most rational thing to do is to swap back again
  11. To be rational I will thus end up swapping envelopes indefinitely
  12. As it seems more rational to open just any envelope than to swap indefinitely we have a contradiction

The puzzle is to find the flaw in the very compelling line of reasoning above. (That includes to be able to say exactly why that step is not correct, and under what conditions it's not correct, so we can be absolutely sure we don't make this mistake in a more complicated situation where the fact that it's wrong isn't this obvious.)

Comment: Because the subjectivistic interpretation of probability is closer to the layman's conception of probability this paradox is understood by almost everybody. However, for a working statistician or probability theorist endorsing the more technical frequency interpretation of probability this puzzle isn't a problem, as the puzzle can't even be stated when imposing those more technical restrictions. Consequently, all published papers below with different ideas on a solution are written from the subjectivistic or Bayesian point of view.

Contents

[edit] Proposed solution

The most common way to explain the paradox is to observe that A isn't a constant in the expected value calculation, step 7 above. In the first term A is the smaller amount while in the second term A is the larger amount. To mix different instances of a variable or parameter in the same formula like this shouldn't be legitimate, so step 7 is thus the proposed cause of the paradox.[23]

[edit] A harder problem

The solution above doesn't explain what's wrong if you are allowed to open the first envelope before you're offered to switch. In this case A in the expected value calculation above is indeed a constant. Hence, the proposed solution in the first case breaks down and we have to find another explanation here.

However, you will not end up switching envelopes indefinitely in this case as you know both contents after the first switch. Imagine instead that your twin brother/sister opens the other envelope without telling you the amount it contains. Now both of you will find that it's better to switch due to the same argument. This is contradictory as you and your twin can't both win when switching envelopes with each other.

[edit] Proposed solution

Once we have looked in the envelope, we have new information—namely the value A. The subjective probability changes when we get new information, so our assessment of the probability that A is the smaller and larger sum changes. Therefore step 2 above isn't always true and is thus the proposed cause of this paradox.

Step 2 can be justified, however, if we can find a prior distribution such that every pair of possible amounts {X, 2X} are equally likely, where X = 2nA, n = 0, ±1, ±2,.... But as this set contains infinitely many elements we can't make a uniform probability distribution over all values in this set. So some values of A must be more likely than others. However, we don't know which values are more likely than others, that is, if we don't happen to know the prior distribution.

[edit] An even harder problem

The solution above doesn't rule out the possibility that there is some prior distribution of sums in the envelopes (but not a uniform distribution, since that's impossible) that makes the paradox work.

Suppose that the envelopes contain the integer sums {2n, 2n+1} with probability 2n/3n+1 where n = 0, 1, 2,.... [6]

Of course there's a sensible strategy which guarantees a win, which is to swap only when the envelope we open contains 1, as we know that the other must contain 2. But we can apparently do better than that, for suppose we open one envelope and find 2. Now we know that there are only two possibilities; the envelope pair in front of us is either {1, 2} or {2, 4}. The conditional probability that it's the {1, 2} pair is

P(\{1,2\} \mid 2)= \frac{P(\{1,2\})}{P(\{1, 2\}) + P(\{2, 4\})} = \frac{1/3}{1/3 + 2/9} = 3/5,

and consequently the probability it's the {2, 4} pair is 2/5 since all other envelope pairs have zero conditional probability.

It turns out that these proportions holds in general unless you find 1 in the first envelope you open. Thus, if we denote by x the amount we find where x = 2n for some n ≥ 1, then the other envelope contains x/2 with probability 3/5 and 2x with probability 2/5. So the expected gain if we switch is

{3 \over 5} {x \over 2} + {2 \over 5} 2x = {11 \over 10}x

which is more than x. This means that in all cases you expect to gain if you switch.

But once again, you can go through this reasoning before you open either envelope, and deduce that you should always choose the other envelope. This conclusion is just as clearly wrong as it was in the first and second cases. But now the flaws noted above doesn't apply; the x in the expected value calculation is a known constant (in every single case) and we get the probabilities in the formula from a specified prior distribution.

[edit] Proposed solution

It can be shown that the distribution producing this variant of the paradox must have an infinite mean, so before you open any envelope the expected gain from switching is "∞ − ∞" which is not defined. In the words of Chalmers this is “just another example of a familiar phenomenon, the strange behaviour of infinity.” [17] This resolves this paradox according to some authors.

But in every actual single instant when you open an envelope the conclusion is justified: you should switch! Not many authors have addressed this case explicitly trying to give a solution. Chalmers, for example, suggests that decision theory generally breaks down when confronted with games having a diverging expectation, and compares it with the situation generated by the classical St. Petersburg paradox.

Suppose that one has a ticket to the St. Petersburg lottery as stated in that problem. Should you be willing to trade it for another ticket to the lottery? Since the St. Petersburg lottery has an infinite expected value, once your lottery ticket value is determined, no matter what (finite) value it is worth, you should be willing to trade it for another ticket to a new, not yet drawn, lottery. However, even before your lottery ticket is drawn, you could reason as follows: "My ticket has some finite value. No matter what it is, given my ticket's finite value I should switch to another ticket, which has an infinite expected value. Therefore I should keep switching tickets indefinitely." This is clearly absurd, and parallel to the envelope problem; conditioned on any particular finite value of a random variable with infinite expectation, we should switch to another random variable with infinite expectation. The extra wrinkle in the envelope problem is that the second envelope's expectation appears to depend on the first. However, it is not clear that we should truly prefer one infinite expected value to another.

However, Clark and Shackel argues that this blaming it all on "the strange behaviour of infinity" doesn't resolve the paradox at all. Neither in the single case nor the averaged case. They show this by providing a simple example of a pair of random variables both having infinite mean but where one is always better to chose than the other. [11] This is the best thing to do at every instant as well as on average, which shows that decision theory doesn't necessarily break down when confronted with infinite expectations.

[edit] The hardest problem

The logician Raymond Smullyan questioned if the paradox has anything to do with probabilities at all. He did this by expressing the problem in a way which doesn't involve probabilities. The following plainly logical arguments lead to conflicting conclusions:

  1. Let the amount in the envelope you chose be A. Then by swapping, if you gain you gain A but if you lose you lose A/2. So the amount you might gain is strictly greater than the amount you might lose.
  2. Let the amounts in the envelopes be Y and 2Y. Now by swapping, if you gain you gain Y but if you lose you also lose Y. So the amount you might gain is equal to the amount you might lose.

[edit] Proposed solution

So far, the only proposed solution of this variant is due to James Chase.[18] His idea is that this problem is an example of a fallacy in the logic of counterfactuals. He claims that while the statement "If you gain you gain A but if you lose you lose A/2" is unproblematic and true (in virtue of the truth of one of its disjuncts) that is no longer the case when we split this into two separate sentences according to the standard analysis of counterfactuals:

  • the nearest possible world in which you gain on the trade is a world where you gain A
  • the nearest possible world in which you lose on the trade is a world where you lose A/2

In each of the relevant ways the actual world can be, only one of the above sentences can be true.

For suppose that the envelope you have chosen actually contains the greater sum, then the second sentence is true since the nearest possible world where you lose money is the actual world. But what is the nearest possible world where you gain on a switch? Is it a world where, contrary to fact, the envelopes were filled with {A, 2A} instead of {A/2, A}, or is it a world where, contrary to fact, you simply chose the other envelope? Chase claims that it must be the latter, because the former involves change earlier in the causal sequence of events. So the first sentence is in this case false both in the actual world as well as in the nearest counterfactual world. By symmetry only the first sentence is true if we suppose that we chose the smaller amount first. This shows that the unproblematic sentence above can't be split into two sentences that are true either factual or counterfactual. This is the fallacy.

[edit] History of the paradox

The envelope paradox dates back at least to 1953, when Belgian mathematician Maurice Kraitchik proposed this puzzle:

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: "Suppose that 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 2A. 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?

Martin Gardner popularized the puzzle in his 1982 book Aha! Gotcha, also in the form of a wallet game. In 1989, Barry Nalebuff presented it in the two-envelope form, and that's the form in which the paradox has been most commonly presented since.

[edit] See also

[edit] External links

[edit] Bibliography

  • Casper Albers, Trying to resolve the two-envelope problem PDF, Chapter 2 of his thesis Distributional Inference: The Limits of Reason, December 2002
  • Randall Barron, The paradox of the money pump: a resolution, in Maximum Entropy and Bayesian Methods, 1988 ed J Skilling
  • Martin Gardner, Aha! Gotcha, 1982
  • Maurice Kraitchik, La mathématique des jeux, 1953
  • Raymond Smullyan, Satan, Cantor and Infinity, November 1992

[edit] Published papers

  1. Barry Nalebuff, Puzzles: the other person's envelope is always greener, Journal of Economic Perspectives 3, 1989
  2. R Christensen and J Utts, Bayesian Resolution of the 'Exchange Paradox', The American Statistician 1992
  3. Jackson, Menzies and Oppy, The Two Envelope 'Paradox', in Analysis, January 1994
  4. Castell and Batens, The Two Envelope Paradox: The Infinite Case, in Analysis, January 1994
  5. Elliot Linzer, The Two Envelope Paradox, American Mathematical Monthly, Volume 101, Number 5, May 1994, p. 417
  6. John Broome, The Two-envelope Paradox, in Analysis, January 1995
  7. A D Scott and M Scott, What’s in the Two Envelope Paradox?, in Analysis, January 1997
  8. McGrew, Shier and Silverstein, The Two-Envelope Paradox Resolved, in Analysis, January 1997
  9. Arntzenius and McCarthy, The two envelope paradox and infinite expectations, in Analysis, January 1997
  10. John Norton, When the Sum of Our Expectations Fails Us: The Exchange Paradox PDF, 1998
  11. Clark and Shackel, The Two-Envelope Paradox, in Mind July 2000 Abstract
  12. Wilfried Hausmann, On The Two Envelope Paradox PDF, August 2000
  13. Terry Horgan, The Two-Envelope Paradox, Nonstandard Expected Utility, and the Intensionality of Probability, 2000
  14. Olav Gjelsvik, Can Two Envelopes Shake The Foundations of Decision Theory? PDF, September 2001
  15. Terry Horgan The Two-Envelope Paradox and the Foundations of Rational Decision Theory, 2001
  16. Jeff Speaks, The two-envelope paradox and inference from an unknown PDF, June 2002
  17. David J. Chalmers, The St. Petersburg Two-Envelope Paradox in Analysis, April 2002
  18. James Chase, The non-probabilistic two envelope paradox Analysis, April 2002
  19. Friedel Bolle, The Envelope Paradox, the Siegel Paradox, and the Impossibility of Random Walks in Equity and Financial Markets PDF, February 2003
  20. Priest and Restall, Envelopes and Indifference PDF, February 2003
  21. Wilton, The Two Envelopes Paradox PDF, June 2003
  22. Meacham and Weisberg, Clark and Shackel on the Two-Envelope Paradox PDF, October 2003
  23. Eric Schwitzgebel and Josh Dever, Using Variables Within the Expectation Formula PDF, February 2004 A Simple Version of Our Explanation
  24. Dov Samet, Iddo Samet, and David Schmeidler, One Observation behind Two-Envelope Puzzles PDF, April 2004
  25. Franz Dietrich and Christian List, The Two-Envelope Paradox: An Axiomatic Approach PDF, May 2004
  26. Bruce Langtry, The Classical and Maximin Versions of the Two-Envelope Paradox PDF, August 2004
  27. Jan Poland, The Two Envelopes Paradox in a Short Story PDF, 2005
  28. Rich Turner and Tom Quilter, The Two Envelopes Problem PDF, 2006
  29. Adom Giffin, The Error in the Two Envelope Paradox PDF, 2006
  30. Nickerson and Falk, The exchange paradox: Probabilistic and cognitive analysis of a psychological conundrum, in Thinking & Reasoning, May 2006
  31. Paul Syverson, Opening Two Envelopes (forthcoming)
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