Two envelopes problem

The Two Envelopes Problem is a puzzle within the subjectivistic interpretation of probability 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 ½, and that it's the larger also ½
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 ½ and A/2 with probability ½
7. So the expected value of the money in the other envelope is ½(2A) + ½(A/2) = 1¼A
8. This is greater than A, so I gain by swapping
9. As A is arbitrary I should swap whatever I see in the first envelope

But as the situation is symmetric this can't be correct.

The puzzle is to find the flaw (or flaws) in the very compelling line of reasoning above.

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.

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.

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 is indeed a constant, for example you might find $512 in the first envelope you pick. Hence, the solution above breaks down and we have to find another explanation in this case.

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 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.

An Even Harder Problem

The solution above doesn't rule out the possibility that there is some 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 non-negative integer sums 2ⁿ and 2ⁿ⁺¹ with probability q(1 − q)ⁿ for some fixed q < ½. (This variation is due to Marcus Moore.)

Now of course there's a sensible strategy which guarantees a win, which is to swap only when the envelope you open contains 1, when you know that the other must contain 2. But you can apparently do better than that, for suppose you open the envelope and find 2ⁿ for n ≥ 1. Then the other envelope contains

2ⁿ⁻¹ with probability q(1 − q)ⁿ⁻¹/R

2ⁿ⁺¹ with probability q(1 − q)ⁿ/R

where

R = q(1 − q)ⁿ⁻¹ + q(1 − q)ⁿ

So your expected gain if you switch is

½(2ⁿq(1 − q)ⁿ − 2ⁿ⁻¹q(1 − q)ⁿ⁻¹)/R

= 2ⁿ⁻²q(1 − 2q)/R

which is positive when q < ½, so 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 paradox, but now the flaw noted in solution 2 doesn't apply, as we have a specified distribution.

Proposed Solution

The distribution in the statement of this paradox has an infinite mean, so before you open any envelope the expected gain from switching is ∞ − ∞, which is not defined. So the a priori conclusion is not in fact justified.

However, once you open one envelope the conclusion is justified; you shall always 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 an diverging expectation, and compares with the situation generated by the classical St. Petersburg game.

The Hardest Problem

In fact, the hardest variant of the problem to solve is the easiest one to state because the problem can be expressed in a way which doesn't use probabilities at all. The following 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.

See Also

• Monty Hall problem
• St. Petersburg paradox
• Bertrand's paradox

External Links

• David J. Chalmers, The Two-Envelope Paradox: A Complete Analysis?
• Keith Devlin, The Two Envelopes Paradox, August 2004
• Emin Martinian, The Two Envelope Problem
• Amos Storkey, Money Trouble
• Brian Weatherson, Two-Envelopes and Variables, August 2004

Bibliography

• 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

Published Papers

• Barry Nalebuff, Puzzles: the other person's envelope is always greener, Journal of Economic Perspectives 3, 1989
• R Christensen and J Utts, Bayesian Resolution of the 'Exchange Paradox', The American Statistician 1992
• Jackson, Menzies and Oppy, The Two Envelope 'Paradox', in Analysis, January 1994
• Castell and Batens, The Two Envelope Paradox: The Infinite Case, in Analysis, January 1994
• John Broome, The Two-envelope Paradox, in Analysis, January 1995
• A D Scott and M Scott, What’s in the Two Envelope Paradox?, in Analysis, January 1997
• McGrew, Shier and Silverstein, The Two-Envelope Paradox Resolved, in Analysis, January 1997
• Arntzenius and McCarthy, The two envelope paradox and infinite expectations, in Analysis, January 1997
• John Norton, When the Sum of Our Expectations Fails Us: The Exchange Paradox PDF, 1998
• Clark and Shackel, The Two-Envelope Paradox, PDF in Mind July 2000 Abstract
• Wilfried Hausmann, On The Two Envelope Paradox PDF, August 2000
• Terry Horgan, The Two-Envelope Paradox, Nonstandard Expected Utility, and the Intensionality of Probability, 2000
• Olav Gjelsvik, Can Two Envelopes Shake The Foundations of Decision Theory? PDF, September 2001
• Terry Horgan The Two-Envelope Paradox and the Foundations of Rational Decision Theory, 2001
• Jeff Speaks, The two-envelope paradox and inference from an unknown PDF, June 2002
• David J. Chalmers, The St. Petersburg Two-Envelope Paradox in Analysis, April 2002
• James Chase, The non-probabilistic two envelope paradox Analysis, April 2002
• Friedel Bolle, The Envelope Paradox, the Siegel Paradox, and the Impossibility of Random Walks in Equity and Financial Markets PDF, February 2003
• Priest and Restall, Envelopes and Indifference PDF, February 2003
• Wilton, The Two Envelopes Paradox PDF, June 2003
• Meacham and Weisberg, Clark and Shackel on the Two-Envelope Paradox PDF, October 2003
• Eric Schwitzgebel and Josh Dever, Using Variables Within the Expectation Formula PDF, February 2004 A Simple Version of Our Explanation
• Dov Samet, Iddo Samet, and David Schmeidler, One Observation behind Two-Envelope Puzzles PDF, April 2004
• Franz Dietrich and Christian List, The Two-Envelope Paradox: An Axiomatic Approach PDF, May 2004
• Bruce Langtry, The Classical and Maximin Versions of the Two-Envelope Paradox PDF, August 2004
• Jan Poland, The Two Envelopes Paradox in a Short Story PDF, 2005