# Two envelopes problem

The two envelopes problem, also known as the exchange paradox, is a brain teaser, puzzle, or paradox in logic, probability, and recreational mathematics. It is of special interest in decision theory, and for the Bayesian interpretation of probability theory. Historically, it arose as a variant of the necktie paradox.

The problem
You have two indistinguishable envelopes that each contain money. One contains twice as much as the other. You may pick one envelope and keep the money it contains. You pick at random, but before you open the envelope, you are offered the chance to take the other envelope instead.
Explanation

Assume the amount in a selected envelope is $20. If the envelope happens to be the larger of the two envelopes ("larger" meaning the one with the larger amount of money), that would mean that the amount in the envelope is twice the amount in the other envelope. So in this case the amount in the other envelope would be$10.

However if the selected envelope is the smaller of the two envelopes, that would mean that the amount in the other envelope is twice the amount in the selected envelope. In this second scenario the amount in the other envelope would be $40. The probability of either of these scenarios is one half, since there is a 50% chance that the larger envelope was selected and a 50% chance that the smaller envelope was selected. The expected value calculation for how much money is in the other envelope would be the amount in the first scenario times the probability of the first scenario plus the amount in the second scenario times the probability of the second scenario, which is$10 * 1/2 + $40 * 1/2. The result of this calculation is that the expected value of money in the other envelope is$25. Since this is greater than the selected envelope, it would appear to the person selecting the envelope's advantage to switch envelopes. But a completely analogous argument would hold if the other envelope was selected in the first place instead, also leading to the advice to switch envelopes. This means that even before you open your selected envelope you know that you will want to take the other envelope instead.

A large number of solutions have been proposed. The usual scenario is that one writer proposes a solution that solves the problem as stated, but then another writer discovers that altering the problem slightly revives the paradox. In this way, a family of closely related formulations of the problem have been created, which are discussed in the literature.

No proposed solution is widely accepted as correct.[1] Despite this it is common for authors to claim that the solution to the problem is easy, even elementary.[2] However, when investigating these elementary solutions they often differ from one author to the next. Since 1987 new papers have been published every year.[3]

## Problem

Basic setup: 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 are given the chance to take the other envelope instead.[4]

The switching argument: 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 is the larger amount is also 1/2.
3. The other envelope may contain either 2A or A/2.
4. If A is the smaller amount, then the other envelope contains 2A.
5. If A is the larger amount, then 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

${1 \over 2} (2A) + {1 \over 2} \left({A \over 2}\right) = {5 \over 4}A$

8. This is greater than A, so I gain on average by swapping.
9. After the switch, I can denote that content by 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: The puzzle is to find the flaw in the very compelling line of reasoning above.

## Common resolution

A common way to resolve the paradox, both in popular literature and academic literature, is to observe that A stands for different things at different places in the expected value calculation, step 7 above.[5] 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 in the same formula like this is said to be illegitimate, so step 7 is incorrect, and this is the cause of the paradox.

According to this analysis, a correct argument runs on the following lines. By definition the amount in one envelope is twice as much as in the other. Denoting the lower of the two amounts by X, we write the expected value calculation as

${1 \over 2} 2X + {1 \over 2} X = {3 \over 2}X$

Here X stands for the same thing in every term of the equation. We learn that 1.5X, is the average expected value in either of the envelopes. Being less than 2X, the value in the greater envelope, there is no reason to swap the envelopes.

### Mathematical details

Let us rewrite the preceding calculations in a more detailed notation that explicitly distinguishes random from not-random quantities (a different distinction from the usual in ordinary, deterministic mathematics between variables and constants). This is a useful way to compare with the next, alternative, resolution. So far, we were thinking of the two amounts of money in the two envelopes as fixed. The only randomness lies in which one goes into which envelope. We called the smaller amount X, let us denote the larger amount by Y. Given the values x and y of X and Y, where y = 2x and x > 0, the problem description tells us (whether or not x and y are known)

$P(A=x|X=x) = P(A=y|X=x) = \frac{1}{2}$

for all possible values x of the smaller amount X; there is a corresponding definition of the probability distribution of B given X and Y. In our resolution of the paradox, we guessed that in Step 7 the writer was trying to compute the expected value of B given X=x. Splitting the calculation over the two possibilities for which envelope contains the smaller amount, it is certainly correct to write

\begin{align} E(B|X=x) &= E(B | B > A, X=x) \cdot P(B > A | X=x) \\ &+ E(B | B < A, X=x) \cdot P(B < A | X=x). \end{align}

At this point, the writer correctly substitutes the value 1/2 for both of the conditional probabilities on the right hand side of this equation (Step 2). At the same time he correctly substitutes the random variable B inside the first conditional expectation for 2A, when taking its expectation value given B > A and X = x, and he similarly correctly substitutes the random variable B for A/2 when taking its expectation value given B < A and X = x (Steps 4 and 5). He would then arrive at the completely correct equation

\begin{align} E(B|X=x) &= \frac{1}{2} \cdot E(2 A | B > A, X=x) + \frac{1}{2} \cdot E( \frac{A}{2} | B < A, X=x) \\ &= E( A | B > A, X=x) + \frac{1}{4} \cdot E( A | B < A, X=x). \end{align}

However he now proceeds, in the first of the two terms on the right hand side, to replace the expectation value of A given that Envelope A contains the smaller amount and given that the amounts are x and 2x, by the random quantity A itself. Similarly, in the second term on the right hand side he replaces the expectation value of A given now that Envelope A contains the larger amount and given that the amounts are x and 2x, also by the random quantity A itself. The correct substitutions would have been, of course, x and 2x respectively, leading to a correct conclusion

$E(B|X=x) = \frac{3}{2}x$.

Naturally this coincides with the expectation value of A given X=x.

Indeed, in the two contexts in which the random variable A appears on the right hand side, it is standing for two different things, since its distribution has been conditioned on different events. Obviously, A tends to be larger, when we know that it is greater than B and when the two amounts are fixed, and it tends to be smaller, when we know that it is smaller than B and the two amounts are fixed, cf. Schwitzgebel and Dever (2007, 2008). In fact, it is exactly twice as large in the first situation as in the second situation.

The preceding resolution was first noted by Bruss in 1996.[6] Falk gave a concise exposition in 2009.[7]

## Alternative interpretation

The first solution above does not explain what is wrong if the player is allowed to open the first envelope before offered the option to switch. In this case, A stands for the value that is then seen throughout all subsequent calculations. The mathematical variable A stands for any particular amount he might see there (it is a mathematical variable, a generic possible value of a random variable). The reasoning appears to show that whatever amount he would see there, he would decide to switch. Hence, he does not need to look in the envelope at all: he knows that if he looks, and goes through the calculations, they will tell him to switch, whatever he saw in the envelope.

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.

Thus, the proposed "common resolution" above breaks down, and another explanation is needed.

This interpretation of the two envelopes problem appears in the first publications in which the paradox was introduced, Gardner (1989) and Nalebuff (1989). It is common in the more mathematical literature on the problem.

The "common resolution" above depends on a particular interpretation of what the writer of the argument is trying to calculate: namely, it assumes he is after the (unconditional) expectation value of what's in Envelope B. In the mathematical literature on Two Envelopes Problem (and in particular, in the literature where it was first introduced to the world), another interpretation is more common, involving the conditional expectation value (conditional on what might be in Envelope A). To solve this and related interpretations or versions of the problem, most authors use the Bayesian interpretation of probability.

### Introduction to resolutions based on Bayesian probability theory

Here the ways the paradox can be resolved depend to a large degree on the assumptions that are made about the things that are not made clear in the setup and the proposed argument for switching.[8][9] The most usual assumption about the way the envelopes are set up is that a sum of money is in one envelope, and twice that sum is in another envelope. One of the two envelopes is randomly given to the player (envelope A). It is not made clear exactly how the smaller of the two sums is determined, what values it could possibly take and, in particular, whether there is a maximum sum it might contain.[10][11] It is also not specified whether the player can look in Envelope A before deciding whether or not to switch. A further ambiguity in the paradox is that it is not made clear in the proposed argument whether the amount A in Envelope A is intended to be a constant, a random variable, or some other quantity.

If it assumed that there is a maximum sum that can be put in the first envelope, then a simple and mathematically sound resolution is possible within the second interpretation. Step 6 in the proposed line of reasoning is not always true, since if the player holds more than the maximum sum that can be put into the first 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. This resolves all practical cases of the problem, whether or not the player looks in their envelope.[12]

It can be envisaged, however, that the sums in the two envelopes are not limited. This requires a more careful mathematical analysis, and also uncovers other possible interpretations of the problem. If, for example, the smaller of the two sums of money is considered equally likely to be one of infinitely many positive integers, without upper limit—the probability that it is any given number is always zero. This absurd situation exemplifies an improper prior, and this is generally considered to resolve the paradox in this case.

It is possible to devise a distribution for the sums possible in the first envelope such that the maximum value is unlimited, computation of the expectation of what is in B given what is in A seems to dictate you should switch, and the distribution constitutes a proper prior.[13] In these cases it can be shown that the expected sum in both envelopes is infinite. There is no gain, on average, in swapping.

The first two resolutions we present correspond, technically speaking, first to A being a random variable, and secondly to it being a possible value of a random variable (and the expectation being computed is a conditional expectation). At the same time, in the first resolution the two original amounts of money seem to be thought of as being fixed, while in the second they are also thought of as varying. Thus there are two main interpretations of the problem, and two main resolutions.

### Proposed resolutions to the alternative interpretation

Nalebuff (1989), Christensen and Utts (1992), Falk and Konold (1992), Blachman, Christensen and Utts (1996),[14] Nickerson and Falk (2006), 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 2a. Thus step 6 of the argument, which leads to always switching, is a non-sequitur.

#### Mathematical details

According to this interpretation, the writer is carrying out the following computation, where he is conditioning now on the value of A, the amount in Envelope A, not on the pair amounts in the two envelopes X and Y:

\begin{align} E( B | A =a ) &= E( B | A =a , A < B) \cdot P(A < B | A=a) \\ &+ E( B | A =a , A > B) \cdot P(A > B | A=a). \end{align}

Completely correctly, and according to Step 5, the two conditional expectation values are evaluated as

$E( B | A =a , A < B) = E( 2A | A =a , A < B) = 2a,$
$E( B | A =a , A > B) = E( A/2 | A =a , A > B) = a/2.$

However in Step 6 the writer is invoking Steps 2 and 3 to get the two conditional probabilities, and effectively replacing the two conditional probabilities of Envelope A containing the smaller and larger amount, respectively, given the amount actually in that envelope, both by the unconditional probability 1/2: he makes the substitutions

$P(A < B | A=a) = 1/2,$
$P(A > B | A=a) = 1/2.$

But intuitively, we would expect that the larger the amount in A, the more likely it is the larger of the two, and vice-versa. And it is a mathematical fact, as we will see in a moment, that it is impossible that both of these conditional probabilities are equal to 1/2 for all possible values of a. In fact, for step 6 to be true, whatever a might be, the smaller amount of money in the two envelopes must be equally likely to be between 1 and 2, as between 2 and 4, as between 4 and 8, ... ad infinitum. But there is no way to divide total probability 1 into an infinite number of pieces that are not only all equal to one another, but also all larger than zero. Yet the smaller amount of money in the two envelopes must have probability larger than zero to be in at least one of the just mentioned ranges.

To see this, suppose that the chance that the smaller of the two envelopes contains an amount between 2n and 2n+1 is p(n), where n is any whole number, positive or negative, and for definiteness we include the lower limit but exclude the upper in each interval. It follows that the conditional probability that the envelope in our hands contains the smaller amount of money of the two, given that its contents are between 2n and 2n+1, is

$(p(n)/2)/(p(n)/2 +p(n-1)/2).$

If this is equal to 1/2, it follows by simple algebra that

$2p(n)=p(n)+p(n-1),$

or p(n)=p(n-1). This must be true for all n, an impossibility.

### New variant

Though Bayesian probability theory can resolve the alternative interpretation of the paradox above, it turns out that examples can be found of proper probability distributions, such that the expected value of the amount in the second envelope given that in the first does exceed the amount in the first, whatever it might be. The first such example was already given by Nalebuff (1989). See also Christensen and Utts (1992)[15][16][17][18]

Denote again the amount of money in the first envelope by A and that in the second by B. We think of these as random. Let X be the smaller of the two amounts and Y=2X be the larger. Notice that once we have fixed a probability distribution for X then the joint probability distribution of A,B is fixed, since A,B = X,Y or Y,X each with probability 1/2, independently of X,Y.

The bad step 6 in the "always switching" argument led us to the finding E(B|A=a)>a for all a, and hence to the recommendation to switch, whether or not we know a. Now, it turns out that one can quite easily invent proper probability distributions for X, the smaller of the two amounts of money, such that this bad conclusion is still true. One example is analysed in more detail, in a moment.

It cannot be true that whatever a, given A=a, B is equally likely to be a/2 or 2a, but it can be true that whatever a, given A=a, B is larger in expected value than a.

Suppose for example (Broome, 1995)[19] that the envelope with the smaller amount actually contains 2n dollars with probability 2n/3n+1 where n = 0, 1, 2,… These probabilities sum to 1, hence the distribution is a proper prior (for subjectivists) and a completely decent probability law also for frequentists.

Imagine what might be in the first envelope. A sensible strategy would certainly be to swap when the first envelope contains 1, as the other must then contain 2. Suppose on the other hand the first envelope contains 2. In that case there are two possibilities: the envelope pair in front of us is either {1, 2} or {2, 4}. All other pairs are impossible. The conditional probability that we are dealing with the {1, 2} pair, given that the first envelope contains 2, is

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

and consequently the probability it's the {2, 4} pair is 2/5, since these are the only two possibilities. In this derivation, $P(\{1,2\})/2$ is the probability that the envelope pair is the pair 1 and 2, and Envelope A happens to contain 2; $P(\{2,4\})/2$ is the probability that the envelope pair is the pair 2 and 4, and (again) Envelope A happens to contain 2. Those are the only two ways that Envelope A can end up containing the amount 2.

It turns out that these proportions hold in general unless the first envelope contains 1. Denote by a the amount we imagine finding in Envelope A, if we were to open that envelope, and suppose that a = 2n for some n ≥ 1. In that case the other envelope contains a/2 with probability 3/5 and 2a with probability 2/5.

So either the first envelope contains 1, in which case the conditional expected amount in the other envelope is 2, or the first envelope contains a > 1, and though the second envelope is more likely to be smaller than larger, its conditionally expected amount is larger: the conditionally expected amount in Envelope B is

$\frac{3}{5} \frac{a}{2} + \frac{2}{5} 2a = \frac{11}{10}a$

which is more than a. This means that the player who looks in Envelope A would decide to switch whatever he saw there. Hence there is no need to look in Envelope A to make that decision.

This conclusion is just as clearly wrong as it was in the preceding interpretations of the Two Envelopes Problem. But now the flaws noted above do not apply; the a in the expected value calculation is a constant and the conditional probabilities in the formula are obtained from a specified and proper prior distribution.

### Proposed resolutions

Some writers think that the new paradox can be defused.[20] Suppose $E(B|A=a)>a$ for all a. As remarked before, this is possible for some probability distributions of X (the smaller amount of money in the two envelopes). Averaging over a, it follows either that $E(B)>E(A)$, or alternatively that $E(B)=E(A)=\infty$. But A and B have the same probability distribution, and hence the same expectation value, by symmetry (each envelope is equally likely to be the smaller of the two). Thus both have infinite expectation values, and hence so must X too.

Thus if we switch for the second envelope because its conditional expected value is larger than what actually is in the first, whatever that might be, we are exchanging an unknown amount of money whose expectation value is infinite for another unknown amount of money with the same distribution and the same infinite expected value. The average amount of money in both envelopes is infinite. Exchanging one for the other simply exchanges an average of infinity with an average of infinity.

Probability theory therefore tells us why and when the paradox can occur and explains to us where the sequence of apparently logical steps breaks down. In this situation, Steps 6 and Steps 7 of the standard Two Envelopes argument can be replaced by correct calculations of the conditional probabilities that the other envelope contains half or twice what's in A, and a correct calculation of the conditional expectation of what's in B given what's in A. Indeed, that conditional expected value is larger than what's in A. But because the unconditional expected amount in A is infinite, this does not provide a reason to switch, because it does not guarantee that on average you'll be better off after switching. One only has this mathematical guarantee in the situation that the unconditional expectation value of what's in A is finite. But then the reason for switching without looking in the envelope, $E(B|A=a)>a$ for all a, simply cannot arise.

Many economists prefer to argue that in a real-life situation, the expectation of the amount of money in an envelope cannot be infinity, for instance, because the total amount of money in the world is bounded; therefore any probability distribution describing the real world would have to assign probability 0 to the amount being larger than the total amount of money on the world. Therefore the expectation of the amount of money under this distribution cannot be infinity. The resolution of the second paradox, for such writers, is that the postulated probability distributions cannot arise in a real-life situation. These are similar arguments as used to explain the St. Petersburg Paradox.

### Foundations of mathematical economics

In mathematical economics and the theory of utility, which explains economic behaviour in terms of expected utility, there remains a problem to be resolved.[21] In the real world we presumably would not indefinitely exchange one envelope for the other (and probability theory, as just discussed, explains quite well why calculations of conditional expectations might mislead us). Yet the expected utility based theory of economic behaviour assumes that people do (or should) make economic decisions by maximizing expected utility, conditional on present knowledge. If the utility function is unbounded above, then the theory can still predict infinite switching.

Fortunately for mathematical economics and the theory of utility, it is generally agreed that as an amount of money increases, its utility to the owner increases less and less, and ultimately there is a finite upper bound to the utility of all possible amounts of money. We can pretend that the amount of money in the whole world is as large as we like, yet the utility that the owner of all that money experiences, while rising further and further, will never rise beyond a certain point no matter how much is in his possession. For decision theory and utility theory, the two envelope paradox illustrates that unbounded utility does not exist in the real world, so fortunately there is no need to build a decision theory that allows unbounded utility, let alone utility of infinite expectation.[citation needed]

### Controversy among philosophers

As mentioned above, any distribution producing this variant of the paradox must have an infinite mean. So before the player opens an envelope the expected gain from switching is "∞ − ∞", which is not defined. In the words of David Chalmers this is "just another example of a familiar phenomenon, the strange behaviour of infinity".[22] Chalmers 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.

However, Clark and Shackel argue that this blaming it all on "the strange behaviour of infinity" does not resolve the paradox at all; neither in the single case nor the averaged case. They provide a simple example of a pair of random variables both having infinite mean but where it is clearly sensible to prefer one to the other, both conditionally and on average.[23] They argue that decision theory should be extended so as to allow infinite expectation values in some situations.

## Non-probabilistic variant

The logician Raymond Smullyan questioned if the paradox has anything to do with probabilities at all.[24] He did this by expressing the problem in a way that does not involve probabilities. The following plainly logical arguments lead to conflicting conclusions:

1. Let the amount in the envelope chosen by the player be A. By swapping, the player may gain A or lose A/2. So the potential gain is strictly greater than the potential loss.
2. Let the amounts in the envelopes be X and 2X. Now by swapping, the player may gain X or lose X. So the potential gain is equal to the potential loss.

### Proposed resolutions

A number of solutions have been put forward. Careful analyses have been made by some logicians. Though solutions differ, they all pinpoint semantic issues concerned with counterfactual reasoning. We want to compare the amount that we would gain by switching if we would gain by switching, with the amount we would lose by switching if we would indeed lose by switching. However, we cannot both gain and lose by switching at the same time. We are asked to compare two incompatible situations. Only one of them can factually occur, the other is a counterfactual situation—somehow imaginary. To compare them at all, we must somehow "align" the two situations, providing some definite points in common.

James Chase (2002) argues that the second argument is correct because it does correspond to the way to align two situations (one in which we gain, the other in which we lose), which is preferably indicated by the problem description.[25] Also Bernard Katz and Doris Olin (2007) argue this point of view.[26] In the second argument, we consider the amounts of money in the two envelopes as being fixed; what varies is which one is first given to the player. Because that was an arbitrary and physical choice, the counterfactual world in which the player, counterfactually, got the other envelope to the one he was actually (factually) given is a highly meaningful counterfactual world and hence the comparison between gains and losses in the two worlds is meaningful. This comparison is uniquely indicated by the problem description, in which two amounts of money are put in the two envelopes first, and only after that is one chosen arbitrarily and given to the player. In the first argument, however, we consider the amount of money in the envelope first given to the player as fixed and consider the situations where the second envelope contains either half or twice that amount. This would only be a reasonable counterfactual world if in reality the envelopes had been filled as follows: first, some amount of money is placed in the specific envelope that will be given to the player; and secondly, by some arbitrary process, the other envelope is filled (arbitrarily or randomly) either with double or with half of that amount of money.

Byeong-Uk Yi (2009), 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.[27] 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 probability ingredient, there is no way to argue that one action is better than another, anyway.

In perhaps the most recent paper on the subject, 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.[28] If, for example, the envelopes contained $5.00 and$10.00 respectively, a player who opened the $10.00 envelope would expect the possibility of a$20.00 payout that simply does not exist. Were that player to open the $5.00 envelope instead, he would believe in the possibility of a$2.50 payout, which constitutes a smaller deviation from the true value.

Albers, Kooi, and Schaafsma (2005) consider that without adding probability (or other) ingredients to the problem, 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.

## Extensions to the problem

Since the two envelopes problem became popular, many authors have studied the problem in depth in the situation in which the player has a prior probability distribution of the values in the two envelopes, and does look in Envelope A. One of the most recent such publications is by McDonnell and Douglas (2009), who also consider some further generalizations.[29]

If a priori we know that the amount in the smaller envelope is a whole number of some currency units, then the problem is determined, as far as probability theory is concerned, by the probability mass function $p(x)$ describing our prior beliefs that the smaller amount is any number x = 1,2, ... ; the summation over all values of x being equal to 1. It follows that given the amount a in Envelope A, the amount in Envelope B is certainly 2a if a is an odd number. However, if a is even, then the amount in Envelope B is 2a with probability $p(a)/(p(a/2)+p(a))$, and a/2 with probability $p(a/2)/(p(a/2)+p(a))$. If one would like to switch envelopes if the expectation value of what is in the other is larger than what we have in ours, then a simple calculation shows that one should switch if $p(a/2) < 2p(a)$, keep to Envelope A if $p(a/2) > 2p(a)$.

If on the other hand the smaller amount of money can vary continuously, and we represent our prior beliefs about it with a probability density $f(x)$, thus a function that integrates to one when we integrate over x running from zero to infinity, then given the amount a in Envelope A, the other envelope contains 2a with probability $2f(a)/(f(a/2)+2f(a))$, and a/2 with probability $f(a/2)/(f(a/2)+2f(a))$. If again we decide to switch or not according to the expectation value of what's in the other envelope, the criterion for switching now becomes $f(a/2) < 4f(a)$.

The difference between the results for discrete and continuous variables may surprise many readers. Speaking intuitively, this is explained as follows. Let h be a small quantity and imagine that the amount of money we see when we look in Envelope A is rounded off in such a way that differences smaller than h are not noticeable, even though actually it varies continuously. The probability that the smaller amount of money is in an interval around a of length h, and Envelope A contains the smaller amount is approximately $f(a) \cdot h \cdot (1/2)$. The probability that the larger amount of money is in an interval around a of length h corresponds to the smaller amount being in an interval of length h/2 around a/2. Hence the probability that the larger amount of money is in a small interval around a of length h and Envelope A contains the larger amount is approximately $f(a/2) \cdot (h/2) \cdot (1/2)$. Thus, given Envelope A contains an amount about equal to a, the probability it is the smaller of the two is roughly $f(a) \cdot h \cdot (1/2)/(f(a) \cdot h \cdot (1/2)+f(a/2) \cdot (h/2) \cdot (1/2)) = 2f(a)/(2f(a)+f(a/2))$.

If the player only wants to end up with the larger amount of money, and does not care about expected amounts, then in the discrete case he should switch if a is an odd number, or if a is even and $p(a/2) < p(a)$. In the continuous case he should switch if $f(a/2) < 2f(a)$.

Some authors prefer to think of probability in a frequentist sense. If the player knows the probability distribution used by the organizer to determine the smaller of the two values, then the analysis would proceed just as in the case when p or f represents subjective prior beliefs. However, what if we take a frequentist point of view, but the player does not know what probability distribution is used by the organiser to fix the amounts of money in any one instance? Thinking of the arranger of the game and the player as two parties in a two person game, puts the problem into the range of game theory. The arranger's strategy consists of a choice of a probability distribution of x, the smaller of the two amounts. Allowing the player also to use randomness in making his decision, his strategy is determined by his choosing a probability of switching $q(a)$ for each possible amount of money a he might see in Envelope A. In this section we so far only discussed fixed strategies, that is strategies for which q only takes the values 0 and 1, and we saw that the player is fine with a fixed strategy, if he knows the strategy of the organizer. In the next section we will see that randomized strategies can be useful when the organizer's strategy is not known.

## Randomized solutions

Suppose as in the previous section that the player is allowed to look in the first envelope before deciding whether to switch or to stay. We'll think of the contents of the two envelopes as being two positive numbers, not necessarily two amounts of money. The player is allowed either to keep the number in Envelope A, or to switch and take the number in Envelope B. We'll drop the assumption that one number is exactly twice the other, we'll just suppose that they are different and positive. On the other hand, instead of trying to maximize expectation values, we'll just try to maximize the chance that we end up with the larger number.

In this section we ask the question, is it possible for the player to make his choice in such a way that he goes home with the larger number with probability strictly greater than half, however the organizer has filled the two envelopes?

We are given no information at all about the two numbers in the two envelopes, except that they are different, and strictly greater than zero. The numbers were written down on slips of paper by the organiser, put into the two envelopes. The envelopes were then shuffled, the player picks one, calls it Envelope A, and opens it.

We are not told any joint probability distribution of the two numbers. We are not asking for a subjectivist solution. We must think of the two numbers in the envelopes as chosen by the arranger of the game according to some possibly random procedure, completely unknown to us, and fixed. Think of each envelope as simply containing a positive number and such that the two numbers are not the same. The job of the player is to end up with the envelope with the larger number. This variant of the problem, as well as its solution, is attributed by McDonnell and Abbott, and by earlier authors, to information theorist Thomas M. Cover.[30]

Counter-intuitive though it might seem, there is a way that the player can decide whether to switch or to stay so that he has a larger chance than 1/2 of finishing with the bigger number, however the two numbers are chosen by the arranger of the game. However, it is only possible with a so-called randomized algorithm: the player must be able to generate his own random numbers. Suppose he is able to produce a random number, let's call it Z, such that the probability that Z is larger than any particular quantity z is exp(-z). Note that exp(-z) starts off equal to 1 at z=0 and decreases strictly and continuously as z increases, tending to zero as z tends to infinity. So the chance is 0 that Z is exactly equal to any particular number, and there is a positive probability that Z lies between any two particular different numbers. The player compares his Z with the number in Envelope A. If Z is smaller he keeps the envelope. If Z is larger he switches to the other envelope.

Think of the two numbers in the envelopes as fixed (though of course unknown to the player). Think of the player's random Z as a probe with which he decides whether the number in Envelope A is small or large. If it is small compared to Z he switches, if it is large compared to Z he stays.

If both numbers are smaller than the player's Z, his strategy does not help him. He ends up with the Envelope B, which is equally likely to be the larger or the smaller of the two. If both numbers are larger than Z his strategy does not help him either, he ends up with the first Envelope A, which again is equally likely to be the larger or the smaller of the two. However if Z happens to be in between the two numbers, then his strategy leads him correctly to keep Envelope A if its contents are larger than those of B, but to switch to Envelope B if A has smaller contents than B. Altogether, this means that he ends up with the envelope with the larger number with probability strictly larger than 1/2. To be precise, the probability that he ends with the "winning envelope" is 1/2 + P(Z falls between the two numbers)/2.

In practice, the number Z we have described could be determined to the necessary degree of accuracy as follows. Toss a fair coin many times, and convert the sequence of heads and tails into the binary representation of a number U between 0 and 1: for instance, HTHHTH... becomes the binary representation of u=0.101101.. . In this way, we generate a random number U, uniformly distributed between 0 and 1. Then define Z = − ln (U) where "ln" stands for natural logarithm, i.e., logarithm to base e. Note that we just need to toss the coin long enough to verify whether Z is smaller or larger than the number a in the first envelope—we do not need to go on for ever. We only need to toss the coin a finite (though random) number of times: at some point we can be sure that the outcomes of further coin tosses would not change the outcome.

The particular probability law (the so-called standard exponential distribution) used to generate the random number Z in this problem is not crucial. Any probability distribution over the positive real numbers that assigns positive probability to any interval of positive length does the job.

This problem can be considered from the point of view of game theory, where we make the game a two-person zero-sum game with outcomes win or lose, depending on whether the player ends up with the higher or lower amount of money. The organiser chooses the joint distribution of the amounts of money in both envelopes, and the player chooses the distribution of Z. The game does not have a "solution" (or saddle point) in the sense of game theory. This is an infinite game and von Neumann's minimax theorem does not apply.[31]

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

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 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?

In 1988 and 1989, Barry Nalebuff presented two different two-envelope problems, each with one envelope containing twice what's in the other, and each with computation of the expectation value 5A/4. The first paper just presents the two problems, the second paper discusses many solutions to both of them. The second of his two problems nowadays the most 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.

## Notes and references

1. ^ Markosian, Ned (2011). "A Simple Solution to the Two Envelope Problem". Logos & Episteme II (3): 347–57.
2. ^ McDonnell, Mark D; Grant, Alex J; Land, Ingmar; Vellambi, Badri N; Abbott, Derek; Lever, Ken (2011). "Gain from the two-envelope problem via information asymmetry: on the suboptimality of randomized switching". Proceedings of the Royal Society A. doi:10.1098/rspa.2010.0541.
3. ^ A complete list of published and unpublished sources in chronological order can be found in the talk page.
4. ^ Falk, Ruma (2008). "The Unrelenting Exchange Paradox". Teaching Statistics 30 (3): 86–88. doi:10.1111/j.1467-9639.2008.00318.x.
5. ^ Eckhardt, William (2013). "The Two-Envelopes Problem". Paradoxes in Probability Theory. Springer. pp. 47–48.
6. ^ Bruss, F.T. (1996). "The Fallacy of the Two Envelopes Problem". The Mathematical Scientist 21 (2): 112–119.
7. ^ Falk, Ruma (2009). "An inside look at the two envelope paradox". Teaching Statistics 31 (2): 39–41. doi:10.1111/j.1467-9639.2009.00346.x.
8. ^ Albers, Casper (March 2003), "2. Trying to resolve the two-envelope problem", Distributional Inference: The Limits of Reason (thesis).
9. ^ Albers, Casper J; Kooi, Barteld P; Schaafsma, Willem (2005), "Trying to resolve the two-envelope problem", Synthese 145 (1): 91.
10. ^ Falk, Ruma; Nickerson, Raymond, "An inside look at the two envelopes paradox", Teaching Statistics 31 (2): 39–41, doi:10.1111/j.1467-9639.2009.00346.x.
11. ^ Chen, Jeff, The Puzzle of the Two-Envelope Puzzle—a Logical Approach (online ed.), p. 274.
12. ^ Nalebuff, Barry, "Puzzles: The Other Person’s Envelope is Always Greener", Journal of Economic Perspectives 3 (1): 171–81, doi:10.1257/jep.3.1.171.
13. ^ Broome, John, "The Two-envelope Paradox", Analysis 55 (1): 6–11, doi:10.1093/analys/55.1.6.
14. ^ Blachman, NM; Christensen, R; Utts, J (1996). The American Statistician 50 (1): 98–99.
15. ^ Christensen, R; Utts, J (1992), The American Statistician 46 (4): 274–76.
16. ^ Binder, DA (1993), "Letter to editor and response", The American Statistician 47 (2): 160.
17. ^ Ross (1994), "Letter to editor and response", The American Statistician 48 (3): 267.
18. ^ Blachman, NM; Christensen, R; Utts, JM (1996), "Letter with corrections to the original article", The American Statistician 50 (1): 98–99.
19. ^ Broome, John (1995). "The Two-envelope Paradox". Analysis 55 (1): 6–11. doi:10.1093/analys/55.1.6. A famous example of a proper probability distribution of the amounts of money in the two envelopes, for which $E(B|A=a)>a$ for all a.
20. ^ Binder, D. A. (1993). The American Statistician 47 (2): 160. (letters to the editor, comment on Christensen and Utts (1992)
21. ^ Fallis, D. (2009). "Taking the Two Envelope Paradox to the Limit". Southwest Philosophy Review 25 (2).
22. ^ Chalmers, David J. (2002). "The St. Petersburg Two-Envelope Paradox". Analysis 62 (2): 155–157. doi:10.1093/analys/62.2.155.
23. ^ Clark, M.; Shackel, N. (2000). "The Two-Envelope Paradox". Mind 109 (435): 415–442. doi:10.1093/mind/109.435.415.
24. ^ Smullyan, Raymond (1992). Satan, Cantor, and infinity and other mind-boggling puzzles. Alfred A. Knopf. pp. 189–192. ISBN 0-679-40688-3.
25. ^ Chase, James (2002). "The Non-Probabilistic Two Envelope Paradox". Analysis 62 (2): 157–160. doi:10.1093/analys/62.2.157.
26. ^ Katz, Bernard; Olin, Doris (2007). "A tale of two envelopes". Mind 116 (464): 903–926. doi:10.1093/mind/fzm903.
27. ^ Byeong-Uk Yi (2009). The Two-envelope Paradox With No Probability.
28. ^ Bliss (2012). A Concise Resolution to the Two Envelope Paradox.
29. ^ McDonnell, M. D.; Abott, D. (2009). "Randomized switching in the two-envelope problem". Proceedings of the Royal Society A 465 (2111): 3309–3322. doi:10.1098/rspa.2009.0312.
30. ^ Cover, Thomas M (1987). "Pick the largest number". In Cover, T; Gopinath, B. Open Problems in Communication and Computation. Springer-Verlag.
31. ^ Martinian, Emin, The Two Envelope Problem, archived from the original on 2007-11-14.