The two envelopes problem, also known as the exchange paradox, is a
paradox
A paradox is a logically self-contradictory statement or a statement that runs contrary to one's expectation. It is a statement that, despite apparently valid reasoning from true premises, leads to a seemingly self-contradictory or a logically u ...
in
probability theory
Probability theory is the branch of mathematics concerned with probability. Although there are several different probability interpretations, probability theory treats the concept in a rigorous mathematical manner by expressing it through a set o ...
. It is of special interest in
decision theory
Decision theory (or the theory of choice; not to be confused with choice theory) is a branch of applied probability theory concerned with the theory of making decisions based on assigning probabilities to various factors and assigning numerical ...
, and for the
Bayesian interpretation of
probability theory
Probability theory is the branch of mathematics concerned with probability. Although there are several different probability interpretations, probability theory treats the concept in a rigorous mathematical manner by expressing it through a set o ...
. It is a variant of an older problem known as the
necktie paradox
The necktie paradox is a puzzle and paradox with a subjective interpretation of probability theory describing a paradoxical bet advantageous to both involved parties. The two-envelope paradox is a variation of the necktie paradox.
Statement of ...
.
The problem is typically introduced by formulating a
hypothetical
A hypothesis (plural hypotheses) is a proposed explanation for a phenomenon. For a hypothesis to be a scientific hypothesis, the scientific method requires that one can test it. Scientists generally base scientific hypotheses on previous obser ...
challenge like the following example:
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.
Introduction
Problem
Basic setup: A person is given two indistinguishable envelopes, each of which contains a sum of money. One envelope contains twice as much as the other. The person may pick one envelope and keep whatever amount it contains. They pick one envelope at random but before they open it they are given the chance to take the other envelope instead.
The switching argument: Now suppose the person reasons as follows:
The puzzle: ''The puzzle is to find the flaw in the line of reasoning above.'' 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. In particular, the puzzle is ''not'' solved by finding another way to calculate the probabilities that does not lead to a contradiction.
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.
No proposed solution is widely accepted as definitive; despite this, it is common for authors to claim that the solution to the problem is easy, even elementary. However, when investigating these elementary solutions they often differ from one author to the next.
Example resolution
The total amount in both envelopes is a constant
, with
in one envelope and
in the other.
If you select the envelope with
first you gain the amount
by swapping. If you select the envelope with
first you lose the amount
by swapping. So you gain on average
by swapping.
Swapping is not better than keeping. The expected value
is the same for both the envelopes. Thus no contradiction exists.
The famous mystification is evoked by the mixing up of two different circumstances and situations, giving wrong results. The so-called ''"paradox"'' presents two already appointed and already locked envelopes, where one envelope is already locked with twice the amount of the other already locked envelope. Whereas step 6 boldly claims "Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2.", in the given situation, that claim can never apply to ''any A'' nor to ''any average A''.
This claim is never correct for the situation presented, this claim applies to the ''Nalebuff asymmetric variant'' only (see below). In the situation presented, the other envelope cannot ''generally'' contain 2A, but can contain 2A only in the very specific instance where envelope A, by chance contains the ''smaller'' amount of
, but nowhere else. The other envelope cannot ''generally'' contain A/2 but can contain A/2 only in the very specific instance where envelope A, by chance, actually contains
, but nowhere else. The difference between the two already appointed and locked envelopes is always
. No ''"average amount A"'' can ever form any initial basis for any ''expected value'', as this does not get to the heart of the problem.
Other simple resolutions
A common way to resolve the paradox, both in popular literature and part of the academic literature, especially in philosophy, is to assume that the 'A' in step 7 is intended to be the
expected value
In probability theory, the expected value (also called expectation, expectancy, mathematical expectation, mean, average, or first moment) is a generalization of the weighted average. Informally, the expected value is the arithmetic mean of a l ...
in envelope A and that we intended to write down a formula for the expected value in envelope B.
Step 7 states that the expected value in B = 1/2( 2A + A/2 )
It is pointed out that the 'A' in the first part of the formula is the expected value, given that envelope A contains less than envelope B, but the 'A', in the second part of the formula is the expected value in A, given that envelope A contains more than envelope B. The flaw in the argument is that the same symbol is used with two different meanings in both parts of the same calculation but is assumed to have the same value in both cases.
A correct calculation would be:
:Expected value in B = 1/2 ( (Expected value in B, given A is larger than B) + (Expected value in B, given A is smaller than B) )
If we then take the sum in one envelope to be x and the sum in the other to be 2x the expected value calculations become:
:Expected value in B = 1/2 (''x'' + 2''x'')
which is equal to the expected sum in A.
In non-technical language, what goes wrong (see
Necktie paradox
The necktie paradox is a puzzle and paradox with a subjective interpretation of probability theory describing a paradoxical bet advantageous to both involved parties. The two-envelope paradox is a variation of the necktie paradox.
Statement of ...
) is that, in the scenario provided, the mathematics use relative values of A and B (that is, it assumes that one would gain more money if A is less than B than one would lose if the opposite were true). However, the two values of money are fixed (one envelope contains, say, $20 and the other $40). If the values of the envelopes are restated as ''x'' and 2''x'', it's much easier to see that, if A were greater, one would lose ''x'' by switching and, if B were greater, one would gain ''x'' by switching. One does not gain a greater amount of money by switching because the total ''T'' of A and B (3''x'') remains the same, and the difference ''x'' is fixed to ''T/3''.
Line 7 should have been worked out more carefully as follows:
:
A will be larger when A is larger than B, than when it is smaller than B. So its average values (expectation values) in those two cases are different. And the average value of A is not the same as A itself, anyway. Two mistakes are being made: the writer forgot he was taking expectation values, and he forgot he was taking expectation values under two different conditions.
It would have been easier to compute E(B) directly. Denoting the lower of the two amounts by ''x'', and taking it to be fixed (even if unknown) we find that
:
We learn that 1.5''x'' is the expected value of the amount in Envelope B. By the same calculation it is also the expected value of the amount in Envelope A. They are the same hence there is no reason to prefer one envelope to the other. This conclusion was, of course, obvious in advance; the point is that we identified the false step in the argument for switching by explaining exactly where the calculation being made there went off the rails.
We could also continue from the correct but difficult to interpret result of the development in line 7:
:
so (of course) different routes to calculate the same thing all give the same answer.
Tsikogiannopoulos presented a different way to do these calculations.
It is by definition correct to assign equal probabilities to the events that the other envelope contains double or half that amount in envelope A. So the "switching argument" is correct up to step 6. Given that the player's envelope contains the amount A, he differentiates the actual situation in two different games: The first game would be played with the amounts (A, 2A) and the second game with the amounts (A/2, A). Only one of them is actually played but we don't know which one. These two games need to be treated differently. If the player wants to compute his/her expected return (profit or loss) in case of exchange, he/she should weigh the return derived from each game by the average amount in the two envelopes in that particular game. In the first case the profit would be A with an average amount of 3A/2, whereas in the second case the loss would be A/2 with an average amount of 3A/4. So the formula of the expected return in case of exchange, seen as a proportion of the total amount in the two envelopes, is:
:
This result means yet again that the player has to expect neither profit nor loss by exchanging his/her envelope.
We could actually open our envelope before deciding on switching or not and the above formula would still give us the correct expected return. For example, if we opened our envelope and saw that it contained 100 euros then we would set A=100 in the above formula and the expected return in case of switching would be:
:
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 their envelope.
Suppose that the amounts in the two envelopes A and B were not determined by first fixing the contents of two envelopes E1 and E2, and then naming them A and B at random (for instance, by the toss of a fair coin
). Instead, we start right at the beginning by putting some amount in envelope A and then fill B in a way which depends both on chance (the toss of a coin) and on what we put in A. Suppose that first of all the amount ''a'' in envelope A is fixed in some way or other, and then the amount in Envelope B is fixed, dependent on what is already in A, according to the outcome of a fair coin. If the coin fell Heads then 2''a'' is put in Envelope B, if the coin fell Tails then ''a''/2 is put in Envelope B. If the player was aware of this mechanism, and knows that they hold Envelope A, but don't know the outcome of the coin toss, and don't know ''a'', then the switching argument is correct and they are recommended to switch envelopes. This version of the problem was introduced by Nalebuff (1988) and is often called the Ali-Baba problem. Notice that there is no need to look in envelope A in order to decide whether or not to switch.
Many more variants of the problem have been introduced. Nickerson and
Falk systematically survey a total of 8.
Bayesian resolutions
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 (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 the concluding section, on history of TEP.
This kind of interpretation is often called "Bayesian" because it assumes the writer is also incorporating a prior probability distribution of possible amounts of money in the two envelopes in the switching argument.
Simple form of Bayesian resolution
The simple resolution depended on a particular interpretation of what the writer of the argument is trying to calculate: namely, it assumed he was after the (unconditional)
expectation value
In probability theory, the expected value (also called expectation, expectancy, mathematical expectation, mean, average, or first moment) is a generalization of the weighted average. Informally, the expected value is the arithmetic mean of a l ...
of what's in Envelope B. In the mathematical literature on Two Envelopes Problem, a different interpretation is more common, involving the
conditional expectation
In probability theory, the conditional expectation, conditional expected value, or conditional mean of a random variable is its expected value – the value it would take “on average” over an arbitrarily large number of occurrences – give ...
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, which means that probability reasoning is not only applied to truly random events like the random pick of an envelope, but also to our knowledge (or lack of knowledge) about things which are fixed but unknown, like the two amounts originally placed in the two envelopes, before one is picked at random and called "Envelope A". Moreover, according to a long tradition going back at least to
Laplace and his
principle of insufficient reason
The principle of indifference (also called principle of insufficient reason) is a rule for assigning epistemic probabilities. The principle of indifference states that in the absence of any relevant evidence, agents should distribute their cre ...
one is supposed to assign equal probabilities when one has no knowledge at all concerning the possible values of some quantity. Thus the fact that we are not told anything about how the envelopes are filled can already be converted into probability statements about these amounts. No information means that probabilities are equal.
In steps 6 and 7 of the switching argument, the writer imagines that envelope A contains a certain amount ''a'', and then seems to believe that given that information, the other envelope would be equally likely to contain twice or half that amount. That assumption can only be correct, if prior to knowing what was in Envelope A, the writer would have considered the following two pairs of values for both envelopes equally likely: the amounts ''a''/2 and ''a''; and the amounts ''a'' and 2''a''. (This follows from
Bayes' rule
In probability theory and statistics, Bayes' theorem (alternatively Bayes' law or Bayes' rule), named after Thomas Bayes, describes the probability of an event, based on prior knowledge of conditions that might be related to the event. For examp ...
in odds form: posterior odds equal prior odds times likelihood ratio). But now we can apply the same reasoning, imagining not ''a'' but ''a/2'' in Envelope A. And similarly, for 2''a''. And similarly, ad infinitum, repeatedly halving or repeatedly doubling as many times as you like.
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
). 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
improper prior distribution: probability calculus breaks down; expectation values are not even defined.
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.
[.]
But the problem can also be resolved mathematically without assuming a maximum amount. Nalebuff,
Christensen and Utts,
Falk and Konold,
Blachman, Christensen and Utts, Nickerson and Falk,
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.
Introduction to further developments in connection with Bayesian probability theory
The first two resolutions discussed above (the "simple resolution" and the "Bayesian resolution") correspond to two possible interpretations of what is going on in step 6 of the argument. They both assume that step 6 indeed is "the bad step". But the description in step 6 is ambiguous. Is the author after the unconditional (overall) expectation value of what is in envelope B (perhaps - conditional on the smaller amount, ''x''), or is he after the conditional expectation of what is in envelope B, given any possible amount ''a'' which might be in envelope A? Thus, there are two main interpretations of the intention of the composer of the paradoxical argument for switching, and two main resolutions.
A large literature has developed concerning variants of the problem.
[.] The standard 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''). The originally proposed problem does not make clear exactly how the smaller of the two sums is determined, what values it could possibly take and, in particular, whether there is a minimum or a maximum sum it might contain. However, if we are using the Bayesian interpretation of probability, then we start by expressing our prior beliefs as to the smaller amount in the two envelopes through a probability distribution. Lack of knowledge can also be expressed in terms of probability.
A first variant within the Bayesian version is to come up with a proper prior probability distribution of the smaller amount of money in the two envelopes, such that when Step 6 is performed properly, the advice is still to prefer Envelope B, whatever might be in Envelope A. So though the specific calculation performed in step 6 was incorrect (there is no proper prior distribution such that, given what is in the first envelope A, the other envelope is always equally likely to be larger or smaller) a correct calculation, depending on what prior we are using, does lead to the result
for all possible values of ''a''.
In these cases, it can be shown that the expected sum in both envelopes is infinite. There is no gain, on average, in swapping.
Second mathematical variant
Though Bayesian probability theory can resolve the first mathematical 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, conditioned on the amount in the first, does exceed the amount in the first, whatever it might be. The first such example was already given by Nalebuff.
See also Christensen and Utts (1992).
[.]
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 analyzed in more detail, in a moment.
As mentioned before, it cannot be true that whatever ''a'', given ''A=a'', ''B'' is equally likely to be ''a''/2 or 2''a'', but it can be true that whatever ''a'', given ''A=a'', ''B'' is larger in expected value than ''a''.
Suppose for example that the envelope with the smaller amount actually contains 2
''n'' dollars with probability 2
''n''/3
''n''+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 or . All other pairs are impossible. The
conditional probability
In probability theory, conditional probability is a measure of the probability of an event occurring, given that another event (by assumption, presumption, assertion or evidence) has already occurred. This particular method relies on event B occur ...
that we are dealing with the pair, given that the first envelope contains 2, is
:
and consequently the probability it's the pair is 2/5, since these are the only two possibilities. In this derivation,
is the probability that the envelope pair is the pair 1 and 2, ''and'' envelope A happens to contain 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'' = 2
''n'' for some ''n'' ≥ 1. In that case the other envelope contains ''a''/2 with probability 3/5 and 2''a'' 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
:
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 through mathematical economics
Most writers think that the new paradox can be defused, although the resolution requires concepts from mathematical economics. Suppose
for all ''a''. It can be shown that this is possible for some probability distributions of ''X'' (the smaller amount of money in the two envelopes) only if
. That is, only if the mean of all possible values of money in the envelopes is infinite. To see why, compare the series described above in which the probability of each ''X'' is 2/3 as likely as the previous ''X'' with one in which the probability of each ''X'' is only 1/3 as likely as the previous ''X''. When the probability of each subsequent term is greater than one-half of the probability of the term before it (and each ''X'' is twice that of the ''X'' before it) the mean is infinite, but when the probability factor is less than one-half, the mean converges. In the cases where the probability factor is less than one-half,