# Two envelopes problem

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 problem

The setup: The player is given two indistinguishable envelopes, each of which contains a positive sum of money. One envelope contains twice as much as the other. The player may select one envelope and keep whatever amount it contains, but upon selection, is offered the possibility to take the other envelope instead.

The switching argument:

1. Denote by A the amount in the 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: $\frac{1}{2} 2A + \frac{1}{2} \frac{A}{2} = \frac{5}{4}A$
8. This is greater than A, so swapping is favored
9. After the switch, reason in exactly the same manner as above, but denote the second envelope's contents as B
10. It follows that the most rational thing to do is to swap back again
11. This line of reasoning dictates that envelopes be swapped indefinitely
12. As it seems more rational to open just any envelope than to swap indefinitely, the player is left with a paradox

The puzzle: The puzzle is to find the flaw, the erroneous step, in the switching argument above. This includes determining exactly why and under what conditions that step is not correct, in order to be sure not to make this mistake in a more complicated situation where the misstep may not be so obvious. In short, the problem is to solve the paradox.

The sting: 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 properly stated when imposing those more technical restrictions.[1] 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 is used inconsistently 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.[2]

For example, if the lower of the two amounts is denoted by C, the expected value calculation may be written as

$\frac{1}{2} C + \frac{1}{2} 2C = \frac{3}{2} C$

Here C is a constant throughout the calculation—as it should be—and we learn that 1.5C is the average expected value in either of the envelopes. So according to this new calculation there is no contradiction between the decisions to keep or to swap, and hence no need to swap indefinitely.

## An even harder problem

The solution above does not rule out the possibility that there is some non-uniform prior distribution of sums in the envelopes to give the paradox force.

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

A sensible strategy that guarantees a win is to swap only when the opened envelope contains 1, as the other must contain 2. Furthermore, suppose the envelope contains 2. Now 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 hold in general unless the first envelope contains 1. Thus, denote by x the amount found 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 by switching is

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

which is more than x. This means that the player should switch in all cases.

But once again, the player may go through this reasoning before opening either envelope, and deduce that the other envelope should always be chosen. This conclusion is just as clearly wrong as it was in the first and second cases. But now the flaws noted above don't apply; the x in the expected value calculation is a known constant (in every single case) and the probabilities in the formula are obtained from a specified and proper prior distribution.

### Proposed solution

In this variant of the paradox, if the player picks any envelope, the expected payoff is infinite. It can be shown that 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 Chalmers this is "just another example of a familiar phenomenon, the strange behaviour of infinity".[4] This resolves this paradox according to some authors.

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 the player be willing to trade it for another ticket to the lottery? Since the St. Petersburg lottery has an infinite expected value, once the lottery ticket value is determined, no matter what (finite) value it is worth, the player should be willing to trade it for another ticket to a new, not yet drawn, lottery. However, even before the lottery ticket is drawn, the player 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, the player 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 the player should truly prefer one infinite expected value to another.

However, Clark and Shackel argue 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 provide a simple example of a pair of random variables both having infinite mean but where one is always better to choose than the other.[5] 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.

## A non-probabilistic variant

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 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 Y and 2Y. Now by swapping, the player may gain Y or lose Y. So the potential gain is equal to the potential loss.

### Proposed solution

So far, the only proposed solution of this variant is due to James Chase.[6] His idea is that this problem is an example of a fallacy in the logic of counterfactuals. He claims that while the statement "The player may gain A or lose A/2" is unproblematic and true (in virtue of the truth of one of its disjuncts), this is no longer the case when split into two separate sentences according to the standard analysis of counterfactuals:

• the nearest possible world in which the player gains on the trade is a world where the player gains A
• the nearest possible world in which the player loses on the trade is a world where the player loses A/2

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

Suppose that the chosen envelope actually contains the greater sum; then the second sentence is true since the nearest possible world where the player loses money is the actual world. But what is the nearest possible world where the player gains 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, the player 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 the smaller amount is chosen first. This shows that the unproblematic sentence above can't be split into two sentences that are true either factual or counterfactual, and therein lies the fallacy.

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.

## References

1. ^ See Priest and Restall, 2003
2. ^ See Schwitzgebel and Dever, 2004
3. ^ This example is from John Broome, 1995
4. ^ See David J. Chalmers, 2002
5. ^ See Clark and Shackel, 2000, p 426
6. ^ See James Chase, 2002

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