Opening Two Envelopes Paul Syverson Center for High Assurance Computer Systems Naval Research Laboratory, Washington, DC 20375, USA
[email protected] www.syverson.org Abstract In the two-envelope problem, one is offered a choice between two envelopes, one containing twice as much money as the other. After seeing the contents of the chosen envelope, the chooser is offered the opportunity to exchange for the other envelope. However, it appears to be advantageous to switch, regardless of what is observed in the chosen envelope. This problem has an extensive literature with connections to probability and decision theory. The literature is roughly divided between those that attempt to explain what is flawed in arguments for the advantage of switching and those that attempt to explain when such arguments can be correct if counterintuitive. We observe that arguments in the literature of the two-envelope problem that the problem is paradoxical are not supported by the probability distributions meant to illustrate the paradoxical nature. To correct this, we present a distribution that does support the usual arguments. Aside from questions about the interpretation of variables, algebraic ambiguity, modal confusions and the like, most of the interesting aspects of the two-envelope problem are assumed to require probability distributions on an infinite space. Our next main contribution is to show that the same counterintuitive arguments can be reflected in finite versions of the problem; thus they do not inherently require reasoning about infinite values. A topological representation of the problem is presented that captures both finite and infinite cases, explicating intuitions underlying the arguments both that there is an advantage to switching and that there is not.
Keywords: Conditional expectation, exchange paradox, epistemic possibility, bounded and unbounded values, decision theory, topology
1
Introduction
You are presented with two envelopes. One contains twice as much money as the other. You choose one at random and open it to find that it contains, e.g., one hundred dollars. You are offered the opportunity to keep the amount you see or to exchange it for the contents of the other envelope. Should you switch? 1
This, along with a few variants, is known as the two-envelope problem or “two-envelope paradox”. Why ‘paradox’ ? Given that the first envelope contains $100, the other envelope must contain either $50 or $200, and either case is equally possible. Thus, to switch is to risk $50 to gain $100. Put another way, the expected value of switching is (1/2 × $50) + (1/2 × $200) = $125 Thus, you have the opportunity to exchange an expected value of $100 for one of $125. Obviously it makes sense to switch. The problem is that you could have employed this reasoning no matter which envelope you opened. So, no matter what value you saw, the expected value of the other envelope is always 1.25 times that value. To underscore the point, suppose that two people are made this offer a thousand times, each of them being offered the same pair at the same play of the game. Suppose they never choose the same envelope. Each always chooses the opposite of the other, opens it, and then switches. They cannot both end up better off than they would have had they stuck with the original choice (Nalebuff 1989). Broome (1995) calls a probability distribution ‘paradoxical’ if, whatever the value in the chosen envelope, x, the expectation of the other envelope, y, conditional on x is greater than x. This by itself is not paradoxical, merely counterintuitive. However, as Broome notes If you have a paradoxical prior probability distribution, then for every value of x [what is in the envelope chosen first], there is a positive expectation of gain from switching envelopes. This appears to be a valid argument for switching. To emphasize the paradox, I should mention that there also appears to be a valid argument that show you have no reason to switch. Your envelope contains either s, the smaller of the two cheques, or 2s, and these possibilites are equally likely. If your envelope contains s and you switch, you will gain s. If your envelope contains 2s and you switch, you will lose s. So, your expected gain from switching, conditional on s, is zero, and this is true for all s. Therefore, you have no reason to switch. We have two apparently valid arguments with conflicting conclusions, then. (Broome 1995, p. 8) Note that ‘paradox’ is a bit strong, even accepting Broome’s conclusions; for they do not actually conflict. There is nothing paradoxical in the dual assertions that, on the one hand, whatever amount you see in the opened envelope, you should switch, but on the other hand, no matter what the total amount is in the envelopes, there is no advantage to switching. (Henceforth we will call these the ‘switch’ and ‘whichever’ arguments respectively.) In the one case the expectation is conditional on the amount in the opened envelope, in the other on the total amount in the two envelopes (or the amount in the smaller one, which is essentially the same thing). The known amount is determined by what is seen in the opened envelope; the (unknown) total amount is fixed by the pair 2
presented. A similar point is made by McGrew et al. (1997). Of course we must still examine if and when the switch and whichever arguments are correct, and, in that examination, we will still sometimes follow tradition in this loose usage of ‘paradox’. This is now a “paradox” with a fairly extensive literature (Albers et al. 2005; Arntzenius and McCarthy 1997; Binder 1993; Blachman et al. 1996; Brams and Kilgour 1995, 1998; Broome 1995; Bruss 1996; Castell and Batens 1994; Chalmers 1994, 2002; Chase 2002; Chihara 1995; Christensen and Utts 1992, 1993; Clark and Shackel 2000; Horgan 2000; Jackson et al. 1994; Jeffrey 2004; Linzer 1994; Malinas 2003; McGrew et al. 1997; Nalebuff 1989; Norton 1998; Rawling 1994, 1997; Ridgway 1993; Ross 1994; Scott and Scott 1997; Schwitzgebel and Dever 2008; Smullyan 1992, 1998; Sobel 1994; Wagner 1999). Indeed if there is anything inherently unbounded about the two-envelope paradox, it is that each search will uncover at least one more reference. Much of the literature hinges on observing that it is possible to have a random variable that always takes a finite value but for which the expected value is infinite. And, it is possible to construe the choices in this problem as two such random variables. There is no paradox in the observation that E(B|A) = 1.25E(A) and E(A|B) = 1.25E(B) if the expected values of A and B are both infinite. There is at most the unintuitive nature of infinity. This is representative of the observations in the literature but is not meant to capture them all. (In particular, Rawling (1997) presents “distributions with infinite means that do not suffer the paradox”.) What is common to all of them is that they see infinitude as necessary for the ‘switch’ argument. In the finite case, it simply fails. There are numerous other subtleties of the infinite that we will not explore. (Rawling (1997) surveys much, though not all, of the previous work. Nalebuff (1989) gives a complementary survey of the work until that point.) We will take a slightly different tack. The primary contributions of this paper are the following: 1. As is typical of paradoxes, in analysis of the two-envelope problem some authors have attributed the paradox to a sleight-of-hand or equivocation in either the algebra or probability arguments of presentations of the paradox, e.g., (Rawling 1994; Jeffrey 2004; Chalmers 2002). And, it is typically with the switch argument that they find fault. We show that if the problem is presented in simple form involving only known constant values, the argument for switching is not subject to such criticism. We later present a general topological framework and treatment of the problem which not only provides resolution but also explains the intuitions both that switching is advantageous and that it is not. 2. Since the argument for switching is the one that appears problematic, most of the literature has focused on how to construe it as rational (or on refuting it as irrational). In defending the switch argument, authors invariably construct distributions which have a distinct infimum. Properly done, this recovers the rationality of the switch argument, but it also
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undermines the argument that switching makes no difference. Thus these distributions justify the validity of the controversial result but only by undermining the intuitively accepted one. So, they do not support the above cited paradoxical result in which two valid arguments yield conflicting conclusions. We present a distribution that maintains both the switch argument and the whichever argument. 3. In the literature, the argument for switching is generally taken as dependent on the unboundedness of expected value. We contend that this is not the central aspect of the switch argument. We construct versions of the two-envelopes game in which expected value may be bounded but in which it is still correct to switch. This allows us to examine the argument for switching in finite terms, not dependent on or distracted by subtle issues that arise in the infinite case. Games with infinite expected value are subcases of the game we construct. We characterize the game topologically and show that the central requirement for the switch argument is related to openness rather than to unboundedness per se. We use this to examine realistic plays of the two-envelopes game.
2
Equivocating the Problem
Care must be taken in how the problem is presented or one can fall victim to what Jeffrey calls “the discharge fallacy” (Jeffrey 2004, p. 71)1 Specifically, one cannot in general argue as follows: 1. Premise: p(Y = .5X) = p(Y = 2X) = .5 2. Then E(Y ) = .5E(Y |Y = .5X) + .5E(Y |Y = 2X) 3. By the Discharge Fallacy: E(Y ) = .5E(.5X) + .5E(2X) 4. E(Y ) = .25E(X) + E(X) = 1.25E(X)
QED
Such reasoning may be valid under special circumstances: for example, if X is constant or if the expectation of X does not change for any occurrence within the right-hand side of the second equation above.2 However, one cannot as easily argue symmetrically, that for each presented pair, whichever envelope had been opened, the expected value of the other would be 1.25 of that. In such 1 This was also raised to me in separate personal communications with Dick Jeffrey and Sjoerd Zwart, both in 2001. 2 Besides Jeffrey, similar points are discussed by others. For example, Schwitzgebel and Dever (2008) introduce the criterion of unchanging expectation. Note that they incorrectly attribute to Jeffrey the position that “one can discharge such X-for-Y substitutions only when X is a true constant”. First, they are referencing a passage in a 1995 draft of (Jeffrey 2004) that did not appear in the final published book. More importantly, even in the draft passage, Jeffrey noted this only as a sufficient condition. He never claimed it was necessary. He was also primarily focused on giving a subjectivist explication of expectation. The two-envelopes problem was simply a convenient example briefly introduced as part of that.
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a case, both X and Y are variables. See (Scott and Scott 1997) and (Chalmers 2002) for related discussion. Similar reliance on an equivocation in the use of variables is noted by Rawling (1994). (Cf. also (McGrew et al. 1997).) Consider an offer of either $4 or a fair chance at either $2 or $8. Here it seems there is agreement that switching is the right way to maximize your expected payoff. But, suppose the results of the gamble on $2 or $8 is placed in an envelope and you can either take $4 or the contents of the envelope. If you now reason that E($4)
= .5(.5(contents of envelope) + .5(2(contents of envelope)) =
1.25(contents of envelope),
you commit the “cardinal sin of algebraic equivocation” that results from using ‘contents of envelope’ to stand for two values in the same equation (Rawling 1994). We will confine our discussion of the switch argument to the basic problem as initially stated above in the introduction. That is, we will simply consider whether a switch makes sense given the (constant) value of the envelope you open, whatever that value happens to be when you see it. We will not condition anything on the (unknown) total amount in the two envelopes; rather we will only condition on the known amount you see when you open the envelope. We will thus take discharge and equivocation issues to be the result of red herrings in the presentation of the problem and direct interested readers to the literature for further discussion of these issues. Some authors have argued that conditioning on the known amount is itself illegitimate because it treats this as fixed, rather than taking the total in the two envelopes to be fixed, e.g. (McGrew et al. 1997). Chase (2002) also argues that the possibility that a different amount was placed in the envelope not chosen is counterfactually more distant from the actual situation than the possibility that the other from the presented pair of envelopes was chosen because this “involves changes earlier in the causal sequence”. It is true that the total is fixed in the sense that the pair of envelopes has been chosen and presented, but this does not automatically preclude treating what is known as fixed for purposes of determining relevant expectation. Suppose a fair coin is flipped and you are to be given $50 or $200 depending on the outcome of this coinflip. (To limit the situation entirely to the point of the example, you have no choices to make here. You will just be given one amount or the other based on the coinflip.3 ) It is possible that sometimes there is another payoff, i.e., some other action happens after the coinflip. Nonetheless, when this is the payoff and you are told it is, you can know (unequivocally) that your expected payoff is $125. It makes no practical difference if you are told 3 This is related to variants of the two-envelope problem presented by many authors in which an amount is placed in one envelope, a fair coin is flipped, and half or twice that amount is placed in the other envelope. Then you are presented the pair and choose one. We discuss similar cases for analysis but consider any such specific causal stories as additional to the basic problem and not necessarily implied by it.
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the payoff possibilities before the coin is flipped or after it is flipped but before you have learned the result. It also makes no practical difference that at other times there are other payoffs associated with the coinflip. For the case where you know that the payoff will be $50 if heads and $200 if tails (or vice versa), then you will be correct to expect an average payoff of $125. This is a practical point and is not meant to address whether or not your payoff for a given coinflip is determined once the coin lands, or whether for that matter it is determined after the coin is tossed but before it lands (or even perhaps before tossing). Now suppose there are three envelopes containing $50, $100, and $200 respectively and you know this. Supppose you are given one of these that you know to contain $100. After that, one of the other two is chosen at random and offered to you as a trade. All authors agree that you should switch, in the sense that you will have a longterm average payoff of $125 if you choose repeatedly to switch whenever you are in this circumstance. But suppose instead that the choice of which of the other two to offer you was made prior to showing you the contents of the $100 envelope and prior to telling you the setup, i.e., that one of the envelopes contains $50, one contains $100, and one contains $200. Thus whether you will gain or lose by switching was fixed prior to your knowing how much was in the $100 envelope. This does not matter for your practical decision making and expectation. Now if we accept the constraints of the stated problem, then you are given no more information than that the envelope you hold contains $100 and that one envelope before you contains twice as much as the other. It does not matter for your practical expectation whether you got to this point by a choice made before or after you knew that the envelope you hold contains $100 or even whether that choice was made by you or not. It may be that you simply do not believe (or believe it likely) that you will be offered a chance at, e.g., more than $100. We will discuss this below. The above analysis is only meant to show that for a given known amount in a chosen envelope, it can be correct to condition expectation in virtue of knowing that amount. Analyses such as Chase’s critique of (Smullyan 1992) are limited to considerations of only alethic possibility and thus are forced to describe and choose between causal accounts of how the amounts came to be placed in the envelopes, when they were placed, which was placed first, etc. But this unnecessarily complicates the problem by forcing an assessment of which of two counterfactual situations is causally closer to the given circumstance. The above should show that such an assessment may not be possible or (fortunately) necessary. By instead focusing on epistemic possibility, we avoid the need to make guesses about which causal history preceded the given circumstance.4 We have not yet given our own analysis of whether or when it makes sense to switch. In this section we have only been looking at some of the criticisms that have been directed at prior analyses. We have examined certain types of faulty reasoning (discharge and equivocating on variables) to understand them 4 For another discussion of epistemic possibility and problems in trying to construe it metaphysically in the context of the two-envelope problem, see (Horgan 2000). For a general discussion of the related dangers of construing epistemic modality in terms of alethic rather than epistemic possible worlds, see (Syverson 2003).
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and make sure to avoid them ourselves. And we have examined reasoning that we will use (such as conditioning on a known constant value) to be sure that it withstands criticisms that have been leveled against it. But we are not quite done considering prior analyses. Some of the other big topics that previous authors have stuffed into the two-envelope problem will also be relevant to our own analysis.
3
The Infinite Case
One way in which people have attempted to resolve the paradox is by pointing out that there can be no uniform probability distribution on the reals. Thus, we cannot simply assume a null hypothesis that all pairs of envelopes are equally likely. Of course there is no particular reason we should assume real values, but similarly there is no uniform distribution on the set {{2n , 2n+1 } : n ∈ N}. Various authors have responded that this is too quick a dismissal: there are reasonable distributions such that, given any value in the first envelope, the expected value of the other envelope is greater. For example, Broome (1995) shows this for the above set using the distribution Pr({2n , 2n+1 }) = 2n /3n+1 for all n ∈ N. Continuous distributions are also possible (Broome 1995; Chalmers 1994). We will call attention to two features of all these distributions. First, they all have an infimum. In the above case, the lowest possible value is 1. Broome uses this distribution to demonstrate the paradox (cf., the quotation in the introduction above). He does not explain why his distribution has an inf, nor does Chalmers. Presumably they were focused on showing that there exist wellformed distributions that support a switch and were not concerned with other features of the constructed distribution. It may be that among the distributions that illustrate the intended point such distributions are easy to construct or are most similar to familiar distributions, but we can only speculate. Rawling reiterates Broome’s argument, casting it in terms of dominance (Rawling 1997, p. 267). But, Broome’s argument is too hasty, as is Rawling’s statement that sticking and switching each weakly dominate the other, i.e., there is no gain in switching. For, if the first envelope contains $1, the expected value of switching is $2 any way you slice it. Your original envelope cannot contain 2s, and switching thus strictly dominates sticking (in Rawling’s terms of expected actuarial value, for n = 0 : [eav(X) | s = 2n ] = 1 6= [eav(Y ) | s = 2n ] = 2). Note that this point is not affected by whether or not a player of the game knows what the infimum is; it is enough to know that it exists for switching to strictly dominate sticking.5 Thus, these “paradoxical” distributions do not show any paradox because in such distributions one should always switch envelopes, which is surprising but not paradoxical. Fortunately, this is easily correctible; we can construct a distribution on {{2n , 2n+1 } : n ∈ Z} with no inf. Before giving the distribution we note a 5 In the terminology introduced in Section 5, switching dominates sticking because the game is shut below and open above.
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second feature of all the paradoxical distributions in the literature: they are monotonically decreasing. (As with the occurrence of an inf, none of the authors explains why such a distribution was chosen or if this feature was accidental or intentional.) A player who knows he is being offered pairs from such a distribution thus always knows that the other envelope is less likely to contains a larger amount than a smaller amount because, according to the distribution, he is less likely to have been offered the larger pair than the smaller pair. For example, if Broome’s distribution mentioned above is used, whatever amount the player sees, the odds are 3 to 2 that the other envelope has half that amount versus twice that amount. This does not undermine the switch argument, but it would be nice if we could construct a distribution that is arbitrarily flat and such that a player of the two-envelopes game typically does not know whether the other envelope is more likely to contain a larger or a smaller amount. Assume that there are infinitely many envelopes laid out in size order. 1 1 1 . . . , , , , 1, 2, 4, 8, . . . 8 4 2 Picking a pair of envelopes thus amounts to picking a point between any two. We give the probability of such an assignment for any {2n , 2n+1 } by Pr({2n , 2n+1 }) =
(1 − )|n| 2
And, the total probability is thus ∞ X 1 ·2 (1 − )n = · =1 2 1 − (1 − ) n=0
Here can be any value 0 < < 1. But, we will typically think of it as an arbitrarily small value: the smaller is, the flatter the distribution. This takes care of one desired feature of the distribution. However, when is it paradoxical? That is, when can we support both the switch argument and the whichever argument? The probability of {2n , 2n+1 }, given 2n or of {2n−1 , 2n }, given 2n are both very close to 1/2. Respectively 1− 2− if n ≥ 0 Pr({2n , 2n+1 } | 2n ) = 1 if n < 0 2− and n−1
Pr({2
n
n
,2 } | 2 ) =
1 2−
if n ≥ 0
1− 2−
if n < 0
Expected value of switching for n ≥ 0 is 2n+1 (
1− 1 2n−1 (5 − 4) ) + 2n−1 ( )= 2− 2− 2− 8
which is greater than 2n whenever < 1/2. Expected value of switching for n < 0 is 1 1− 2n−1 (5 − ) ) + 2n−1 ( )= 2− 2− 2− n which is greater than 2 whenever > −1. So, the distribution is “paradoxical”, whenever 0 < < 1/2. Of course, not only can we make the distribution paradoxical, we can also make it as close to uniform as desired: for any chosen , the distribution will be such that the difference between the probability of any two pairs {2n , 2n+1 } and {2m , 2m+1 } is less than /2. Let the center be the point at which the probability transitions from the lower pair being slightly more likely to the higher being slightly more likely. In the above distribution the center is 1. We can easily shift the center an arbitrary amount to the right or left simply by adding to or subtracting from the exponent |n| of the distribution definition. This does nothing to affect whether or not the switch or whichever arguments are justified, nor where the distribution is paradoxical. But, it does provide an intuitive counter to any intuitive sense that whether the probability favors the larger or smaller pair should play a role. For here, especially if is small, the player cannot even be sure which of the two pairs is slightly more likely if she has not been told the center of the distribution. She could ultimately discern the center, given a very large number of plays of the game in which the center does not change from one play to the next. One more thing to note is that it is trivial to generalize the above result from a single chain of powers of 2 to arbitrarily many such chains. We can always multiply the chains by any constant value k. And, it is easy enough to take such a space of chains, and then normalize the probability distribution around the number of disjoint chains. We will have more to say about these and related chains in Section 5, where we will use them to form the basic open sets of a topology. 2n+1 (
4
Open and Shut Finite Games
Suppose that you are just given pairs between some fixed set of values, e.g., 1 and 2048. You choose a pair and open an envelope. You then play the envelopes game according to the following rules. • If not 1 or 2048, choose to switch or stick. • If 1 or 2048, quit and begin again. This game is interesting because it effectively removes the immediate lower and upper bounds. One can play variants where the bounds are effectively hidden, e.g., by having a large range and a filtering agent that just checks if you
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are OK to proceed before showing you the envelope you picked. This will not matter to the analysis that follows. For this game it would always make sense to switch. One way to view this is that the value of the times you might get {1024, 2048} and open the 1024 envelope first, outweigh that of the times you might get {1, 2} and open the 2 envelope first. More generally, the distribution is uniform, and the potential for loss that is greater than the sum of all the potential for gain is removed. Let us call the above game, an open game, since it is in some sense bounded but does not include its bounds. Similarly, let us call the game where boundary values may occur in the first envelope the shut game. As an even simpler example of an open game, suppose that the only possible pairs of envelopes are {1, 2} and {2, 4} from which you chose a pair uniformly, and that the filtering agent only lets you proceed if the envelope you choose is neither the 1 nor the 4 envelope, i.e., when the envelope you choose is the 2 envelope. If you know this is the game, you do not need to even look in the envelope to know that the expected value of switching is 2.5 and the expected value of sticking is 2. More generally, whether or not you even know the range of pairs in the game, if you know the game is open, you do not need to see the value in the envelope passed to you by the filtering agent to know that it is preferable to switch. Though this may be immediately clear, the remaining sections should illuminate why it is so. Now if, as is realistically the case, a player is given that the game being played is on a bounded space such as just described, and he is presented with the standard choice of the two-envelope problem, there is not enough information to know if he should switch envelopes. In the open case the answer is yes. In the shut case, the answer is that it doesn’t matter (unless we know the bounds or possibly can make reasonable guesses about the bounds). We will return to this scenario below. However, we note at this point that we have already shown something that had not been considered in the previous literature, namely a bounded game with finite expected value for which the switch argument is undoubtedly correct. We next explore how to put the intuitive notions we have been discussing on mathematically firmer ground.
5
Topology on the Back of Two Envelopes
We will now consider chains of the type mentioned in Section 3. However, we will not only be considering infinite chains; we also consider chains of finite length. More specifically, we will be concerned with points, all of which are ordered pairs of the form h{k2n , k2n+1 }, k2m i where n ∈ Z and m = n or m = n + 1. We will generally ignore the set from which k is drawn and will limit our primary discussion to the case k = 1. Let X be the set of such points. We can define the obvious lexicographic ordering on points in X, namely, h{2n , 2n+1 }, 2m i < 0 0 0 h{2n , 2n +1 }, 2m i iff n < n0 or n = n0 and m < m0 (i.e., m0 = m + 1). • A chain is any connected set of such points.
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• A chain C is open above if it contains a point c = h{2n , 2n+1 }, 2n i but not the point c0 = h{2n , 2n+1 }, 2n+1 i, or if c ∈ C and c0 > c implies c0 ∈ C. • A chain C is open below if it contains a point c = h{2n , 2n+1 }, 2n+1 i but not the point c0 = h{2n , 2n+1 }, 2n i, or if c ∈ C and c0 < c implies c0 ∈ C. • A set of points is open above if it is the union or finite intersection of chains that are open above. • A set of points is open below if it is the union or finite intersection of chains that are open below. • A chain is open if it is both open above and open below (or empty). It is easy to see that the finite open chains form the basis of a topology on X. The open game of the last section is an open set in this topology (in fact a basic open set). However, the shut game is neither open nor closed in this topology, which is why we called it ‘shut’ rather than ‘closed’. In fact the open sets are the closed sets, i.e., all open sets are clopen. Note that this topology is strictly coarser than the order topology on X. It would be nice if the finite open sets (or open above sets) characterized those finite games for which switching is the right choice. Switching is right for all such sets, but there are others as well. For example, if a finite game contains no chains open below and even one chain that is open above, players of that game should switch. To pick out those finite games for which switching is correct we must consider the set’s border.6 • A point is on the upper border of a set if it is of the form h{2n , 2n+1 }, 2n+1 i and either the sup of a chain in the set or the next larger element above the sup of a chain in the set. • A point is on the lower border of a set if it is of the form h{2n , 2n+1 }, 2n i and either the inf of a chain in the set or the next smaller element below the inf of a chain in the set. • A point is on the border of a set if it is on either the upper border or the lower border. • A set is shut above if it is bounded above and contains its upper border. • A set is shut below if it is bounded below and contains its upper border. • A set is shut if it is shut above and shut below. Let the value of a point p = h{2n , 2n+1 }, 2m i be given by value(p) = 2n . In other words, the value of the lower element in the pair, regardless of whether the lower or higher is chosen first. 6 The border we define is not to be confused with a set’s (topological) boundary. Recall, the boundary of a set is the intersection of the closure of that set with the closure of its complement. Because all open sets in this topology are clopen they all have empty boundary.
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The finite sets for which it is correct to switch are precisely those where the total value of the upper border that is not in the set is greater than the total value of the lower border that is not in the set. Here and elsewhere, we have been assuming a uniform distribution on the finite games. One way to view the paradox is that, for any finite run, the player is always assumed to be playing a game that is shut. The fact that she doesn’t know which shut game she is playing, and that the union of all the possible shut games is open, makes it seem that an open game is being played, hence, giving it that feel of paradox. But, it’s only open if she never stops playing. This seems to support the dependency on infinite value to validate the switch argument. However, we will shortly see strategies that yield openness even after only finitely many plays of this game. Note that the topological account we have given does not mention probability at all. In this respect, it follows Smullyan (1992; 1998) who analyzed the paradox specifically eschewing probabilistic reasoning. His analysis did not support switching per se; it merely showed that what you gain if you gain by switching is more than what you lose if you lose. One can of course attach probabilities to our chains simply by assuming a null hypothesis. Since infinite games have no uniform distribution, we cannot make this move for them. We can instead assume that there is some distribution reasonably close to uniform, which is actually all that hypothesis testing could establish anyway. But, in practical play, we should not need to consider distributions on infinite sets at all. With the topological conceptual machinery established, we now turn to looking more carefully at reasoning about practical plays of the game.
6
Playing for Real
In this section, we attempt to characterize what would happen in any real iterated play of the two-envelopes game. That is, we assume a human player of the game. The game will thus only be played some finite number of times, say L. Suppose that many orders of magnitude more than L pairs of envelopes are generated and used as the set from which pairs are offered to this player. She will not be able to distinguish this case from one in which the pairs she is offered are based on the distribution set out in Section 3 (with a small ). That is, she will not be able to distinguish the case (1) in which she is offered pairs from a very large finite set of pairs that was generated and given a uniform distribution from the case (2) in which she is offered pairs drawn from a very large set that was generated using the distribution of Section 3. Thus, a real player cannot distinguish whether she is playing a finite or an infinite game.7 7 There is valid concern about the amount of money available in the world and expectations about how much could actually be offered in the presented pair (Jackson et al. 1994) and about the representation on finite sized paper or other media of the values in the envelopes. We can assume wrt money that this is just a game. Or, as is sometimes suggested, we can assume that the amount is in the form of a check. Thus, we separate what is won from any ability
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As noted, many authors have concluded that in any finite play of the game, the paradox does not arise because there is no argument to switch. They essentially assume that the game is shut above. It is easy to see that for a uniform distribution on a finite set of pairs of envelopes, the switching argument is ungrounded when the game is shut above. The expected value for all the possible pairs and choices can easily be calculated. And, in the end, the expected value of switching is the same as that of sticking. The above seems to show that this cannot be prima facie correct, unless there were no argument to switch when the game was based on a paradoxical distribution. But there is, so how can this be? One possible answer is that the original distribution is no longer relevant. Even if the above cases are indistinguishable, what matters is the range of envelope pairs she is actually offered, not the set from which they were drawn. Given that range, it seems that we can again conclude that the case is finite and the switching argument is ungrounded. However, we saw in Section 4, that finiteness does not automatically preclude the correctness of the switching argument. To put it in terms of Section 5, in order to not support the switch argument, the game must contain more value on its upper border than on its lower border. We must consider then not just the offered pairs but also the choices made on them. In the game as usually set out, it is simply assumed in effect that an actual (finite) game contains its entire upper border. This appears to be illustrated by the observation recounted in Section 1 that two players following a dualized strategy on presented pairs cannot both be better off switching envelopes. One of them must always be initially choosing the larger value.8 Rawling notes that this does not mean that they cannot both increase the expected payoff by switching. One still will win exactly to the extent the other loses, but “that is the nature of expectation—loss can accompany an increase in expected utility” (Rawling 1997, p. 259). This appears to be a difference between the infinite and finite unshut games. Notice that in the paradoxical distributions given by Broome and others, there is no upper border on the game. And the disconnect between expected value and real advantage does involve an infinite mean to the distribution. In the infinite case, no matter how much you have lost in prior plays of the game, it is possible that you will win much more in a later play. (In this sense it is much like the St. Petersburg paradox.) But recall that if you ever stop playing, then the situation is indistinguishable from a finite shut game on a set bounded by the largest value pair you ever encountered. In principle winnings are unbounded. In practice they are always less. It thus would appear that switching does not dominate sticking in this situation. (A somewhat related point is made by to collect (sort of like stock options). And, we can assume that whatever representational needs are implied by any pair of values actually chosen to offer the player in a single round of the game, the presented envelopes will be large enough to handle values enough beyond those values that the player will not be able to determine anything from the envelopes actually presented. 8 Cf. also the paradox of the neckties (Kraitchik 1943; Nalebuff 1989).
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Chalmers (2002).) However, recall that there are finite and unshut games for which switching does dominate sticking. We began this section with the indistinguishability of certain infinite and finite two-envelope games for real players. Games that are finite and unshut above in virtue of round cancellations might seem poor illustrators of interesting aspects of the two-envelope problem because a player always has a nontrivial probability of distinguishing whether she is playing a game that is shut above or one that is not shut above by choosing a point that cancels a round of play. However, this need not be the case. Consider a game that is just like the ordinary finite, shut two-envelope game except that, with some probability a play will be cancelled and restarted after the player chooses an envelope but before the player is given an opportunity to switch. In this game the expected value of switching is the same as the expected value of sticking. Consider another game just like this, except that if the player chooses a disallowed point on the upper border, play is also cancelled and restarted. The probability of random cancellation in this case can be reduced commensurate with the size of the disallowed upper border. If as in the initial example of this section, we are in a game that has a range larger than anything the player can detect in the lifetime of the game, then there will be no means by which the player can differentiate the shut game from the unshut game. We thus have a finite game, which, if not paradoxical, is one where a player will never know if there is an advantage to switching.
7
Self-opening envelope play
We have shown that it is possible to be offered games (1) such that a real player will never know whether or not it is shut and based on a finite distribution or open and based on an infinite distribution, (2) such that the game is finite but unshut, and (3) such that the game is finite and the player will not be able to tell if the game is shut or not. Still, in order to have an unshut finite game, we had to allow that rounds might be cancelled. This is an explicatory result, and even without notifying players of the possibility of cancellation it is realistic, in that occasional errors and restarts are to be expected in reality. But, requiring cancellation to be part of the game is outside of what we are taking to be the basic two-envelope scenario. It seems that without cancellation or infinity, the switch argument remains ungrounded. We finish with a description of strategies for the ordinary (no-cancellation) finite two-envelope game that can cause play to be open even within a game that is shut. Several authors, beginning with Christensen and Utts (1992) have described strategies like the following: Choose some value r. If the value in the envelope you open is less than r, switch. Otherwise stick. This strategy is advantageous provided that, for some pair you are offered, r falls between the values in the two envelopes. From our perspective, this strategy creates an open-above chain with a bor-
14
der point h{2n , 2n+1 }, 2n+1 i, where 2n+1 is the least 2i > r.9 If r is not on the inside of any set, i.e., if it is outside the range of any offered pair or occurs only on the border of the set of offered pairs, then play of the game cannot be forced open by this strategy. As some have also pointed out, players may have a prior belief or even a specific distribution for the pairs that will be offered. Put in simple practical terms (assuming you accept that such a choice might be offered to you) you are reasonable to think that a smaller pair is more likely than a larger pair and that you will be offered neither an extremely small nor extremely large pair. Different players in different settings will of course vary in what they consider extremely small or extremely large as well as the fall off of expectation as pair value gets larger, but the choice of r will be affected by such prior expectation. One reason that the problem is typically presented with the open envelope containing $100 is partly to have simple round numbers for the calculation (a power of ten to start, and the smallest such that all numbers including expected value are expressible in whole dollars). But in contemporary middle-class western culture, being offered a choice on a pair of envelopes containing $100 and $200 is about as plausible as being offered a choice on a pair containing $50 and $100.10 Were this not so, I suspect the default example would start with $10, or $1, etc. We have been relying on prior expectation to describe opening the game by this strategy. Even without a given expectation, for a repeated game there is a strategy that will always open the game and thus dominates. Assume a single finite uniform but unknown distribution on a chain of pairs of envelopes. A refinement of the above strategy is possible that will not only be advantageous but will guarantee that you never switch on an upper border point: Always switch for amounts at or below half the largest amount you have seen in previous play. Never switch for amounts larger than that. This strategy will guarantee that the expected value of switching for any play on which you choose to switch, is indeed 1.25 that of sticking on that value. As an aside, note that this strategy cannot in general support dualized players, nor can the one with fixed r above, or any switch-if-below strategy (for example, you could always choose to switch for some number of rounds in order to get some information about the distribution and then adopt the switch-ifbelow-half-previous-maximum strategy ever after). Whether players choose the same or different values below which they will switch and above which they will stick, there will be pairs on which one player would want to switch but the other will refuse. It is possible that a strategy like this could be adopted by each of them and the sequence of offered pairs be such that they never are offered a pair wherein one player is inside his switch range but the other is not. However, it becomes increasingly likely such a conflict pair will arise the longer play is continued. And, this is commensurate with the increasing likelyhood that a switch-if-below strategy will be advantageous as the sequence of rounds 9 Some have also noted that this strategy does not require envelopes with amounts in a two to one ratio. It is enough that they differ. Cf., e.g., (Ross 1994; Bruss 1996). 10 My familiarly assumed currency is US dollars. This point actually holds across most, and possibly all, of the currencies denominated by ‘$’ at the time of writing.
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lengthens. If the set of possible pairs is not a chain, then it is possible to have completely dualized play if both players choose a fixed r from the same ‘gap’ in the set of pairs. Not surprisingly, and as indicated above, in this case the strategy will yield no advantage over sticking. If the set of possible pairs is not a chain, the switch-if-below-half-previous-maximum strategy may not be open above and the expected value of switching may not be 1.25 that of sticking for every play; nonetheless, the strategy is unshut above, and it dominates sticking. In particular, it precludes switching for the largest border point in the set. In the basic two-envelope scenario, however, we are given nothing about the set of pairs nor about the distribution on it. And, if play were repeated, the distribution used in one play need not persist. For that matter we do not know whether or not play is one-off or repeated. Nonetheless, there remains a strategy strictly preferable to sticking even in this setting: just pick an r randomly for each play. But from what range? If we rule out any given information or information acquired through learning about the amounts in the envelopes, there is no guide to what may be in them prior to opening the first envelope. Nonetheless, we introduced a distribution in Section 3 that allows us to pick values from all integer powers of two, arbitrarily close to uniformly. A switch-if-below-r strategy where r is chosen based on this or a similar distribution will be no worse than sticking for all pairs of envelopes and will beat sticking for one pair (actually more since we are not now considering assumptions that could limit envelopes to a single chain of powers of two). Though we cannot say anything about distributions on the set of envelope pairs that lead to being offered a particular pair, we can still say that the set of all possible plays is necessarily unshut. This holds whether the set of pairs from which the offered pair was chosen was infinite, finite, or even just the singleton actually offered.
7.1
Self-opening Envelopes with Real Money
What we have just described is a definitive answer to the question of switching or sticking for the basic two-envelope problem scenario inasmuch as it gives a strictly preferable strategy independent of any assumptions about how one arrives to be in the basic scenario. But, as we have already noted, there are some bounds imposed by what is believable and even just the limited amount of money that exists in the world. Once again it is instructive to crank up the level of reality, regardless of the inherent plausibility of the basic scenario. Assuming that you only care about payments that can actually be made, not just the numbers represented on checks in the envelopes, and that amounts must be rounded to denominations of actual currency for payment (assumed to be US currency for the convenience of the author), the least amount you could receive above nothing is one cent. So the least r you should have is one half of a cent. If the envelope you see has less than that, you will always receive nothing if you stick. In addition, there may be an amount below which you should switch because 16
you don’t care: the amount you would lose if you have picked the larger envelope is just too trivial. (Note that no matter where you place the threshold of minimum significant amount, the amount you would gain if you have picked the smaller envelope will always cross the threshold to significance before the amount you would lose if you have picked the larger envelope. This is similar to, but not the same as, Smullyan’s analysis because the utility of switching is only the same or larger in this case.) You cannot lose if the effective utility of your potential loss is zero. Thus the lower bound on your range for r may also be above one half cent. Shifting now to the other end of the scale, for legitimate checks the maximum amount in either envelope cannot be more than that on deposit in some bank somewhere in the world. Here too there may be amounts above which you simply would not care (although the behavior of some of the very rich might serve to counter this as a general claim). Probably well below that, however, are amounts that exceed what you believe you would be offered. If there is a history to the game and if past performance can serve as an indicator of bounds, then one can at least fix an upper bound on the current range from which r is chosen by using the previously noted strategy of setting the upper bound to half of the highest value seen in all previous play. Unlike the case where the range from which offered pairs are selected forms a chain, this cannot be assumed to provide a payoff of 1.25 times sticking, but it will still give a reasonable upper bound to r (again if past plays of the game indicate anything about the range and distribution of pairs in later play). As we have already argued, if there is no distribution that can be placed on the set of possible pairs of envelopes, then choosing an r randomly from some reasonable distribution is strictly preferred. This remains true if the range of pairs is limited to realistic bounds. The only difference is that now the distribution could be really uniform rather than merely practically uniform. While a uniform distribution is possible, this will not be the most advantageous distribution for a given range. A tight upper bound on what could be in the larger envelope might be hard to pin down exactly; however, you can still use a distribution for r that is highest at the lower end of the range and then falls off with your belief. You still retain a necessary advantage in this case by picking a diminishing distribution for r on a range with such reality-imposed bounds. But what if an adversary intentionally chooses a distribution on possible pairs meant to shut (literally) your advantage? He could of course always choose amounts below one half cent. Let us rule out such cases as trivial. Similarly, since we have shifted focus in this section from distributions on infinite possibilities to a more reality oriented setting, let us ignore as trivial the fact that if the first envelope thus contains between one half and one cent, you will know that you should switch. Now however, in order to reduce the probability that your strategy will yield an r that falls between offered-pair values, he will be required to weight his distribution on pairs higher in the range of realistic possibilities. Which means that your payoff will be higher if you choose such a strategy than if you do not. In sum, whether there is a distribution on possible offered pairs or not, and if there is, whether it is known to you or not, the strategy of picking 17
an r based on a diminishing distribution on a range with realistic bounds is strictly preferable to sticking.
8
Conclusion
In this paper we have examined the two-envelope problem. We have shown that one of the arguments in the literature is not supported by the distributions that are usually taken as supporting it. We have constructed a distribution that does support the argument, as well as having other nice features. We have also looked at the problem from a finite perspective and shown that it is not necessary to have an infinite game in order to support the argument to switch. We showed how to view the problem topologically and used our characterization to illustrate some of the aspects of actual finite reasoning that would apply in non-idealized plays of the game.
Acknowledgements I thank Raymond Smullyan for first introducing me to the two-envelope problem and for many lively discussions back in the 1980s. I thank Michael Jackson for reintroducing me to the problem in 1999 and for the ensuing discussions that enticed me to write this paper. For other helpful comments and discussions I thank Iliano Cervesato, Dick Jeffrey, Sjoerd Zwart, and anonymous commentators.
References Albers, C. J., Kooi, B. P., and Schaafsma, W. (2005). Trying to resolve the two-envelope problem. Synthese, 145:89–109. Arntzenius, F. and McCarthy, D. (1997). The two envelope paradox and infinite expectations. Analysis, 57(1):42–50. Binder, D. A. (1993). Comment on Christensen and Utts (1992). The Amercian Statistician, 47:160. Blachman, N. M., Christensen, R., and Utts, J. (1996). Comment on Christensen and Utts (1992). The American Statistician, 50(1):98–99. Brams, S. J. and Kilgour, D. M. (1995). The box problem: to switch or not to switch. Mathematics Magazine, 68:27–34. Brams, S. J. and Kilgour, D. M. (1998). The fallacy of the two envelopes problem. Mathematical Scientist, 23:58–59. Broome, P. (1995). The two-envelope paradox. Analysis, 55(1):6–11. Bruss, F. T. (1996). The fallacy of the two envelopes problem. Mathematical Scientist, 21:112–119. 18
Castell, P. and Batens, D. (1994). The two envelope paradox: The infinite case. Analysis, 54(1):46–69. Chalmers, D. J. (1994). The two-envelope paradox: A complete analysis? Manuscript. HTML available at http://www.u.arizona.edu/~chalmers/ papers/envelope.html (PostScript at envelope.ps). Chalmers, D. J. (2002). The st. petersburg two-envelope paradox. Analysis, 62(2):155–157. Chase, J. (2002). 62(2):157–160.
The non-probabilistic two envelope paradox.
Analysis,
Chihara, C. (1995). The mystery of julius: A paradox in decision theory. Philosophical Studies, 80:1–16. Christensen, R. and Utts, J. (1992). Bayesian resolution of the “exchange paradox”. The American Statistician, 46(4):274–276. Christensen, R. and Utts, J. (1993). Reply to Ridgway (1992). The American Statistician, 47:311. Clark, M. and Shackel, N. (2000). The two-envelope paradox. Mind, 109:415– 442. Horgan, T. (2000). The two-envelope paradox, nonstandard expected utility, and the intensionality of probability. Noˆ us, 34(4):578–603. Jackson, F., Menzies, P., and Oppy, G. (1994). The two envelope ‘paradox’. Analysis, 54(1):43–45. Jeffrey, R. (2004). Subjective Probability: The Real Thing. Cambridge University Press. Kraitchik, M. (1943). Mathematical Recreations. George Allen & Unwin. Linzer, E. (1994). The two-envelope paradox. The American Mathematical Monthly, 101(5):417–419. Malinas, G. (2003). Two envelope problems and the roles of ignorance. Acta Analytica, 18(30/31):217–225. McGrew, T. J., Shier, D., and Silverstein, H. S. (1997). The two-envelope paradox resolved. Analysis, 57(1):28–33. Nalebuff, B. (1989). Puzzles: The other person’s envelope is always greener. The Journal of Economic Perspectives, 3(1):171–181. Norton, J. D. (1998). When the sum of our expectations fails us: The exchange paradox. Pacific Philosophical Quarterly, 79:34–58.
19
Rawling, P. (1994). A note on the two envelopes problem. Theory and Decision, 36:97–102. Rawling, P. (1997). Perspectives on a pair of envelopes. Theory and Decision, 43:253–277. Ridgway, T. (1993). Comment on Christensen and Utts (1992). The American Statistician, 47:311. Ross, S. M. (1994). Comment on Christensen and Utts (1993). The American Statistician, 48:267. Schwitzgebel, E. and Dever, J. (2008). The two envelope paradox and using variables within the expectation formula. Sorites, (20):135–140. Scott, A. D. and Scott, M. (1997). What’s in the two envelope paradox? Analysis, 57(1):34–41. Smullyan, R. (1992). Satan, Cantor, and Infinity. Alfred A. Knopf. Smullyan, R. (1998). The Riddle of Scheherazade: And Other Amazing Puzzles. Harvest Books. Sobel, J. H. (1994). Two envelopes. Theory and Decision, 36:69–96. Syverson, P. (2003). Logic, Convention, and Common Knowledge: A Conventionalist Account of Logic. CSLI Publications. Wagner, C. G. (1999). Misadventures in conditional expectation: The twoenvelope problem. Erkenntnis, 51:233–241.
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