Peremptory Challenges and Jury Selection Francis X Flanagan Wake Forest University [email protected]

Abstract I examine how peremptory challenges, which are vetoes that attorneys may use on prospective jurors, affect jury composition. The purpose of peremptory challenges is to eliminate biased jurors, however I show that under the two most common rules used in the United States peremptory challenges actually increase the probability of juries composed entirely of members on one extreme or another of some ideological spectrum. I then show that it is not possible to design a peremptory challenge procedure which unambiguously makes such juries less likely. I show that if unanimity is required for conviction, the distribution of juror types is symmetric about some mean type, and each attorney has the same number of challenges, then challenges benefit the prosecution.

JEL classification: K40, C44

Keywords: Jury Selection, Peremptory Challenge, Struck Jury, Strike and Replace

1

2

1

Introduction

A criminal defendant’s right to a trial by an impartial jury is one of the foundations of the United States judicial system. This right is guaranteed by the Sixth Amendment of the Constitution, and every US jurisdiction has a series of checks established to remove biased1 jurors. One such check is that courts allow attorneys a limited number of unconditional vetoes, known as peremptory challenges, to be used on prospective jurors. Ostensibly the use of peremptory challenges removes biased jurors by eliminating them from the juror pool, however I show that peremptory challenges can actually increase the probability of juries composed entirely of members on one extreme or another of some ideological spectrum, regardless of the procedure used to implement the challenges. The main observation of the paper is that allowing attorneys to strategically eliminate prospective jurors introduces correlation among the chosen jurors, which means that even if individual jurors with extreme biases are less likely, juries composed mostly or entirely of jurors with similar characteristics (biases) are more likely.

1.1

Peremptory Challenges in Practice

When selecting a jury, each party to a trial is afforded a predetermined number of peremptory challenges which can be used to veto veniremen, the prospective jurors.2 These challenges are “peremptory” in the sense that they do not need to be justified: attorneys can remove whomever they wish, up to their allotted number of challenges.3 The purpose of challenges is to eliminate biased jurors and encourage confidence in the trial by jury. As the Supreme Court of the United States explained: The function of the challenge is not only to eliminate extremes of partiality on both sides, but to assure the parties that the jurors before whom they try the case will decide on the basis of the evidence placed before them, and not otherwise.4 The modern use of peremptory challenges stems from the English common law, although the practice dates at least to Roman law of 104 B.C.5 Currently there are two types of procedures for the execution of 1 Bias

is the lack of ability to reach a verdict solely based on evidence presented at trial. term for an individual prospective juror is venireman, and the entire pool of prospective jurors is the venire. 3 The only limitations are that challenges cannot be used to discriminate based on race or gender, but it is unclear how often these limitations are heeded. See for example Grosso and O’Brien (2012). 4 Swain v. Alabama, 380 U.S. 202 (1965). 5 W. Forsyth, History of Trial by Jury 175 (1852). 2 The

3 peremptory challenges in the Unites States, the most common of which is known as “strike and replace.” The defining characteristic of this procedure is that prospective jurors are presented and challenged sequentially, rather than all at one time: a single venireman, or small group of veniremen, is questioned and then each side is given the opportunity to either accept or challenge. If both sides accept a venireman then she is impaneled, and if either side challenges then the venireman exits and is replaced by a new candidate.6 The other common method for peremptory challenges is known as “struck jury,” which is defined by having all prospective jurors presented and questioned before challenges are used. Therefore in this method at least enough veniremen to fill the jury plus the number of challenges are all questioned, then challenges are executed, and whoever is left fills the jury. If more than the number of jurors needed are left then the jury is chosen randomly from the remaining pool.7 The choice of procedure, as well as the number of challenges allowed to each side, varies widely by jurisdiction and crime, and to some extent is left to the discretion of judges. For example, many jurisdictions do not specify which selection rule should be used, but do impose a minimum number of challenges allowed per side. These minimums are usually an equal number per side, but sometimes more challenges are given to the defense. The number of challenges also generally rises with the severity of the crime. For minor crimes it may be that each side receives three to six challenges, while for capital crimes it is common that each side receives at least twenty challenges.8 The challenge process also varies in how questions are presented to veniremen. In some courts veniremen are interviewed in front of the entire venire, while in others they are interviewed individually in private quarters. In some courts veniremen merely fill out a standardized questionnaire which can be minimally augmented by attorneys. There is also some variation in who asks the questions. In some courts lawyers may lead questioning, while in others the judge leads. In some courts lawyers and judges share the questioning. Most legal experts, even those who are critical of peremptory challenges, accept that these rules are well designed to reduce biases, however the term is defined, in juries. In fact, although peremptory challenges are not Constitutionally mandated, the Supreme Court has acknowledged that “[there is a] widely held belief that peremptory challenge is a necessary part of trial by jury,”9 and that “the right of challenge is almost 6 Cf.

Jeffrey T. Frederick, Mastering Voir Dire and Jury Selection, ABA (2010).

7 Ibid. 8 Cf.

Federal Rule of Criminal Procedure 24(b). v. Alabama, 380 U.S. 202 (1965).

9 Swain

4 essential for the purpose of securing perfect fairness and impartiality in a trial.”10,11 Therefore the view shared by the Supreme Court, lawyers, and scholars, seems to be that peremptory challenges reduce the bias in juries, and the long history of peremptory challenges is seen as testimony to this ability. The present paper seeks to dissect the theory behind this assertion.

1.2

Main Results

I introduce a simple model, explained in Section 3, of jury selection which abstracts away from many real world concerns but allows us to isolate the effects of peremptory challenges. My main results are as follows: • Giving attorneys the ability to challenge individual veniremen necessarily introduces correlation among the chosen jurors, which means that even if individual jurors with extreme biases are less likely, juries composed mostly or entirely of jurors with similar characteristics, including biases, are more likely. I call this a composition bias. • If unanimity is required for conviction, the distribution of juror types is symmetric about some mean, and each attorney has the same number of challenges, then challenges benefit the prosecution. I call a change in the expected conviction rate a conviction bias. I then show that these results are not unique to the peremptory challenge system used in US courts: I show that there does not exist a peremptory challenge rule which would avoid these issues. The remainder of the paper is organized as follows: Section 2 discusses previous literature. Section 3 introduces the general model and contains some illustrative examples. Section 4 discusses alternative selection rules and extensions of the model, such as strategic voting by the jurors. Section 5 concludes. All proofs are in the Appendix. 10 W.

Forsyth, History of Trial by Jury 175 (1852), quoted in Chief Justice Burger’s dissent in Batson v. Kentucky, 476 U.S. 79 (1986). 11 The most common criticism of peremptory challenges is that the process leads to racial or gender discrimination which violates the right to equal protection under the law. Batson v. Kentucky (1986) established the current precedent that race cannot be a factor in challenges, and later gender was also included. The challenge system has also been criticized over the use of jury selection software, which uses demographic information about veniremen and data from previous cases to determine the optimal strategy for lawyers, which apparently by definition violates the Batson criteria. Nevertheless, the existence of such software is evidence that lawyers put effort into forming their strategies, which helps motivate the present paper.

5

2

Related Literature

While there are very few papers which formally model the use of peremptory challenges, one that is closely related to my results is Brams and Davis (1978). Brams and Davis establish the model for the strike and replace system of peremptory challenges which I use in the present paper. However, Brams and Davis do not explore the implications of the model beyond defining the game and presenting equilibrium strategies. I redevelop the equilibrium strategies and I extend the analysis by analyzing how these strategies affect the expected impaneled jury as the primitives of the model change. Roth et al. (1977) and Kadane et al. (1999) use a similar model to Brams and Davis, but focus mostly on a setting in which agents may have different beliefs about the distribution that jurors are drawn from. Degroot and Kadane (1980) generalize the model of both Roth et al. and Brams and Davis to include a larger class of preferences for the players, which is not relevant for the present paper. Another paper which formally models peremptory challenges is Ford (2010), which models the struck jury system rather than strike and replace. Unlike the other literature, this paper does explore the effect that challenges have on jury composition. The author runs simulations of the jury selection process by drawing veniremen from a specified beta distribution. The result of the simulation is that the struck jury system drives the composition of impaneled jurors to the median of the venireman distribution; in other words, a single juror chosen randomly from the set of all impaneled jurors in all juries is closer to the median. While this result is true, especially for games in which the number of challenges is large, Ford does not examine the distribution of the composition of the full jury. One of the main results of the present paper is that it is possible that the expected composition of the jury is pushed away from the mean and median jury relative to a model with no challenges, since the impaneled jurors are correlated. I refine the results of Ford and provide a theoretical background for the findings of the simulations.12 Schwartz and Schwartz (1996) combine a model of jury voting with a model of peremptory challenges to examine how challenges affect the likelihood of a unanimous verdict. However the peremptory challenge aspect of the model is not a formal game, as the authors do not model the strategic behavior of the attorneys. Rather, the authors assume that for any distribution of veniremen the challenge process will truncate the 12 The thesis of Ford is that peremptory challenges distort the composition of the jury by making juries homogeneous and composed of median type jurors, and that this is against the spirit of a representative jury. While the simulations in Ford appear accurate, the author falsely assumes that the order statistics for the realized veniremen are independent, and thus provides an incorrect theoretical explanation of the findings. In a response to Ford, Stevenson (2012) informally argues that peremptory challenges actually increase uncertainty and that this is a desirable quality in jury selection.

6 distribution at the tails. My results show that strategic play by the attorneys does not lead to a truncation of the venireman distribution. However, like Schwartz and Schwartz, my results do show that peremptory challenges make juries more homogeneous. There is also some empirical literature showing the effect that peremptory challenges can have on the composition and conviction rate of the impaneled jury. Zeisel and Diamond (1978) use an experiment to test the effect that challenges have on verdicts. For a limited number of criminal cases in the United Sates District Court for the Northern District of Illinois, the authors were allowed to observe the challenge process and have access to both the challenged veniremen and the impaneled jury. Zeisel and Diamond conclude that some attorneys were clearly better than others at exercising the challenges, and that in at least one case the use of challenges actually changed the verdict delivered. In a more recent and larger study, Anwar et al. (2012) show empirically that the presence of just one black venireman, even if the jury has no black members, can have a significant impact on the conviction rate. The explanation suggested by Anwar et al. (2012) is exactly the formal result in the present paper: a black venireman may be more likely to favor the defense, and due to the correlation of the order statistics of the venire the presence of such a venireman makes it more likely that the jurors who survive challenges also favor the defense. These studies help motivate the present paper, and give support for the claim that peremptory challenges can have a large impact on the composition of juries. Related to the subject of jury selection, there is also a large and growing literature on equilibrium voting by strategic jurors (for example Austen-Smith and Banks (1996), Feddersen and Pesendorfer (1998), Coughlan (2000), Gerardi and Yariv (2007)). For the most part I assume that jurors do not act strategically, however via a simple example I extend the model to include strategic voting and show that the use of peremptory challenges could exacerbate some of the criticism of unanimous voting.13 Many of the results of the present paper would extend to a model with strategic voting, however the results which directly examine the expected conviction rates of impaneled juries clearly rely on the non-strategic assumption. The problem of selecting a jury from a larger population is also related to other institutional design and implementation theory problems such as selecting arbitrators and committee members from some pool of candidates. de Clippel et al. examine the related arbitrator selection problem, which differs from jury selection in that only one arbitrator needs to be chosen, and that the preferences of the players may be aligned. Similarly, the literature on committee selection, which includes Alpern and Gal (2009) and Alpern et al. 13 Cf.

Feddersen and Pesendorfer (1998).

7 (2010), studies models which are closely related to jury selection, with one the major difference that the players may have aligned preferences. Nevertheless, the results of the present paper can be used to examine how changing the pool of eligible arbitrators or committee members and the selection rules may affect the outcomes of such procedures. Lastly, given the structure of the strike and replace game, there is some similarity between this model other optimal stopping problems, such as a labor market search model in the spirit of McCall (1970), Mortensen (1970), and Gronau (1971), or the secretary problem, in which an employer attempts to choose the best secretary from an applicant pool of known size offered sequentially; cf. Degroot (1970), Ferguson (1989), Bearden (2006). Section 3.3 discusses strike and replace and briefly comments on some of the similarities and differences among these models.

3

Model of Existing Peremptory Challenge Rules

3.1

General Environment

The two strategic players are the defense, D, and the prosecution, P, who are involved in selecting a jury of n > 1 jurors, { j1 , . . . , jn }, according to some selection rule, R. Veniremen, the potential jurors, each have a type, ti , which indicates the venireman’s probability of voting for conviction. Venireman types are independent and distributed according to a known, non-degenerate distribution F with support [t, t] ⊆ [0, 1]. F is conditional on the particular trial, thus it can vary with guilt, strength of physical evidence, credibility of witnesses, et cetera.14 Therefore I abstract away from these concerns, and formally defining bias,15 in order to isolate the effects of peremptory challenges on the jury composition. The defense would like the jury to acquit, and the prosecution would like the jury to convict. A unanimous verdict is required for conviction, and otherwise the result is acquittal; in real-world courts unanimity is also required for acquittal, however hung juries are often viewed as a victory for the defense, therefore for simplicity I assume that anything but conviction is acquittal.16 Thus for a jury J = { j1 , . . . , jn }, the payoffs to the defense, UD , and to the 14 This

is motivated by the practice of a pre-trial discovery, during which each side of a case must share the evidence which can be presented in trial. The implicit assumption is that both attorneys know the information the jury will receive during trial, and the attorneys’ posterior beliefs about the conviction probability of each juror given this information are the same. 15 The interpretation here is that bias will be captured by the probability of conviction, and that this probability is the payoff relevant statistic for the attorneys; ceteris paribus, relative to another juror, a juror is more biased in favor of the prosecution if her type is higher. 16 Empirical evidence from California courts shows that only about one third of cases resulting in hung juries are retried, and it is very rare for a case to go through more than two trials. The majority of hung jury cases result in plea bargains at reduced sentences or are dismissed (P. Hannaford-Agor, V.P. Hans, N.L. Mott, G.T. Munsterman, Are Hung Juries a Problem?, Natl. Cent. State

8 prosecution, UP , are defined as: UD = 1 − ∏in=1 ti , UP = ∏in=1 ti . D and P each have a finite number of peremptory challenges, d and p respectively, which, subject to the selection rule, can be used to veto veniremen. A game Γ is described by p, d, n, F, and R; Γ = Γ( p, d, n, F, R).17 Assuming the game has a unique equilibrium, the conviction rate is the probability that the jury votes for conviction given equilibrium play. If F and R are fixed, I write the conviction rate as V ( p, d, n). Let µ be the mean of the distribution F, so that a randomly chosen jury of size n would have a conviction rate V (0, 0, n) = µn . I say the game favors P if V ( p, d, n) > µn , and the game favors D if V ( p, d, n) < µn .

3.2

A Static Procedure: Struck Jury

This section formally states the model and results for the struck jury selection rule, which I denote SJ. The rule is as follows: n + p + d veniremen are drawn and all types are revealed before challenges are used. P and D then simultaneously challenge p and d veniremen, respectively, and the remainder are impaneled.18 The unique equilibrium of Γ( p, d, n, F, SJ ) is for P to challenge the p veniremen with the lowest types, and for D to challenge the d veniremen with the highest types.19 Illustrative Examples Suppose there are two types of jurors, { L, H }, L < H, each venireman has probability

1 2

of being either

type, and a jury is composed of two members. If there were no peremptory challenges, and juries were drawn randomly from the population, then there would be three possible outcomes for the composition of the impaneled jury, with the distribution shown in Table 1. Half of the impaneled juries would contain both types of jurors, and half would be what I call an extreme jury, containing only one type of juror. However, when each attorney has one peremptory challenge the venire must consist of four members: two to be challenged and two to sit on the jury. This results in the venire distribution shown in Table 2. Courts, Natl. Inst. Justice, 2002). 17 The players and their payoff functions will be held constant throughout the analysis. I often refer simply to Γ when it is clear that I am speaking of a specific game. 18 Given the information structure of this game the assumption of simultaneous challenges is without loss of generality. The assumption that all challenges are used is also without loss of generality, since if one side does not use all of its challenges then all remaining veniremen are of the same type. 19 To see why this is true in general, suppose D has some positive number of challenges, d, but does not use them on the d highest value veniremen that P is not challenging. Then the conviction rate of the jury would be strictly lower, and minimized conditional on P’s challenges, if D changes strategies and challenges the highest d veniremen not being challenged by P. The analogous argument shows that P would always challenge the lowest p veniremen that D does not challenge.

9

Table 1: Jury distribution with no challenges Impaneled Jury Composition Percentage of Impaneled Juries

LL 25%

LH 50%

HH 25%

Table 2: Venire distribution with one challenge per side Venire Composition Percentage of Venires

LLLL 6.25%

LLLH 25%

LLHH 37.5%

LHHH 25%

HHHH 6.25%

Since the defense will challenge the highest valued venireman, and the prosecution will challenge the lowest valued venireman, the only way to impanel a jury of one L and one H juror is if there are two of each type in the venire. Therefore, with one challenge per side, extreme juries become more likely than when there are no challenges; the distribution of impaneled juries is shown in Table 3. Table 3: Jury distribution with one challenge per side Impaneled Jury Composition Percentage of Impaneled Juries

LL 31.25%

LH 37.5%

HH 31.25%

This effect of pushing mass towards the tails of the jury distribution is generally true for any distribution of veniremen, as long as the number of challenges are low relative to the size of the jury, as will be discussed later. I will call this change in the expected composition of the jury a composition bias. To see that this distortion will generally be true notice that we can reinterpret the previous example as drawing veniremen from a continuous population, with type L and H veniremen representing those below and above the median type, respectively. It is immediate that the resulting impaneled juries would more often contain only jurors from one side of the median, relative to a game with no challenges. In fact, this is true for any distribution with a median, regardless of symmetry or continuity. Thus it is always the case that peremptory challenges cause some correlation among the impaneled jury members. The reason this correlation occurs is that the peremptory challenges are not eliminating the tails of the potential juror distribution, but rather they are eliminating the tails of a sample from that overall distribution. Despite the fact that each individual venireman is drawn independently from the population, the order statistics of the resulting venire are not independent. For example, suppose potential jurors are

10 distributed uniformly on [0, 1] and we draw four jurors, t1 < t2 < t3 < t4 . Given t1 , the expected values of the remaining three veniremen are uniformly distributed between t1 and 1; all three of these expected values are increasing in t1 . Thus even if an attorney can eliminate the most extreme venireman in a sample, in expectation the remaining veniremen will still be correlated. Peremptory challenges will also affect the expected conviction rate. For instance, suppose in the first example that type L and H jurors correspond to conviction rates of 1/4 and 3/4, respectively. Then the mean type would be 1/2 and the expected conviction rate of a randomly drawn two person jury would be 1/4. However, after allowing one challenge to each attorney, the expected conviction rate becomes 17/64 > 1/4. This is because, despite the fact that the new jury distribution is symmetric, there is asymmetry inherent in the expected conviction rate. Since the conviction rate is the product of all of the juror types it is convex, which makes it more sensitive to the extreme juries composed of type H jurors relative to those composed of type L. In general I will call any change in the expected conviction rate, increase or decrease, a conviction bias. To see the two biases in a more complex example, let there be three types of prospective jurors,

{ L, M, H } = { 41 , 21 , 43 }, with each venireman having probability 1/3 of being any specific type. Figure 1 reflects both types of biases through the empirical distribution functions of the impaneled jury conviction rate for three different combinations of challenges.20 It is clear that for the highest value juries, the V(p,d,12) ECDFs 1 0.8 0.6 0.4 V (0, 0, 12) V (2, 2, 12) V (6, 6, 12)

0.2 0

2

6

10 14 Value of Jury

18 ·10−4

Figure 1: Empirical CDFs for V ( p, d, 12), Struck Jury 20 The

data is from a simulation of 100,000 juries generated from the distribution used in the example.

11 distributions generated when agents have more challenges lie below the distributions when there are fewer challenges: higher value juries are more likely when challenges are present. Although it cannot be seen in Figure 1 due to the scale, a somewhat similar effect is present for the low value juries: more challenges increases the probability of low value juries. Allowing challenges is pushing some mass towards both tails of the jury distribution. However, these effects do not have an equal impact on the conviction rate, and thus the conviction rate will change as the number of challenges changes. In this example, when each side receives the same number of challenges, the conviction rate will again be biased in favor of the prosecution.

3.2.1

Composition Bias Under Struck Jury

Recall that the unique equilibrium of the the SJ game is for P to challenge the p lowest veniremen, and for D to challenge the d highest. For the distributions in the examples we saw that this equilibrium moved weight into the tails of the impaneled jury distribution. Here I show that for large n, this is universally true, however we define the “tails.” T HEOREM 1. If D is given at least one challenge, then for any interval [t, t + e] ⊂ [t, t), there exists an n0 such that for all juries larger than n0 , the probability that all impaneled jury members fall in the interval

[t, t + e] is greater in the game Γ( p, d, n, F, SJ ) than in a game with no challenges. Likewise, if P is given at least one challenge, then for any interval [t − e, t] ⊂ (t, t], there exists an n0 such that for all juries larger than n0 , the probability that all impaneled jury members fall in the interval [t − e, t] is greater in the game Γ( p, d, n, F, SJ ) than in a game with no challenges. Thus for any interval at either endpoint of the juror distribution, if the size of the jury is large enough the presence of challenges push enough mass towards the tails of the jury distribution that it becomes more likely that all impaneled jurors come from this interval. It is also immediate that for m < n we can find an n0 such that it is more likely that at least m jurors fall in the targeted interval. For intervals that have very small measure the jury may have to be quite large before this happens, but it is also possible to fix the size of the jury and solve for the minimal measure for which it is more likely that all of the jurors fall in the extreme intervals.21 T HEOREM 2. If D has at least one challenge then there exists an x ∈ (0, 1) such that if the measure of the 21 For

the remainder of this section I will restrict attention to the case of impaneled jurors being in the interval [t, t + e], with the understanding that the analogous results hold for the case of all impaneled jurors being in the interval [t − e, t].

12 interval [t, t + e] ⊂ [t, t) is at least x, it is more likely that all jurors fall in the interval [t, t + e] in the game Γ( p, d, n, F, SJ ) than in a game with no challenges. C OROLLARY 1. If each player receives the same number of challenges, then it is always more likely that the impaneled jury is composed entirely of jurors above the median, or entirely of jurors below the median, in game Γ( p, d, n, F, SJ ) than in a game with no challenges. Thus, given a fixed jury size and number of challenges, for an arbitrary distribution it is possible to calculate the probability that all jurors fall in some extreme upper or lower quantile of the distribution. Corollary 1 shows that the shift in probability towards more extreme juries can occur in small juries. In fact, this effect can be quite drastic. For example, in Florida all juries in non-capital felony cases are composed of six members and each side is given six peremptory challenges.22 If no peremptory challenges were given, the fraction of juries composed entirely of jurors above the median type is approximately 1.6%. However, when each side has six challenges, if the selection rule is struck jury, then this fraction increases to approximately 11.9% of all impaneled juries. Thus, in such courts, we would expect that almost a quarter of all juries are composed entirely of individuals on one side of the median juror. There is also empirical evidence that challenges can have large effects on the composition of six member juries. Diamond et al. (2009) use voire dire data for juried civil trials in the First Municipal District of the Circuit Court of Cook County, Illinois. The jury pool population was 25% black, which would lead to 17.8% of juries containing no black jurors if the juries were chosen randomly. However the authors show that 28.1% of the six member juries had zero black jurors.23 This is despite the fact that Diamond et al. find that there was no discernible difference in the impaneled jury population after challenges: overall black and white24 veniremen were challenged at rates equal to their population shares, but black veniremen were challenged more often by the prosecution, while white veniremen were challenged more often by the defense. This suggests that peremptory challenges were in fact pushing some mass of the jury distribution towards the tails of the distribution.25 The following section shows that such a change in the composition of the jury can also affect the conviction rate. 22 Florida

Rules of Criminal Procedure, Rule 3.270, Rule 3.350.(a)(2). were 89 six member juries, of which 25 had zero black members. Given the demographics, the probability of this event occurring randomly is 1.1%. 24 “White” was the largest demographic, accounting for 63.4% of the venire. 25 Diamond et al. (2009) do not include data for juries with a black majority. 23 There

13

3.2.2

Conviction Bias Under Struck Jury

As mentioned previously, the unique equilibrium of this game is for P to challenge the p lowest veniremen, and for D to challenge the d highest. Therefore the conviction rate of the game is equal to the expected product of the p + 1 to p + n order statistics of the venire. L EMMA 1. For the venire J 0 with | J 0 | = p + d + n, let t(k) denote the kth order statistic of the sample, so that t(1) ≤ t(2) ≤ . . . ≤ t( p+d+n) . Then the conviction rate of the game Γ( p, d, n, F, SJ ) is: " V ( p, d, n) = E

#

p+n

∏

t(k)

k = p +1

And if F is absolutely continuous this value can be expressed as:

V ( p, d, n) =

(n + p + d)! p!d!

n

Z

∏ (t p+k ) F p (t p+1 )(1 − F(t p+n ))d dF(t p+1 )dF(t p+2 ) . . . dF(t p+n )

(1)

W k =1

Where W = {(t p+1 , . . . , t p+n ) : 0 ≤ t p+1 ≤ . . . ≤ t p+n ≤ 1}. The conviction rate given by (1) merely uses the well known expression for the joint pdf of order statistics drawn from an absolutely continuous distribution.26,27 Although this conviction rate is simple to describe, it is not immediately transparent how changing the number of challenges or the size of the jury will affect its value for general distributions. However if agents are distributed uniformly on [0, 1] then the expression simplifies into a form which makes it easy to perform comparative statics on p, d, and n. L EMMA 2. For F = U [0, 1], Γ( p, d, n, F, SJ ),

VΓ =

(n + p + d)! n ( p + 2k − 1) (2n + p + d)! k∏ =1

(2)

This result, which is due to a more general result of David and Johnson (1954), makes analysis of the struck jury method very tractable when veniremen are distributed uniformly, which can be used as a 26 Cf.

David and Nagaraja (2003). this expression the F p (t p+1 ) inside the integral accounts for the probability of having p veniremen below the middle p + 1 to p + n veniremen, while the (1 − F (t p+n ))d accounts for the probability of having d veniremen above this range. The fraction 27 In

(n+ p+d)!

outside of the integral, p!d! draws from the distribution.

, accounts for the number of ways the veniremen can be split into these three ranges given n + p + d

14 reference to examine how this selection rule is affected by changes in the primitives. The main result of this section is that under SJ, if veniremen are distributed uniformly on [0, 1], then if each player has the same number of challenges the game always favors the prosecution. T HEOREM 3. For Γ( p, d, n, F, SJ ), if p = d > 0 and F = U [0, 1], then Γ favors P. This result is due to the same phenomenon as in the example in Section 3.2: allowing challenges pushes some of the mass in the jury distribution towards the tails, and this benefits the prosecution because of the asymmetry of the payoffs. It turns out that because of the symmetry of the distribution, this happens immediately. This is the same insight that suggests that peremptory challenges under a struck jury rule may increase the probability of juries which are composed entirely of relatively extreme jurors. Compared to a game with no peremptory challenges, Theorem 1 and Theorem 2 show that under a struck jury rule the relative likelihood of extreme juries increases with the size of the jury. This suggests that if we hold n fixed but increase challenges we may be able to produce the opposite effect. The following theorem shows that indeed it is possible to reduce the probability of extreme juries if there are enough challenges awarded to each side. T HEOREM 4. If R = SJ and F (c) =

p d

is constant for all n, then lim V ( p, d, n) = cn , where c is such that p→∞

p p+d .

This result supports the common view that peremptory challenges may eliminate biased veniremen. However the counter effect described in Theorem 2 is often ignored, and Corollary 1 shows that this force can be significant: it may take a long time before the limit described in Theorem 4 is reached. Figure 2 shows the conviction rate for a twelve member jury when F = U [0, 1] as the number of challenges remains equal for both sides but increases. Note that in practice the maximum number of challenges allowed is generally below thirty per side. This figure illustrates that the conviction rate does not change monotonically with the number of challenges, and significantly reducing the variance of the conviction rate can take more challenges than is realistically implementable. There is also empirical evidence that peremptory challenges can affect the conviction rate. As mentioned, in a sample of twelve criminal cases Zeisel and Diamond (1978) find at least one in which the use of peremptory challenges seems to have changed the verdict, and in a larger sample Anwar et al. (2012) show that the presence of just one black member in the venire, not necessarily on the jury, can have a significant

15

7

·10−4 V ( x, x, 12) V (0, 0, 12)

6

Conviction Rate

5

4

3

2

1

0

10

20

30 40 Challenges

50

60

70

Figure 2: Conviction Rates for F = U [0, 1] impact on the expected conviction rate: in felony cases in two Florida counties between 2000 and 2010, the presence of one black venireman caused the conviction rate to fall to 73% from 81% for black defendants, and rise to 74% from 66% for white defendants. One interpretation of this result is that a black venireman represents the realization of a low order statistic for black defendants, and a high order statistic for white defendants, and thus the respective impaneled juries are more likely to contain lower and higher types. These results show that the struck jury rule may have the perverse effect of increasing the probability that juries are composed of entirely of more extreme jurors. Furthermore, although the results are dependent on the number of challenges relative to the size of the jury, it appears quite reasonable that the number of challenges used in US courts shift the probabilities of extreme juries in undesirable ways. The following section shows that the sequential, strike and replace rule leads to similar results.

3.3

A Sequential Procedure: Strike and Replace

In this section I develop the model of the sequential, strike and replace game. This rule, denoted SR, is as follows: an individual venireman j1 is drawn and her type t1 is revealed. Each agent, P and D, simultaneously chooses to accept or challenge.28 If both agents accept then j1 joins the jury. If either P or D challenges then j1 does not join the jury. After this round the next venireman is drawn, and the process continues. The selection ends when either the jury reaches its capacity, or when both sides run out of challenges, in which case the remaining jurors are chosen randomly. 28 I

will show that this timing is without loss of generality.

16 Suppose we reach round k, in which P and D have p0 and d0 challenges left, there are n0 places to fill on the jury, and currently there are m jurors impaneled. If we let the product of the types of the impaneled juries be π = ∏im=1 ti , then the decision for agent P to use his challenge against jk is equivalent to choosing between the payoffs πV ( p0 − 1, d0 , n0 ) and πtk V ( p0 , d0 , n0 − 1). Therefore the values of the jurors already impaneled do not affect P’s decision. Thus the relevant information to determine players’ strategies is the number of challenges they each have, p0 and d0 , and the number of jurors left to impanel, n0 . I will refer to this as point ( p0 , d0 , n0 ) in the game. Using this insight it is possible to solve the game for any distribution using backward induction.

Characterizing the Equilibrium L EMMA 3. A subgame-perfect equilibrium for the game Γ( p, d, n, F, SR) exists and the unique value of the conviction rate is V ( p, d, n), defined recursively as follows for p, d, n > 0:

V ( p − 1, d − 1, 0) ≡ 1

(3)

V (0, 0, n) ≡ µn

(4) V ( p−1,0,n) V ( p,0,n−1)

Z

!

V ( p, 0, n) = V ( p, 0, n − 1)

µ+

V (0, d, n) = V (0, d, n − 1)

V (0, d − 1, n) − V (0, d, n − 1)

Z

V ( p, d, n) = V ( p, d, n − 1)

V ( p, d − 1, n) − V ( p, d, n − 1)

Z

t

F ( x )dx V (0,d−1,n) V (0,d,n−1)

t V ( p,d−1,n) V ( p,d,n−1) V ( p−1,d,n) V ( p,d,n−1)

(5) ! F ( x )dx

(6) !

F ( x )dx

(7)

Identity (3) is used to make the full characterization well defined, and (4) is the conviction rate of a randomly drawn jury of n members. When there are empty spots on the jury we can see that a player with challenges has a unique threshold strategy determined by the ratios of the conviction rates of the two subgames he can choose to enter. P’s threshold at point ( p, d, n) will sometimes be denoted t P ( p, d, n), and likewise D’s threshold will be t D ( p, d, n). For example, as mentioned earlier, P will accept juror jk at point

( p, d, n) if V ( p − 1, d, n) < tk V ( p, d, n − 1). This relationship is represented in the lower limit of the integral in (7). Likewise, D’s threshold strategy is represented by the upper limit of this integral. One way to interpret (7) is that adding an extra juror while holding the number of challenges constant will change the conviction rate by the factor within the parentheses. The first term within the parentheses,

V ( p,d−1,n) , V ( p,d,n−1)

17 is D’s threshold strategy, which is the maximum value of any juror who would be accepted. Thus we can see that D can keep the expected value of the accepted juror below this threshold as long as d ≥ 1, but depending on the distribution and the values of p, d, and n this term may be greater than or less than the mean type. From expression (5) it is apparent that when P is the only player with challenges, adding an extra member to the jury will affect the conviction rate by a factor larger than the mean: being the only player with challenges allows P to shift the conviction rate in his favor. It is less transparent, but (6) shows a similar effect when D has the only challenges: adding one more juror affects the conviction rate by a factor less than the mean type. In general though, the value V ( p, d, n) requires the computation of the conviction rates of ( p + 1)(d + 1)n subgames,29 and it is not always immediately clear how changing the size of the jury or the number of challenges affects this rate. However, as we increase the size of the jury, in the limit the analysis of the game becomes much simpler. T HEOREM 5. Fix the number of challenges each side receives and increase the size of the jury. If t = 0 and   p  d µ V ( p,d,n) µ V ( p,d,n) p > 0 then limn→∞ µn = ∞, otherwise limn→∞ µn = t . t Theorem 5 shows the impact that each player’s challenges can have on the expected conviction rate. In µ

the limit, every challenge that P receives scales up the conviction rate by a factor of t , while every challenge µ

D receives scales down the conviction rate by a factor of t . Holding the mean fixed, increasing the maximum of the support actually helps D because it gives D’s challenges more influence over the conviction rate, and likewise P benefits from having a lower minimum of the support. However D’s challenge can never scale down the conviction rate by more than a factor of µ, and it can only approach this limit if the maximum of the support is t = 1. Yet as the minimum of the support decreases there is no limit to the impact that P’s challenge can have on the conviction rate. This asymmetry comes from the requirement that anything except unanimous conviction is acquittal. This result is of particular relevance in cases where the range of juror types is largest. It is well documented that in capital cases in which the death penalty may be imposed potential jurors often enter the venire with existing beliefs either in favor of or against the death penalty, and that these beliefs can have a large and significant impact on the verdict.30,31 There are also numerous studies showing that the crime committed, as well as the demographics of the defendant, victim, and jurors, can interact to lead to changes in probability 29 This

is the number of subgames in which the jury is not full, including those in which both agents have zero challenges. Devine et al. (2001). 31 Another source of conviction bias in these cases is that jurors often must be “death-qualified,” meaning all veniremen who oppose the death penalty are removed for cause. 30 Cf.

18 of conviction, length of sentences, and more.32 These results show that certain types of cases may lead to a larger range of juror types than others. Theorem 5 suggests that changing the number of challenges awarded to each side may have a large impact on the conviction rate in such cases. The proof of Theorem 5 also shows that as the size of the jury increases, D’s threshold strategy monotonically approaches t, the maximum of the support of the venireman distribution. Similarly, P’s threshold monotonically decreases towards t. T HEOREM 6. Under strike and replace, as n increases, t D ( p, d, n) is strictly increasing and approaching t, and t P ( p, d, n) is strictly decreasing and approaching t. As the size of the jury grows players have a stronger incentive to save their challenges for a more extreme venireman, since the probability of drawing an extreme type increases with the number of jurors.33 Thus as the number of jurors grows, eventually it is very unlikely that either player will use a challenge on the first venireman presented, which means adding an extra member to the jury will scale the conviction rate by a factor close to µ. Interestingly though, the convexity of the conviction rate and the asymmetry of the payoffs cause these threshold strategies to change at different rates even for symmetric distributions. This also means that under symmetric distributions, if each player starts with the same number of challenges, then the game always favors the prosecution, which is formalized in the following section. Challenges Favor the Prosecution T HEOREM 7. If F is symmetric and p = d > 0, then Γ( p, d, n, F, SR) favors P. C OROLLARY 2. For symmetric distributions and x > 0, at point ( x, x, n) D accepts a larger set of veniremen than P; µ − t P ( x, x, n) < t D ( x, x, n) − µ. With a symmetric distribution and an equal number of challenges, the asymmetry of the payoffs forces D to be less aggressive than P with his challenges. This means that, conditional on neither player challenging, the expected value of an impaneled juror at point ( x, x, n) is above the mean type. This has the effect of shifting the conviction rate higher than if there were no challenges. It is also important to note that the symmetry is defined over the true support of F, [t, t], not over the possible support [0, 1]. Thus these results 32 Cf.

Mustard (2001), Mitchell et al. (2005). is similar to the dynamics of the cardinal secretary problem, a variant of a classic optimal stopping game. However in the cardinal secretary problem the actual types of the secretaries (veniremen) presented are not known to the employer (attorney), only whether or not the current candidate is the best one so far. Cf. Ferguson (1989), Bearden (2006). 33 This

19 do not restrict the mean to be 1/2, but show that any time there is a symmetric distribution around some type in the population, the strike and replace rule favors the prosecution if each attorney is given the same number of challenges.34 As noted, peremptory challenge procedures vary widely across jurisdictions. While it is often required that an equal number of challenges are awarded to each side, in some cases more challenges are given to the defense.35 The results of this section suggest that, under certain conditions, the latter procedure results in a more equal influence over the conviction rate by each side. However it is unclear when the juror distribution will be symmetric, although it is often the case in the literature that baseline models of peremptory challenges assume symmetric distributions.36

Lemma 3 and the following results make it possible to do precise comparative statics on p, d, and n, however it may also be of interest to do comparative statics on the distribution of veniremen. It is possible that changes in laws or judicial practice affect the juror pool, and different shifts in the distribution may have varying effects on the conviction rate.37 It is possible to show that in general a distribution which first order stochastically dominates another will be better for the prosecution. This result is unsurprising, but we may also want to know how an increase in variability of jurors affects the value of the game. As the following example shows, a second order stochastic shift in the form of a mean preserving spread has an ambiguous effect on the conviction rate. E XAMPLE 1. Let U be the uniform distribution on [0, 1], and let F be a mean preserving spread of U created by taking all of the mass on the interval [ 58 − e, 85 ], e < 18 , and placing it on the point

5 8

− e, and taking all

of the mass on the interval [1 − e, 1] and placing it on 1, as in Figure 3. The mean is unchanged, µ F =

1 2,

and the expected value of a venireman conditional on being above

or below the mean is also unchanged: E[t|t > 12 , F ] = E[t|t > 12 , U ], E[t|t < 12 , F ] = E[t|t < 12 , U ]. Therefore in both games we have that V (1, 0, 1) =

5 8

and V (0, 1, 1) =

3 8.

This implies that at point

(1, 1, 1) the threshold strategies for each player remain the same, with D challenging if t > 34 Also,

5 8,

and P

since this is a strict change in the conviction rate, the result would also hold for “almost” symmetric distributions. that in all non-capital criminal cases in Florida each side is awarded 6 peremptory challenges. In federal court each side receives 20 challenges in a capital case and 3 in a misdemeanor case, however more challenges are awarded to the defense in non-capital felony cases (Federal Rules of Criminal Procedure Rule 24(b).) 36 Cf. Brams and Davis (1978), Schwartz and Schwartz (1996), Anwar et al. (2012). 37 As mentioned, in capital cases juries often must be “death-qualified.” This may result in a shift to a distribution which first order stochastically dominates the general population. Also, since the distribution is conditional on all information available at trial, a change in admissible evidence could have a predictable impact on the distribution. 35 Recall

20 1

A

A

0

0

3 8

1 2

5 8

1

Figure 3: F, Mean Preserving Spread of U challenging if t < 83 . The probabilities of reaching subgames Γ(1, 0, F, SR) and Γ(0, 1, F, SR) also remain the same as the game with the uniform distribution, since U ( 38 ) = F ( 38 ) and U ( 85 ) = F ( 58 ). However, when the distribution is F the expected value of an unchallenged veniremen at point (1, 1, 1) decreases: E[t| 38 < t < 58 , F ] < E[t| 83 < t < 85 , U ]. Therefore the overall value of the game decreases, and P is worse off. An analogous example would show that a mean preserving spread can hurt D in the same manner. This example highlights how this model differs form a one-sided employment search model. In such a search model increased variance in the wage distribution would always weakly help the searcher,38 but since this game is constant sum it is of course not possible to help both players. Once one player has used all of his challenges then the game does resemble a more traditional search model, and increased juror variance does weakly help the player who has challenges remaining. Yet when both players have challenges it is possible to prevent the opposition from exploiting the shift in the distribution, and thus the overall effect on the value of the game is ambiguous for general distributions and general shifts. Nevertheless, the characterizations in this section provide a framework to precisely examine the strike and replace method for any set of primitives. 38 Cf.

Mortensen (1986).

21

4 4.1

Extensions Impossibility of a Perfect Selection Rule

The results so far have not ruled out the existence of an optimal peremptory challenge rule. We may think it is possible to design a rule which unambiguously reduces the probability of more extreme juries; but we would be wrong. It turns out that there is no rule which allows challenges of individual veniremen which avoids these problems. To motivate this result I will define a selection rule R to be unbiased if it does not favor either side. T HEOREM 8. There does not exist a rule which is unbiased for every distribution F. The proof of this theorem uses the fact that if there are only two possible types of veniremen, then any selection rule will result in the same impaneled jury. When this is the case the results from the previous sections carry through exactly to any other selection rule. This result is quite intuitive, and it highlights the limitations inherent in a peremptory challenge system which attempts to control the composition of the jury by allowing agents to veto individual veniremen. Peremptory challenges give each player some influence over the impaneled jurors, which removes the statistical independence of the jury members, regardless of the distribution or the specific selection rule. Even if the distribution of veniremen were known to a central planner it may not be possible to design a practical challenge procedure which reduces the variance of the expected jury.

4.2

Improving Peremptory Challenges

The question of how to improve peremptory challenges may not have a satisfying answer. As Theorem 1 and Theorem 8 show, there may be no way to allow challenges without significantly altering the expected composition of the jury. However this result is partly due to the assumption that challenges attempt to preserve the mean jury. If society wants to minimize jury variance while targeting the median jury, or some point between the mean and the median, then it is possible to design a system which assists this goal. One such solution would be if challenges were used to veto potential juries, rather than individual jurors. The selection rule could be as follows: P and D are given p and d peremptory challenges, respectively, and n + k jurors are drawn from F and interviewed. A set of J = { J1 , J2 , . . . , Jm } of m ≥ p + d + 1 juries is produced via random draws of n jurors from the set of veniremen. Attorneys can then use challenges on the

22 resulting juries, as in a struck jury selection rule. As k and m grow, the expected impaneled jury approaches the median jury. The appeal of this is that the median jury is not necessarily a homogenous jury; it may be composed of a wide range of jurors.39 This is also appealing since for the same size venire it requires no more information than the struck jury method; once attorneys have interviewed the entire venire it would take minimum effort to rank juries. Nevertheless, in order to improve the peremptory challenge system society must have a well defined notion of what it desires in a jury.

4.3

Strategic Voting

Feddersen and Pesendorfer (1998) famously criticize unanimous voting as a requirement for conviction, pointing out that in equilibrium each juror will rationally condition her vote on the event of it being pivotal, since under unanimity a vote affects a juror’s payoff only if all other jurors are also voting to convict. This conditioning bounds the probability of false conviction strictly away from zero, even though the requirement of unanimity would seem to protect defendants from such an outcome. Although fully incorporating a model of strategic voting for general selection rules is outside the scope of the present paper, I will show that my model of the struck jury selection rule extends quite naturally to the setting of Feddersen and Pesendorfer. Under the struck jury rule, since all information about veniremen is revealed to players before challenges are executed, even with strategic jurors the equilibrium strategies will remain the same. I will illustrate this point using a very simple transformation of the model. Instead of t representing a juror’s likelihood of conviction, let it denote 1 − q, where q is what Feddersen and Pesendorfer call a juror’s threshold of reasonable doubt. This number represents the threshold for the probability of guilt at which the juror is willing to vote for conviction. A juror ji ’s utility is given by u( A| I ) = u(C | G ) = 0, u( A| G ) = −(1 − qi ), and u(C | I ) = −qi , where A and C stand for acquit and convict, and G and I stand for guilty and innocent; thus each juror maximizes her utility by voting correctly. During the trial, each juror receives a signal as to whether the defendant is innocent or guilty. Each defendant is equally likely to truly be guilty or innocent, and the signal is correct with known probability s ∈ (.5, 1]. Feddersen and Pesendorfer show that in this game there is no Nash equilibrium in which all jurors vote to convict when the signal is guilty and acquit when the signal is innocent. Jurors must play a mixed strategy when the signal is innocent, which leads to a lower bound on the probability that a convicted defendant is 39 For example, for the distribution in the leading example with three types, the median two member jury falls between J = { L, H } and J = { M, M}.

23   1− q 0 innocent of min .5, 1+q0 2s−1 , where q0 = maxi {qi }. (1− s )2

In this model with strategic voting, P would want to minimize the maximum qi on the jury, and likewise D would want to maximize the minimum. Thus for the struck jury rule this does not alter the strategies of the attorneys: P still wishes to challenge the p lowest types, and D wishes to challenge the d highest. Therefore the shift in jury composition discussed in Section 3.2 would still manifest, which may exacerbate the problems associated with unanimous voting rules in this model. For example, it is possible that the expected lower bound of false conviction increases when peremptory challenges are allowed. Suppose T = {t L , t H }, t L < t H , and there is probability x that a venireman is of type t H .40 The probability that a jury is composed entirely of type t H jurors is given p and d peremptory challenges is: p

Pr (| L| ≤ p| p + d + n veniremen) =

∑

k =0



 p+d+n (1 − x ) k x p + d + n − k k

Whereas in a random selection the probability of a jury composed entirely of type t H jurors is x n . The ratio of the likelihood of an extreme jury when challenges are allowed relative to when there are no challenges  p  p+d+n is given by ∑ (1 − x )k x p+d−k . This ratio is unbounded and monotonically increasing as n k k =0 grows, thus there is some n0 such that for all n > n0 , an all t H jury is more likely when p and d challenges are allowed. This means that the expected lower bound of the false conviction rate will rise for a large enough jury. For example, if x =

1 2

and s = 34 , then if we give each side one challenge, p = d = 1, the

expected lower bound of the false conviction rate relative to the expected lower bound from a game with no challenges is

2n +2 + 6 ( n + 3 ) , 2n+2 +24

which is greater than one for n > 1.41

Clearly this is a stylized example, but it emphasizes the point that the composition of the jury will affect the outcome of the jury’s vote, and allowing attorneys to alter that composition can have significant effects. This also acts as somewhat of a robustness check for the more general results of the present paper; regardless of whether jurors vote strategically, the strategies for P and D under a struck jury rule would always be the same, and therefore the distortions that peremptory challenges cause would remain. 40 Given Lemma 4 all selection rules are equivalent for this distribution, thus this example also works for a strike and replace rule.

However, in general the strategies under strike and replace would be affected by strategic voting. 41 The lower bounds implied by the game both with and without challenges are less than 1 , and therefore are the correct bounds. 2

24

5

Conclusion

In this paper I present a simple model of jury selection which explores whether or not peremptory challenges, under ideal conditions, can decrease the probability of biased juries without affecting the expected conviction rate. The answer is that peremptory challenge procedures may fail both criteria. By allowing players to strategically remove veniremen without reason, impaneled jurors are no longer statistically independent, and thus mass is pushed towards the tails of the distribution of impaneled juries. The effects of peremptory challenges can be quite large, and the distortions are robust to a variety of selection rules and a variety of interpretations of juror types. These results suggest that the common perception that challenges unambiguously limit the prevalence of extreme juries is incorrect, and that more thought should be given as to how the larger society should be represented via an “impartial” jury.

Appendix: Proofs Proof of Lemma 2. The proof is by direct computation given Lemma 1. Fix p, d, n and let F = U [0, 1]. Then, by Lemma 1, we have

V ( x, x, n) =

(n + p + d)! p!d!

Z

n

∏ (t p+k )(t p+1 ) p (1 − t p+n )d dt p+1 dt p+2 . . . dt p+n

W k =1

Direct integration of this expression is possible after a change of variables. Let t p+n = yn , t p+n−1 = yn−1 yn , . . ., t p+1 = ∏nk=1 yk . Then the Jacobian matrix for the substitution function is:       J=     

∏ k 6 =1 y k

∏ k 6 =2 y k

∏ k 6 =3 y k

···

0

∏k6=1,2 yk

∏k6=1,3 yk

···

0 .. .

0 .. .

0

0

∏k6=1,2,3 yk · · · .. .. . . 0

···

∏k6=n yk



  y ∏k6=1,n k    ∏k6=1,2,n yk    ..   .  1

25 So the Jacobian determinant is | J | = ∏nk=2 ykk−1 . Substituting out the t’s and noting that the yk ’s are independent42 gives us:

( n + p + d ) ! n −1 VΓ = ∏ p!d! k =1 =

Z 1 p+2k −1 0

yk

dyk

Z 1 2n+ p−1 0

yn

(1 − yn )d dyn

(n + p + d)! n ( p + 2k − 1) (2n + p + d)! k∏ =1

This comes from integrating all of the terms and realizing that the last integration term is a beta function.

 Proof of Theorem 1. I will prove the first part of the theorem for d > 0; the proof of the second statement is analogous. Let the measure of the interval [t, t + e] be x ∈ (0, 1). Therefore in a game with no challenges the probability that all n jurors fall in the interval is x n . When there are challenges and the selection rule is SJ, then P will challenge the p lowest type veniremen. Therefore in order for there to be n impaneled jurors in the interval, there must be at least n + p veniremen in that interval from the n + p + d veniremen drawn. Equivalently, there can be at most d veniremen outside of the interval. The probability of this occurring is: d

Pr (at most d veniremen fall outside of [t, t + e]) =

∑

k =0



 n+p+d (1 − x ) k x n + p + d − k k

We want to know when this is greater than the probability under a random draw: d



 n+p+d (1 − x ) k x n + p + d − k > x n ∑ k k =0  d  n+p+d (1 − x ) k x p + d − k > 1 ⇐⇒ ∑ k k =0 For fixed p and d, with d > 0, the lefthand side of the last inequality is unbounded and monotonically increasing as n grows, which proves the theorem.

 42 For

k = 1, . . . , n − 1, the limits of integration for t p+k were 0 and t p+k+1 , which under the change of variables gives the limits of integration for yk as 0 and 1, since t p+k = yk t p+k+1 .

26 Proof of Theorem 2. For fixed p, d, and n, define the polynomial P( x ) as follows: d

P( x ) =

∑

k =0



 n+p+d (1 − x ) k x n + p + d − k − x n k

We want to know when P( x ) > 0. For any p, d, and n we have that P(1) = 0, since all elements in the summation with k > 0 are equal to zero at x = 1. The derivative of P( x ) at x = 1 is: ∂ ∑dk=0 (n+ kp+d)(1 − x )k x n+ p+d−k − x n ∂P( x ) = ∂x 1 ∂x 1  = (n + p + d) x n+ p+d−1 + (n + p + d)((n + p + d − 1)(1 − x ) x n+ p+d−2 ! n+ p+d d k x n+ p+d−k − x n 1 − x ) ∂ ( )( ∑ k = 2 k − x n+ p+d−1 ) − nx n−1 + ∂x 1

= (n + p + d) − (n + p + d) − n = −n The continuity of P( x ) together with the facts P(1) = 0 and P0 (1) < 0 are sufficient to complete the proof.

 Proof of Theorem 3. The proof will be by induction on n. Let c = p = d and n = 2. Then the conviction rate is given by:

V (c, c, 2) =

(2 + 2c)! 2 (c + 1)(c + 3) 1 (c + 2k − 1) = > ∏ (4 + 2c)! k=1 (4 + 2c)(3 + 2c) 4

Now assume that for Γ(c, c, n, U, SJ ), VΓ > 2−n . Increasing the size of the jury to n + 1 changes the expected value by a factor of

(n + 1 + 2c)(2n + 1 + c) 1 > (2n + 2 + 2c)(2n + 1 + 2c) 2 Thus the value of the game Γ(c, c, n + 1, U, SJ ) is greater than 2−(n+1) .

 Proof of Theorem 4. The set of veniremen J is the empirical distribution of F with n + p + d draws; denote this Fn+ p+d . Thus as n + p + d → ∞, Fn+ p+d → F, and

n n+ p+d

→ 0. For any fixed game Γ( p, d, n, F, SJ ),

27 P challenges the p veniremen with the lowest types and D challenges the d veniremen with the highest types. Therefore, in the limit the measure of the set of veniremen that P challenges is value of each impaneled juror is c, where F (c) =

p p+d ,

and thus the

p p+d .

 Proof of Lemma 3. First I will derive equation (5), which is the value of the game when P is the only agent with challenges. Take the game Γ(1, 0, 1, F, SR). When the venireman j1 is drawn, the optimal strategy is for P to accept if t1 > µ and to challenge otherwise.43,44 Thus the unique value of V (1, 0, 1) is given by the following expression:

Rt µ

V (1, 0, 1) = F (µ) µ + (1 − F (µ))

= µ+

Z µ t

xdF ( x )

1 − F (µ)

F ( x )dx

(8)

Line (8) comes from integrating by parts and the relation µ = t −

Rt t

F ( x )dx. If we define V ( p, d, 0) and

V (0, 0, n) as in the statement of the theorem, then (8) is consistent with equation (5). Now, for p ≥ 1 assume that V ( p − 1, 0, 1) is well defined according to (5). Then in the game Γ( p, 0, 1, F, SR), P wishes to challenge j1 if t1 < V ( p − 1, 0, 1). This gives us a similar equation to when P has one challenge, and we R V ( p−1,0,1) see that V ( p, 0, 1) = µ + t F ( x )dx, which is also consistent with (5). The same argument shows that V ( p, 0, 2) is well defined by (5). Therefore, if we assume that V ( p − 1, 0, n) and V ( p, 0, n − 1) are well defined, then we get, letting a = V ( p − 1, 0, n) and b = V ( p, 0, n − 1), V ( p, 0, n) is given by the following expression:  V ( p, 0, n) = F

V ( p − 1, 0, n) V ( p, 0, n − 1)



= V ( p, 0, n − 1) µ +

V ( p − 1, 0, n) + V ( p, 0, n − 1) V ( p−1,0,n) V ( p,0,n−1)

Z t

Z t V ( p−1,0,n) V ( p,0,n−1)

xdF ( x )

! F ( x )dx

A similar argument shows that in the game Γ(0, 1, 1, F, SR), D’s optimal strategy is to challenge j1 if t1 > µ. Rµ Thus V (0, 1, 1) = µ − t F ( x )dx. Therefore V (0, 1, 1) is consistent with the definition in the statement of faces a discrete choice between t1 and E[t2 | challenge] = µ. Thus P maximizes his payoff by accepting when t1 > µ and challenging otherwise. 44 For continuous distributions this strategy is unique up to a set of measure zero. For discrete distributions there may be positive probability that t1 = µ, but the conviction rate of the game is unaffected by assuming P challenges j1 when he is indifferent. 43 P

28 the theorem. The analogous induction arguments as for V ( p, 0, n) show that V (0, d, n) is well defined. Lastly I will derive equation (7). The derivation is almost analogous to the previous steps, except now there is a possible interaction between the strategies of the two players. Begin with the game Γ =

(1, 1, 1, F, SR). For now assume that P and D do not challenge the same venireman. Then P challenges Rµ Rµ j1 if t1 < V (0, 1, 1) = µ − t F ( x )dx, and D challenges j1 if t1 > V (1, 0, 1) = µ + t F ( x )dx. This confirms the assumption that P and D do not challenge the same venireman in this game. Thus there are unique threshold strategies for P and D, and given these strategies it is possible to define V (1, 1, 1) via the expectation of reaching the subgames V (0, 1, 1), V (1, 0, 1), or t1 V (1, 1, 0). Induction on p,d and n produces the general expression (7).

 Proof of Theorem 5. The proof of Theorem 5 is broken into three steps: first for p > 0, d = 0, then d > 0, p = 0, and lastly p, d > 0. Each step uses an induction argument. S TEP 1. First I will prove that as n grows, if t > 0 then

V ( p,0,n) µn

→

 p µ t

, and if t = 0 then

V ( p,0,n) µn

→ ∞.

From Lemma 3, V (1, 0, n) can be expressed as follows:

V (1, 0, n) = V (1, 0, n − 1)

µ+

µn V (1,0,n−1)

Z t

! F ( x )dx

(9)

This recursive representation in turn implies that the upper limit of the integral above can be written as n −1 µn =µ∏ V (1, 0, n − 1) k =1

µ µ+

R

µk V (1,0,k −1)

t

(10) F ( x )dx

This shows that the limit of the integral in (9) is decreasing as n grows, since each element of the product term is less than 1. It is also the case that the ratio is bounded below by t, and therefore it converges to some limit L; I want to show that L = t.45 For the purpose of contradiction, suppose that L > t. This would imply that: n −1 µn µ ≤ lim µ ∏ =0 RL n→∞ V (1, 0, n − 1) n→∞ F ( x )dx i =1 µ +

lim

0

45 The lower bound is t because V (1, 0, 1) > µ and (9) imply that V (1, 0, n ) > µV (1, 0, n − 1) for all n > 0. If L < t then for some finite n we would have that V (1, 0, n) = µV (1, 0, n − 1), a contradiction.

29 However, for t > 0 this contradicts the assumption that L > t, thus we have that limn→∞

µn V (1,0,n−1)

= t.

Rearranging this term and iterating the expression forward one step gives us the desired result for p = 1. Now for the inductive step assume that

V ( p−1,0,n) V ( p,0,n−1)

→ t for some p ≥ 1. By Lemma 3 we have that:

R V ( p−1,0,n) µ + t V ( p,0,n−1) F ( x )dx V ( p, 0, n) V ( p, 0, n − 1) = · R V ( p,0,n−1) V ( p + 1, 0, n − 1) V ( p + 1, 0, n − 2) µ + t V ( p+1,0,n−2) F ( x )dx By assumption,

V ( p−1,0,n) V ( p,0,n−1)

→ t, therefore:

Rt R V ( p−1,0,n) µ + t F ( x )dx µ + t V ( p,0,n−1) F ( x )dx lim = lim −1) n→∞ n→∞ R VV( p(+p,0,n R V ( p,0,n−1) 1,0,n−2) µ+ t F ( x )dx µ + t V ( p+1,0,n−2) F ( x )dx µ

= lim

n→∞

µ+

R

V ( p,0,n−1) V ( p+1,0,n−2)

t

(11) F ( x )dx

Once again, since (11) is decreasing and bounded below by t, the limit L exists. Similar to the previous argument, if L < 1, meaning the upper limit of the integral in (11) is bounded away from t, then V ( p,0,n) V ( p+1,0,n−1)

→ 0, contradicting the lower bound of t. Therefore,

V ( p, 0, n) =t n→∞ V ( p + 1, 0, n − 1) lim

(12)

Together, (12) and Lemma 3 show that in the limit, adding one more juror will affect the conviction rate by a factor of µ, since P is unlikely to challenge: limn→∞

V ( p,0,n) V ( p,0,n−1)

= µ. Along with (12), this implies µ

that giving one more challenge to P affects the conviction rate by a factor of t : limn→∞ Iterating iterating this expression from p = 0 generates the desired result: limn→∞ Taking the limit of t to zero shows that, if t = 0, then limn→∞

V ( p,0,n) µn

V (0,d,n) µn

that for d = 1 we have:

V (0, 1, n) = V (0, 1, n − 1)

µn V (0,1,n−1)

Z t

V ( p,0,n) V (0,0,n)

µ

= t  p = µt .

= ∞.

S TEP 2. Next I will prove the theorem for p = 0. That is, as n grows,

µn − V (0, 1, n − 1)

V ( p+1,0,n−1) V ( p,0,n−1)

! F ( x )dx

→

 d µ t

. Lemma 3 shows

30 The term in the parentheses above is strictly less than µ. To see this note that this is true for n = 1, which by µn

induction implies that V (0, 1, n) < µV (0, 1, n − 1) for all n > 0. This in turn implies that V (0,1,n−1) < t, Rt Rt µn since for any t < t, we have that t − 0 F ( x )dx < t − 0 F ( x )dx = µ. Therefore V (0,1,n−1) is bounded above by t. From Lemma 3 we can rewrite this ratio as n −1 µn =µ∏ V (0, 1, n − 1) k =1

µ µk V (0,1,k−1)

−

R

µk V (0,1,k−1)

t

(13) F ( x )dx

This is increasing as n grows, since each element of the product term is strictly greater then one. Thus, since it is bounded above by t, the limit exists. If this limit were strictly less than t then the ratio would be unbounded, which is a contradiction, therefore limn→∞

V (0, 1, n − 1) = lim lim n→∞ n→∞ µn

V (0, 1, n − 2)

µn V (0,1,n−1)

µ n −1 V (0,1,n−2)

−

R

= t. This also shows that, µ n −1 V (0,1,n−2)

t

! F ( x )dx

µn

V (0, 1, n − 2) n→∞ µn

= µ lim

(14)

The analogous argument as in the previous step generates the desired result: limn→∞ inductive step assume that for some d ≥ 1,

V (0,d−1,n) V (0,d,n−1)

V (0,1,n) µn

= µt . For the

→ t. Then we have:



V (0,d−1,n) V (0,d,n−1)

 F ( x ) dx V (0, d, n) t  V (0, d, n − 1)  = lim  · lim  V (0,d,n−1) n→∞ n→∞ V (0, d + 1, n − 1) R V (0,d+1,n−2) V (0, d + 1, n − 2) V (0,d,n−1) − t F ( x )dx V (0,d+1,n−2)   Rt t − F ( x ) dx   V (0, d, n − 1) t = lim  ·  V (0,d,n−1) n→∞ R V (0,d+1,n−2) V (0, d + 1, n − 2) V (0,d,n−1) − t F ( x )dx V (0,d+1,n−2)   V (0,d−1,n) V (0,d,n−1)

 V (0, d, n − 1) = lim  · n→∞ V (0, d + 1, n − 2)

Once again, since

lim

n→∞

V (0,d,n) V (0,d+1,n−1)

−

R

µ V (0,d,n−1) V (0,d+1,n−2)

−

R

V (0,d,n−1) V (0,d+1,n−2)

t

< t for all n and is increasing with n, it must be that

V (0, d, n − 1) − V (0, d + 1, n − 2)

V (0,d,n−1) V (0,d+1,n−2)

Z t

! F ( x )dx

= µ,

  F ( x )dx

31 otherwise

V (0,d,n) V (0,d+1,n−1)

would be unbounded. This implies that limn→∞

V (0,d,n−1) V (0,d+1,n−2)

= t. Once again we

have that in the limit adding an extra jury member changes the conviction rate by a factor of µ, and giving µ

D one more challenge changes the conviction rate by a factor of t . Iterating this implies the desired result:  d µ V (0,d,n) limn→∞ µn = t .

S TEP 3. Lastly I must prove the result for p, d > 0. From Lemma 3, V (1, 1, n) can be represented as

V (1, 1, n) = V (1, 1, n − 1)

Assume that limn→∞

V (1,1,n) µn

V (1, 0, n) − V (1, 1, n − 1)

exists. The limits of

Z

V (1,0,n) V (1,1,n−1) V (0,1,n) V (1,1,n−1)

V (1,0,n) V (1,1,n−1)

! F ( x )dx

and

V (0,1,n) , V (1,1,n−1)

which are the endpoints of

the integral, are respectively bounded above by t and below by t,46 thus if t > 0,

lim

n→∞

V (1, 1, n − 1) V (1, 0, n) V (1, 0, n) 1 µ ≥ lim ≤ t ⇒ lim = · n→∞ n→∞ V (1, 1, n − 1) µn tµn t t

And, V (0, 1, n) 1 µ V (0, 1, n) V (1, 1, n − 1) ≤ lim = · ≥ t ⇒ lim n n n→∞ n→∞ V (1, 1, n − 1) n→∞ µ tµ t t lim

These together imply limn→∞

V (1,1,n−1) µn

=

µ , tt

which shows that limn→∞

V (1,1,n) µn

exists and is equal to

µ2 . tt

Now for the inductive step assume that for some p ≥ 1 we have that, V ( p, 1, n) lim = n→∞ µn

 p   µ µ t t

Then through an analogous squeeze argument as in the previous stage we can see that,   p +1   p +1 µ 1 V ( p + 1, 1, n − 1) µ 1 ≤ lim ≤ n n→∞ t µ t t t 46 If this were not true then we would have, for example, lim V (1,0,n)

V (1,0,n) n→∞ V (1,1,n−1)

which would in turn imply that limn→∞ V (1,1,n−1) = 0, a contradiction. t.

V (0,1,n) > t, V (1,1,n−1) V (0,1,n) The analogous argument shows that limn→∞ V (1,1,n−1) ≥

> t. This would imply that limn→∞

32

Therefore the limit exists and is equal to

  p +1 µ t

1 . t

The analogous induction can be done on d to show the

desired result that for any 0 < p, d < ∞, if t > 0, then, V ( p, d, n) = lim n→∞ µn

 p   µ µ d t t

Taking the limit of t to zero gives the final claim that the limit is unbounded if this is the case.

 Proof of Theorem 7. First I will show that the theorem is true for c = p = d = 1 and n = 2, and the proof will continue via induction. The following facts about a symmetric distribution about µ will be useful: 1. F ( x ) = 1 − F (2µ − x ) + Pr ( x ) 2. E[ x | x < y] = 2µ − E[ x | x > 2µ − y] for y ∈ (t, t) Fix any symmetric distribution F and let the selection rule be SR. From Lemma 3 we have that:

V (1, 0, 1) = µ + V (0, 1, 1) = µ −

Z µ t

Z µ t

F ( x )dx F ( x )dx

I claim that V (1, 0, 2) − µ2 > µ2 − V (0, 1, 2). To see this first note that P’s threshold for challenging at point (1, 0, 2) is t P (1, 0, 2) = µ−

2 R µµ . F ( x )dx t

t∗ = 2µ − µ2

2 R µµ , µ+ t F ( x )dx

while D’s threshold at point (0, 1, 2) is t D (0, 1, 2) =

Suppose instead that at point (1, 0, 2), P adopts the suboptimal strategy of vetoing if t < 2 R µµ , µ− t F ( x )dx

thus t∗ = 2µ − t D (0, 1, 2). Then F (t∗ ) = 1 − F (2µ − t∗ ) + Pr (t∗ ) = 1 − µ2

F ( µ−R µ F(x)dx ) + Pr ( µ−R µ F(x)dx ); ex ante, P and D are equally likely to challenge the venireman t. Let t

t

V ∗ (1, 0, 2) denote the conviction rate of the game when P uses the suboptimal cutoff. Then the values V ∗ (1, 0, 2) and V (0, 1, 2) are given by: ∗

∗

∗

∗

V (1, 0, 2) = F (t )µ + (1 − F (t ))E[t|t > t ](µ + 2

Z µ t

F ( x )dx )

V (0, 1, 2) = F (t∗ )µ2 + (1 − F (t∗ ))E[t|t < 2µ − t∗ ](µ −

Z µ t

F ( x )dx )

33 Now I will show that V ∗ (1, 0, 2) − µ2 > µ2 − V (0, 1, 2), or equivalently: E[t|t > t∗ ](µ +

Rµ t

F ( x )dx ) + E[t|t < 2µ − t∗ ](µ −

Rµ t

F ( x )dx )

2

> µ2

Since F is symmetric, I can let b = E[t|t > t∗ ] − µ = µ − E[t|t < 2µ − t∗ ], and a =

(15)

Rµ t

F ( x )dx. Then

(15) becomes

(µ + b)(µ + a) + (µ − b)(µ − a) > µ2 ⇐⇒ µ2 + ab > µ2 2 The last inequality is true since a, b > 0. This proves the original claim that V (1, 0, 2) − µ2 > µ2 − V (0, 1, 2). This result implies that V (1, 1, 2) > µ2 . The logic for this is the same: P could adopt a suboptimal strategy in which his threshold for challenging the venireman at the point (1, 1, 2) is t = 1 − t D (1, 1, 2), but play optimally at all other points. Then there would be equal probability of reaching points

(1, 0, 2) and (0, 1, 2). With the remaining probability both sides would accept the veniremen, who would have expectation µ since the cutoff strategies are symmetric, and we would reach point (1, 1, 1). For the inductive step on p and d, assume that for some c > 1, V (c − 1, c − 1, 2) > µ2 . By the symmetry of the game with only one juror, it must be that V (c, c − 1, 1) and V (c − 1, c, 1) are equidistant from µ, and thus the analogous argument as in the previous step implies that V (c, c, 2) > µ2 . All that is left is to prove the induction on n. This argument is again analogous to the original step. Starting with game Γ(1, 1, 3, F ), we can show that V (1, 0, 3) − µ3 > µ3 − V (0, 1, 3): P can guarantee this by adopting the “symmetric” strategy as D’s optimal strategy, and the convexity of the functions leads to the result. Therefore V (c, c, n) > µn for all c, n > 1.

 Proof of Theorem 8. The proof is by counterexample. I propose a simple family of distributions for which all selection rules are equivalent, and then show that one of these selection rules is biased. L EMMA 4. If T = {tl , t H }, t L < t H , all selection rules for which P always wishes to challenge t and D always wishes to challenge t result in the same value of the game. Proof. Take any ordered set of veniremen, J 0 = { j1 , . . . , j p+d+n }. Under any rule R the unique optimal strategy for D is to challenge any ji if ti = t H and d > 0, and not to challenge otherwise. Likewise, it is

34 the unique optimal strategy for P to challenge any ji if ti = t L and p > 0, and not to challenge otherwise. Therefore, if l is the number of veniremen of type t L in J 0 , and h is the number of type t H , then under any l− p

rule, if l − p ≥ 0 and h − d ≥ 0, then the value of the game is V ( p, d, n) = t L thH−d , and if l − p < 0 or h − d < 0, then the value of the game is V ( p, d, n) = tnH or V ( p, d, n) = tnL , respectively. Next I will show that under a rule of struck jury, as long as at least one side has a challenge, there exists a distribution F such that SJ is biased, which, given Lemma 4, completes the proof. First let veniremen be distributed uniformly on T = {0, 1}. For this distribution µ = 12 , but for any draw of jurors the expected probability of conviction is only positive if all jurors are of type t H = 1. The value of the game when there are no challenges is µn = 2−n , and under a rule of struck jury, the value when there are p > 0, d ≥ 0, challenges is: p

V ( p, d, n) = Pr (|t L | ≤ p) =

∑



k =0

 n + p + d −(n+ p+d) 2 k

This rule is unbiased only if V ( p, d, n) = 2−n , which is satisfied if and only if: p

∑

k =0



 n + p + d −(n+ p+d) 2 = 2− n k  p  n+p+d ⇐⇒ ∑ = 2 p+d k k =0

(16)

For fixed p and d the righthand side of (16) is constant, while the lefthand side is unbounded and monotonically increasing as n → ∞, thus there can be at most one value of n for which the rule is unbiased. If the game happens to be unbiased for some p, d, n combination, then there exists e ∈ (0, 21 ) such that perturbing F so that there is

1 2

− e mass on 0 and

1 2

+ e mass on 1 makes the rule biased.47 An analogous argument

proves the theorem for d > 0, p ≥ 0.

 p

47 The

conviction rate of the game, V ( p, d, n) =

∑

k =0





n+p+d 1 1 ( − e)(n+ p+d−k) ( + e)k , and the expected value of k 2 2

a random draw, µn = ( 12 + e)n , are different polynomials as a function of e, and thus can have only finitely many points of intersection on the interval (0, 12 ).

35

References Alpern, S. and Gal, S. (2009). Analysis and design of selection committees: A game theoretic secretary problem. International Journal of Game Theory, 38(3):377 – 394. Alpern, S., Gal, S., and Solan, E. (2010). A sequential selection game with vetoes. Games and Economic Behavior, 68(1):1 – 14. Anwar, S., Bayer, P., and Hjalmarsson, R. (2012). The impact of jury race in criminal trials. Quarterly Journal of Economics, 127(2):1017 – 1055. Austen-Smith, D. and Banks, J. S. (1996). Information aggregation, rationality, and the condorcet jury theorem. The American Political Science Review, 90(1):pp. 34–45. Bearden, J. N. (2006). A new secretary problem with rank-based selection and cardinal payoffs. Journal of Mathematical Psychology, 50(1):58 – 59. Brams, S. J. and Davis, M. D. (1978). Optimal jury selection: A game-theoretic model for the exercise of peremptory challenges. Operations Research, 26(6):pp. 966–991. Coughlan, P. J. (2000). In defense of unanimous jury verdicts: Mistrials, communication, and strategic voting. The American Political Science Review, 94(2):pp. 375–393. David, F. N. and Johnson, N. L. (1954). Statistical treatment of censored data part i. fundamental formulae. Biometrika, 41(1/2):pp. 228–240. David, H. A. and Nagaraja, H. N. (2003). Order Statistics. John Wiley. de Clippel, G., Eliaz, K., and Knight, B. (forthcoming). On the selection of arbitrators. The American Economic Review. Degroot, M. H. (1970). Optimal Statisical Decisions. Degroot, M. H. and Kadane, J. B. (1980). Optimal challenges for selection. Operations Research, 28(4):pp. 952–968. Devine, D. J., Clayton, L. D., Dunford, B. B., Seying, R., and Pryce, J. (2001). Jury decision making: 45 years of empirical research on deliberating groups. Psychology, Public Policy, and Law, 7(3):622 – 727.

36 Diamond, S. S., Peery, D., Dolan, F. J., and Dolan, E. (2009). Achieving diversity on the jury: Jury size and the peremptory challenge. Journal of Empirical Legal Studies, 6(3):425 – 449. Feddersen, T. and Pesendorfer, W. (1998). Convicting the innocent: The inferiority of unanimous jury verdicts under strategic voting. American Political Science Review, 92(1):23 – 35. Ferguson, T. S. (1989). Who solved the secretary problem? Statistical Science, 4(3):pp. 282–289. Ford, R. A. (2010). Modeling the effects of peremptory challenges on jury selection and jury verdicts. George Mason Law Review, 17(2):377–422. Gerardi, D. and Yariv, L. (2007). Deliberative voting. Journal of Economic Theory, 134(1):317 – 338. Gronau, R. (1971). Information and frictional unemployment. The American Economic Review, 61(3):pp. 290–301. Grosso, C. M. and O’Brien, B. (2012). A stubborn legacy: The overwhelming importance of race in jury selection in 173 post-batson north carolina capital trials. Iowa Law Review, 97(5):1531 – 1559. Kadane, J. B., Stone, C. A., and Wallstrom, G. (1999). The donation paradox for peremptory challenges. Theory and Decision, 47(2):139–155. McCall, J. J. (1970). Economics of information and job search. The Quarterly Journal of Economics, 84(1):pp. 113–126. Mitchell, T. L., Haw, R. M., Pfeifer, J. E., and Meissner, C. A. (2005). Racial bias in mock juror decision-making: A meta-analytic review of defendant treatment. Law and Human Behavior, 29(6):pp. 621–637. Mortensen, D. T. (1970). A Theory of Wage and Employment Dynamics. Mortensen, D. T. (1986). Job search and labor market analysis. volume 2 of Handbook of Labor Economics, chapter 15, pages 849 – 919. Elsevier. Mustard, D. B. (2001). Racial, ethnic, and gender disparities in sentencing: Evidence from the u.s. federal courts. Journal of Law and Economics, 44(1):pp. 285–314.

37 Roth, A., Kadane, J. B., and Degroot, M. H. (1977). Optimal peremptory challenges in trials by juries: A bilateral sequential process. Operations Research, 25(6):pp. 901–919. Schwartz, E. P. and Schwartz, W. F. (1996). The challenge of peremptory challenges. Journal of Law, Economics, & Organization, 12(2):pp. 325–360. Stevenson, D. (2012). The function of uncertainty within jury systems. George Mason Law Review, 19(2):513–548. Zeisel, H. and Diamond, S. S. (1978). The effect of peremptory challenges on jury and verdict: An experiment in a federal district court. Stanford Law Review, 30(3):pp. 491–531.