Likelihood of voting outcomes with generalized IAC probabilities

Accepted Manuscript Likelihood of voting outcomes with generalized IAC probabilities Tomas J. McIntee, Donald G. Saari PII: DOI: Reference:

S0165-4896(17)30029-X http://dx.doi.org/10.1016/j.mathsocsci.2017.01.003 MATSOC 1921

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Mathematical Social Sciences

Received date: 13 January 2016 Revised date: 18 January 2017 Accepted date: 20 January 2017 Please cite this article as: McIntee, T.J., Saari, D.G., Likelihood of voting outcomes with generalized IAC probabilities. Mathematical Social Sciences (2017), http://dx.doi.org/10.1016/j.mathsocsci.2017.01.003 This is a PDF file of an unedited manuscript that has been accepted for publication. As a service to our customers we are providing this early version of the manuscript. The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its final form. Please note that during the production process errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.

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Generalizes the three-candidate, IAC probability distribution to a new, wide class of octahedral distributions. Computes the probability of a majority vote cycle for this wide class of probabilities Determines the likelihood that the Borda and Condorcet winners agree for this class of probabilities. New approach to determine all possible profiles supporting specified threecandidate majority vote outcomes.

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LIKELIHOOD OF VOTING OUTCOMES WITH GENERALIZED IAC PROBABILITIES TOMAS J. MCINTEE, DONALD G. SAARI

Abstract. After determining all supporting profiles with any number of voters for any specified three-candidate pairwise majority vote outcome, a new, large class of “octahedral” probability distributions, motivated by and including IAC, is introduced to examine various three-candidate voting outcomes involving majority vote outcomes. Illustrating examples include computing each distribution’s likelihood of a majority vote cycle and the likelihood that the Borda Count and Condorcet winners agree. Surprisingly, computations often reduce to a simple exercise of finding the volumes of tetrahedrons.

1. Introduction A new way to analyze three candidate relationships connecting pairwise and positional election outcomes is developed in Saari and McIntee [16] (denoted by S&M). This structure identifies all profiles that satisfy any specified pairwise and positional criteria, such as where the pairwise majority votes define a cycle (“Condorcet paradox”), or the plurality and Condorcet winners (the candidate who beats all others in majority vote comparisons) differ, or the Borda Count (i.e., tally a ballot by assigning 2, 1, and 0 points, respectively, to the first, second, and third place candidate) and Condorcet winners agree. By knowing all supporting profiles, we can determine a behavior’s likelihood. For instance, although the Condorcet and Borda Count winners can differ, some results (Saari [12], Gehrlein and Lepelley [7]) show that they often agree. But these assertions involve special assumptions (geometry and Impartial Anonymous Culture (IAC)), so it is reasonable to question whether agreement holds for a wider array of probability distributions. Answers for these three-candidate concerns (with large numbers of voters) follow by using a new class of “octahedral” probability distributions. To indicate how our distributions complement other choices, some distributions (such as the Gaussian) impose assumptions where pairwise majority votes are close to ties. There may be, however, an interest in decisive pairwise differences. Conversely, one might worry whether IAC places undue emphasis on profiles with extreme, perhaps unrealistic majority vote outcomes; what happens by emphasizing midrange outcomes? These concerns sample what can be addressed with the continuum of octahedral distributions introduced here. Beyond introducing the octahedral distributions, our main contribution (Thm. 3) is to • extend known conclusions about the likelihood of a Condorcet paradox and • show that Condorcet and Borda winners agree with a surprisingly high probability. Our thanks to the referees for useful comments. 1

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Our easily used techniques are intuitive, yield robust results, and can be applied to several concerns beyond the illustrating issues used here. When computing likelihoods of voting behavior, only a limited number of approaches tend to be used. Saari and Valognes [17, 18] created a geometric method to resolve certain problems, and Saari and Tataru [15] showed how to handle Gaussian distributions. (Others, such as Merlin, Tataru and Valognes [9], used the Saari–Tataru approach to address related topics.) The octahedral probability distributions introduced here differ; they are motivated by Gehrlein’s influential and pioneering work about IAC done with his coauthors. This body of research includes, for instance, where Gehrlein and Fishburn [6] compute the IAC likelihood of a Condorcet paradox (Sect. 2.1); also see Gehrlein and Lepelley [8]. As IAC is a special case of our distributions, their Condorcet paradox result is extended by computing its likelihood with each of our octahedral distributions. In this manner, our conclusions address concerns about IAC. For instance, IAC assumes all profiles are equally likely, so it treats a 100 voter profile where everyone has the A  B  C ranking as likely as the 100 voter profile where 20 prefer A
 B  C and the rest prefer B  A  C. But with 100 voters, there is precisely 100 100
= 1 way to assign the one hundred voters to 100 create the first profile, while there are 20 = 5.36 × 1020 ways to create the second one. This mind boggling difference makes it arguable that IAC assigns unrealistic distinction to certain profiles; e.g., the first one. The approach developed here resolves such concerns by allowing adjustments to be made to compensate for these significant differences. We also compute the likelihood that the Borda and Condorcet winners agree. Others who have examine this kind of question include Gehrlein with his coauthors such as Lepelley [7] and others [2, 3, 4]. (As some of these results compare majority vote and positional outcomes, they also follow from conclusions (Saari [11]) that identify and explain how and why these methods relate to one another. Also see Chap. 4 of [13].) What simplifies our analysis is that our approach adopts the geometric flavor of [17, 18] where probability computations reduce to finding volumes of standard geometric objects. Other results developed here and in S&M use the three-candidate representation cube (reviewed next) to extract majority vote consequences. For instance, with most N -voter profiles, can some candidate beat another with over 70% of the vote, or do most profiles prevent anyone from having more than 60% of the vote? These issues are addressed with Thm. 2, which describes how profiles are distributed relative to majority vote outcomes. Theorem 1 determines how many profiles support specified pairwise tallies and then identifies all of them. As a sample of the kind of questions that now can be answered: Example 1. Profiles exist where Ann beats Barb by 65:35, Barb beats Connie by 55:45 and Ann beats Connie by 60:40, and where Ann beats Barb by 65:35, Barb beats Connie by 51:49 and Ann beats Connie by 53:47. Which setting, if either, is supported by the larger number of profiles? Another sample issue is whether there can be majority vote elections where Example 2. Ann beats Barb by 35:15, Barb beats Connie by 40:10, and Ann beats Connie by 27:23?

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Example 3. Or the closely related setting where Ann beats Barb by 35:15, Barb beats Connie by 40:10, and Connie beats Ann by 26:24? If either or both choices can occur, how many supporting profiles are there? Can all of them be identified? v4........................................................................................v5 v4......................................................................................v5 . ........ ...... ..... ........................ ..... ........... . . . . . .. .. ..... .............................. .. ..... ...................................... .... . . . . . . . . . . . . . . .... . . . . . . . . . . . . . . ..... ............................................ .. ..... ............. . . . . . . . . . . . . . . ... .. ..... .................................................... ... ............................................................................................ .... ... ................................................................................ ... ... ................................................. ... .......................................... ... .......................................... ... ... . ...... . . . . . . . . . . . . . . .. ... ................................... ... .................................. ..... ... ...... . . . . . . . ... . ... ............................. ... ... . .......................... . . ... . ...... . . . . . . . ... ... . . . ... . ....................... ..... .................... ... ...... . . ... ... ......... ... ................. ..... ............. ..... ... ...... ... .... ... . .. .. .............................................................................................

v3

v7

v6

v2 v1 a. Representation cube

. ...... . ... ..... ......... . . ...... ..... ....................... .. ..... .......... . . . . . . ... .. ............................................ .... ..... . . . . . . . . . . . . . . . . ... . ..... . . . . . . . . . . . ..... ................................................ .. ..... ............ . . . ................... . . . . . . . ... .. ..... .............................................................. ... ........................................................................................................................... .... ... ... ...................................................................................... ... ... ............................................................................... ... ............................................................................ ... ....................................................................... ... ... . ...... . . . ........................... ....................... ... ... .................................................. ...... ... ....................................................... ..... ... ...... ........ ...... . . . . . . . ... ................................... ... ... . ............................ . . ... . ...... . . . . . . . . .. ... . . . ... . ........................ ..... ...................... ... ...... . . . . ... ......... ... ................. ..... .............. ..... ... ...... . ... .... ... .. . .. .............................................................................................

v3

v6

v2 v1 b. Cyclic region

Figure 1 Representation Cube 2. Geometry of profiles Central to our development is the “representation cube” (Saari [10]), which geometrically represents all possible majority vote tallies. For candidates A, B, C, let P (X, Y ) be the difference between X’s and Y ’s tallies in a majority vote election. If Example 2 represents a N = 50 voter election, for instance, then P (A, B) = 35 − 15 = 20 = −P (B, A), P (B, C) = 40 − 10 = 30 = −P (C, B), P (A, C) = 27 − 23 = 4 = −P (C, A). As known (e.g., [10], S&M), the parity of each of P (A, B), P (A, C), P (B, C) and N (the number of voters) is the same: either all four values are even integers, or all are odd. The geometry uses a scaling where p(X, Y ) = N1 P (X, Y ). So with P (A, B) = 20, 1 p(A, B) = 50 (20) = 0.4. In general, −1 ≤ p(X, Y ) ≤ 1. (Recall, p(X, Y ) = −p(Y, X).) Positive values of p(A, B), p(B, C), p(C, A) are, respectively, on the positive x, y, z axis of R3 . Thus a normalized (p(A, B), p(B, C), p(C, A)) majority vote tally defines a point in the Fig. 1a cube [−1, 1] × [−1, 1] × [−1, 1]. These coordinates are selected so that the cube’s positive and negative orthants represent cyclic rankings. The positive orthant, for instance, is where all (p(A, B), p(B, C), p(C, A)) terms are positive, so A  B, B  C, C  A. The cube’s faces represent unanimous votes; e.g., the front, right-side, and top faces require, respectively, an unanimous p(A, B) = 1, p(B, C) = 1, and p(C, A) = 1 vote. Thus, each of the cube’s vertices represents an unanimous vote for the particular ranking. Let vj be the vertex for j th ranking, j = 1, . . . , 6, as listed in Table 1 and illustrated in Fig. 1. Point Vertex Ranking Point Vertex Ranking (1, 1, −1) v1 A  B  C (1, −1, −1) v2 ACB (1, −1, 1) v3 C  A  B (−1, −1, 1) v4 CBA (−1, 1, 1) v5 B  C  A (−1, 1, −1) v6 BAC 1. Names of transitive ranking vertices

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The two remaining vertices represent unanimous votes for the cyclic A  B, B  C, C  A and B  A, A  C, C  B rankings located, respectively, at v7 = (1, 1, 1) and v8 = (−1, −1, −1). But unanimous cyclic outcomes are impossible with voters who have transitive preferences, which requires identifying all admissible tallies. If Nj is the number of voters of the j th type, the possible pairwise outcomes are given by the sum (1)

q = (q1 , q2 , q3 ) =

6 X j=1

n j vj ,

nj =

Nj ≥ 0, N

6 X

nj = 1.

j=1

An Eq. 1 convex combination is an election tally; conversely, any majority vote tally can be so represented [10]. Consequently, these tallies reside in the convex hull defined by the six Table 1 vertices: The geometry of tallies, then, requires removing from the cube the two tetrahedrons with the inadmissible v7 and v8 . (One detached tetrahedron has the vertices {v1 , v3 , v5 , v7 }; its base is the Fig. 1 slanted, shaded region. The other one has vertices {v2 , v4 , v6 , v8 }.) What remains is an octahedron called the representation cube (RC), where its name is chosen to emphasize the starting structure. The set is defined by (2)

RC = {q = (q1 , q2 , q3 ) ∈ R3 | − 1 ≤ q1 + q2 + q3 ≤ 1, −1 ≤ qj ≤ 1}

where summation equalities define the top and bottom RC slanted faces. Thus (Eq. 1) any RC point with common denominator N and numerators with the same parity as N (called an N -point) is an N -voter majority vote outcome. Conversely, any N voter majority vote outcome defines an N -point in RC. As such, RC identifies all possible three-candidate majority vote tallies. (Garvey’s seminal result [5] identifies all pairwise rankings that can arise, but not the tallies. Thus the RC structure extends Garvey’s result for three candidates by also identifying all possible pairwise tallies.) Example 2 satisfies Eq. 2, so its tallies define election outcomes supported by 50-voter profiles. Example 3, however, cannot be an election outcome because the sum of the P (B,C) P (C,A) 20 30 2 ( P (A,B) 50 , 50 , 50 ) = ( 50 , 50 , 50 ) components exceeds unity to violate Eq. 2. Similarly, when expressing (1, −0.4, 0.3) in a fractional form with the common denomi3 numerator has an odd value where the parities do not match; nator of 10, the p(C, A) = 10 this point cannot represent a 10 voter profile. With the denominator of 100, however, the numerators and common denominator of 100 are even integers, so the point represents a N = 100 voter election outcome. As this point is on the front face (p(A, B) = 1), it is a convex combination of v1 , v2 , v3 , or (1, −0.4, 0.3) = n1 v1 +n2 v2 +n3 v3 where n1 +n2 +n3 = 1. Solving these algebraic equations leads to n1 = 0.30, n2 = 0.05, n3 = 0.65, with the unique supporting profile of N1 = 30, N2 = 5, N3 = 65, N4 = N5 = N6 = 0. We show in Sect. 2.1 how to recover all supporting profiles for an N -point. The RC geometry identifies all possible pairwise outcomes; e.g., RC points in the positive and negative orthants have cyclic rankings. (One cyclic region is the heavier shaded Fig. 1b sector.) To illustrate how this cube identifies what can, or cannot, happen, notice how the geometry prevents an outcome from being in the positive orthant’s cyclic region without involving all {v1 , v3 , v5 }vertices; similarly, it is impossible to have a negative orthant cyclic

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outcome without including all {v2 , v4 , v6 }. Thus, if a profile is missing at least one preference ranking from each set, the outcome cannot be cyclic [10]. This property includes Ward’s conditions [19] and, as a special case, the widely used Black’s single-peaked condition [1]. As it is easily computed, the volume of these two cyclic tetrahedrons constitutes 1 1 16 of RC’s total volume. 2.1. Counting profiles. By assuming that all profiles are equally likely, IAC is a counting approach; it defines, for N voters, the ratio of the number of profiles satisfying a specified condition with the total number of profiles. Using IAC, Gehrlein and Fishburn [6] showed 1 that the probability of a Condorcet paradox approaches 6.25%, or 16 , as N → ∞. As 1 proved here (Thm. 3), it is not a coincidence that this 16 value agrees with the relative volume of RC cyclic regions (previous paragraph). While our method is motivated by the IAC counting argument, our distributions offer significantly more flexibility by allowing certain classes of profiles to be emphasized, or deemphasized; in this way, adjustments can be made to handle regions that IAC overly emphasizes. Our method requires finding how many profiles support each N -point: An N -point on a RC surface represents a unique profile determined (as above) by the face’s three vertices, but an N -point in the RC interior has many supporting profiles. To find this number, first link an N -point with a RC face; i.e., draw a line from the RC origin through the N -point until it meets a RC face; this is the point’s associated face. To convert this face-point geometry into a computational tool, in S&M we introduced the strongly non-cyclic condition where if U beats V and V beats W (so, p(U, V ) ≥ 0, p(V, W ) ≥ 0), then (3)

p(U, W ) ≥ min(p(U, V ), p(V, W )).

As a cyclic outcome requires p(U, W ) = −p(W, U ) < 0, it always fails Eq. 3. With a transitive ranking, Eq. 3 requires the tally difference between the Condorcet winner and loser (here, U and W , respectively) to be at least as large as one other tally difference. If Eq. 3 is satisfied, the N -point is linked with a face of the original cube; i.e., if p(X, Y ) is the largest of the six values, then the face is p(X, Y ) = 1. The vertices of this face play a central role in computations. If Eq. 3 is not satisfied, the point is associated with the closest slanted face, where its vertices are used in computations. This face can be found with a line from the RC origin through the point and hitting the face. Alternatively, the sign of the sum of components determines the face; a positive value represents the top face (with q1 + q2 + q3 = 1) while a negative value has the bottom face. For instance, 25 15 , 50 , − 10 ( 50 50 ) fails to satisfy Eq. 3, so it is associated with a slanted face; as the sum of components is positive, the point is linked with the top face with vertices v1 , v3 , v5 . Definition 1. (Saari [11, 13]) A “reversal pair” consists of two complete transitive rankings where each ranking reverses the other. The three pairs are (4)

(A  B  C, C  B  A), (A  C  B, B  C  A), (C  A  B, B  A  C).

1This is given in [10] and derived again, for the reader’s convenience, in Sect. 4.1, Eq. 4.1.

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The following theorem asserts that an N -point has a unique v-voter essential profile (v is defined in the theorem) determined by the associated face’s vertices. These v voters have specified rankings, so the remaining 2η = N − v voters need to be assigned preferences. As proved in [11, 13], and repeated in Thm. 1, the only way to do so is to assign each pair of voters a reversal pair of preferences. (This does not change P (X, Y ) values as the majority vote of a reversal pair is a complete tie.) Theorem 1. Each N -point has a unique essential profile defined by the three vertices of the associated RC face. A strongly non-cyclic point (Eq. 3) is associated with the p(X, Y ) = 1 face’s vertices where P (X, Y ) has the largest value; its essential profile has v = P (X, Y ) voters. If the point does not satisfy Eq. 3, it is associated with the closest slanted face; the essential profile has v = |P (A, B) + P (B, C) + P (C, A)| voters. The parity of v and N agree, so η = N 2−v is a non-negative integer. The set of all profiles supporting the N -point consists of adding η reversal pairs to the essential profile. This can be done in     (η + 2)(η + 1) η+2 η+2 = = (5) δ(η) = 2 η 2 different ways, so the N -point is supported by precisely δ(η) different profiles.

Theorem 1 provides a comparison with the IAC uniform distribution of profiles over the probability simplex as N tends to infinity. According to Thm. 1, this uniform distribution fails to hold for the RC. To see the interesting RC symmetry, notice how the number of profiles supporting an N -point is determined by η, where η reversal pairs are added to the essential profile. This structure can be done in δ(η) ways, so it generates δ(η) different supporting profiles. Of importance (Thm. 2), the η value corresponds to the N -point’s distance from the RC surface; a point closer to the RC origin has a larger η value. Indeed, all N -points with the same η value are the same distance (defining a shell) from the RC face. We use this octahedral shell structure to define our octahedral probability distributions. Theorem 1 and Eq. 5 admit all sorts of conclusions. For instance, δ(η) is an increasing function of η, so, for a given N , the closer the majority vote outcome is to a complete tie, the larger the number of supporting profiles. This makes sense; a nearly tied outcome must include many reversal pairs. (A reversal pair’s pairwise outcome is a complete tie.) For a strongly non-cyclic outcome and an odd value of N , the largest η value is η = N 2−1 , where v = P (X, Y ) = 1 (the winning candidate has one more vote than the loser). According +1) to Eq. 5, this outcome is supported by δ( N 2−1 ) = (N +3)(N profiles. Illustrating with 8 N = 21, the A  B, B  C, A  C outcomes with P (A, B) = P (B, C) = P (A, C) = 1 (which are strongly non-cyclic) are supported by 66 profiles. Stronger results follow after we determine the number of N -points with a given ν value (Thm. 2). It turns out, for instance, that v ≈ N 2+3 has the largest number of profiles, which include outcomes with a landslide 75% victory. 2.1.1. Near a cube face. A strongly non-cyclic N -point is associated with a non-slanting 8 2 −4 RC face (one of the original cube faces). For instance, ( 10 , 10 , 10 ) is in the RC by satisfying

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Eq. 2. As the numerators and common denominator are even integers, the point represents outcomes from 10-voter profiles, which, by satisfying Eq. 3, is strongly non-cyclic. Because P (A, B) = 8 is the largest numerator, the point’s essential profile is determined by the p(A, B) = 1 face’s vertices {v1 , v2 , v3 }. (The face is determined by the largest, positive P (X, Y ) value; e.g., if P (X, Y ) = −20, it would be expressed by P (Y, X) = 20 to place emphasis on the p(Y, X) = 1 face.) This 8-voter essential profile (from v = P (A, B) = 8) is found by solving (0.8, 0.2, −0.4) = 8 v = 10 . The unique solution has 5 voters pren1 v1 + n2 v2 + n3 v3 , where n1 + n2 + n3 = 10 ferring A  B  C, 1 preferring A  C  B, and 2 preferring C  A  B. Preferences must be assigned to the remaining 2η = N − v = 2 voters without affecting 8 2 −4 ( 10 , 10 , 10 ). These preferences must use η = 1 reversal pair from Eq. 4 (Thm. 1). Adding any of the three reversal choices to the essential profile creates a supporting profile, so 8 2 −4 , 10 , 10 ) represents outcomes for δ(1) = (3)(2) = 3 different 10-voter profiles. ( 10 2 Of importance (Sect. 3), Nv remains fixed, so v → ∞ as N → ∞. For instance, replacing 800 200 −400 8 2 −4 , 10 , 10 ) with ( 1000 , 1000 , 1000 ) creates an outcome with 100 times more voters; the v ( 10 value jumps from 8 to v = 8 × 100, and η = 1 to η = 100, but with δ(100) = 5151. 24 4 44 Point (− 100 , − 100 , 100 ) represents the transitive rankings C  A, C  B, B  A where C and A are, respectively, the Condorcet winner and loser. The parity condition is satisfied, so the point is supported by 100-voter profiles. Also, Eq. 3 is satisfied because p(C, A) = .44 > min(p(C, B) = −p(B, C) = .04, p(B, A) = −p(A, B) = .24). With the largest numerator of P (C, A) = 44, the v = 44 voter essential profile (Thm. 1) is found from the p(C, A) = 1 face’s vertices {v3 , v4 , v5 } by solving the equations (−

24 4 44 ,− , ) = n 3 v3 + n 4 v4 + n 5 v5 , 100 100 100

n3 + n4 + n5 =

v 44 = . 100 100

Thus 10 voters prefer C  A  B, 14 prefer C  B  A, and 20 prefer B  C  A. To = 28 reversal pairs (Eq. 4) to this essential find all supporting profiles, add η = 100−44 2 profile (Thm. 1), which can be done in δ(28) = (30)(29) = 435 different ways. Thus this 2 point is supported by 435 different 100-voter profiles. Borda Count rankings never are affected by reversal pairs ([11, 13]), so the Borda ranking for these 435 profiles remains C  B  A. But all possible differences between the Borda ranking and those of other positional methods (i.e., tally ballots by assigning specific points to candidates depending on their ballot position) are caused by reversal pairs, and only reversal terms! [11, 13] Indeed, the different reversal terms in these 435 profiles generate different plurality rankings such as C  A  B, or C  B  A, or B  C  A. 20 30 10 To answer the Example 1 question, the first profile is the N -point ( 100 , 100 , − 100 ), while 30 2 6 the second is ( 100 , 100 , − 100 ). Both satisfy Eqs. 2 and 3 along with the parity condition, so both are supported by 100-voter profiles. Both have a v = 30 voter essential profile, )= so both outcomes are supported (Thm. 1) by the devilishly large number of δ( 100−30 2 (37)(36) 30 δ(35) = = 666 profiles. The v value for both is determined by p(A, B) = 100 , so 2 both points are the same distance from the RC surface.

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4 4 4 , 10 , − 10 ), can be assigned to two faces; e.g., p(A, B) = 1 Some N -points, such as ( 10 and p(B, C) = 1. Here the essential profile’s rankings are from Table 1 vertices that are 4 4 4 on both faces. With ( 10 , 10 , − 10 ) and using either face, the essential profile has only one ranking with N1 = 4, which is consistent with the geometry where only v1 is on both faces.

2.1.2. The slanted faces. Points that are not strongly non-cyclic are associated with the nearest RC slanted face. This includes cyclic and transitive outcomes where the Condorcet winner and loser election is tighter than any other paired comparison. An example 2 14 6 , − 40 , 40 ) with transitive rankings A  B, C  B, C  A; here, C and B are, is ( 40 respectively, the Condorcet winner and loser. By satisfying Eq. 2 and the parity condition, this point is supported by 40-voter profiles. But p(C, B) = −p(B, C) = 0.05, while p(C, A) = 0.35 and p(A, B) = 0.15, so Eq. 3 is not satisfied. With the positive sum of components, the essential profile is determined by top slanted face’s vertices {v1 , v3 , v5 }. The essential profile (Thm. 1) has v = |P (A, B) + P (B, C) + P (C, A)| = |6 − 2 + 14| = 18 voters, so the defining equations are 6 2 14 v 18 ( , − , ) = n1 v1 + n3 v3 + n5 v5 , n1 + n3 + n5 = = . 40 40 40 40 40 The essential profile has 2 voters preferring A  B  C, 10 preferring C  A  B and 6 preferring B  C  A. Because η = N 2−v = 40−18 = 11, to find all supporting profiles add 2 = 78 different ways eleven reversal pairs to the essential profile; there are δ(11) = (13)(12) 2 to do this, so this point is supported by 78 different profiles. As shown above, Example 2 is a 50-voter election outcome where the v = |20 + 30 − 4| = 46-voter essential profile has 25 voters preferring A  B  C, 8 preferring C  A  B, and =2 13 preferring B  C  A. All supporting profiles are found by adding η = N 2−v = 50−46 2 (4)(3) reversal pairs to the essential profile, which can be done in δ(2) = 2 = 6 ways. Thus Example 2 has six supporting profiles. (For three of them, the two reversal pairs are the same, for the last three they differ.) 2.1.3. Geometry of N -points with the same number of profiles. For a given v, all N -points supported by δ(η) profiles, η = N 2−v , are on the boundary surface of an object in the interior of RC, which geometrically resembles RC. For 0 < α ≤ 1, the object is given by (6)

RC α = {q = (q1 , q2 , q3 ) ∈ R3 | − α ≤ q1 + q2 + q3 ≤ α, −α ≤ qj ≤ α}

Call the boundary surface the α-octahedral shell. For α = Nv , call the boundary surface the v-octahedral shell. It follows from the above that an N -point is on the v-octahedral shell if and only if its essential profile has v voters if and only if the essential profile is accompanied by δ( N 2−v ) reversal pairs. This means that all N -points with the same number of supporting profiles are on the same v-shell. The next statement specifies both the number of N -points and the total number of supporting profiles on a v-shell.
Theorem 2. N voters define N 5+5 different profiles. The number of N -points on a voctahedral shell (the pairwise tallies define an essential profile with v > 0 voters) is (7)

4v 2 + 2

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9

Thus the number of supporting profiles for N -points on the v-shell for v > 0 is 1 N −v ) = [4v 2 + 2][v 2 − (2N + 6)v + (N + 2)(N + 4)] (8) [4v 2 + 2]δ( 2 8 For v = 0, N is an even number and the essential profile is the origin representing a complete tie. A supporting profile consists of η = N2 reversal pairs; there are δ( N2 ) = (N +4)(N +2) different profiles of this type. 8 For large N values, the v-octahedral shell with the maximum number of profiles is near v ≈ N 2+3 . For a strongly non-cyclic N -point on this shell, the winner of at least one pairwise competition has a landslide victory by winning approximately three-fourths of the vote. About half of all profiles are between v = N2 and the v = N octahedral shells, where strongly non-cyclic points have at least one pairwise winner with N 2+v ≥ 34 of the vote.

The Thm. 2 final paragraph describes strongly non-cyclic terms where computations are easier. This is because the v value identifies a particular pair where P (X, Y ) = v. Thus, with v ≈ N 2+3 , we have x − y ≈ N 2+3 , x + y = N for x and y representing X’s and Y ’s tallies, so x ≈ 3N4+3 , where a candidate receives 75% of the vote! But if the strongly non-cyclic conditions is not satisfied, then v = |P (A, B) + P (B, C) + P (C, A)|, so no particular paired comparison is identified. This requires analyzing different cases, such as the cyclic setting where P (A, B) = P (B, C) = P (C, A) and 3P (A, B) = v. Thus with v ≈ N 2+3 , x − y ≈ N 6+3 , x + y = 1, so x ≈ 7N12+3 , or a 58.3% victory! All remaining settings have a more extreme tally; e.g., should P (A, B) = P (B, C) > 0 and P (A, C) = 0 (on the boundary of the Fig. 1 cyclic region), the largest outcome increases to x ≈ 5N8+3 , where the winner receives about 62.5% of the vote. With v ≈ N 2+3 , all outcomes outside of the Fig. 1 cyclic regions have a
winner with at least 62.5% of the vote. With N = 10, the 10+5 = 3003 different profiles are distributed as given below. Re5 member, the v value is the number of voters in the essential profile and, for strongly non-cyclic N -points, where some pair has the N 2+v to N 2−v tally. Thus larger v values represent more decisive election outcomes that may be unrealistic. v Number of essential profiles η = 10−v 2 0 1 5 2 18 4 66 3 4 6 146 2 8 258 1 10 402 0 Total 891

δ(η) Number of profiles 21 21 15 270 10 660 6 876 3 774 1 402 3003

2. Counting profiles For large N values, the maximum number of profiles on a shell is associated with v ≈ Even with just N = 10, this v ≈ 10+3 = 6.5 value is supported by Table 2 with 2

N +3 2 .

10

TOMAS J. MCINTEE, DONALD G. SAARI

v = 6. For a strongly non-cyclic N -point, this v = 6 value requires the winner of at least tally, or 80% of the vote! one pairwise election to have a landslide victory with the 10+6 2 Over half of the Table 2 profiles are on just the two shells (v = 6, 8 shells) near the RC surface; this means, with large N values, that the largest number of profiles have at least one landslide pairwise outcome. 1 The profile structure raises questions about IAC predictions. Is the 16 likelihood for cycles biased by the preponderance of profiles with unrealistically large tally differences? (According to the above, over half of the profiles have a landslide outcome.) Would the conclusion differ by modifying the IAC philosophy to emphasize profiles where, say, a majority vote winner receives no more than 55% of the vote? Questions of this type motivate, and are answered by, the distributions introduced next. ................................................................................................................................................... ... .......................................................................................................................................................... ... ... ...................................................................................................... ... ... ... ............................................................................................................... ... .... ... ... ... .............................................................. ... .... ... ... ... ... ...................................................... ... ... ... ... ... ... ... ..................................... ... .... ... .... ... ... ... ... ... ........... ... .. ... .. ... ... ... ... ... ............................. ... ... ... ... ..... ... .. .. .. ... ... ... .......... . . . ... ... ... .......................................... .... .... .... ... ... ............................................................ .. ... ........ ... ... .......... ... ....................................................................... .... ..................................................................................

............................................................................................................. ... .................................................................................................................................. ... ... ........................................................................................ ... ... ....................................................... ... ................................... ... ... ....................................... ... .............................. ... ...................... ..... ... ............ . ... . ... . ......... . ... . . ... . ... ....... . . . .... ... . ..... ... . . . ... . . ... . ..... ... . . .. . . . ... . ..... .. ..... .... ... ......... ..... .. ... ...... . . .....................................................................................

B

a. Square

.................................................................................................................................................. ... ....................................................................................................................................................... ... ........................................................................................................ ... .................................................................. .. ... ............................................................ .... ................................................... ... ... .............................................. ... ...................................... ..... ... .......................... ......... ... ... .. ..... . ... . . ... . ... ....... . . . .... ... . ..... ... . . . ... . . ... . ..... ... . . .. . . . ... . ..... .. ..... .... ... ......... ..... .. ... ...... . . .....................................................................................

.................................................................................................................................................... ... ........................................................................................................................................................... ... ... ...................................................................................................... ... ... ... .................................................................................................. ... .... ... .. ............................................... ... ... ...................................... ... ... ... ... ............................. ................... .... .... ... ... ........... .. ... ... ... ........ . ... ... . . ... ... . . ... ....... . . . ... ... .. ... . ..... ... . . . . . ... ... . ..... .... .... .. ..... .. ... ... ... ......... ..... .. .. ... .......... .. . ... ....................................................................... .... . . ................................................................................

c. Symmetry b. Square section rings Figure 2. Ilustrating examples

d. Lack of symmetry

3. Octahedral probability distributions R To introduce our approach by integrating S f (x) dA over the Fig. 2a square S = [−1, 1]× [−1, 1], let the Riemann integrable f be constant valued on the perimeter of each square about the center. (That is, f (x) has the same value on {x = (x1 , x2 ) | max(|x1 |, |x2 |) = s} for each 0 ≤ s ≤ 1.) Using the definition of the integral as a limit of sums of step functions, (9)

Z

a

R

b

g(x) dx = lim

K→∞

K X i=1

g(xi ) ∆x,

∆x =

b−a , K

S f (x) dA can be approximated, as closely as desired, by step functions over sections of squares. The approach mimics how onion rings partition an onion; here (Fig. 2b) “square rings” partition S. Namely, for a given positive integer K, partition S into the regions suggested by Fig. 2b. As with Fig. 2c, start with S and then create a smaller square in S’s interior that also is centered about the origin. The difference between these two squares is the first square ring SR(1). Continue this process to construct K square rings. Starting from the outside and working inwards, the exterior boundary of first square ring, SR(1), is S’s boundary; SR(1)’s interior boundary is {x = (x1 , x2 ) | max(|x1 |, |x2 |) = 1 − ∆} where ∆ = K1 . The j th square ring, SR(j), has its exterior boundary given by {x | max(|x1 |, |x2 |) = 1 − (j − 1)∆} and its interior boundary by {x | max(|x1 |, |x2 |) = 1 − j∆}. Because f is constant valued on squares, then, as in Eq. 3, the integral is

LIKELIHOOD OF VOTING OUTCOMES WITH GENERALIZED IAC PROBABILITIES

11

approximated by step functions over square rings as Z

(9)

f (x) dA = lim

K→∞

S

Similarly, (9)

Z

R

B

f (xj ) Area(SR(j)),

j=1

xj ∈ SR(j).

f (x) dA for region B ⊂ S (e.g., the shaded region in Fig. 2a) is

f (x) dA = lim

B⊂S

K X

K→∞

K X j=1

 Area(B ∩ SR(j)) Area(SR(j)), f (xj ) Area(SR(j)) 

xj ∈ B ∩ SR(j).

If, R for instance, f (x) is R a probability distribution function over S, then f (x) ≥ 0 and S f (x) dA = 1, where B⊂S f (x) dA is the probability that B occurs.

3.1. Octahedral rings. The same approach is used with RC where, rather than square rings, RC is partitioned into K octahedral rings centered about the origin. The first K−1 th ring, OR(1), is the region between the K K and K octahedral shells. In general, the j K−(j−1) K−j octahedral ring, j = 1, . . . , K, is the region between the and K octahedral shells. K Start with step functions and mimic the above. The µi values will be used to adjust the value attached to certain shells. Definition 2. For integer K > 1, sequence µK = {µ1 , . . . , µK } is a K-octahedral probability distribution if each µj ≥ 0 and K X

(9)

µj Vol(OR(j)) = 1.

j=1

For B ⊂ RC, if the volume of B ∩ OR(j) is defined for each j, the probability of B is (9)

ProbµK (B) =

K X j=1

µj



 Vol(B ∩ OR(j)) Vol(OR(j)) Vol(OR(j))

For N > 1, the probability that N -points are in B is given by (9)

ProbµK (B, N ) =

K X j=1

µj



 Number of N -points in (B ∩ OR(j)) Vol(OR(j)) Number of N -points in OR(j))

With a Riemann integrable probability distribution on RC approximated by the µK step functions, the probability of N -points satisfying the condition B as N → ∞ is   (9) Prob(B) = lim lim ProbµK (B, N ) . K→∞ N →∞

12

TOMAS J. MCINTEE, DONALD G. SAARI

3.2. Examples and results. The considerable flexibility offered by Eq. 2 permits addressing concerns such as those raised about IAC. The approach is immediate: select smaller µj values for the shell regions to be de-emphasized, and larger µj values for regions of emphasis. In this manner, a wide variety of issues can be considered. Of value, if the concern defines an appropriate RC symmetry, then computations reduce to simple geometry! This is illustrated by computing the likelihood of a Condorcet winner and the likelihood of a Borda-Condorcet agreement: Many other issues can be similarly analyzed. 1 for A natural choice of a K-octahedral probability distribution is µj = KVol(OR(j)) j = 1, . . . , K where Eq. 2 determines the relative volume of B to that of RC. This µK choice yields the fact (paragraph preceding Sect. 2.1) that by using either the step functions 1 the volume of RC. (Eq. 2) or the limit as K → ∞, the volume of RC cyclic regions is 16 It follows (Thm. 2) that if an N -point with precisely η reversal pairs is in the interior of an OR(j), then all N -points that are supported with precisely η reversal pairs are in the same ring. Thus the probability distributions captured by Def. 2 include choices that depend on the number of supporting profiles for each N -point, such as IAC. Indeed, IAC would count the number of N -points supported by η reversal pairs and multiply this number by δ(η)—the number of distinct profiles—to determine the µj choice. In the limit, these number of ways to create distinct probabilities approaches a function, so Eq. 3.1 applies to find the likelihood of, say, cyclic outcomes. As another example, it may be impossible for the winner of a pairwise election to receive −0.4N = 0.2. To over 60% of the vote, so all p(X, Y ) outcomes are bounded above by 0.6N N capture this setting, the likelihood of something occurring in an earlier octahedral rings (closer to the RC boundary) is zero, so set µj = 0 for j ≤ 0.8K. Similarly, perhaps at least one candidate will receive at least 55% of the vote. This −0.45N = 0.1, so set µj = 0 for j > 0.9K. defines a p(U, V ) value of 0.55N N A fourth possibility is to find the likelihood of Borda and Condorcet winners agreeing if at least one pairwise outcome has more than, say, 55% of the vote, but less than 65%. −0.35N Here the largest p(U, V ) values are bounded below by 0.1 and above by 0.65N N = 0.3, so octahedral distributions would select µj = 0 for j < 0.7K and j > 0.9K. The µj choices for 0.7 ≤ j ≤ 0.9 are free to be selected as long as Eq. 2 is satisfied. As indicated, K-octahedral distributions admit a surprisingly wide selection of settings. For this reason, and because there are an uncountable number of octahedral probability distributions, the following results are unexpected and surprising. Theorem 3. For three candidates and any K-octahedral probability distribution (Eq. 2), as N → ∞, the following are true: 1 (1) The likelihood of a cyclic outcome is 16 . (2) The likelihood that a Borda and Condorcet winner agree is

41 45

= 0.911 . . . .

If the step functions from a sequence µK , as K → ∞, approximates a probability distribution over RC, then the above three likelihood assertions hold for the probability Eq. 3.1.

An immediate, surprising consequence of Thm. 3 is that the Gehrlein-Fishburn [6] IAC 1 result that cycles have a likelihood of 16 , and the Gehrlein-Lepelley [7] likelihood result

LIKELIHOOD OF VOTING OUTCOMES WITH GENERALIZED IAC PROBABILITIES

13

41 for Borda and Condorcet agreement (indicating a Borda efficiency of the Condorcet of 45 method) as N → ∞ may be biased by profiles with unrealistic pairwise tallies, but the conclusion remains the same for an uncountable number of different weightings! The settings described at the start of this section define, in the limit, a probability distribution over RC, so the theorem’s last statement applies. In particular, both the volume integral over RC and IAC are octahedral probability distributions, which is why 1 both assign the probability of 16 for cyclic behavior. Surprisingly, the same conclusion holds even if points near indifference are excluded, or if points near the RC boundaries (with landslide outcomes) are emphasized, deemphasized, or whatever choice is made. A way to appreciate this surprising result is to recall that the IAC probability of an event B is the number of profiles satisfying B divided by the number of profiles. With K-octahedral distributions, we compute the ratio of profiles in each shell that satisfy B to the number of profiles in the shell (Eq. 2). Dividing the space into shells creates the opportunity to assign different weights to different shells, which leads to our distributions. The two events selected to illustrate this approach, cycles and Borda-Condorcet agreement, are events B where the RC geometry has the same ratio in each shell as N → ∞. Thus, the weighting cancels, and Thm. 3 follows. (See Prop. 1.) More generally, the likelihood of any issue B with a symmetric RC geometry (e.g., union of cones emerging from the RC origin) becomes easy to compute with a Thm. 3 conclusion of having the same K-octahedral likelihood. For instance, one might wish to find the K-octahedral likelihoods of a strongly non-cyclic point (Eq. 3). According to Fig. 1b, all N -points violating Eq. 3 are in two RC tetrahedrons; one is defined by the origin and the vertices v1 , v3 , v5 ; the other by the origin and vertices v2 , v4 , v6 . Using the techniques and arguments of Sect. 4.1, it turns out that the probability of a strongly non-cyclic profile is 3 4 . Or, motivated by the “strongly transitive condition” [11, 14], let a “strong Condorcet winner” be where the difference in tallies between the Condorcet winner U and loser W is at least the sum of the tally difference between U and second ranked candidate V , and V and W ; i.e., P (U, W ) ≥ P (U, V ) + P (V, W ). As this event defines a cone, likelihoods can be computed. But if B does not admit a symmetric RC setting, as in Fig. 2d, nor reduced to combinations of symmetric settings, then the likelihood computations become more difficult, and a Thm. 3 conclusion need not apply. Still other results follow; e.g., the likelihood of a plurality and Condorcet winner agreeing 41 . The basic argument is simple (but many details must be carried out); it is less than 45 involves reversal terms that never affect the Borda ranking, but can change the plurality ranking to where the plurality and Condorcet winners disagree. Thus, for any K-octahedral probability distribution emphasizing profiles with a larger predominance of reversal terms (i.e., points away from the RC boundary), a shell has a smaller percentage of plurality— pairwise agreement than with the Borda Count. The conclusion follows.

4. Proofs Several proofs depend on the standard combinatorics of placing b indistinguishable balls into c cups. (These expressions are in most undergraduate textbooks on probability.)

14

TOMAS J. MCINTEE, DONALD G. SAARI


(1) If some cups can be empty, this can be done in b+c−1 c−1
ways. b−1 (2) If each cup cannot be empty, this can be done in c−1 ways.

Proof of Thm. 1. Although stated in different terms, most of Thm. 1 is in S&M; only Eq. 5 remains to be proved. This is equivalent to
placing b = η reversal pairs into c = 3
η+2 = choices, so (from #1) the answer is η+3−1 2 , which is Eq. 5. 3−1 Proof of Thm. 2. To count the number of profiles, each of the N voters can
select any
+6−1 of the six rankings. So (#1), where b = N and c = 6, there are N 6−1 = N 5+5 profiles. To count the number of essential profiles (N -points) on a v-shell, recall that an essential profile consists of the three rankings associated with the appropriate RC face’s vertices.

• All v voters could have the same preference ranking. There are six preference rankings, so this can be done in six ways. • The preferences involve two different selected rankings. These rankings cannot be reversals because essential profiles do not allow reversal pairs; this follows immediately by observing that a reversal pair cannot be on the same RC face. – There are three pairs of rankings with the same candidate top-ranked. – There are three pairs of rankings with the same candidate bottom-ranked. – The (X  Y  Z, Z  Y  X) reversal pairs are not admitted. – What remains are six pairs of rankings of the X  Y  Z, Z  X  Y form. Thus there are 12 pairs of rankings. Each must have at least one voter, so #2
applies showing there are 12 v−1 = 12(v − 1) essential profiles of this type. 2−1 • What remains is where the v voters
are split among all three rankings. For each triplet, this can be done in v−1 3−1 ways for v ≥ 3. (As this cannot be done for v < 3, here the number is zero.) Each of the eight faces defines a triplet, so this
v−1 case defines 8 2 essential profiles.
The sum of the values in the three bulleted points is 6 + 12(v − 1) + 8 v−1 2 , which, after collecting terms, equals 4v 2 + 2 that is Eq. 7. Each N -point is identified with a unique essential profile. Each essential profile with v voters is supported by δ(η) = δ( N 2−v ) profiles. Thus the product of Eq. 7 and δ( N 2−v ) is Eq. 8, which is the number of profiles with this v value. The number of profiles on a v-shell (Eq. 8) is the product of the number of N -points and δ( N 2−v ), so it has a f (v)g(v) form. To find the v value with the maximum number of profiles, set the first derivative equal to zero, which yields f (v)g 0 (v) + f 0 (v)g(v) = 0, or f 0 (v) g 0 (v) 8v 2v f (v) = − g(v) . The first term is 4v 2 +2 = v 2 +0.5 , while the second is v2

2[(N + 3) − v] 2[(N + 3) − v] = . − 2(N + 3)v + (N + 2)(N + 4) [(N + 3) − v]2 − 1

Rather than finding the exact v value, an approximation for large N values suffices. The +3)−v] expression becomes 2v ≈ 2[(N , or v ≈ (N + 3) − v leading to the v ≈ N 2+3 conclusion. v2 [(N +3)−v]2 Finally, summing the number of profiles (Eq. 8) over the v values is closely approximated by the integral over this range. To simplify the integration, let v = tN , 0 ≤ t ≤ 1, to convert

LIKELIHOOD OF VOTING OUTCOMES WITH GENERALIZED IAC PROBABILITIES

15

the integrand to 1 2 2 2 6 2 4 N (4t + 2 )[t2 − 2t(1 + ) + (1 + )(1 + )], 8 N N N N which, for large values of N is essentially a multiple of t2 (t2 − 2t + 1) = t4 − 2t3 + t2 . Therefore, about half of the profiles are in the region [0, aN ] where a is defined by Z 1 Z a 4 3 2 [t4 − 2t3 + t2 ] dt, [t − 2t + t ] dt = a

0

or where

1 5

−

1 2

+

1 3

=

1 30

= 2[

a5 5

−

a4 2

+

a3 3

2

] = 23 a3 [ 3a5 −

3a 2

+ 1]. Here a = 12 .

Proof of Thm. 3. By the nature of their construction, the N -points are uniformly distributed in RC. Thus, for an open region B ⊂ RC,     Number of N -points in (B ∩ OR(j)) Vol(B ∩ OR(j)) (9) lim = N →∞ Number of N -points in OR(j)) Vol(OR(j)) The basic idea is captured by the following Proposition.

Proposition 1. Let B be a RC region with a finite number of boundaries that are planes passing through the origin. All octahedral probability distributions have the same value, Vol(B) . Similarly, as N → ∞, the likelihood of N -points being in B also is which is Vol(RC) Vol(B) Vol(RC) .

As shown next, both Thm. 3 conditions define RC regions where the boundaries are planes passing through the origin. According to Prop. 1, each likelihood value can be determined by computing the relative volume of the appropriate region to that of RC. Incidentally, the Prop. 1 constraint imposed on the boundaries of B was selected only because it suffices to prove Thm. 3. As long as B is the finite union of cones emanating from the origin, the assertion holds. Intuition about the proof for Prop. 1 comes from Fig. 2c and the construction of the first square ring SR(1). The shaded region B in the original square constitutes one-fourth of the total area, or 41 Area(S). By geometric similarity, the shaded region in the inner square, S1 , also has the same proportion, or 14 Area(S1 ). As such, the shaded region in 1 1 1 Area(S) − Area(S1 ) = Area(SR(1)). 4 4 4 The same argument holds for all SR(j), which means that the summation in Eq. 3 is   K K X Area(B ∩ SR(j)) 1X 1 Probµj (B) = µj Area(SR(j)) = µj Area(SR(j)) = . Area(SR(j)) 4 4 Area(B ∩ SR(1)) =

j=1

j=1

The bracketed value equals 14 independent of the choices of µj . As such, as long as the µj satisfy the square ring version of Eq. 2 (so the last summation equals unity), all of the distributions have the same likelihood value.

16

TOMAS J. MCINTEE, DONALD G. SAARI

By using Eq. 4 as applied to the square, if M points are uniformly distributed in S, then the Eq. 3.1 version becomes   1 Prob(B) = lim lim Probµj (B, M ) = . K→∞ M →∞ 4

The Fig. 2d shaded region, however, does not have the Prop. 1 symmetry structure; certain square rings have a larger portion of B than in other rings. For this reason, the Prop. 1 conclusion does not hold; instead, different square ring distributions can have different probability values. Proof of Proposition: The proof mimics the discussion for the square. Partition B so that each partition set is in one orthant and the exterior boundary is a single RC face. (To do so, intersect B with the orthant planes and each plane that passes through the origin and a slanted face edge.) The analysis is done on each partition set. In a partition set, the intersection of two boundary planes defines a line passing through the origin. Each of these boundary planes intersects another plane (another boundary plane, or the orthant’s boundary plane). These three lines define a tetrahedron with vertices at the origin and the RC surface. This tetrahedron has β of the total RC volume. By geometric similarity, the portion of the tetrahedron in the contracted version of RC, or RC 1−∆ , is β of that volume. Thus, the portion of the tetrahedron in the octahedral ring OR(1) is βVol(RC) − βVol(RC 1−∆ ) = βVol(OR(1)).

This means that the bracketed term in Eq. 2 for the first ring equals β. As the same similarity argument holds for each ring, each bracketed term in Eq. 2 equals β, so the summation equals β. This completes the proof. 4.1. Cyclic outcomes. The first assertion of Thm. 3 concerns cyclic outcomes. The boundaries for the cycle region in the positive orthant are the three coordinate planes where x = 0, y = 0, z = 0. The same is true for the cyclic region in the negative orthant. According to Prop. 1, all that remains is to compute the relative volume of these cyclic regions to that of RC. To find Vol(RC), start with the volume of the [−1, 1] × [−1, 1] × [−1, 1] cube, which is 23 = 8. Two tetrahedrons are removed from this cube (Fig. 1a); the edges of the tetrahedron defined by {v7 , v1 , v3 , v5 } emerging from v7 are v1 − v7 = (1, 1, −1) − (1, 1, 1) = (0, 0, −2), v3 − v7 = (0, −2, 0), and v5 − v7 = (−2, 0, 0). This tetrahedron volume is given by the determinant 0 0 −2 1 4 0 = . Vol(tetrahedron) = 0 −2 6 −2 3 0 0 The tetrahedron defined by {v8 , v2 , v4 , v6 } has the same volume, so it follows that (9)

Vol(RC) = 8 −

8 16 = . 3 3

LIKELIHOOD OF VOTING OUTCOMES WITH GENERALIZED IAC PROBABILITIES

17

The cyclic region in the positive orthant (Fig. 1b) is the tetrahedron defined by the origin and the unit vectors on the three positive axis, {(0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 0, 1)}, so, using the three vectors with their base at the origin, its volume is 1 0 0 1 1 (9) Vol(cyclic region in positive orthant) = 0 1 0 = . 6 0 0 1 6 Thus the volume of the two cyclic regions is 13 . This means that Vol(cyclic regions) = Vol(RC)

(9)

1 3 16 3

which is the sought after conclusion.

1 , 16

v4

z=p(C, A) .. ................................................

... ..... .. ... . ... ................... .................... ........ . .. .. ........................ ... .. ..................................... .. .. .................... .. .... . . . . . . . . . ............................. ... .. .......... ....... ... ... ..... ............ .. ..... ..... .. ........ ........ . . . . . ..... . .... ...... .. ... ........................................................................................................... ..... ... ..... .... . . . . ... ..... ..... ..... .... ... ..... ... .

=

v5

........................................................................................................... ..... .... ... ......... ....... ..... ..... ............... ..... ... .. .... ............ .... . .. ... . .................... . . .. .... .................................... ... . . . . . . . . . . . . . . . . .. ..... .......................................... ..... ........... ... ... ..... .......... ... .... ..................................... .................. ... .................................................... ... .. ... .. ........................................................................... .... ......... . ... ....................................... .. ... ....... ................................................................................................................................ ..... ..... ... ..... ... .. ..... ..... ... ... ........................................................................................................................................... . . . . . . ... . . . . . . . . . . . . . . . . . . . . . . . . . . ... . . . . . . . . . . . . ................ .... . . .. . . ...... .. . ......... ............................................................................................................... ... . ..... .......... . ... . . ..... . ...... ...... . ... . . . ..... .. ........ ......... ..... ... . ..... ... ... ......... ..... ..... ... ... ..... ..... ... ........ ... ..... ....... ... . . ...................................................................................

v3

y=p(B, C)

x=p(A, B)

v6

v2 v1 a. Disagreement in B  C  A orthant b. Disagreement in three of six orthants Figure 3. Comparison of Borda, Condorcet winners 4.2. Borda and Condorcet. The result needed to prove the assertion about the Borda and Condorcet winners agreeing is Thm. 4 from S&M (also see [11]). It asserts that for candidates U, V, W , if U and W are, respectively, the Condorcet winner and loser, then a necessary and sufficient condition for Borda and Condorcet winners to agree is (9)

2P (U, V ) + P (U, W ) > P (V, W ).

In each of the six Fig. 1a transitive regions, the regions with agreement, and disagreement, can be plotted. Figure 3a depicts this region for the B  C  A orthant, where B and A are, respectively, the Condorcet winner and loser: The shaded portion represents disagreement. The shaded Fig. 3b portions identify where there is disagreement in three of the six transitive ranking orthants. Starting at the top and rotating in a clockwise manner, these orthants are B  C  A, A  B  C, C  A  B. (Regions in the remaining three orthants are hidden from this perspective.) In the B  C  A orthant, Eq. 4.2 becomes 2p(B, C) + p(B, A) > p(C, A). When expressed in terms of (x, y, z) coordinates, it is 2y − x > z. This means that the (9)

2y = x + z

plane separates N -points with the same Condorcet and Borda winners from those N -points with disagreement. As the region is defined by a plane passing through the origin, and the planes of the orthants and surface of RC, Prop. 1 applies. Thus, only the relative volumes need to be computed.

18

TOMAS J. MCINTEE, DONALD G. SAARI

The octahedral distribution giving the probability of agreement between the Borda and Condorcet winners is the geometric volume of the Fig. 3a unshaded region to the volume of the full orthant. To find this volume, the Eq. 4.2 plane intersects the two RC surfaces z = 1 and x + y + z = 1, along with the A ∼ B and B ∼ C surfaces of x = 0 and y = 0. Thus the plane passes through the four points 1 1 1 2 (0, 0, 0), (0, , ), (− , , 1), (−1, 0, 1) 3 3 3 3 This means that the region of the shaded Fig. 3 region, where the winners disagree, is defined by the two tetrahedrons with vertices 1 2 1 1 1 1 {(0, 0, 0), (0, 0, 1), (0, , ), (− , , 1)} and {(0, 0, 0), (0, 0, 1), (− , , 1), (−1, 0, 1)} 3 3 3 3 3 3 with respective volumes of 0 0 1 0 0 1 1 1 1 1 1 1 0 13 23 = , − 3 3 1 = 6 1 1 54 6 18 −1 0 1 −3 3 1

1 1 4 for a total volume of 54 + 18 = 54 . Therefore, the volume of the region of agreement in any 4 41 5 transitive region is 6 − 54 = 54 . 1 The volume of RC is 16 3 (Eq. 4.1), and the volume of each cyclic region (Eq. 4.1) is 6 . Thus the volume of each of the six transitive regions is one-sixth [the volume of RC minus 1 5 the volume of the cyclic regions], or 16 [ 16 3 − 3 ] = 6 . Thus, the fraction of Borda-Condorcet

agreement in this region is the desired value of

41 54 5 6

=

41 45

= 0.911 . . . .

While Prop. 1 extends immediately to the Fig. 1a geometry, it is stated for the full RC. The volume of agreement within each transitive region is 41 54 , so the volume of agreement 41 41 in RC is 6 × 54 = 9 . Thus, probability of Condorcet winner agreeing with Borda winner is 41 9 16 3

=

41 48 .

This probability, however, includes sections where a Condorcet winner does not

exist. To complete the proof, we need to find the probability of agreement subject to the condition that the Condorcet winner exists. Let Bj and Cj be where j = A, B, C is, respectively, the Borda and the Condorcet winner. Thus the conditional probability that the Borda and Condorcet winners agree subject to the condition that there is a Condorcet winner is 41 Prob(∪j (Bj ∩ Cj )) 41 = 48 15 = 45 = 0.911 . . . , Prob(∪j Cj ) 16 which completes the proof.

References [1] Black, D., The Theory of Committees and Elections, Cambridge University Press, New York, 1958. (Reissued edition, 2011.) [2] Cervone, D., W.V. Gehrlein, and W. Zwicker. Which scoring rule maximizes Condorcet efficiency under IAC? Theory and Decision, 58 (2005), 145-185.

LIKELIHOOD OF VOTING OUTCOMES WITH GENERALIZED IAC PROBABILITIES

19

[3] Diss, M. and W.V. Gehrlein. Borda’s paradox with weighted scoring rules. Social Choice and Welfare, 38 (2012), 121-136. [4] Diss, M. and W.V. Gehrlein. The true impact of voting rule selection on Condorcet efficiency. Economics Bulletin, 35 (2015), 2418-2426. [5] Garvey, D., A theorem on the construction of voting paradoxes, Econometrica 21 (1953), 608-610. [6] Gehrlein, W. V, and P. C. Fishburn, Condorcet’s paradox and anonymous preference profiles, Public Choice 26 (1976), 1 - 18. [7] Gehrlein, W. V. and D. Lepelley, The Condorcet Efficiency of Borda Rule with Anonymous Voters, Mathematical Social Sciences, 41 (2001), 39-50. [8] Gehrlein, W. V. and D. Lepelley, Voting Paradoxes and Group Coherence, Springer, New York, 2011. [9] Merlin, V., M. Tataru, M. and F. Valognes, On the probability that all decision rules select the same winner, Journ. Math. Econ., 33 (2000), 183-207. [10] Saari, D. G., Basic Geometry of Voting, Springer, New York, 1995. [11] Saari, D. G., Explaining All Three-Alternative Voting Outcomes, Jour. Econ. Theory 87 (1999), 333355. [12] Saari, D. G., Analyzing a nail-biting election, Social Choice & Welfare 18 (2001), 415-430. [13] Saari, D. G., Disposing Dictators, Demystifying Voting Paradoxes Cambridge, New York, 2008. [14] Saari, D. G., Unifying voting theory from Nakamura’s to Greenberg’s Theorems, Mathematical Social Sciences, 69 (2014), 1-11. [15] Saari, D. G., and M. Tataru, The likelihood of dubious election outcomes, Econ. Theory 13 (1999), 345-363. [16] Saari, D. G., and T. J. McIntee, Connecting pairwise and positional election outcomes, Mathematical Social Sciences, 66 (2013), 140-151. [17] Saari, D. G., and F. Valognes, Geometry, Voting, and Paradoxes, Math Magazine 4 (1998), 243-259. [18] Saari, D. G., and F. Valognes, The geometry of Black’s single peakedness condition, Jour. Math. Econ. 32 (1999), 429-456. [19] Ward, B. Majority voting and the alternative forms of public enterprise, pp 112126, in J. Margolis (Ed.) The Public Economy of Urban Communities. Baltimore MD, Johns Hopkins University Press, 1965 E-mail address: [email protected], [email protected] TJM is at the Dept. of Mathematics, Virginia Tech, Blacksburg, VA 24061-0123; DGS is at the Institute for Mathematical Behavioral Sciences, University of California, Irvine, CA. 92697-5100