An Analysis of Random Elections with Large Numbers of Voters
AAn Analysis of Random Electionswith Large Numbers of Voters ∗ Matthew Harrison-TrainorSeptember 8, 2020
Abstract
In an election in which each voter ranks all of the candidates, we consider the head-to-head results between each pair of candidates and form a labeled directed graph,called the margin graph, which contains the margin of victory of each candidate overeach of the other candidates. A central issue in developing voting methods is that therecan be cycles in this graph, where candidate A defeats candidate B , B defeats C , and C defeats A . In this paper we apply the central limit theorem, graph homology, andlinear algebra to analyze how likely such situations are to occur for large numbers ofvoters. There is a large literature on analyzing the probability of having a majoritywinner; our analysis is more fine-grained. The result of our analysis is that in electionswith the number of voters going to infinity, margin graphs that are more cyclic in acertain precise sense are less likely to occur. The Condorcet paradox is a situation in social choice theory where every candidate in anelection with three or more alternatives would lose, in a head-to-head election, to some othercandidate. For example, suppose that in an election with three candidates A , B , and C , andthree voters, the voter’s preferences are as follows:First choice Second choice Third choiceVoter 1 A B C
Voter 2
B C A
Voter 3
C A B
There is no clear winner, for one can argue that A cannot win as two of the three votersprefer C to A , that B cannot win as two of the three voters prefer A to B , and that C cannotwin as two of the three voters prefer B to C .More formally, fix a set V = { v , . . . , v n } of voters and a set C = { c , . . . , c (cid:96) } of candidates.Let L = L ( C ) be the set of all linear orders on C ; we think of such a linear order as a ranking ∗ We thank Wesley Holliday and Eric Pacuit for extensive comments and discussions. a r X i v : . [ ec on . T H ] S e p f the candidates. A profile is a map P : V → L , mapping each voter v to a ranking P ( v ) ofthe candidates; we call P ( v ) voter v ’s ballot . So a profile is exactly the data we might getfrom an election. We write c > P v d if voter v prefers candidate c to d in the profile P .Given a profile P , the margin of one candidate c over another d is the margin of vic-tory/loss of c over d in a direct comparison:Margin P ( c , d ) = { v ∈ V : c > P v d } − { v ∈ V : d > P v c } . If Margin P ( c , d ) > c is majority preferred to d . We will always consider the numberof voters to be odd so that for every pair of candidates, one is majority preferred to theother. We can construct a labeled directed margin graph M ( P ) whose vertices are thecandidates, and with an edge from c to d , labeled with Margin P ( c , d ), exactly when c ismajority preferred to d . For example, the margin graph of the profile described above is A (cid:127) (cid:127) B (cid:47) (cid:47) C (cid:95) (cid:95) When we forget the margins of victory, we obtain a tournament which we call the majoritygraph G ( P ) of P . Theorem 1.1 (Debord [Deb87]) . For any labeled tournament M , such that all weights ofedges have the same parity, there is a profile P such that M is the margin graph of P . The Condorcet paradox occurs when there is a cycle in the margin graph of a profile, sothat there are candidates c , . . . , c k such that c is majority preferred to c , c to c , and soon, and c k is majority preferred to c . If there is no cycle in the margin graph of a profile P ,then the margin graph is just a linear ordering of the candidates, and it is plausible that thewinner should be the greatest candidate according to this ordering. A central problem ofvoting theory is to come up with a voting method—formally a function mapping each profile P to a set of winning candidates (or sometimes to a ranking of all the candidates)—whichdeals as well as possible with cycles in the margin graph.Thus an important area of research has been to identify how often there occur cycles in themargin graph, both in historical situations and theoretically under various assumptions onthe voters. Riker [Rik58] argues that various amendments in the Agricultural AppropriationAct of 1953 in the US House of Representatives formed a cycle; Bjurulf and Niemi [BN78]found similar situations in the Swedish parliament; Stensholt [Ste99] found a cycle in a de-cision by the Norwegian national assembly; see also Van Deemen [Dee14], Kurrild-Klitgaard[KK01], and Truchon [Tru98].In this paper we take the more theoretical point of view where we assume that voters fillout their ballot randomly according to some probability distribution, and we consider theprobability of a paradox occurring There is also a large literature of results here. The bookby Gehrlein [Geh06] is an excellent reference for what is known.To begin, we must make an assumption about the probability distribution of ballots foreach voter. We assume for the rest of the paper that each voter is equally likely to pick any2able 1: Probability of a majority winner for n voters and (cid:96) candidates (cid:96)n ∞ (cid:96) ! linear orders on the (cid:96) candidates, an assumption referred to in the literature as theImpartial Culture (IC) condition. An important probability is the probability P W ( n, (cid:96) ) of having a Condorcet winner—asingle candidate who is majority preferred to each other candidate. Such a candidate can beargued to be a clear winner, and an important class of voting methods, the Condorcet meth-ods, select such a candidate as the winner. Using increasingly sophisticated methods, Sevcik[Sev69], DeMeyer and Plott [DP70], Niemi and Weisberg [NW68], Garman and Kamien,[GK68], and Gehrlein and Fishburn [GF76, GF79, Geh99] calculated these probabilities forsmall numbers of voters and candidates. These values are included in Table 1. One can alsocalculate other related probabilities, such as the probability P T ( n, (cid:96) ) of having a transitivemajority graph. Gehrlein [Geh88, Geh89] calculated these values for various n and (cid:96) .As the number of voters becomes large, it is known that due to the central limit theo-rem, the distribution of margin graphs approaches a multivariate normal distribution. Guil-baud [Gui52] was the first to compute for three candidates the probability P W ( ∞ ,
3) =lim n →∞ P W ( n,
3) = 0 . P W ( ∞ , (cid:96) ) = lim n →∞ P W ( n, (cid:96) ) for various numbers (cid:96) of candidates. These probabilitiesare shown in Table 1 as well. Gehrlein and Fishburn [GF78] computed other probabilitiessuch as that of being transitive or having a Hamiltonian cycle. Most of these are numericalapproximations.We quickly notice that the probability of avoiding a paradox is higher than one mightnaively expect. Given three candidates, there are eight possible majority graphs: six linearorders and two cycles. If these all occurred with equal likelihood, then we would expect theprobability P W ( ∞ ,
3) = 3 / P W ( ∞ ,
3) is much higherthan this. Intuitively, this is because each voter’s ballot is linearly ordered, and this makes We note that there is a wide range of work on what happens under assumptions other than the ImpartialCulture condition. One can apply the same sort of analysis in this paper to these other assumptions, thoughthe covariance matrix Σ obtained will be different. It would be very interesting to compare the resultsobtained. We leave this for future work. A beats B is positively correlated with the event that A beats C .So far most research has been on the probability of particular events occurring, suchas having a Condorcet winner, being transitive, having a Hamiltonian cycle, etc. In thispaper we will look at individual tournaments T , and the probability Pr( T ) of having thattournament as the majority graph of a random election with a large number of voters n → ∞ .We also have some results on the relative weighting of the edges in a majority graph.As remarked above, as the number n of voters goes to infinity, the distribution of margingraphs approaches a multivariate normal distribution. Such a distribution can be analyzedusing the covariance matrix which appears as a quadratic form in the probability densityfunction. This matrix is computed in Theorem 4.1. One can view the space of labeleddirected graphs as a vector space acted on by the covariance matrix. The vector space oflabeled directed graphs is known to split as a direct sum of the cycle space and cut space .We show in Theorem 5.1 that these two spaces are exactly the eigenspaces of the covariancematrix of our distribution, and that moreover, the eigenvalue of the former is smaller thanthat of the latter. What this means is that: The more cyclic a margin graph is, the less likely it is to arise.
By this we mean that the probability density function takes lower values on margin graphswhich are more cyclic.Margin graphs are labeled, but to compute Pr( T ) for a tournament T we need to in-tegrate the probability density function over all of the margin graphs compatible with T .Unfortunately, such probabilities are not well-understood for five or more candidates, butwe formulate a conjecture which we check holds for up to five candidates. Definition 1.2.
Given a tournament T on (cid:96) vertices, we can assign a number to T whichwe call the linearity of T :lin( T ) = 12 (cid:88) v deg − ( v ) + deg + ( v ) = (cid:88) v deg − ( v ) = (cid:88) v deg + ( v ) . This value differs by a constant (depending on (cid:96) ) from the following value: for each pair ofedges meeting at a common vertex v , add 1 / v , and subtract 1 / Conjecture 6.3. If lin( T ) < lin( T (cid:48) ) , then Pr( T ) < Pr( T (cid:48) ) . In Sections 5, 7, and 8 we compute the probability of obtaining each possible majority graphin the cases of three, four, and five candidates respectively, and find that the probabilitieswe compute agree with this conjecture.This paper was motivated by a question of Holliday and Pacuit posed in Holliday’s 2019seminar on Voting and Democracy at the University of California, Berkeley. Many voting4ethods are margin based in the sense that the set of winners of an election depends only onthe margin graph. Of these, some voting methods are majority based ; they chose a winnerbased solely on the majority graph, that is, depending only on which candidates defeat whichother candidates head-to-head.Holliday and Pacuit [HPa] define an intermediate category of qualitative-margin basedmethods. We first need the following definition:
Definition 1.4.
A qualitative margin graph is a pair ( M, ≺ ) where M is a majority graphand ≺ is a strict weak order on the set of edges of M . The qualitative margin graph of aprofile P is the pair ( G ( P ); ≺ P ) such that for any edges ( a, b ) and ( c, d ) in G ( P ), we have( a, b ) ≺ P ( c, d ) if Margin P ( a, b ) < Margin P ( c, d ).A voting method is said to be qualitative-margin based if it selects the set of winners basedonly on the qualitative margin graph. There are examples of voting methods which arequalitative-margin based but not majority based, such as the Simpson-Kramer Minimaxmethod [Sim69, Kra77], Ranked Pairs [Tid87], weighted covering solutions [DL99, DBT00],Beat Path [Sch11], and Split Cycle [HPc, HPa].Holliday and Pacuit asked whether any qualitative margin graph, with the ordering onthe edges being total, is obtained with positive probability even as the number of voters goesto infinity. We require that the ordering on edges be total because it is very unlikely, withlarge numbers of voters, to have two different margins be equal. We show that the answerto this question is positive: Theorem 4.2.
Let T be a tournament on a set of candidates C and let ≺ be an ordering ofthe edges of T . There is a number N and a positive probability p > such that: Given a set V of voters with |V| ≥ N , the probability is at least p that the qualitative margin graph of arandomly chosen profile P : V → L ( C ) is ( T, ≺ ) . Of course, this also means that any tournament is obtained as the majority graph withpositive probability.
Corollary 1.6.
Let T be a tournament on a set of candidates C . There is a number N anda positive probability p > such that: Given a set V of voters with |V| ≥ N , the probabilityis at least p that the majority graph of a randomly chosen profile P : V → L ( C ) is T . As an immediate consequence of Theorem 4.2, most behaviours that a qualitative-marginbased voting method could have with a small number of voters will also happen with posi-tive probability even with large numbers of voters. (The only exception is that with smallnumbers of voters, there might be ties or two margins might be equal, and both of thesesituations are unlikely with large numbers of voters.) There are too many applications ofthis to list them all, but we will give two examples.Holliday and Pacuit [HPc] were interested in whether their new voting method
Split Cycle satisfies a certain property of asymptotic resolvability for k >
Theorem 1.7.
Split Cycle does not satisfy asymptotic resolvability for k > candidates:As the number n of voters approaches infinity, there is a positive probability that Split Cycleselects more than one winner. D is the Condorcet loser, but its greatest loss is only by 5. Eachother candidate is defeated by a margin of 7, 9, or 11, and so D is selected as the winner. A (cid:127) (cid:127) (cid:15) (cid:15) B (cid:47) (cid:47) (cid:31) (cid:31) C (cid:95) (cid:95) (cid:127) (cid:127) D . This margin graph corresponds to the qualitative margin graph A γ (cid:127) (cid:127) σ (cid:15) (cid:15) B β (cid:47) (cid:47) τ (cid:31) (cid:31) C α (cid:95) (cid:95) ρ (cid:127) (cid:127) D with α (cid:31) β (cid:31) γ (cid:31) ρ (cid:31) σ (cid:31) τ . We can then prove: Theorem 1.8.
As the number of voters approaches infinity, Minimax has a positive proba-bility of choosing as the unique winner the Condorcet loser.
The final section of this paper includes brief remarks on performing Monte Carlo simula-tions. Given a voting method, one might want to know, for example, how often it chooses aunique winner, or how often it deviates from some other voting method. Such calculationsare performed for example by Holliday and Pacuit in [HPc], where for n voters they pick n ballots at random and compute the margin graph. For large numbers of voters, one caninstead pick a random margin graph from the multinomial distribution on margin graphswith large numbers of voters; this simply requires applying a linear transformation to inde-pendent normal random variables. We describe this method and give sample computationsin Section 9. This method is faster than a voter-by-voter simulation with a number of votersmuch larger than the number of candidates, and has already been used in [HPb]. We begin with a quick review of multivariate distributions and the central limit theorem.6et X = ( X , . . . , X k ) be a k -dimensional random vector. We will define what it meansfor X to have normal distribution X ∼ N k ( µ , Σ ) with mean µ = ( µ , . . . , µ k ) = E [ X ]and k × k covariance matrix with entriesΣ i,j = E [( X i − µ i )( X j − µ j )] = Cov[ X i , X j ] . The covariance matrix is always positive semi-definite, and it is positive definite if and onlyif it is invertible.If Σ is positive definite then we are in what is called the non-degenerate case, and thedistribution has probability density function f X ( x ) = e − ( x − µ ) T Σ − ( x − µ ) (cid:112) (2 π ) k | Σ | where | Σ | is the determinant of Σ . Note that the level sets of f X are ellipses.The book by Tong [Ton90] is a good reference for many properties of the multivariatenormal distribution distribution.The importance of the multivariate normal distribution is due to the central limit theoremfor multivariate random variables. Let X , X , X , . . . be i.i.d. k -dimensional random vectorswith mean µ and covariance matrix Σ . Define S n = 1 n n (cid:88) i =1 X i . The multivariate central limit theorem says that √ n ( S n − µ ) converges in distribution tothe multivariate normal distribution with mean 0 and covariance matrix Σ : √ n ( S n − µ ) D −→ N k ( , Σ ) . What we mean by convergence in distribution is: For a sequence of random vectors X , X , . . . ∈ R k , we say that this sequence converges in distribution to a random k -vector X if for each A ⊆ R k which is a continuity set of X (i.e., A is a Borel and Pr( X ∈ ∂A ) = 0),lim n →∞ Pr( X n ∈ A ) = Pr( X ∈ A ) . See the book by van der Vaart [vdV98] as a reference on the central limit theorem.
There is a homology theory for graphs in which one constructs for a graph G and field F aquite simple chain complex · · · → → E F ( G ) ∂ → V F ( G ) → → · · · .
7o knowledge of homology will be necessary as all of the objects we use are easily definablein purely graph-theoretic terms. This material is contained in most books covering algebraicgraph theory, e.g., [Big93].Fix for the rest of this section a graph G , which in this paper will always be the completegraph K n on n vertices, and the field F = R . As is usual in homology, we must fix anorientation for each edge—for K n the complete graph on v , . . . , v n , we can think of an edgebetween v i and v j as a directed edge v i → v j for i < j .The edge space of G = ( V, E ) is the R -vector space E ( G ) = R E . We can think of eachelement of E ( G ) as a formal sum (cid:80) e ∈ E r e e , i.e., a labeling of each edge of the graph witha real number. The orientation of an edge ( u, v ) is essentially just a choice to say that apositive number assigned to the edge is going from u to v , and a negative number from v to u ; if we had chose the opposite orientation ( v, u ), then a positive number assigned to theedge would be going from v to u , and a negative number from u to v . So if an edge hasorientation ( u, v ), we can define ( v, u ) = − ( u, v ).The vertex space of G = ( V, E ) is similarly defined using formal sums of vertices: it is the R -vector space V ( G ) = R V , and we think of each element of V ( G ) as a formal sum (cid:80) v ∈ V r v v .There is also a boundary operator ∂ : E ( G ) → V ( G ) which takes ( u, v ) to u − v .The homology group H ( G ) = ker ∂ is easily seen to be the cycle space of G . This is thesubspace C ( G ) of E ( G ) consisting of all of those elements of the edge space with the propertythat, for each vertex, the sum of the numbers assigned to each incoming edge is equal to thesum of the numbers assigned to each outgoing edge; i.e., those elements (cid:80) e ∈ E r e e such thatfor every vertex u , (cid:88) ( u,v ) ∈ E r ( u,v ) − (cid:88) ( v,u ) ∈ E r ( u,v ) = 0 . Using the convention ( v, u ) = − ( u, v ) described above, we can just write (cid:88) ( u,v ) ∈ E r ( u,v ) = 0 . If u → u → · · · → u n → u is a cycle in G , then ( u , u ) + ( u , u ) + · · · + ( u n , u ) is anelement of the cycle space. In fact, the cycle space is generated by all such elements.The cut space of G is the subspace C (cid:48) ( G ) of E ( G ) generated by the edge cuts of G . A cut of G is a partition of the vertices of G into two sets S and T ; the cut-set of the cut is theset of edges with one end in S and the other end in T . The cut space is generated by theelements (cid:88) ( u,v ) ∈ E, u ∈ S, v ∈ T ( u, v )where ( S, T ) is a cut, and we again use the convention that ( v, u ) = − ( u, v ).Let T be a spanning tree of G . We can use T to compute bases for the cycle space andcut space. First, for each edge e = ( u, v ) not in T , there is a unique path in T from u to v ;together with e , this forms a cycle. We call the set of all such cycles the fundamental systemof cycles associated with T . It is not hard to see that these cycles are linearly independent,because each of them contains an edge not contained by any of the others. In fact, they forma basis for the cycle space. By counting, we see thatdim C ( G ) = | E | − | V | + c ( G )8here c ( G ) is the number of connected components of G . For the cut set, it will be easiestto think about the case when G is connected, and this will be the only case we use in thepaper. Given an edge e of the spanning tree T , T − { e } splits into two connected components S and S which partition V , forming a cut of G containing e (and no other edge of T ) inits edge set. The set of all such cuts forms a basis for the cycle space, anddim C (cid:48) ( G ) = | V | − c ( G ) . Note that dim C ( G ) + dim C (cid:48) ( G ) = dim E ( G ). In fact, we can equip the edge space with thenatural inner product, taking the edges as an orthonormal basis, and we have: Theorem 3.1.
The cycle space and the cut space are orthogonal complements, so that E ( G ) = C ( G ) ⊕ C (cid:48) ( G ) . As mentioned above, in this paper we will always take G to be the complete graph on n vertices v , . . . , v n and the orientation of each edge to be from v i to v j for i < j . As aspanning tree, we can take the tree v (cid:118) (cid:118) (cid:126) (cid:126) (cid:15) (cid:15) (cid:40) (cid:40) v v v · · · v n Then as a basis for the cycle space we can take the elements( v , v i ) + ( v i , v j ) + ( v j , v ) = ( v , v i ) + ( v i , v j ) − ( v , v j )for i < j . For the cut space, we take for each i ≥ (cid:88) j (cid:54) = i ( v i , v j ) , i.e., the sum of all outgoing edges from v i . We begin by fixing some notation for the rest of the paper. Earlier we fixed a set of (cid:96) candidates C = { c , . . . , c (cid:96) } . Let G be the complete graph whose vertices are the candidates C , and choose the natural orientation for the edges: c i → c j if i < j . We often write i for c i when we are thinking of it as a vertex of G and write c i only when we need to be especiallyclear.Given a voter v , under the IC assumption, v chooses a ballet, i.e., a ranking of thecandidates, uniformly at random. We can associate with this a random vector X v = ( X v i,j ) i We have, for i, j, k, (cid:96) all distinct: Σ ( i,j );( i,j ) = Cov( X v i,j , X v i,j ) = Var( X v i,j ) = 1Σ ( i,j );( j,k ) = Cov( X v i,j , X v j,k ) = − ( i,j );( k,(cid:96) ) = Cov( X v i,j , X v k,(cid:96) ) = 0 . Using our convention that X v i,j = − X v j,i , we see thatΣ ( i,j );( i,k ) = Cov( X v i,j , X v i,k ) = 13 and Σ ( i,k );( j,k ) = Cov( X v i,k , X v j,k ) = 13 . Proof of Lemma 4.1. We omit the superscript v for this proof.First, Var( X i,j ) = E ( X i,j ) − E ( X i,j ) = 1since X i,j is always 1, and in exactly half of the linear orders L ∈ L ( C ) we have c i > L c j , sothat E ( X i,j ) = 0.For i , j , r , and s all distinct, we haveCov( X i,j , X r,s ) = E [ X i,j · X r,s ] − E [ X i,j ] · E [ X r,s ] = E [ X i,j · X r,s ] . Then X i,j and X r,s are independent and have covariance 0; indeed, among all of the linearorders L ∈ L ( C ) with c i > L c j , exactly half of them have c r > L c s , and the other half have c s > L c r .Finally, given i, j, k distinct, we haveCov( X i,j , X j,k ) = E [ X i,j · X j,k ] − E [ X i,j ] · E [ X j,k ] = E [ X i,j · X j,k ] . Half of the linear orders have c i > c j . Of these, have c i > c j > c k , have c i > c k > c j ,and have c k > c i > c j . Similarly, half the linear orders have c j > c i , and of these, have c j > c i > c k , have c j > c k > c i , and have c k > c j > c i . SoCov( X i,j , X j,k ) = E [ X i,j · X j,k ] = 16 (1 − − − − − . P using n voters V = { v , . . . , v n } . Ifeach voter chooses a ballot uniformly at random, then the random margin graph we obtainis a random variable equal to X v + · · · + X v n ; this is again an element of the integral edgespace E Z ( G ). The X v i are i.i.d. with mean µ = and covariance matrix Σ . As we see belowin Theorem 4.3 (where we compute the inverse) or in Theorem 5.1 (where we compute theeigenvalues), Σ is invertible and hence positive semi-definite; we leave the proofs to lateras they are heavily computational. Let S n = n (cid:80) ni =1 X v i . By the central limit theorem, √ n · S n converges in distribution to the multivariate normal distribution N ( , Σ ) with thesame mean and covariance matrix, which has probability density function f ( x ) = e − x T Σ − x (cid:112) (2 π ) k | Σ | . That is, if P n is a random variable which represents a random profile given by n voterschoosing a ballot under the IC assumption, then √ n P n converges in distribution to N ( , Σ ).Let Y ∼ N ( , Σ ). Then we think of Y as an element of the real-valued edge space E R ( G )which represents the margin graph of a random profile with a very large number of voters,with the edge weights normalized by a factor of √ n . From this, without even making anyfurther analysis of Σ , we can already make several conclusions. For example, the margin ofvictory of one candidate over another is usually on the order of √ n .The fact that f ( x ) is always positive for every x means that, given Y ∼ N ( , Σ ) andany region R ⊆ E R ( G ) with positive measure, Pr( Y ∈ R ) > 0. As a consequence, for anytournament, we get a positive probability of that tournament being the majority graph of arandom profile. Moreover, we can prove Theorem 4.2 which says that the same is true forqualitative margin graphs. Theorem 4.2. Let T be a tournament on a set of candidates C and let ≺ be an ordering ofthe edges of T . There is a number N and a positive probability p > such that: Given a set V of voters with |V| ≥ N , the probability is at least p that the qualitative margin graph of arandomly chosen profile P : V → L ( C ) is ( T, ≺ ) .Proof. For a given set V of n voters, randomly choose a profile P : V → L ( C ) by choosing,for each voter v ∈ V , a ballot P ( v ) ∈ L ( X ) uniformly among all possible ballots with eachballot being chosen by a particular voter with probability n ! . Taking the margin graph of P , and dividing by the number n of voters, we generate a random vector S n from E R ( G ) asabove. As n → ∞ , √ n · S n converges in distribution to the multinormal distrubtion withprobability density function f ( x ) = exp( − ( x − µ ) T Σ − ( x − µ )) (cid:112) (2 π ) k | Σ | . Let Y = ( Y c i > c j ) i The inverse of Σ is the matrix Γ = Σ − with entries Γ ( i,j ) , ( i,j ) = 3( (cid:96) − (cid:96) + 1Γ ( i,j ) , ( i,k ) = − (cid:96) + 1 , and for i, j, r, s all distinct, Γ ( i,j ) , ( r,s ) = 0 . We again use the convention that Γ ( i,j ) , ( r,s ) = − Γ ( j,i ) , ( r,s ) = − Γ ( i,j ) , ( s,r ) = Γ ( j,i ) , ( s,r ) , so that,e.g., Γ ( i,j ) , ( j,k ) = 3 (cid:96) + 1 . Proof. We check that ΓΣ is the identity matrix by checking entry by entry. For i < j ,( ΓΣ ) ( i,j ) , ( i,j ) = (cid:88) ( r,s ) Γ ( i,j ) , ( r,s ) · Σ ( r,s ) , ( i,j ) = Γ ( i,j ) , ( i,j ) · Σ ( i,j ) , ( i,j ) + (cid:88) k (cid:54) = i,j Γ ( i,j ) , ( i,k ) · Σ ( i,k ) , ( i,j ) + (cid:88) k (cid:54) = i,j Γ ( i,j ) , ( j,k ) · Σ ( j,k ) , ( i,j ) = 1 · (cid:96) − (cid:96) + 1 − · ( (cid:96) − · · (cid:96) + 1= 1 . Note that for k < i , the middle terms on the second line are really Γ ( i,j ) , ( k,i ) · Σ ( k,i ) , ( i,j ) butthis is equal to Γ ( i,j ) , ( i,k ) · Σ ( i,k ) , ( i,j ) , and similarly for the last terms when k < j . We willmake the same kind of adjustments throughout the proof.12or i, j, k distinct, we have( ΓΣ ) ( i,j ) , ( i,k ) = (cid:88) ( r,s ) Γ ( i,j ) , ( r,s ) · Σ ( r,s ) , ( i,k ) = Γ ( i,j ) , ( i,k ) · Σ ( i,k ) , ( i,k ) + Γ ( i,j ) , ( i,j ) · Σ ( i,j ) , ( i,k ) + Γ ( i,j ) , ( j,k ) · Σ ( j,k ) , ( i,k ) + (cid:88) t (cid:54) = i,j,k Γ ( i,j ) , ( i,t ) · Σ ( i,t ) , ( i,k ) = − · (cid:96) + 1 + 13 · (cid:96) − (cid:96) + 1 + 13 · (cid:96) + 1 − ( (cid:96) − · (cid:96) + 1 · 13= 0One of course must note, from the first line, that our normal convention still holds in thesense that, e.g., ( ΓΣ ) ( i,j ) , ( i,k ) = − ( ΓΣ ) ( j,i ) , ( i,k ) .Now given i, j, r, s all distinct, we have:( ΓΣ ) ( i,j ) , ( r,s ) = (cid:88) ( u,v ) Γ ( i,j ) , ( u,v ) · Σ ( u,v ) , ( r,s ) = Γ ( i,j ) , ( i,r ) · Σ ( i,r ) , ( r,s ) + Γ ( i,j ) , ( i,s ) · Σ ( i,s ) , ( r,s ) + Γ ( i,j ) , ( j,r ) · Σ ( j,r ) , ( r,s ) + Γ ( i,j ) , )( j,s ) · Σ ( j,s ) , ( r,s ) = 3 (cid:96) + 1 · − (cid:96) + 1 · − (cid:96) + 1 · 13 + 3 (cid:96) + 1 · 13= 0So ΓΣ is the identity matrix, and Γ = Σ − . We know that each tournament has a positive probability of being obtained as the majoritygraph of a random election. But which tournaments are more likely than others?Let P n be a random variable which represents a random margin graph given by n voterschoosing a ballot under the IC assumption. We know that √ n P n converges in distributionto N ( , Σ ), with probability density function f ( x ) = e − x T Σ − x (cid:112) (2 π ) k | Σ | . So to understand what a random margin graph looks like with a large number of voters,we want to understand this distribution. We view this distribution as taking values in thereal-valued edge space E R ( G ). In particular, we view Σ and its inverse Σ − as acting on E R ( G ).The geometry of the probability density function is determined by the quadratic form x T Σ − x . We analyse this by computing its eigenvalues and eigenvectors. This is where thecycle space and cut space make an appearance.Covariance matrices are always positive semidefinite, and as Σ is invertible it is positivedefinite. So the eigenvalues of Σ are all positive. The inverse Σ − has the same eigenspaceswith reciprocal eigenvalues. Because Σ is symmetric, its eigenspaces will be orthogonal.13 heorem 5.1. Σ has two eigenvalues: • λ = with eigenspace the cycle space C R ( G ) ; and • λ = + (cid:96) − with eigenspace the cut space C (cid:48) R ( G ) . Recall that the cycle space and the cut space are orthogonal. The inverse Σ − has thesame eigenspaces with eigenvalues 1 /λ and 1 /λ respectively. Proof of Theorem 5.1. Consider a cycle ( c i , c j ) + ( c j , c k ) + ( c k , c i ) ∈ E R ( G ). Then Σ acts onthis as Σ · (( c i , c j ) + ( c j , c k ) + ( c k , c i )) = (cid:88) r 13 = 13 . A similar argument works for ( c j , c k ) and ( c k , c i ). If r and s are distinct from i, j, k , thenthe coefficient of ( c r , c s ) is clearlyΣ ( i,j );( r,s ) + Σ ( j,k );( r,s ) + Σ ( k,i );( r,s ) = 0 + 0 + 0 = 0 . If, say, r = i and s is distinct from i, j, k , thenΣ ( i,j );( i,s ) + Σ ( j,k );( i,s ) + Σ ( k,i );( i,s ) = 13 + 0 − 13 = 0 . So we conclude that Σ · (( c i , c j ) + ( c j , c k ) + ( c k , c i )) = 13 (( c i , c j ) + ( c j , c k ) + ( c k , c i )) . So one eigenvalue of Σ is , and each element of the cycle space is an eigenvector.Now consider for a fixed i the cut (cid:80) j (cid:54) = i ( c i , c j ). Σ acts on this as Σ · (cid:32)(cid:88) j (cid:54) = i ( c i , c j ) (cid:33) = (cid:88) r 13 13 − with inverse Σ − = 32 34 − − − − 34 32 . The cycle space has dimension one and is generated by the cycle ( A , B ) + ( B , C ) + ( C , A ).This is the eigenspace of Σ corresponding to the eigenvalue λ = , and the correspondingeigenvalue of Σ − is 12. The cut space has dimension two and contains the elements ( A , B ) +( A , C ), ( B , A )+( B , C ) , and ( C , A )+( C , B ). It is the eigenspace corresponding to the eigenvalue λ = + (cid:96) − = of Σ and 3 of Σ − .Now consider the probability density function f ( x ) = e − x T Σ − x (cid:112) (2 π ) k | Σ | . x as a linear combination of cycles and cuts, say x = y + z with y ∈ C ( G ) an elementof the cycle space and z ∈ C (cid:48) ( G ) an element of the cut space. The vectors y and z areorthogonal, and so x T Σ − x = 1 λ || y || + 1 λ || z || = 12 · || y || + 3 · || z || . The larger x T Σ − x is, the smaller f ( x ) is; so, given that || x || = || y || + || z || , for a fixedvalue of || x || , f ( x ) is maximized for x in the cut space and minimized for x in the cyclespace; and in general, the closer x is to being a cut, and the further it is from being a cycle,the larger f ( x ) is.Now let us consider the geometry of the level sets of f ( x ). These level sets are ellipsoids,and the lengths of the axes are 1 / √ λ = 1 / √ 12 and √ λ = 1 / √ 3, with λ < λ . So thecycle space corresponds to the minor axis of the ellipsoid, and the cut space corresponds tothe major axis. We plot one of these level sets, x T Σ − x = 1, in the three-candidate case,in Figure 1. What we see is that the ellipsoid is shorter in two of the octants, as the minoraxis of the ellipsoid is along the vector ( A , B ) + ( B , C ) + ( C , A ) which is the diagonal vectorof an octant.Now an octant corresponds to choosing a direction for each edge of the graph G , i.e., toa tournament on the candidates. There are eight tournaments on three candidates, the firstsix of which are the linear orders, and last two of which are cycles.Linear orders Cycles A (cid:127) (cid:127) (cid:31) (cid:31) A (cid:31) (cid:31) A (cid:127) (cid:127) A (cid:127) (cid:127) B (cid:47) (cid:47) C B (cid:63) (cid:63) (cid:47) (cid:47) C B C (cid:111) (cid:111) (cid:95) (cid:95) B (cid:47) (cid:47) C (cid:95) (cid:95) A (cid:127) (cid:127) (cid:31) (cid:31) A A A (cid:31) (cid:31) B C (cid:111) (cid:111) B (cid:63) (cid:63) (cid:47) (cid:47) C (cid:95) (cid:95) B (cid:63) (cid:63) C (cid:111) (cid:111) (cid:95) (cid:95) B (cid:63) (cid:63) C (cid:111) (cid:111) The minor axis of the ellipsoid is within the two octants corresponding to the cycles, and themajor axis (which has dimension two) is orthogonal to this. Now this means that the volumeof the ellipsoid within each of the two cycle octants is less than its volume within each ofthe other six. As f ( x ) decreases as one gets further away, this means that the integral of f ( x ) over each of the six octants will be larger than the integral of f ( x ) over the two cycleoctants.This means that for a fixed tournament T , the probability that T is the majority graphof a random profile is larger when T is a linear order than it is when T is a cycle. Since thereare six linear orders and two cycles, the probability that the majority graph is a linear order(i.e., that there is a Condorcet winner) is at least , and the probability that the majoritygraph is a cycle (i.e., that the paradox of voting occurs) is at most .16igure 1: x T Σ − x = 1.In the case of three candidates, there are exact expressions for the probabilities of lying ina particular octant and hence of obtaining each of these tournaments [NW68, GK68]. E.g.,for Σ = a ba cb c the probability of being in the positive orthant is p = arccos( − a )4 π + arccos( − c )4 π − arccos( b )4 π . p = arccos( − / π = 0 . . . . and for the each of the two cycles, the probability is p = arccos(1 / π − arccos( − / π = 0 . . . . . With more candidates, the same sort of analysis works, except that there are more possibletournaments, and the ellipsoid sits in a higher-dimensional space, e.g., with four candidatesit sits in six dimensional space and with five candidates in ten dimensional space. Unfortu-nately there are exact expressions for orthant probabilities in higher dimensions only in veryparticular special cases. Recall that if we write x ∈ E R ( G ) as a linear combination of cycles and cuts, say x = y + z with y ∈ C ( G ) an element of the cycle space and z ∈ C (cid:48) ( G ) and element of the cut space, wehave x T Σ − x = 1 λ || y || + 1 λ || z || . So it is natural to try to compute || y || and || z || from x .Given x = (cid:80) ( c i , c j ) x ( c i , c j ) ( c i , c j ) ∈ E R ( G ) and a candidate c i , we introduce the valueflow x ( c i ) = (cid:88) ( c i , c j ) x ( c i , c j ) . We think of this as the net flow out of the vertex c i , counting flow out of c i positively andflow into c i negatively. It is not hard to see that for a fixed c i , flow x ( c i ) is a linear functionof x . Moreover, if x is part of the cycle space, then flow x ( c i ) = 0; one can easily see this bychecking that it is true for the basic cycles ( c i , c j ) + ( c j , c k ) + ( c k , c i ) and extends linearly tothe whole cycle space.Then we can compute: Theorem 6.1. Let x ∈ E R ( G ) be x = (cid:80) ( c i , c j ) x ( c i , c j ) ( c i , c j ) . Write x = y + z with y ∈ C ( G ) an element of the cycle space and z ∈ C (cid:48) ( G ) an element of the cut space. Then z = (cid:88) ( c i , c j ) flow x ( c i ) − flow x ( c j ) (cid:96) ( c i , c j ) . Moreover, || z || = 2 (cid:96) · (cid:88) ( c i , c j ) x c i , c j ) + (cid:88) ( c i , c j ) , ( c i , c k ) x ( c i , c j ) x ( c i , c k ) . where the convention x ( c i , c j ) = − x ( c j , c i ) handles the sign of the terms x ( c i , c j ) x ( c i , c k ) . roof. Since flow is a linear operator, and y is in the cycle space, we have for each vertex c i that flow x ( c i ) = flow y ( c i ) + flow z ( c i ) = flow z ( c i ) . Now writing z = (cid:80) ( c i , c j ) z ( c i , c j ) ( c i , c j ) we argue thatflow z ( c i ) − flow z ( c j ) = (cid:96)z ( c i , c j ) . It suffices by linearity to show that this is true for a generating set of the cut space, i.e., thatit is true for each of the cuts (cid:80) c k ( c i , c k ), (cid:80) c k ( c i , c k ), and for a fixed k ∗ (cid:54) = i, j , (cid:80) c k ( c k ∗ , c k ).For z = (cid:80) c k ( c i , c k ), we haveflow z ( c i ) − flow z ( c j ) = ( (cid:96) − − ( − 1) = (cid:96), and the coefficient of ( c i , c j ) in z is 1. A similar argument works for z = (cid:80) c k ( c j , c k ). For z = (cid:80) c k ( c k ∗ , c k ), we have flow z ( c i ) − flow z ( c j ) = − − ( − 1) = 0 , and the coefficient of ( c i , c j ) in z is 0. Thus we haveflow x ( c i ) − flow x ( c j ) = (cid:96)z ( c i , c j ) and so z = (cid:88) ( c i , c j ) flow x ( c i ) − flow x ( c j ) (cid:96) ( c i , c j ) . This completes the first part of the theorem.Now we must compute || z || . We have || z || = (cid:88) c i , c j z c i , c j ) = (cid:88) ( c i , c j ) (flow x ( c i ) − flow x ( c j )) (cid:96) = 1 (cid:96) (cid:88) ( c i , c j ) (cid:32)(cid:88) c r x ( c i , c r ) − (cid:88) c s x ( c j , c s ) (cid:33) . After expanding this out, we will get a linear combination of terms of the form x ( c i , c r ) · x ( c j , c s ) .First consider terms of the form x ( c t , c u ) · x ( c v , c w ) with t, u, v, w all distinct. Such a termshows up four times after expanding, for ( c i , c j ) equal to each of ( c t , c v ), ( c t , c w ), ( c u , c v ), and( c u , c w ). It shows up with the opposite sign for t as compared to u , and v as compared to w . So these all cancel out and the coefficient is zero.Next, consider terms of the form x ( c u , c v ) · x ( c u , c w ) . Such a term appears for: • ( c i , c j ) = ( c u , c v ): For c r = c v we get x ( c i , c r ) = x ( c u , c v ) , and for c r = c w we get x ( c i , c r ) = x ( c u , c w ) ; and for c s = c u we get x ( c j , c s ) = x ( c v , c u ) = − x ( c u , c v ) . After squaringwe get a term 4 · x ( c u , c v ) · x ( c u , c w ) . 19 ( c i , c j ) = ( c u , c w ): Similarly to the above, after squaring we get a term 4 · x ( c u , c v ) · x ( c u , c w ) . • c i = c u , c j (cid:54) = c v , c w : For c r = c v we get x ( c i , c r ) = x ( c u , c v ) , and for c r = c w we get x ( c i , c r ) = x ( c u , c w ) ; after squaring, we get 2 · x ( c u , c v ) · x ( c u , c w ) . • ( c i , c j ) = ( c v , c w ): For c r = c u we get x ( c i , c r ) = x ( c v , c u ) = − x ( c v , c u ) , and for c s = c u weget x ( c j , c s ) = x ( c w , c u ) = − x ( c w , c u ) ; after squaring, we get − · x ( c u , c v ) · x ( c u , c w ) .There are (cid:96) − x ( c u , c v ) · x ( c u , c w ) is 4 + 4 + 2 (cid:96) − − (cid:96) . There are four other similar cases,such as x ( c u , c v ) · x ( c v , c w ) . In this case for example, the coefficient will be − (cid:96) . The signis positive if the edges are either both into or both out of the same common vertex, andnegative otherwise.Finally, consider terms of the form x c u , c v ) . This term appears, with a coefficient of 4, for( c i , c j ) = ( c u , c v ), and with a coefficient of 1 for each ( c i , c j ) which has exactly one vertex incommon with ( c u , c v ), of which there are 2( (cid:96) − x c u , c v ) has a coefficient of 2 (cid:96) .So: || z || = 2 (cid:96) · (cid:88) ( c i , c j ) x c i , c j ) + (cid:88) ( c i , c j ) , ( c i , c k ) x ( c i , c j ) x ( c i , c k ) . Now let us apply this to our scenario. We have f ( x ) = e − x T Σ − x (cid:112) (2 π ) k | Σ | , where x T Σ − x = 1 λ || y || + 1 λ || z || . Then we can write f ( x ) = e − λ || y || − λ || z || (cid:112) (2 π ) k | Σ | = e − λ || x || + (cid:16) λ − λ (cid:17) || z || (cid:112) (2 π ) k | Σ | . Since λ − λ ≥ 0, this makes it clear that for a specific value of || x || , the larger || z || is, thelarger f ( x ) is. Using Theorem 6.1, we have || z || = 2 (cid:96) · (cid:88) ( c i , c j ) x c i , c j ) + (cid:88) ( c i , c j ) , ( c i , c k ) x ( c i , c j ) x ( c i , c k ) . Given a tournament T on the candidates C , the probability that we obtain T as the majoritygraph of a random profile X ∼ N ( , Σ ) isPr( T ) = (cid:90) R T f ( x ) d x where R T is the orthant corresponding to T ; R T is the octant which has x ( c i , c j ) positive ifand only if c i → c j in T . (Note that by our convention that x ( c i , c j ) = − x ( c j , c i ) , if i < j and20 → i in T , then the region R T has x ( c i , c j ) negative which means that x ( c j , c i ) is positive.) Sowe can write || z || = 2 (cid:96) · (cid:16) (cid:88) c i → c j x c i , c j ) + (cid:88) c i → c j , c i → c k x ( c i , c j ) x ( c i , c k ) + (cid:88) c j → c i , c k → c i x ( c j , c i ) x ( c k , c i ) − (cid:88) c i → c j , c k → c i x ( c i , c j ) x ( c k , c i ) (cid:17) where → is interpreted as in T . In the region R T , all of the terms x ( · , · ) appearing in thisexpression will be positive. Intuitively, on might expect that the more positive terms showup, the greater the integral Pr( T ) = (cid:90) R T f ( x ) d x should be. Moreover, our geometric intuition agrees. Suppose that we have an ellipsoidsuch that all of the axes take on one of two values (dividing the axes into major axes andminor axes). Suppose also that we have an orthant. It seems plausible that the closer thediagonal vector in the orthant is to the major axes of the ellipsoid, the greater the volumewithin that orthant is. (Unfortunately, volumes of octants of ellipsoids are directly relatedto the measurement of solid angles of simplicial cones, and this is not understood in higherdimensions. See, e.g., [Rib06].)To this effect, we assign a number to each tournament: Definition 6.2. Given a tournament T on (cid:96) vertices, we can assign a number to T whichwe call the linearity of T :lin( T ) = 12 (cid:88) v deg − ( v ) + deg + ( v ) = (cid:88) v deg − ( v ) = (cid:88) v deg + ( v ) . Then the number of positive terms in || z || on the octant R T is exactly lin( T ) − (cid:96) ( (cid:96) − 1) as (cid:88) v deg − ( v ) · (deg − ( v ) − 1) + deg + ( v ) · (deg + ( v ) − 1) = lin( T ) − (cid:96) ( (cid:96) − . So we have a guess at what is going on: The greater the linearity of T , the greater theprobability Pr( T ) should be. Conjecture 6.3. Let T and T (cid:48) be tournaments on (cid:96) candidates. If lin( T ) < lin( T (cid:48) ) , then Pr( T ) < Pr( T (cid:48) ) . This conjecture has already been verified for (cid:96) = 3 (as the linearity of a linear order is higherthan the linearity of a cycle on three candidates). In the next two sections we verify theconjecture for (cid:96) = 4 and (cid:96) = 5 as well. 21 Four Candidates Consider the case of four candidates A , B , C , and D . As usual we fix by convention the edgedirections of the linear order A > B > C > D . A (cid:47) (cid:47) (cid:15) (cid:15) (cid:31) (cid:31) B (cid:127) (cid:127) (cid:15) (cid:15) C (cid:47) (cid:47) D Then we compute Σ = AB AC AD BC BD CD AB / / − / − / AC / / / − / AD / / / / BC − / / / − / BD − / / / / CD − / / − / / . For four candidates, there are four possible isomorphism types: T : A (cid:47) (cid:47) (cid:15) (cid:15) (cid:31) (cid:31) B (cid:127) (cid:127) (cid:15) (cid:15) T : A (cid:47) (cid:47) (cid:15) (cid:15) (cid:31) (cid:31) B (cid:127) (cid:127) T : A (cid:47) (cid:47) (cid:31) (cid:31) B (cid:15) (cid:15) (cid:127) (cid:127) T : A (cid:47) (cid:47) (cid:31) (cid:31) B (cid:15) (cid:15) (cid:127) (cid:127) C (cid:47) (cid:47) D C (cid:47) (cid:47) D (cid:79) (cid:79) C (cid:79) (cid:79) (cid:47) (cid:47) D C (cid:79) (cid:79) D (cid:111) (cid:111) There are twenty-four different labeled copies of the first one, eight of the second and third,and twenty-four of the last.We have a new phenomenon that shows up here, which is that T and T are dual in thesense that one is obtained from the other by reversing the directions of every arrow. Thusthe probability of obtaining T will be the same as the probability of obtaining T . T and T are self-dual in the sense that by reversing arrows, we keep the same isomorphism type.Dual tournaments have the same linearity.To compute the probabilities of obtaining each of these tournaments directly, one wouldhave to find orthant probabilities in six dimensions, and there is no known general wayof doing this. However, a clever argument by Gehrlein and Fishburn [GF78] allows us tocompute exact probabilities.First, the probability of having A as a majority winner is the probability that A beats B , C , and D ; thus we have reduced ourselves to a three-dimensional space and can computethe probability exactly. By multiplying by four, we get the probability of having a majoritywinner, which is the same as the probability that the margin graph of a random tournamentis isomorphic to either T or T . (This value is 0.82452.)Second (and this is the clever part) the probability that the margin graph of a randomelection is transitive (i.e., isomorphic to T ) can be computed as follows. First, consider theprobability that A beats C and D and the probability that B beats C and D . This puts uswithin a four-dimensional space, and exact orthant probabilities are known in certain special22ases (including this one) in four dimensions [DM61]. This allows Gehrlein and Fishburn tocompute an exact expression for the probability that A beats C and D and that B beats C and D . The are four different majority graphs satisfying this (depending on whether A beatsor loses to B , and C with D ), each of which is transitive, and each of these is equally likely,so we can compute the probability 0.7395 of obtaining a transitive graph as the outcome ofa random election.From these values we can compute the value of Pr( T ) and the value of Pr( T ) + Pr( T ).Using the fact that T and T are dual and so Pr( T ) = Pr( T ), we can compute all ofthese values. They are displayed in the table below. The score sequence is the sequenceof out-degrees of the nodes of the tournament. The number of labelings is the number oflabeled tournaments with the isomorphism type. The labeled probability is the probability ofobtaining T i specifically as the outcome of an election, while the probability is the probabilityof obtaining the isomorphism type of T i .Label Score sequence Linearity Num Labelings Labeled Probability Probability T . . T . . T . . T . . T is less probable than T ,because there are so many more isomorphic copies of T the isomorphism type of T on thewhole is more likely. Consider the case of five candidates A , B , C , D , and E . As always we fix a standard orientationof the edges between candidates, corresponding to the linear order A > B > C > D > E : A (cid:127) (cid:127) (cid:7) (cid:7) (cid:31) (cid:31) (cid:23) (cid:23) B (cid:39) (cid:39) (cid:47) (cid:47) (cid:15) (cid:15) EC (cid:47) (cid:47) (cid:55) (cid:55) D (cid:79) (cid:79) Σ = AB AC AD AE BC BD BE CD CE DE AB / / / − / − / − / AC / / / / − / − / AD / / / / / − / AE / / / / / / BC − / / / / − / − / BD − / / / / / − / BE − / / / / / / CD − / / − / / / − / CE − / / − / / / / DE − / / − / / − / / . Up to isomorphism, there are twelve different tournaments on the five candidates. Recall thatthe score sequence of a tournament is the sequence of outdegrees. These twelve isomorphismtypes have between them nine different score sequences; one score sequence, 3 , , , , 1, hastwo different isomorphism types, and another, 3 , , , , 1, has three different isomorphismtypes. We number these tournaments from 1 to 12. T : A (cid:127) (cid:127) (cid:7) (cid:7) (cid:31) (cid:31) (cid:23) (cid:23) T : A (cid:127) (cid:127) (cid:7) (cid:7) (cid:31) (cid:31) (cid:23) (cid:23) T : A (cid:127) (cid:127) (cid:7) (cid:7) (cid:31) (cid:31) (cid:23) (cid:23) B (cid:39) (cid:39) (cid:47) (cid:47) (cid:15) (cid:15) E B (cid:47) (cid:47) (cid:15) (cid:15) E B (cid:39) (cid:39) (cid:47) (cid:47) (cid:15) (cid:15) E (cid:119) (cid:119) C (cid:47) (cid:47) (cid:55) (cid:55) D (cid:79) (cid:79) C (cid:47) (cid:47) (cid:55) (cid:55) D (cid:103) (cid:103) (cid:79) (cid:79) C (cid:47) (cid:47) D (cid:79) (cid:79) T : A (cid:127) (cid:127) (cid:31) (cid:31) (cid:23) (cid:23) T : A (cid:127) (cid:127) (cid:7) (cid:7) (cid:31) (cid:31) (cid:23) (cid:23) T : A (cid:127) (cid:127) (cid:7) (cid:7) (cid:31) (cid:31) B (cid:39) (cid:39) (cid:47) (cid:47) (cid:15) (cid:15) E B (cid:39) (cid:39) (cid:15) (cid:15) E (cid:111) (cid:111) (cid:119) (cid:119) B (cid:39) (cid:39) (cid:47) (cid:47) (cid:15) (cid:15) EC (cid:47) (cid:47) (cid:55) (cid:55) (cid:71) (cid:71) D (cid:79) (cid:79) C (cid:47) (cid:47) D (cid:79) (cid:79) C (cid:47) (cid:47) (cid:55) (cid:55) D (cid:79) (cid:79) (cid:87) (cid:87) T : A (cid:127) (cid:127) (cid:7) (cid:7) (cid:23) (cid:23) T : A (cid:127) (cid:127) (cid:7) (cid:7) (cid:23) (cid:23) T : A (cid:127) (cid:127) (cid:7) (cid:7) (cid:23) (cid:23) B (cid:39) (cid:39) (cid:47) (cid:47) (cid:15) (cid:15) E (cid:95) (cid:95) (cid:119) (cid:119) B (cid:39) (cid:39) (cid:47) (cid:47) (cid:15) (cid:15) E (cid:95) (cid:95) B (cid:39) (cid:39) (cid:15) (cid:15) E (cid:111) (cid:111) (cid:95) (cid:95) C (cid:47) (cid:47) D (cid:79) (cid:79) C (cid:47) (cid:47) (cid:55) (cid:55) D (cid:79) (cid:79) C (cid:47) (cid:47) (cid:55) (cid:55) D (cid:79) (cid:79) : A (cid:127) (cid:127) (cid:23) (cid:23) T : A (cid:127) (cid:127) (cid:7) (cid:7) (cid:23) (cid:23) T : A (cid:127) (cid:127) (cid:7) (cid:7) B (cid:39) (cid:39) (cid:15) (cid:15) E (cid:111) (cid:111) (cid:95) (cid:95) B (cid:47) (cid:47) (cid:15) (cid:15) E (cid:95) (cid:95) B (cid:39) (cid:39) (cid:15) (cid:15) E (cid:95) (cid:95) (cid:111) (cid:111) C (cid:47) (cid:47) (cid:55) (cid:55) (cid:71) (cid:71) D (cid:79) (cid:79) C (cid:47) (cid:47) (cid:55) (cid:55) D (cid:79) (cid:79) (cid:103) (cid:103) C (cid:47) (cid:47) (cid:55) (cid:55) D (cid:79) (cid:79) (cid:87) (cid:87) In the table below we list: their score sequence; linearity; the number of different labeledtournaments having that isomorphism type; the probability of obtaining that isomorphismtype as the majority graph of a random profile under IC; and the probability of obtaining afixed labeling of that isomorphism type. There are no known closed form solutions for orthantprobabilities in such high dimensions, so the probabilities given are numeric approximationscomputed using the R [R C13] package orthant [CM12], which contains an implementationof Craig’s algorithm from [Cra08].Label Score sequence Linearity Num Labelings Labeled Probability Probability T . . T . . T . . T . . T . . T . . T . . T . . T . . T . . T . . T . . T , T , T , or T ; our valuesgive 0.748612396 which closely agrees with 0.74869 from Gehrlein and Fishburn [GF76].Gehlrein [Geh88] used a Monte-Carlo simulation to obtain an estimate of the probability ofobtaining a transitive tournament as 0.529; we computed 0.527.Once again we verify that Conjecture 6.3 is true for five candidates. We have an interest-ing new phenomenon, which is that there are non-dual tournaments with the same linearity.For example, T , T , and T all have the same linearity, and T and T are dual, but T isself-dual. We have computed Pr( T ) > Pr( T ) = Pr( T ). ( T and T , and T and T , are theonly two non-self-dual tournaments here.) If Conjecture 6.3 is true, we must then ask whatmakes one tournament more probable than another tournament with the same linearity? Suppose that we want to perform a Monte Carlo simulation of a margin based voting method,for example to see how often it selects multiple winners. Given a set V of n voters, a Monte25arlo simulation might repeatedly generate for each voter a random ballot and compute themargin graph from all of these ballots. The problem is that if we want to do this with a largenumber of voters, we have to generate a random ballot for each voter and tally them all, andit might take a long time to generate enough random values to get an accurate Monte Carlosimulation.Using the results of this paper we get a more efficient method. In the notation of Section4, the random margin graph generated by n voters will be (cid:80) ni =1 X v i . By the central limittheorem, √ n · S n = √ n (cid:80) ni =1 X v i converges in distribution to the multivariate normal distri-bution N ( , Σ ) with the same mean and covariance matrix, which has probability densityfunction f ( x ) = e − x T Σ − x (cid:112) (2 π ) k | Σ | . So instead of generating random profiles, we can instead use a random variable Y ∼ N ( , Σ ).We can easily compute from Y the majority graph, the qualitative margin graph, and (es-sentially) the margin graph. This is enough to determine the winning set by any qualitativemargin-based voting method, and by most margin-based voting methods such as Bordacount.To generate values of Y , one can either simply use a software package, or one can useindependent normal random variables Z = ( Z i,j ) i 00% in the table below are non-zero entries which wererounded down.) 26ize of winning set (cid:96) . 80% 3 . 15% 0 . 06% — — — — — — 3 . . 19% 7 . 38% 0 . 42% 0 . 01% — — — — — 7 . . 26% 13 . 13% 1 . 52% 0 . 09% 0 . 00% — — — — 14 . . 80% 24 . 22% 6 . 65% 1 . 17% 0 . 15% 0 . 01% 0 . 00% 0 . 00 — 32 . . 58% 28 . 47% 11 . 24% 3 . 01% 0 . 62% 0 . 09% 0 . 01% 0 . 00% — 43 . . 35% 29 . 95% 16 . 55% 6 . 88% 2 . 62% 0 . 51% 0 . 11% 0 . 03% — 56 . . 68% 26 . 71% 16 . 73% 8 . 77% 3 . 54% 1 . 21% 0 . 49% 0 . 05% 0 . 01% 67 . References [Big93] Norman Biggs. Algebraic graph theory . Cambridge Mathematical Library. Cam-bridge University Press, Cambridge, second edition, 1993.[BN78] Bo H. Bjurulf and Richard G. Niemi. Strategic voting in scandinavian parliaments. Scandinavian Political Studies , 1(1):5–22, 1978.[CM12] Peter Craig and Tetsuhisa Miwa. orthants: Multivariate normal orthant probabili-ties , 2012. R package version 1.5.[Cra08] Peter Craig. A new reconstruction of multivariate normal orthant probabili-ties. Journal of the Royal Statistical Society: Series B (Statistical Methodology) ,70(1):227–243, 2008.[DBT00] Philippe De Donder, Michel Le Breton, and Michel Truchon. Choosing from aweighted tournament. Mathematical Social Sciences , 40:85–109, 2000.[Deb87] Bernard Debord. Caractrisation des matrices des prfrences nettes et mthodesd’agrgation associes. Mathmatiques et Sciences Humaines , 97:5–17, 1987.[Dee14] Adrian Van Deemen. On the empirical relevance of condorcet’s paradox. PublicChoice , 158(3/4):311–330, 2014.[DL99] Bhaskar Dutta and Jean-Francois Laslier. Comparison functions and choice corre-spondences. Social Choice and Welfare , 16:513–532, 1999.[DM61] F. N. David and C. L. Mallows. The variance of Spearman’s rho in normal samples. Biometrika , 48:19–28, 1961.[DP70] Frank DeMeyer and Charles R. Plott. The probability of a cyclical majority. Econo-metrica , 38(2):345–354, 1970.[Geh88] William V. Gehrlein. Probability calculations for transitivity of the simple majorityrule. Economics Letters , 27(4):311–315, 1988.27Geh89] William V. Gehrlein. The probability of intransitivity of pairwise comparisons inindividual preference. Mathematical Social Sciences , 17(1):67–75, 1989.[Geh99] William V. Gehrlein. Approximating the probability that a condorcet winner exists.In Proceedings of International Decision Sciences Institute , pages 626–628, 1999.[Geh06] William V. Gehrlein. Condorcet’s Paradox . Theory and Decision Library C.Springer, 2006.[GF76] William V. Gehrlein and Peter C. Fishburn. The probability of the paradox ofvoting: A computable solution. Journal of Economic Theory , 13(1):14–25, 1976.[GF78] William V Gehrlein and Peter C Fishburn. Probabilities of election outcomes forlarge electorates. Journal of Economic Theory , 19(1):38–49, 1978.[GF79] William V. Gehrlein and Peter C. Fishburn. Proportions of profiles with a majoritycandidate. Computers & Mathematics with Applications , 5(2):117–124, 1979.[GK68] Mark B. Garman and Morton I. Kamien. The paradox of voting: Probabilitycalculations. Behavioral Science , 13(4):306–316, 1968.[Gui52] Georges-Thodule Guilbaud. Les thories de l’intrt gnral et le problme logique del’agrgation. Economie Applique , 5:501–584, 1952.[HPa] Wesley H. Holliday and Eric Pacuit. Axioms for defeat in democratic elections.https://arxiv.org/abs/2008.08451.[HPb] Wesley H. Holliday and Eric Pacuit. Probabilistic stability for winners in votingwith even-chance tiebreaking. preprint.[HPc] Wesley H. Holliday and Eric Pacuit. Split cycle: A new condorcetconsistent voting method independent of clones and immune to spoilers.https://arxiv.org/abs/2004.02350.[KK01] Peter Kurrild-Klitgaard. An empirical example of the condorcet paradox of votingin a large electorate. Public Choice , 107(1/2):135–145, 2001.[Kra77] Gerald H. Kramer. A dynamical model of political equilibrium. Journal of EconomicTheory , 16(2):310–334, 1977.[NW68] Richard G. Niemi and Herbert F. Weisberg. A mathematical solution for the prob-ability of the paradox of voting. Behavioral Science , 13(4):317–323, 1968.[R C13] R Core Team. R: A Language and Environment for Statistical Computing . RFoundation for Statistical Computing, Vienna, Austria, 2013. ISBN 3-900051-07-0.[Rib06] Jason M. Ribando. Measuring solid angles beyond dimension three. Discrete andComputational Geometry , 36(3):479–487, 2006.28Rik58] William H. Riker. The paradox of voting and congressional rules for voting onamendments. The American Political Science Review , 52(2):349–366, 1958.[Sch11] Markus Schulze. A new monotonic, clone-independent, reversal symmetric, andcondorcet-consistent single-winner election method. Social Choice and Welfare ,36(2):267–303, 2011.[Sev69] K. E. Sevcik. Exact probabilities of a voter’s paradox through seven issues and sevenjudges. University of Chicago Institute for Computer Research Quarterly Report 22 ,1969. Sec. III-B.[Sim69] Paul B. Simpson. On defining areas of voter choice: Professor Tullock on stablevoting. The Quarterly Journal of Economics , 83(3):478–490, 1969.[Ste99] Eivind Stensholt. Voteringens kvaler: flyplassaken i stortinget 8. oktober 1992. Sosialøkonomen , 4:28–40, 1999.[Tid87] T. Nicolaus Tideman. Independence of clones as a criterion for voting rules. SocialChoice and Welfare , 4:185–206, 1987.[Ton90] Y. L. Tong. The multivariate normal distribution . Springer Series in Statistics.Springer-Verlag, New York, 1990.[Tru98] M. Truchon. Figure Skating and the Theory of Social Choice. Technical Report,Laval-Recherche en Politique Economique, 1998.[vdV98] A. W. van der Vaart. Asymptotic statistics , volume 3 of