Probabilistic Inference of Winners in Elections by Independent Random Voters
PProbabilistic Inference of Winners in Elections byIndependent Random Voters
Aviram Imber
Technion, Israel
Benny Kimelfeld
Technion, Israel
Abstract
We investigate the problem of computing the probability of winning in an election where voterattendance is uncertain. More precisely, we study the setting where, in addition to a total ordering ofthe candidates, each voter is associated with a probability of attending the poll, and the attendancesof different voters are probabilistically independent. We show that the probability of winning can becomputed in polynomial time for the plurality and veto rules. However, it is computationally hard( k -approval and k -veto for k >
1, Borda, Condorcet,and Maximin. For some of these rules, it is even hard to find a multiplicative approximationsince it is already hard to determine whether this probability is nonzero. In contrast, we devise afully polynomial-time randomized approximation scheme (FPRAS) for the complement probability,namely the probability of losing, for every positional scoring rule (with polynomial scores), as wellas for the Condorcet rule.
Theory of computation → Algorithmic game theory and mechan-ism design
Keywords and phrases
Social choice, voting rules, probabilistic voters, approximation.
Digital Object Identifier
The theory of social choice targets the question of how voter preferences should be aggregatedto arrive at a collective decision. It spans centuries of research, from law making in AncientRome to the 20th century’s fundamental theory and the recent study of the algorithmic andcomputing aspects—computational social choice. (See [7] for an overview.) In the commonformal setting, there are voters and candidates, each voter has an individual preference, namelyordering over the candidates, and a voting rule is applied to the preference profile (collectionof orderings) in order to elect a winner. Past research has investigated various ways ofincorporating situations of uncertainty in this setting, including incomplete and probabilisticpreferences [20, 1, 16, 19, 14], probabilistic candidate validity [28, 6], and probabilistic voterparticipation [24, 28, 27, 11, 23]. Further settings include bribery with probabilistic votercooperation [10, 9] and random elicitation for efficient outcome determination [8, 22] .In this work, we focus on the basic setting where voters are randomly drawn. Inparticular, the election outcome is probabilistic and, therefore, each candidate has a marginalprobability of being elected. More formally, each voter v i is associated with a total orderover the candidates, and she casts a vote with probability p i (and steps outside the voter listwith probability 1 − p i ). Importantly, we assume that different voters are probabilisticallyindependent. For a candidate c , the probability of winning is the probability that the randomset of casting voters elects c . The aggregation of the preferences of these voters is done by avoting rule, and we consider several well-known alternatives: positional scoring rules, the Condorcet rule, and the
Maximin rule (also known as the
Simpson rule). We investigate thecomputational problem of calculating the probability that a given candidate c is a winner. © Aviram Imber, Benny Kimelfeld;licensed under Creative Commons License CC-BYLeibniz International Proceedings in InformaticsSchloss Dagstuhl – Leibniz-Zentrum für Informatik, Dagstuhl Publishing, Germany a r X i v : . [ c s . G T ] J a n X:2 Probabilistic Inference of Winners in Elections by Independent Random Voters
Table 1
Overview of the complexity results.
Problem plurality, veto k -{approv., veto} Borda R ( f, ‘ ) other positional Condorcet MaximinPr[win] P P -h P -h P -h for ( f, ‘ ) = (1 ,
1) ? P -h P -h[Thm. 2] [Thm. 3] [Thm. 6] [Thm. 7] [Thm. 8] [Thm. 9]Pr[win] > P P NP -c NP -c for ( f, ‘ ) = (1 ,
1) ? NP -c NP -c[Thm. 2] [Thm. 15] [Thm. 10] [Cor. 13] [Thm. 14] [Thm. 14]Approximate P FPRAS FPRAS FPRAS FPRAS for poly. scores FPRAS ?Pr[lose] [Thm. 2] [Thm. 16] [Thm. 16] [Thm. 16] [Thm. 16] [Thm. 16]
Our contribution is summarized in Table 1. We first consider the complexity of exactevaluation (the first row of the table). We start with positional scoring rules; recall thatin such a rule, each voter assigns to each candidate a score based on the position of thecandidate in the voter’s ranking, and the vector of scores is shared among all voters. Forexample, in k -approval the top k positions get the score 1 and the rest get 0, and in k -vetothe bottom k positions get 0 and the rest get 1. The probability of winning can be computedin polynomial time for k = 1 (i.e., the plurality and veto rules), but k > additive approximation, we can get a FullyPolynomial-Time Randomised Approximation Scheme (FPRAS) by straightforward samplingand averaging. So, we focus on a multiplicative approximation. Yet, in some of the ruleswe consider it is intractable to get any approximation guarantee of a sub-exponential ratio,since it is already hard to decide whether the probability of winning is nonzero (as shownin the second row of Table 1). Exceptional are k -approval and k -veto where this decisionproblem is solvable in polynomial time, but the existence of an efficient approximationremains an open problem. Nevertheless, we can also consider another kind of a multiplicativeapproximation, where we consider the complement probability of losing . (In the additivevariant, approximating the probability is the same as approximating its complement, but itis not so in the case of a multiplicative approximation.) Inspired by the work of Kenig andKimelfeld [19] on probabilistic preferences, we can apply the Karp-Luby-Madras technique [18]to establish a multiplicative FPRAS for the probability of losing. This applies to every positional scoring rule (with the mild assumption that the scores are all polynomial in thenumber of voters and candidates) and the Condorcet rule; the problem remains open forMaximin.While it is important to understand richer frameworks with more complex probabilisticmodeling of voters, little has been known about the computational aspects of our basicsetting, to the best of our knowledge. The closest studied problem is lot-based voting , wherewe uniformly select k voters for a pre-determined number k [27], and its generalization thatentails a preprocessing step of choosing k itself randomly from a given distribution [28].Lot-based polling relates to, and can adopt complexity results on, the problem of control inelections, where the goal is to detect a small subset of voters of whom elimination can leadto a desired outcome [28]. Yet, it is not clear how to draw conclusions from lot-based votingto out setting. The positive results in this model are based on counting, so the algorithms donot immediately apply to our setting where voters can have individual probabilities. As forthe hardness results, in some cases we adapt proofs from that work to our setting, while inother cases the proofs are heavily based on the assumption that the cardinality of the voter . Imber and B. Kimelfeld XX:3 set is given. Moreover, that work focused on the exact (but not approximate ) computationof probabilities where our problem is We first give definitions, notation, and terminology that we use throughout the paper.
We denote by C = { c , . . . , c m } the set of candidates and V = { v , . . . , v n } the set of voters .A voting profile T = ( T , . . . , T n ) consists of n linear orders on C , where each T i is theranking of (i.e., linear order over) C by v i . A voting rule is a function that maps every profileon C to a set of winners from C .A positional scoring rule r is a series { ~s m } m ∈ N + of m -dimensional score vectors ~s m =( ~s m (1) , . . . , ~s m ( m )) of natural numbers where ~s m (1) ≥ · · · ≥ ~s m ( m ) and ~s m (1) > ~s m ( m ).We make the conventional assumption that ~s m ( j ) is computable in polynomial time in m .Examples of positional scoring rules include the plurality rule (1 , , . . . , k -approval rule (1 , . . . , , , . . . ,
0) that begins with k ones followed by zeroes, the veto rule (1 , . . . , , k -veto rule (1 , . . . , , , . . . ,
0) that starts with ones and ends with k zeros, and the Borda rule ( m − , m − , . . . , f and ‘ , we denote by R ( f, ‘ ) the three-valued rule with scoringvector ~s m = (2 , . . . , , , . . . , , , . . . ,
0) that begins with f occurrences of two and ends with ‘ zeros. For example, the scoring vector for R (1 ,
1) is ~s m = (2 , , . . . , , T = ( T , . . . , T n ), the score s ( T i , c, r ) that the voter v i contributesto the candidate c is ~s m ( j ) where j is the position of c in T i . The score of c in T is s ( T , c, r ) = P ni =1 s ( T i , c, r ) that we may denote simply by s ( T , c ) if r is clear from context.A candidate c is a winner (also referred to as co-winner ) if s ( T , c ) ≥ s ( T , c ) for all candidates c . For two candidates c, c ∈ C , denote by N T ( c, c ) the number of voters that prefer c to c . A candidate c is a Condorcet winner if N T ( c, c ) > N T ( c , c ) for all c = c . (Note that aCondorcet winner does not necessarily exist; when exists, it is unique.) Under the Maximin rule, the score of c is s ( T , c ) = min { N T ( c, c ) : c ∈ C \ { c }} and a winner is a candidatewith a maximal score. In the setting we study, voters participate by being drawn randomly from V . We are givenas part of the input a vector ( p , . . . , p n ) ∈ [0 , n of probabilities. Define a random variable I ⊆ [ n ], where every i ∈ [ n ] is in I with probability p i and different indices are probabilistically All of our results apply to, and can be easily adapted to, the unique winner semantics, where c is aunique winner if s ( T , c ) > s ( T , c ) for all candidates c = c . X:4 Probabilistic Inference of Winners in Elections by Independent Random Voters independent. (Note our notation of [ n ] = { , . . . , n } .) The random set of voters thatparticipate in the election is V = { v i : i ∈ I } , and the resulting random profile is T = { T i } i ∈ I .The probability of I being a subset U ⊆ [ n ] is Pr[ I = U ] = Q i ∈ U p i Q i ∈ [ n ] \ U (1 − p i ).Denote by Win( c ; T ) the event that c is a winner of T according to a positional scoringrule r . Similarly, denote by Lose( c ; T ) the event that c is not a winner. We will investigatethe evaluation of the probabilities Pr[Win( c ; T )] and Pr[Lose( c ; T )].A Fully-Polynomial Randomized Approximation Scheme (FPRAS) for a probabilityfunction p ( x ) is a randomized algorithm A ( x, (cid:15), δ ) that given x for p and (cid:15), δ ∈ (0 , (cid:15) -approximation of p ( x ) with probability at least 1 − δ , in time polynomial in the size of x , 1 /(cid:15) , and log(1 /δ ). Formally, an FPRAS, satisfies:Pr[(1 − (cid:15) ) p ( x ) ≤ A ( x, (cid:15), δ ) ≤ (1 + (cid:15) ) p ( x )] ≥ − δ . Note that this notion of FPRAS refers to a multiplicative approximation, and we adopt thisnotion implicitly, unless stated otherwise.
Let r be a voting rule. In the problem of Constructive Control by Adding Voters (CCAV) weare given a set C of candidates, a voting profile M of voters who are already registered, avoting profile Q of yet unregistered voters, a preferred candidate c ∈ C , and a bound k ∈ N .The goal is to test whether we can choose a sublist Q ⊆ Q of size at most k such that c is awinner of M ◦ Q , where M ◦ Q is the concatenation of the two profiles M and Q . (Theunderlying rule r will be clear from the context.) (cid:73) Theorem 1 ([21, 25, 2, 13]) . The problem of CCAV is solvable in polynomial time forplurality, veto and k -approval for k ≤ . CCAV is NP -complete for Borda, Condorcet, and k -approval for k ≥ . In the corresponding counting problem Q ⊆ Q of size at most k such that c is a winner of M ◦ Q .They showed that FP under plurality, and is P -complete under Condorcet,Maximin and k -approval for k ≥ r , we define the problem of Constructive Controlby Adding an Unlimited number of Voters (CCAUV) as follows: We are given a set C ofcandidates, a voting profile M of registered voters, a voting profile Q of yet unregisteredvoters, and a preferred candidate c ∈ C . We ask whether we can select a sublist Q ⊆ Q (of any cardinality) such that c is a winner of M ◦ Q . We also denote the correspondingcounting problem by For a set A and a partition A , . . . , A t of A , we use O ( A , . . . , A t ) to denote an arbitrarylinear order on A that satisfies a (cid:31) · · · (cid:31) a t for every a ∈ A , . . . , a t ∈ A t . A linear order a (cid:31) · · · (cid:31) a t is also denoted as a vector ( a , . . . , a t ). The concatenation of two linear orders( a , . . . , a t ) ◦ ( b , . . . , b ‘ ) is ( a , . . . , a t , b , . . . , b ‘ ). In this section, we study the complexity of computing the probability that a given candidateis a winner. Recall that the input consists of a voting profile T , a vector of voter probabilities . Imber and B. Kimelfeld XX:5 and a candidate c , and the goal is to compute Pr[Win( c ; T )]. For plurality and veto, if the probabilities of all voters are identical, we can easily computePr[Win( c ; T )] in polynomial time by reducing the case of independent voters to theprobability distribution studied by Wojtas and Faliszewski [28]. For the general case, wherethe probabilities can differ, we need a different algorithm. (cid:73) Theorem 2.
For the plurality and veto rules,
Pr[Win( c ; T )] is computable in polynomialtime. Proof.
We begin with plurality. Let T = ( T , . . . , T n ) be a voting profile and let ( p , . . . , p n )be the probabilities. In the plurality rule, every voter increases the score of a single candidateby 1. Since different voters are independent, the scores of different candidates are independent.Therefore, we get the following (due to the law of total probability).Pr[Win( c ; T )] = n X s =0 Pr[Win( c ; T ) ∧ s ( T , c ) = s ]= n X s =0 Pr[ s ( T , c ) = s ] · Y c = c Pr[ s ( T , c ) ≤ s ]We show that for all c ∈ C and s ∈ { , . . . , n } , we can compute Pr[ s ( T , c ) = s ] in polynomialtime via dynamic programming. Then, by summing and multiplying these values, we cancompute Pr[Win( c ; T )].For a candidate c , a number t ≤ n of voters and an integer score 0 ≤ y ≤ n , define N ( t, y ) := Pr[ P i ∈ I ∩ [ t ] s ( T i , c ) = y ]. In particular, N ( n, s ) = Pr[ s ( T , c ) = s ]. For t = 0 wehave N ( t, y ) = 1 if y = 0 and N ( t, y ) = 0 otherwise.Let t ≥
1. If c is not ranked first in T t , then the appearance of the voter v t does not affectthe score of c , hence N ( t, y ) = N ( t − , y ). If c is ranked first in T t , then we consider twocases. If t ∈ I , then for the event P i ∈ I ∩ [ t ] s ( T i , c ) = y we need P i ∈ I ∩ [ t − s ( T i , c ) = y − P i ∈ I ∩ [ t − s ( T i , c ) = y . Hence, we have the following. N ( t, y ) = p t · N ( t − , y −
1) + (1 − p t ) · N ( t − , y )For the veto rule, denote by b ( T , c ) the (random variable that holds the) number of votersin T who place c at the bottom position. Note that b ( T , c ) and b ( T , c ) are independentfor c = c . We can write the following.Pr[Win( c ; T )] = n X b =0 Pr[ b ( T , c ) = b ] · Y c = c Pr[ b ( T , c ) ≥ b ]For c ∈ C and b ∈ { , . . . , n } , we can compute Pr[ b ( T , c ) = b ] in polynomial time similarlyto Pr[ s ( T , c ) = s ] under plurality. (cid:74) In the remainder of this section, we show that computing Pr[Win( c ; T )] is P -hard forseveral other voting rules. For every voting rule we consider, we have a reduction from c ; T )], as follows. Let C , M , Q and c ∈ C be an instanceof M participate X:6 Probabilistic Inference of Winners in Elections by Independent Random Voters with probability 1, and every voter of Q participates with probability 1 /
2. Denote by k thenumber of voters of Q and by α ( Q , M ) the number of subsets Q ⊆ Q such that c is a winnerof M ◦ Q . We get that Pr[Win( c ; T )] = 2 − k α ( Q , M ). We conclude that P -completenessof P -hardness of computing Pr[Win( c ; T )]. Consequently, in this sectionthe proofs of our hardness results are essentially showing that P -completeunder the voting rule in consideration. k -Approval and k -Veto We begin with the hardness of k -approval for k > k = 1 is the plurality rulediscussed in Section 3.1). (cid:73) Theorem 3.
For every fixed k > , computing Pr[Win( c ; T )] is P -hard under k -approval. Proof.
We show a reduction from counting the (not necessarily perfect) matchings in a graphto k -approval. Given a graph G = ( U, E ), we wish to compute the numberof subsets E ⊆ E such that every vertex u ∈ U is incident to at most one edge of E . Thisproblem is known to be P -complete [26]. Given a graph G = ( U, E ) where E = { e , . . . , e m } ,define a set C of candidates by C = U ∪ { c, d } ∪ F where F = F ∪ F ∪ · · · ∪ F m and F i = { f i, , . . . , f i,k − } . The voting profile is T = M ◦ Q that we define next.The first part, M , consists of a single voter O ( { c, d } ∪ F , C \ ( { c, d } ∪ F )). Observe thatthe candidates of { c, d } ∪ F receive a score of 1 from M , and the other candidates receive0. The second part is Q = ( Q , . . . , Q m ). For i ∈ [ m ], define Q i = O ( e i ∪ F i , C \ ( e i ∪ F i )).Observe that the candidates of e i ∪ F i receive a score of 1 from Q i , and the other candidatesreceive 0. Also observe that here we are using the assumption that k >
1, as both endpointsof e i need to gain 1 from Q i .Let Q ⊆ Q and denote T = M ◦ Q . Since only the voter of M contributes to the scoreof c and d , we have s ( T , c ) = s ( T , d ) = 1. Every f ∈ F can get a positive score only froma single voter, hence s ( T , f ) ≤
1. If c is a winner, then s ( T , u ) ≤ u ∈ U , andthen { e i : Q i ∈ Q } is a matching in G because each voter Q i contributes a score of 1 to thevertices of e i . Conversely, if E ⊆ E is a matching in G , then c is a winner of M ◦ { Q e } e ∈ E .Overall, the number of subsets Q ⊆ Q such that c is a winner of M ◦ Q is the numberof matchings in G , as required. (cid:74) Next, by a similar reduction to the proof of Theorem 3, we obtain hardness of k -veto for k > k = 1 is the veto rule discussed in Section 3.1). (cid:73) Theorem 4.
For every fixed k > , computing Pr[Win( c ; T )] is P -hard under k -veto. Proof.
We show a reduction from the problem of counting the edge covers in a graph to k -veto. Formally, given a graph G = ( U, E ), the goal is to compute thenumber of subsets E ⊆ E such that every vertex u ∈ U is incident to at least one edge of E . This problem is known to be P -complete [5].Given a graph G = ( U, E ) where E = { e , . . . , e m } , define a set C of candidates by C = U ∪ { c, d } ∪ F where F = { f , . . . , f k − } . The voting profile T = M ◦ Q consists of m + 1 voters. The first part, M , consists of a single voter O ( U, { c, d } ∪ F ). Note that thecandidates of { c, d } ∪ F receive a score of 0 from M , and the other candidates receive 1.The second part is Q = ( Q , . . . , Q m ) where Q i = O ( C \ ( e i ∪ F ) , e i ∪ F ). Note that thecandidates of e i ∪ F receive a score of 0 from Q i , and the other candidates receive 1.Let Q ⊆ Q and denote T = M ◦ Q . For every c ∈ C , let b ( T , c ) be the numberof voters in T that rank c among the bottom k positions. Note that c is a winner in T . Imber and B. Kimelfeld XX:7 if and only if b ( T , c ) ≤ b ( T , c ) for all c = c . We know that b ( T , c ) = b ( T , d ) = 1 and b ( T , f ) ≥ f ∈ F . For c to be a winner, we need b ( T , u ) ≥ u ∈ U . Eachvoter Q i ranks the candidates of e i ∪ F at the bottom k positions, hence b ( T , u ) ≥ u ∈ U if and only if { e : Q e ∈ Q e } is an edge cover in G .Conversely, if E ⊆ E is an edge cover, then c is a winner of M ◦ { Q e } e ∈ E . Overall, thenumber of subsets Q ⊆ Q such that c is a winner of M ◦ Q is the number of edge covers in G . (cid:74) Next, we consider additional positional scoring rules: the Borda rule ( m − , m − , . . . , R ( f, ‘ ). We use a technique of Dey and Misra [12]. (cid:73) Lemma 5 ([12]) . Let C = { c , . . . , c m } ∪ D be a set of candidates, where D is nonempty,and ~s | C | a normalized scoring vector (i.e., the greatest common divisor of the scores is one).For every ~x = ( x , . . . , x m ) ∈ Z m , there exists λ ∈ N and a voting profile M such that s ( M , c i ) = λ + x i for i ∈ [ m ] and s ( M , d ) < λ for all d ∈ D . Moreover, the number of votesin M is polynomial in | C | · P mi =1 | x i | . We first show hardness for the Borda rule, using Lemma 5. (cid:73)
Theorem 6.
Computing
Pr[Win( c ; T )] is P -hard under Borda. Proof.
We show a reduction from counting the matchings in a graph, as defined in the proofof Theorem 3, to G = ( U, E ) where U = { u , . . . , u n } ,define a set C of candidates by C = U ∪ { c } ∪ F where F = { f , . . . , f n − n +1 } . Note that | C | = n + 2, hence the scoring vector is ( n + 1 , n , . . . , T = M ◦ Q ,as explained next.The first part, M , is the profile that exists by Lemma 5 such that (for some constant λ > f ∈ F we have s ( M , f ) < λ .For every u ∈ U we have s ( M , u ) = λ + n . s ( M , c ) = λ + n + 2 n − Q = { Q e } e ∈ E . For every edge e = { u, w } ∈ E , define Q e := O ( e, F, U \ e, { c } ). From the construction we can see the following. First, the scores of the candidates of e satisfy s ( Q e , u ) ≥ n . Second, for every other vertex u ∈ U \ e we have s ( Q e , u ) ≤ n − f ∈ F we have s ( Q e , f ) < n . Finally, s ( Q e , c ) = 0.Let E ⊆ E be a set of edges. For every u ∈ U , denote by deg E ( u ) the number ofedges of E incident to u . Define Q = { Q e } e ∈ E and T = M ◦ Q . Since every voter of Q contributes a score of 0 to c , we have s ( T , c ) = s ( M , c ) = λ + n + 2 n −
1. For every f ∈ F ,we have the following since | E | ≤ n : s ( T , f ) < λ + | E | · n ≤ λ + n < s ( T , c )For u ∈ U , if deg E ( u ) ≤ u gains at most n + 1 from edges that cover it, and at most | E | ( n −
2) from the other edges of E . Overall, s ( T , u ) ≤ n + 1 + | E | ( n −
2) + s ( M , u ) ≤ n + 1 + n ( n −
2) + λ + n < s ( T , c ) . Otherwise, if deg E ( u ) ≥
2, then we have s ( T , u ) ≥ n + s ( M , u ) = λ + n + 2 n > s ( T , c ) . X:8 Probabilistic Inference of Winners in Elections by Independent Random Voters
We can deduce that c is a winner of T if and only if deg E ( u ) ≤ u ∈ U , that is, E is a matching. Since there is a bijection between the subsets E ⊆ E and the sub-profiles T = M ◦ Q , we get the correctness of the reduction. (cid:74) The next theorem states hardness for all positional scoring rules of the form R ( f, ‘ ),except for the rule R (1 ,
1) with the scoring vector (2 , , . . . , , R (1 ,
1) rule, which gota considerable attention in the context of the possible-winner problem [3], remains an openproblem. The proof again uses Lemma 5. (cid:73)
Theorem 7.
Computing
Pr[Win( c ; T )] is P -hard under R ( f, ‘ ) whenever ( f, ‘ ) = (1 , . Proof.
First, consider the case where f >
1. We show a reduction from computing Pr[Win( c ; T )] under f -approval, which is P -hard by Theorem 3, to computing Pr[Win( c ; T )] under R ( f, ‘ ). Let T = ( T , . . . , T n ), be an instance for f -approval with probabilities ( p , . . . , p n )over a set C of candidates. Define a instance under R ( f, ‘ ) with candidate set C = C ∪ D where D = { d , . . . , d ‘ } . The voters are T = ( T , . . . , T n ) where T i = T i ◦ ( d , . . . , d ‘ ) forevery i ∈ [ n ]. The probabilities are ( p , . . . , p n ).Observe that for every i ∈ [ n ], the candidates of D receive a score of 0 from T i , and forevery c ∈ C the score is s ( T i , c ) = s ( T i , c ) + 1. Since the probabilities are the same as in theinstance under f -approval, we can deduce for every c ∈ C , the probability that c is a winnerof T under f -approval is the same as the probability that c is a winner of T under R ( f, ‘ ).Next, assume that ‘ >
1. For this case we show a reduction from computing Pr[Win( c ; T )]under ‘ -veto, which is P -hard by Theorem 4, to computing Pr[Win( c ; T )] under R ( f, ‘ ).Let T = ( T , . . . , T n ) be an instance for ‘ -veto with probabilities ( p , . . . , p n ) over a set C of candidates. Define a instance under R ( f, ‘ ) with the candidate set C = C ∪ D where D = { d , . . . , d f } .The voting profile T = ( T , . . . , T n ) ◦ M consists of two parts. For the first part, forevery i ∈ [ n ] define T i = ( d , . . . , d f ) ◦ T i and the probability is p i . The second part, M , isobtained by applying Lemma 5, where (for some constant λ > s ( M , d ) < λ for every d ∈ D and s ( M , c ) = λ + 3 n for every c ∈ C . All voters of M appearwith probability 1.For every subset A ⊆ [ n ], define a profile T A = (cid:8) T i (cid:9) i ∈ A ◦ M . For every d ∈ D the scoresatisfies s ( T A , d ) < | A | + λ ≤ λ + 2 n and for every c ∈ C we get s ( T A , c ) ≥ λ + 3 n .Therefore, every candidate in C always defeats all candidates in D . Furthermore, for every i ∈ [ n ] and c ∈ C we have s ( T i , c ) = s ( T i , c ). Overall, for every c ∈ C , the probabilitythat c is a winner of T under ‘ -veto equals the probability that c is a winner of T under R ( f, ‘ ). (cid:74) So far, we considered the complexity of computing the probability of winning only forpositional scoring rules. Next, we show hardness of two rules of a different type: Condorcetand Maximin. We begin with the former. (cid:73)
Theorem 8.
Computing
Pr[Win( c ; T )] is P -hard under Condorcet. Proof.
We show a reduction from U = { u , . . . , u q } and a collection E of 3-element subsets of U , and thegoal is to count the q -element subsets of E that cover U using pairwise-disjoint sets. Thisproblem is known to be P -complete [17]. . Imber and B. Kimelfeld XX:9 Our reduction is an adaptation of the proof of Wojtas and Faliszewski [28] that Condorcet- P -complete. The reduction is as follows. Let U = { u , . . . , u q } and E be aninstance of C = U ∪ { c } and the voting profile is T = M ◦ Q .The first part, M , consists of q − u , . . . , u q , c ). The secondpart, Q = { Q e } e ∈ E , contains a voter for every set e in E , where Q e = O ( e, { c } , U \ e ). Theyshowed a bijection between two collections: The sub-profiles Q ⊆ Q such that | Q | ≤ q and c is a Condorcet winner of M ◦ Q , and the subsets E ⊆ E that are exact covers.We change the reduction as follows to show P -hardness of d , so now C = U ∪ { c, d } . Second, the voting profile is T = M ◦ Q .The first part, M , consists of q − u , . . . , u q , c, d ), and twovoters with the preferences ( c, d, u , . . . , u q ). We have the following for all u ∈ U . N M ( u, c ) = q − N M ( u, c ) + 2 N M ( c, u ) = 2 = N M ( c, u ) + 2For c, d we have N M ( c, d ) = q + 1 and N M ( d, c ) = 0. The second part, Q = { Q e } e ∈ E ,contains the voter Q e for each e ∈ E , where Q e = O ( e, { d } , { c } , U \ e ).To prove the correctness of our reduction, we will show a one-to-one correspondencebetween our witnesses and those of Wojtas and Faliszewski [28], that is, between the sub-profiles Q ⊆ Q such that c is a Condorcet winner of M ◦ Q , and the sub-profiles Q ⊆ Q such that | Q | ≤ q and c is a Condorcet winner of M ◦ Q .Let Q ⊆ Q , denote T = M ◦ Q . Also define the sub-profiles Q = { Q e : Q e ∈ Q } ⊆ Q and T = M ◦ Q . Observe that the following holds for all u ∈ U : N T ( u, c ) = N M ( u, c ) + N Q ( u, c ) = N M ( u, c ) + 2 + N Q ( u, c )= N T ( u, c ) + 2Similarly, we have N T ( c, u ) = N T ( c, u ) + 2, N T ( c, d ) = q + 1 and N T ( d, c ) = | Q | . Hence,for all Q ⊆ Q , if | Q | > q then N T ( c, d ) ≤ N T ( d, c ) and c is not a Condorcet winner of T . Otherwise, | Q | ≤ q . In this case we have N T ( c, d ) > N T ( d, c ) and for every u ∈ U , N T ( u, c ) = N T ( u, c ) + 2 and N T ( c, u ) = N T ( c, u ) + 2.Similarly, given a sub-profile Q ⊆ Q such that | Q | ≤ q and c is a Condorcet winnerof M ◦ Q , we can define a sub-profile Q = { Q e : Q e ∈ Q } ⊆ Q and get that c is aCondorcet winner of M ◦ Q .We conclude the claimed correspondence, and hence, we get a polynomial-time reductionfrom (cid:74) Next, we show hardness for the Maximin rule. (cid:73)
Theorem 9.
Computing
Pr[Win( c ; T )] is P -hard under Maximin. Proof.
We show a reduction from NP -complete under Maximin. Let U = { u , . . . , u q } and E be an instance of C = U ∪ { c, d, w } and voting profile T = M ◦ Q , as detailed next.The M part consists of 4 q voters as follows: q voters with the preference O ( c, d, U, w ),then q − O ( c, U, w, d ), then a single voter with O ( U, c, w, d ), and lastly 2 q voterswith O ( d, w, U, c ). The second part is Q = { Q e } e ∈ E where Q e = O ( w, U \ e, c, e, d ). X:10 Probabilistic Inference of Winners in Elections by Independent Random Voters
Let Q ⊆ Q , define T = M ◦ Q and E = { e ∈ E : Q e ∈ Q } . For d we have that N M ( d, c ) = 2 q and, for all u ∈ U , that N M ( d, u ) = 3 q and N M ( d, w ) = 3 q . Since the votersof Q rank d at the bottom position, Q does not affect the score of d and s ( T , d ) = 2 q .For w it holds that every voter of Q ranks w at the top position and, hence, we havethat N T ( w, c ) = 2 q + | Q | , that N T ( w, d ) = q + | Q | , and that N T ( w, u ) = 2 q + | Q | for all u ∈ U . Therefore s ( T , w ) = q + | Q | . For u ∈ U we have N T ( u, d ) = q + | Q | , therefore, s ( T , u ) ≤ q + | Q | .Finally, for c we have N T ( c, d ) = 2 q + | Q | and N T ( c, w ) = 2 q . For every u ∈ U , letdeg E ( u ) be the number of sets of E incident to u , we get that N T ( c, u ) = 2 q − E ( u ).We complete the proof by showing that the number of subsets Q ⊆ Q wherein c is a winnerof M ◦ Q is the number of exact covers.First, suppose that E ⊆ E is an exact cover, that is, | E | = q and deg E ( u ) = 1 for every u ∈ U . Define Q = { Q e : e ∈ E } and T = M ◦ Q . From the above we have: s ( T , d ) = s ( T , w ) = 2 q ; s ( T , u ) ≤ q for all u ∈ U ; s ( T , c ) = 2 q since N T ( c, u ) = 2 q for all u ∈ U .Hence, c is a winner of T .Conversely, let Q ⊆ Q be such that c is a winner of T = M ◦ Q , and let E be thecorresponding subset of E . If | Q | > q then s ( T , w ) > q and s ( T , c ) ≤ N T ( c, w ) = 2 q ;hence, we get a contradiction. If there exists u ∈ U such that deg E ( u ) = 0, then s ( T , c ) ≤ N T ( c, u ) = 2 q − s ( T , d ) = 2 q , so we again get a contradiction. We conclude that | E | ≤ q and deg E ( u ) ≥ u ∈ U , therefore E is an exact cover. (cid:74) To summarize the section, we established the row Pr[win] of Table 1 on the exactevaluation of the probability of winning. In the next two sections, we discuss approximateevaluation.
Observe that there is an additive FPRAS for Pr[Win( c ; T )] whenever we can test in polynomialtime whether c is a winner of a sampled (fully deterministic) profile. Such an FPRAS isobtained by taking the ratio of successes in trials wherein we sample voters (according totheir distribution) and test whether c is a winner; then, an FPRAS can be shown in standardways (e.g., the Hoeffding’s Inequality).In this and the next section, we study the complexity of multiplicative approximation.Note that a multiplicative FPRAS implies an additive FPRAS, but not vice versa: ifthe probability is exponentially small, then 0 is already an additive FPRAS, but not amultiplicative one. In probability estimation, it is often important to get a multiplicativeapproximation since, unlike the additive approximation, it allows for approximating ratios ofprobabilities (that are needed, e.g., for conditional probabilities) and for comparison betweenthe likelihood of rare events. In the remainder of the paper, we restrict the discussion tomultiplicative approximations, unless explicitly stated otherwise.In this section, we argue that for most of the rules considered in the previous section,(multiplicative) approximation is also intractable and an FPRAS does not exist underconventional complexity assumptions. We show it by proving that it is NP-hard to determinewhether Pr[Win( c ; T )] = 0 and, therefore, a multiplicative FPRAS for Pr[Win( c ; T )] under r implies that NP ⊆ BPP .As a general technique, observe that deciding whether Pr[Win( c ; T )] > . Imber and B. Kimelfeld XX:11 is a polynomial-time reduction from the decision of Pr[Win( c ; T )] > M , and each remaining voter is put in Q . In the seconddirection, every voter of M participates with probability 1 and every voter of Q participateswith probability 1/2 (or any other probability in (0 , We begin by showing hardness for the Borda rule. (cid:73)
Theorem 10.
For the Borda rule, CCAUV is NP -complete; hence, under Borda it is NP -complete to decide whether Pr[Win( c ; T )] > . Proof.
We show a reduction to Borda-CCAUV from Borda-CCAV, which is known to be NP -complete [25]. Let M , Q , c ∗ and k be an input for CCAV under Borda over a set C of m candidates, where Q = ( Q , . . . , Q n ). By the proof of Russell [25] that CCAV is NP -complete for Borda, we can assume that all voters of Q rank c ∗ at the top position. Weconstruct an instance of CCAUV under Borda, where the candidate set is C = C ∪ { d , d } .Let M be the profile of Lemma 5 such that (for some λ > s ( M , d ) < λ and s ( M , d ) = λ + 2 mn + s ( M , c ∗ ) − k .For every c ∈ C we have s ( M , c ) = λ + 2 mn + s ( M , c ).The second profile Q = ( Q , . . . , Q n ) consists of n voters, where Q i = ( d , d ) ◦ Q i for every i ∈ [ n ].Let Q be a sublist of Q , define Q = (cid:8) Q i : Q i ∈ Q (cid:9) , and let T = M ◦ Q , T = M ◦ Q . For c ∈ C we have the following. s ( T , c ) = s ( M , c ) + s ( Q , c ) = λ + 2 mn + s ( M , c ) + s ( Q , c )= λ + 2 mn + s ( T , c ) (1)Since we assume that all voters of Q rank c ∗ at the top position, for all i ∈ [ n ] we have s ( Q i , c ∗ ) = s ( Q i , c ∗ ) = m −
1, and overall s ( T , c ∗ ) = λ + 2 mn + s ( M , c ∗ ) + ( m − | Q | Finally, every voter of Q contributes the scores m + 1 and m to d and d , respectively,therefore s ( T , d ) < λ + ( m + 1) | Q | ≤ λ + ( m + 1) n (2) s ( T , d ) = λ + 2 mn + s ( M , c ∗ ) − k + m | Q | = s ( T , c ∗ ) − k + | Q | (3)From Equation (1) we can deduce that for every c ∈ C \ { c ∗ } , c ∗ defeats c in T if and onlyif c ∗ defeats c in T . From Equations (1) and (2) we can deduce that d is defeated by allcandidates of C . From Equation (3) we can deduce that c ∗ defeats d in T if and only if | Q | ≤ k .We show that there exists Q ⊆ Q such that c ∗ is a winner of M ◦ Q if and only if thereexists Q ⊆ Q of size at most k such that c ∗ is a winner of M ◦ Q . Let Q ⊆ Q suchthat c ∗ is a winner of M ◦ Q . In particular, c ∗ defeats d , hence as we said | Q | ≤ k . Let Q = (cid:8) Q i : Q i ∈ Q (cid:9) , we have | Q | ≤ k . For every c ∈ C \ { c ∗ } , c ∗ defeats c in M ◦ Q ,hence c ∗ defeats c in M ◦ Q . Therefore | Q | ≤ k and c ∗ is a winner of M ◦ Q .Conversely, let Q ⊆ Q of size at most k such that c ∗ is a winner of M ◦ Q , define Q = (cid:8) Q i : Q i ∈ Q (cid:9) . By the same arguments as before, c ∗ defeats every candidates of C \ { c ∗ } in M ◦ Q , all candidates of C defeat d , and c ∗ defeats d since | Q | ≤ k . Hence, c is a winner of M ◦ Q . (cid:74) X:12 Probabilistic Inference of Winners in Elections by Independent Random Voters
The following two theorems show that CCAUV is NP -complete under R ( f, ‘ ) whenever( f, ‘ ) = (1 , (cid:73) Theorem 11.
For every fixed f ≥ and ‘ ≥ , CCAUV is NP -complete under R ( f, ‘ ) . Proof.
We show a reduction from the problem of (3DM): Giventhree disjoint sets X = { x , . . . , x q } , Y = { y , . . . , y q } and Z = { z , . . . , z q } of the same size,and a set E ⊆ X × Y × Z , is there a subset E ⊆ E consisting of q pairwise-disjoint triples?This problem is know to be NP -complete [15]. Given X , Y , Z and E = { e , . . . , e m } , weconstruct an instance of CCAUV under R ( f, ‘ ). Denote U = X ∪ Y ∪ Z . The candidate setis C = U ∪ W ∪ W ∪ { c, d } where W = { w , , . . . , w ,f − } and W = { w , , . . . , w ,‘ − } .Let M the profile of Lemma 5 such that (for some λ > s ( M , c ) = λ + 2 m and s ( M , d ) < λ ; s ( M , w ) = λ for all w ∈ W ∪ W ; s ( M , x ) = s ( M , y ) = λ + 2 m − x ∈ X and y ∈ Y ; s ( M , z ) = λ + 2 m + 1 for all z ∈ Z .The second profile Q = { Q e } e ∈ E consists of a voter for every triplet in E . For every e = ( x, y, z ) ∈ E define Q e = O ( { x, y } ∪ W , ( U \ e ) ∪ { c, d } , { z } ∪ W ) . Note that the candidates of { x, y } ∪ W receive a score of 2 from Q e , the candidates of { z } ∪ W receive a score of 0 from Q e , and the remaining candidates receive 1.We state some observations regarding the profile. Let Q ⊆ Q , define E = { e ∈ E : Q e ∈ Q } .Every voter of Q contributes a score of 1 to c and d , hence their scores are: s ( M ◦ Q , c ) = λ + 2 m + | Q | s ( M ◦ Q , d ) < λ + | Q | ≤ λ + m < s ( M ◦ Q , c )Similarly, for every w ∈ W ∪ W the score satisfies s ( M ◦ Q , w ) ≤ λ + 2 m , hence c alwaysdefeats the candidates of W ∪ W ∪ { d } .For every u ∈ U , let deg E ( u ) be the number of triplet in E that are incident to u . Forevery u ∈ X ∪ Y we have s ( M ◦ Q , u ) = λ + 2 m − | Q | + deg E ( u ) = s ( M ◦ Q , c ) − E ( u ) . (4)and for every z ∈ Z we have s ( M ◦ Q , z ) = λ + 2 m + 1 + | Q | − deg E ( z ) = s ( M ◦ Q , c ) + 1 − deg E ( z ) . (5)We show that there exists Q ⊆ Q such that c is a winner of M ◦ Q if and only if thereis a 3DM. Let E ⊆ E be a 3DM, that is, | E | = q and deg E ( u ) = 1 for all u ∈ U . Define Q = { Q e : e ∈ E } . For every u ∈ X ∪ Y , by Equation (4), we have s ( M ◦ Q , u ) = s ( M ◦ Q , c ),and for every z ∈ Z , by Equation (5) we have s ( M ◦ Q , z ) = s ( M ◦ Q , c ). Since c alwaysdefeats the candidates of W ∪ W ∪ { d } , we can deduce that c is a winner of M ◦ Q .Conversely, let Q ⊆ Q such that c is a winner of M ◦ Q , and let E = { e ∈ E : Q e ∈ Q } .For every z ∈ Z , since c defeats every z , by Equation (5) we have deg E ( z ) ≥
1. Every z ∈ Z is covered by at least one set of E , therefore | E | ≥ q . For every u ∈ X ∪ Y , since c defeatsevery u , by Equation (4) we have deg E ( u ) ≤
1. If | E | > q then there exists u ∈ X ∪ Y forwhich deg E ( u ) >
1, hence a contradiction.Overall, | E | = q , each z ∈ Z is covered by at least one set of E , and each u ∈ X ∪ Y iscovered by at most one set of E . Therefore, E is a 3DM. (cid:74) . Imber and B. Kimelfeld XX:13 By a reduction similar to the proof of Theorem 11, we obtain the following. (cid:73)
Theorem 12.
For every fixed f ≥ and ‘ ≥ , CCAUV is NP -complete under R ( f, ‘ ) . Proof.
We again show a reduction from 3DM as defined in the proof of Theorem 11. Giventhe input X = { x , . . . , x q } , Y = { y , . . . , y q } , Z = { z , . . . , z q } and E = { e , . . . , e m } , weconstruct an instance of CCAUV under R ( f, ‘ ). Denote U = X ∪ Y ∪ Z . The candidate setis C = U ∪ W ∪ W ∪ { c, d } where W = { w , , . . . , w ,f − } and W = { w , , . . . , w ,‘ − } .The voting profiles M , Q are constructed in the same way as in the proof of Theorem 11,except that for every e = { x, y, z } ∈ E , the voter Q e contributes the score 2 to x and 0 to y, z (instead of contributing 2 to x, y and 0 to z ), and the scores of M are modified accordingly.Formally, let M the profile of Lemma 5 such that (for some λ > s ( M , c ) = λ + 2 m and s ( M , d ) < λ ; s ( M , w ) = λ for all w ∈ W ∪ W . s ( M , x ) = λ + 2 m − x ∈ X , and s ( M , y ) = s ( M , z ) = λ + 2 m + 1 for all y ∈ Y and z ∈ Z .The second profile is Q = { Q e } e ∈ E . For every e = { x, y, z } ∈ E define Q e = O ( { x } ∪ W , ( U \ e ) ∪ { c, d } , { y, z } ∪ W ) . By the same arguments as in the proof of Theorem 11, there exists Q ⊆ Q such that c isa winner of M ◦ Q if and only if there is a 3DM. (cid:74) From Theorems 11 and 12 we conclude the following. (cid:73)
Corollary 13.
For all fixed ( f, ‘ ) = (1 , , it is NP -complete to determine whether Pr[Win( c ; T )] > under R ( f, ‘ ) . For the rules Condorcet and Maximin, the proofs of Theorem 8 and Theorem 9, respect-ively, show a reduction from Q ⊆ Q such that c is a winner of M ◦ Q . Since it is NP -hard to decide whether there is an exact cover, it isalso NP -hard to decide whether there exists Q ⊆ Q such that c is a winner of M ◦ Q . (cid:73) Theorem 14.
Under Condorcet and Maximin, CCAUV (and deciding whether
Pr[Win( c ; T )] > ) is NP -complete. What about the positional scoring rules that are not covered by the previous section? Inparticular, is there an FPRAS for k -approval and k -veto for k > P -hard (Theorem 3 and 4)? The question remains open. We do know, however, that theproof technique of this section fails on them since it turns out that zeroness can be decidedin polynomial time for these rules (and, in fact, every binary positional scoring rule). (cid:73) Theorem 15.
For every binary positional scoring rule, CCAUV is solvable in polynomialtime and, hence, whether
Pr[Win( c ; T )] > can be decided in polynomial time. Proof.
Let ( C, M , Q , c ) be an instance of CCAUV. Let Q ∗ ⊆ Q be the set of all voters ofwho contribute 1 to c . We claim that if any Q ⊆ Q is such that c is a winner of M ◦ Q ,then c is a winner of M ◦ Q ∗ . Hence, it suffices to test whether c is a winner of M ◦ Q ∗ .Let Q ⊆ Q such that c is a winner of M ◦ Q . For every voter of Q ∗ \ Q , if we add itto M ◦ Q then the score of c increases by 1, and the score of every other candidate increasesby at most 1. Hence, we can add all voters of Q ∗ \ Q and c remains a winner, that is, c is a X:14 Probabilistic Inference of Winners in Elections by Independent Random Voters
Algorithm 1
Sampling from the posterior distribution conditioned on L( c, d ; T ). Define J := ∅ for i = 1 , . . . , n do Compute q i := Pr[ i ∈ I | L( c, d ; T ) ∧ I i − = J i − ] With probability q i define J i := J i − ∪ { i } , otherwise J i := J i − return J := J n winner of M ◦ Q where Q = Q ∪ Q ∗ . Now, for every voter in Q \ Q ∗ , if we remove it from M ◦ Q then the score of c is unchanged (since these voters contribute 0 to the score of c )and the score of the other candidates cannot increase. Therefore c is a winner of M ◦ Q ∗ . (cid:74) In the previous section, we showed that a multiplicative approximation of Pr[Win( c ; T )]is often intractable since it is hard to determine whether this probability is zero. In caseswhere it is tractable to determine whether it is zero, the existence of an approximationscheme (and even a constant-ratio approximation) remains an open problem. In thissection we show that, in contrast, we can often get an FPRAS for the probability of losing :Pr[Lose( c ; T )] = 1 − Pr[Win( c ; T )]. (cid:73) Theorem 16.
There is a multiplicative FPRAS for estimating
Pr[
Lose ( c ; T )] = 1 − Pr[Win( c ; T )] under every positional scoring rule with polynomial scores, and under theCondorcet rule. In the remainder of this section, we prove Theorem 16. We adapt the technique ofKarp-Luby-Madras [18] for approximating the number of satisfying assignments of a DNFformula.For a positional scoring rule, c is not a winner if there exists another candidate d suchthat s ( T , d ) > s ( T , c ). For a pair of candidates c = d , let L( c, d ; T ) be the event that s ( T , c ) > s ( T , d ). We can write Lose( c ; T ) = ∨ d = c L( d, c ; T ). Under Condorcet, c is not awinner if there exists d = c such that N T ( d, c ) ≥ N T ( c, d ). Therefore the event L( c, d ; T )is N T ( c, d ) ≥ N T ( d, c ) and we again have Lose( c ; T ) = ∨ d = c L( d, c ; T ).As before, let I ⊆ [ n ] be the random variable that represents the random set of voterswho cast their vote. For the Karp-Luby-Madras algorithm to estimate Pr[ ∨ d = c L( d, c ; T )],we need to perform the following tasks in polynomial time: Test whether L( c, d ; T ) is true in a given sample; Compute Pr[L( c, d ; T )] for every c and d ; Sample a random set I of voters from the posterior distribution conditioned on the eventL( c, d ; T ).The first task is straightforward. For the other two tasks, we assume that we can computePr[L( c, d ; T ) | I ∩ S = S ] in polynomial time, given a profile T , a pair of candidates c, d and voter sets S ⊆ [ n ] and S ⊆ S . Later, we will show how this can be done for particularrules. This solves the second task because Pr[L( c, d ; T )] = Pr[L( c, d ; T ) | I ∩ ∅ = ∅ ].For the task of sampling from the posterior distribution conditioned on L( c, d ; T ), forevery i ≤ n we define a random variable I i = I ∩ { , . . . , i } . Algorithm 1 presents thesampling procedure, which constructs a random set of voters iteratively. It begins with J := ∅ . Then, for every i ∈ [ n ], at the i th iteration, the algorithm defines J i := J i − ∪ { i } . Imber and B. Kimelfeld XX:15 with probability Pr[ i ∈ I | L( c, d ; T ) ∧ I i − = J i − ], and J i := J i − otherwise. The outputis J := J n .We show that the algorithm is correct: for all U ⊆ [ n ] we havePr[ J = U ] = Pr[ I = U | L( c, d ; T )] . (6)For that, we show by induction on i that for every U ⊆ [ i ] we havePr[ J i = U ] = Pr[ I i = U | L( c, d ; T )] . (7)Then, for i = n we get Pr[ J = U ] = Pr[ I = U | L( c, d ; T )].For the base i = 1, the equality (7) holds by the definition of the algorithm. Let i > i −
1, and let U ⊆ [ i ]. If i ∈ U thenPr[ J i = U ] = Pr[ J i − = U \ { i } ] · Pr[ i ∈ J i | J i − = U \ { i } ]= Pr[ I i − = U \ { i } | L( c, d ; T )] · Pr[ i ∈ I | L( c, d ; T ) ∧ I i − = U \ { i } ]= Pr[ I i = U | L( c, d ; T )] . Otherwise, i / ∈ U , similarly,Pr[ J i = U ] = Pr[ J i − = U \ { i } ] · Pr[ i / ∈ J i | J i − = U \ { i } ] = Pr[ I i = U | L( c, d ; T )] . This concludes the correctness of the algorithm.We now show that the algorithm can be realized in polynomial time. In the the i thiteration, we need to compute q i = Pr[ i ∈ I | L( c, d ; T ) ∧ I i − = J i − ], that is: q i = Pr[L( c, d ; T ) ∧ I i = J i − ∪ { i } ]Pr[L( c, d ; T ) ∧ I i − = J i − ] . For the denominator, observe thatPr[L( c, d ; T ) ∧ I i − = J i − ] = Pr[ I i − = J i − ] × Pr[L( c, d ; T ) | I i − = J i − ] . We can compute Pr[ I i − = J i − ] directly from the definition:Pr[ I i − = J i − ] = Y j ∈ [ i − ∩ J i − p j Y j ∈ [ i − \ J i − (1 − p j ) . For the second part, we havePr[L( c, d ; T ) | I i − = Ji −
1] = Pr[L( c, d ; T ) | I ∩ [ i −
1] = J i − ] , which can be computed in polynomial time by our assumption that Pr[L( c, d ; T ) | I ∩ S = S ]is computable in polynomial time. We can similarly compute the numerator Pr[L( c, d ; T ) , I i = J i − ∪ { i } ].It remains to show that Pr[L( c, d ; T ) | I ∩ S = S ] can indeed be calculated in polynomialtime for S ⊆ [ n ] and S ⊆ S . We start with positional scoring rules with polynomial scores. (cid:73) Lemma 17.
For every positional scoring rule with polynomial scores, we can compute
Pr[ L ( c, d ; T ) | I ∩ S = S ] in polynomial time given a voting profile T over a set C ofcandidates, a pair c = d of candidates, probabilities ( p , . . . , p n ), and sets S ⊆ S . X:16 Probabilistic Inference of Winners in Elections by Independent Random Voters
Proof.
For every voter v i define x i = s ( T i , c ) − s ( T i , d ) and define D = P i ∈ S x i . ObservethatPr[L( c, d ; T ) | I ∩ S = S ] = Pr "X i ∈ I x i > (cid:12)(cid:12) I ∩ S = S = Pr X i ∈ I ∩ S x i + X i ∈ I \ S x i > (cid:12)(cid:12) I ∩ S = S = Pr X i ∈ I \ S x i > − D . We show that we can compute Pr hP i ∈ I \ S x i > − D i in polynomial time. Assume,without loss of generality, that the set of voters outside S is [ n ] \ S = { , . . . , k } .Let A be the sum of the negative values in x , . . . , x k and B the sum of the positivevalues. For every t ≤ k and y ∈ Z , define N ( t, y ) = Pr[ P i ∈ I ∩ [ t ] x i > y ], our goal is tocompute N ( k, − D ). If y < A then N ( t, y ) = 1 and if y > B then N ( t, y ) = 0. Using dynamicprogramming, we compute N ( t, y ) for all t ≤ k and y ∈ { A, . . . , B } in polynomial time. Notethat B − A = poly( n, m ) since the scores are polynomial.For t = 0 we have N ( t, y ) = 1 if y < N ( t, y ) = 0 otherwise. Let t ≥
1, consider twocases. If t ∈ I then for the event P i ∈ I ∩ [ t ] x i > y we need P i ∈ I ∩ [ t − x i > y − x t , otherwisewe need P i ∈ I ∩ [ t − x i > y . Hence, N ( t, y ) = p t · N ( t − , y − x t ) + (1 − p t ) · N ( t − , y ) . This concludes the proof. (cid:74)
Using a similar analysis to the proof of Lemma 17, we obtain the same result for theCondorcet rule. (cid:73)
Lemma 18.
For the Condorcet rule, we can compute
Pr[ L ( c, d ; T ) | I ∩ S = S ] inpolynomial time given a voting profile T over a set C of candidates, a pair c = d ofcandidates, probabilities ( p , . . . , p n ) and sets S ⊆ S . Proof.
Recall that for the Condorcet rule, the event L( c, d ; T ) is N T ( c, d ) ≥ N T ( d, c ). Let V c be the set of voters which rank c higher than d , and let V d be the set of voters which rank d higher than c . Define D = | S ∩ V c | − | S ∩ V d | , observe thatPr[L( c, d ; T ) | I ∩ S = S ] = Pr [ | I ∩ V c | ≥ | I ∩ V d | | I ∩ S = S ]= Pr (cid:2) | ( I ∩ S ) ∩ V c | + | ( I \ S ) ∩ V c | ≥ | ( I ∩ S ) ∩ V d | + | ( I \ S ) ∩ V d | | I ∩ S = S (cid:3) = Pr [ | ( I \ S ) ∩ V c | − | ( I \ S ) ∩ V d | ≥ − D ] . We show that we can compute this probability in polynomial time. Assume w.l.o.g.that the voters which are not in S are [ n ] \ S = { , . . . , k } . Let A = −| [ k ] ∩ V d | and let B = | [ k ] ∩ V c | . For every t ≤ k and y ∈ Z , define N ( t, y ) = Pr [ | ( I ∩ [ t ]) ∩ V c | − | ( I ∩ [ t ]) ∩ V d | ≥ y ] . Our goal is to compute N ( k, − D ). If y < A then N ( t, y ) = 1 and if y > B then N ( t, y ) = 0.Using dynamic programming, we compute N ( t, y ) for all t ≤ k and y ∈ { A, . . . , B } inpolynomial time. Note that B − A = O ( n ).For t = 0 we have N ( t, y ) = 1 if y ≤ N ( t, y ) = 0 otherwise. Let t ≥
1, if t ∈ V c then N ( t, y ) = p t · N ( t − , y −
1) + (1 − p t ) · N ( t − , y ) . . Imber and B. Kimelfeld XX:17 Otherwise, t ∈ V d , then N ( t, y ) = p t · N ( t − , y + 1) + (1 − p t ) · N ( t − , y ) . This concludes the proof. (cid:74)
Note that Theorem 16 covers all voting rules that we discuss in the paper, except forMaximin. The correctness of Algorithm 1 holds for Maximin, and in fact for every votingrule. However, to obtain an FPRAS for estimating Pr[Lose( c ; T )] under Maximin, we needto be able to compute the probabilities q i of Algorithm 1 in polynomial time for Maximin.Whether this is the case remains an open problem. We studied the complexity of evaluating the marginal probability of wining in an electionwhere voters are drawn independently at random, each with a given probability. The exactprobability is computable in polynomial time for the plurality and veto rules, and P -hardfor the other rules we considered, including classes of positional scoring rules, Condorcetand Maximin. In some of these cases, it is also intractable to compute a multiplicativeapproximation of the probability of winning, since it is NP -complete to determine whetherthis probability is nonzero. For k -approval and k -veto, the zeroness problem is solvable inpolynomial time, but the complexity of the multiplicative approximation remains unknown.In contrast, we devised a multiplicative FPRAS for the probability of losing in all positionalscoring rules with polynomial scores, in addition to Condorcet.Several problems are left open for future investigation. First, can we establish a fullclassification of the (pure) positional scoring rules in terms of their complexity for ourproblem, as in the case of the possible-winner problem over incomplete preferences [3, 4, 29]?It might be the case the same classification holds, since our results so far are consistentwith the possible winner (tractability for plurality and veto, and hardness for the rest). Tobegin with, a specific rule of which absence stands out in Table 1 is (2 , , . . . , , R (1 , / / References Yoram Bachrach, Nadja Betzler, and Piotr Faliszewski. Probabilistic possible winner determ-ination. In Maria Fox and David Poole, editors,
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