Sèbastien Courtin

Boniface Mbih
email: boniface.mbih@unicaen.fr

Issofa Moyouwou

Thomas Sennè

The reinforcement axiom under sequential positional rules

The reinforcement axiom roughly states that when an alternative is selected by two di §erent constituencies, it must also be selected by their union. Hare and Coombs rules are special cases of sequential positional voting rules which are known to violate this axiom. In this paper, we Örst show that reinforcement can be violated by all such rules. We then evaluate, by the use of Monte Carlo simulations and the Fishburn-Gehrlein technique, the proportion of proÖles at which this phenomenon occurs.

Introduction

In the theoretical debate on the attractiveness of Condorcet consistent social choice methods on the one hand, and positional (or scoring) systems -voting ‡l aBorda -on the other hand, the two strongest arguments in favor of the second type of mechanisms are the participation axiom and the Youngís reinforcement axiom (YRA). Roughly, according to the participation axiom, no voter can take advantage of not taking part to the election. For more details, see [START_REF] Nurmi | A responsive voting system[END_REF] or [START_REF] Mbih | On the responsiveness of parliamentary social choice functions[END_REF] among others. In this contribution, we are only interested in YRA.

The notion of reinforcement was Örst introduced by [START_REF] Smith | Aggregation of Preferences with Variable Electorate[END_REF], under the name of separability. The intuition is as follows: if two di §erent blocks of individuals rank an alternative at least as high as some other alternative, then so does the combination of the blocks with regard to these two alternatives.

In the same way -and indeed at the same period - [START_REF] Young | An Axiomatization of Bordaís Rule[END_REF][START_REF] Young | So cial Choice Scoring Functions[END_REF] proposes a variant of the notion of separability, which he calls consistency. Following this idea, if two disjoint groups of individuals separately select two non-disjoint subsets of alternatives, then the union of these groups should exactly select the intersection of the two subsets. In fact, the di §erence between the two contributions lies in the type of social choice mechanisms studied: Young is concerned with social choice correspondences (selecting possibly more than a single alternative), while Smith is interested in aggregation functions (selecting a social ranking of alternatives from preferences reported by individual voters).

And Önally, in the special context of social choice functions (correspondences selecting a unique outcome), Moulin (1988) introduces the phrase Youngís reinforcement axiom.

Actual situations where this phenomenon could have arisen can also be found. In France, President Georges Pompidou and more recently President Jacques Chirac both abandoned the idea of submitting to the vote of the CongrËs de Versailles, the bills on the reduction of the presidentís term of o¢ce from seven to Öve years in 1973 and the status of French Polynesia 1999, respectively, although those bills had been voted in both AssemblÈe Nationale and SÈnat.

In the Condorcet-Borda debate, arguments in favor of positional rules (PR), based on the notion of reinforcement, are summarized in Youngís theorem (1975). This theorem shows that all PR satisfy YRA whereas there is no Condorcet consistent social choice procedure satisfying YRA.

There is no doubt about the importance of this result. However it does not take into consideration the potentially sequential aspects of positional rules. This paper is speciÖcally concerned with sequential positional rules (see DeÖnition 6), whose famous examples are the well-known Hare and Coombs methods.

Our goal is to examine precise conditions at which violations of Youngís reinforcement axiom are susceptible to arise under sequential positional rules. Our main contribution in this context states that all sequential positional rules violate this axiom. This normative contribution is completed by a measure of the quantitative signiÖcance of the violation of YRA.

The remainder of this paper is organized as follows: Section 2 is a presentation of the general framework with deÖnitions, assumptions and some examples. Section 3 provides some general results on the behavior of sequential positional rules vis- ‡-vis the reinforcement axiom. Then, Section 4 studies the precise conditions at which a proÖle may violate the axiom, and frequencies are given in Section 5; Önally, Section 6 discusses and concludes the paper.

Notations and deÖnitions

Consider a Önite set N of n individuals or voters, with n ! 2 and a Önite set A of m alternatives. And suppose two disjoint groups (or constituencies)

T 1 and T 2 of individuals, and T 1 [ T 2 = N: Let 2 A be the set of nonempty subsets of A. Assume that the preference relation R i of individual i, i 2 N , is a complete, antisymmetric and transitive binary relation (or simply a linear order) on A and let L = fR k :1% k % m!g be the set of all linear orders on A.

A proÖle is an n-tuple R N =(R i ) i2N of individual preference relations, one for each individual. The set of all proÖles on N will be denoted by L N . Similarly proÖles on T 1 and T 2 will be denoted R T1 and R T2 , respectively.

DeÖnition 1 Given T ' N , a social choice function ( SCF) is a mapping f from L T to A.
In other words, an SCF assigns a single alternative to each proÖle. We now present the YRA property more formally.

DeÖnition 2 Given T 1 and T 2 , an SCF f satisÖes YRA if for any R T1 2 L T1 and R T2 2 L T2 and for all x 2 A,

! f (R T1 )=x and f (R T2 )=x " =) f (R N )=x
Alternatively,

DeÖnition 3 Given T 1 and T 2 , an SCF f violates YRA if there exist R T1 2 L T1 , R T2 2 L T2 and fx; yg'A, x 6 = y, such that f (R T1 )=x, f (R T2 )=x and f (R N )=y
An illustration of this notion will be presented in Example 1. Now, we need additional notations to present the family of SCF s under study in the present paper. For each R k in L, let n k be the total number of individuals in N with preference relation R k . In particular with A = fx; y; zg, the set of linear orders on A is: R 1 : xyz; R 2 : xzy; R 3 : yxz; R 4 : yzx; R 5 : zxy; R 6 : zyx Given B 2 2 A such that jBj!2 and x 2 B, let r (B; x; k) be the rank, according to R k , of x among alternatives in B. A scoring vector is a jBjtuple v = # v 1 ; :::; v r ; :::; v jBj $ of real numbers such that v 1 =1, v jBj =0and for all r =1 ; :::; jBj+1, v r ! v r+1 . When B is the issue, each individual gives v r points to the alternative in B he ranks at the r th position.

DeÖnition 4 Given B 2 2 A , T ' N , R T 2 L T ;x2 B and a scoring vector v 2 R jBj : i) Sc # B; x; R T ;v $ , the score of x is deÖned as follows: Sc # B; x; R T ;v $ = jBj! 5 k=1 v r(B;x;k) , n k ii) with b B = & x 2 B : Sc # B; y; R T ;v $ ! Sc # B; x; R T ;v $ for all y 2 B '
as the set of alternatives with the smallest score, l # B; R T ;v $ is the losing alternative and is such that l

# B; R T ;v $ = 8 > < > : b B if b B is a singleton, g( b B)
otherwise, where g is a mechanism used to break ties among alternatives in b B.

Note that the mechanism g introduced in the deÖnition above is still quite general and can take di §erent forms: a chance mechanism, the power given to a chairman, the lexicographic order of the alternatives, etc. We will later give precision about the type of mechanism we use in this paper.

In social choice processes using positional rules, the winning alternative is the one with the highest score, as stated in the following deÖnition.

DeÖnition 5 Let B 2 2 A , T ' N and a scoring vector v 2 R jBj , a posi- tional rule ( PR) is a social choice function f such that 8R T 2 L T , 8x 2 B, f (R T )=x if [Sc # B; x; R T ;v $ ! Sc # B; y; R T ;v $ 8y 2 B, x 6 = y],
with ties broken by some mechanism.

From the deÖnitions above, it appears that given some issue B,aPR is deÖned by a vector v 2 R jBj . We can then express three usual procedures: plurality rule if v = (1; 0; :::; 0); antiplurality rule if v = (1; :::; 1; 0) and Borda rule if v =(1; m#2 m#1 ; :::; m#r m#1 ; :::; 1 m#1 ; 0).

We now introduce the sequential positional rules (SPR) under consideration all along our study.

Let V A = & v m ; :::; v jBj ; :::; v 2 ' is a collection of scoring vectors, each vector v jBj being associated with each possible cardinality jBj of the subset B ' A, jBj!2. At the Örst step of the sequential process, scores are computed using the vector v m and the losing alternative is eliminated. In the next step, vector v m#1 is used to compute the scores and again the losing alternative is eliminated. The sequential process is repeated until a simple majority winner is obtained (see [START_REF] Lepelley | Constant Scoring rules, Condorcet criteria and single-peaked preferences[END_REF].

More formally, DeÖnition 6 A sequential positional rule ( SPR) is a social choice function f such that given B ' A, R N 2 L N and V A = & v m ; :::; v jBj ; :::; v 2 ' , f (R N )=A m , with A m sequentially deÖned in the following way:

A 1 = A A 2 = A + l # A 1 ;R N ;v m $ . . . A jBj+1 = A jBj + l # A jBj ;R N ;v m#jBj+1 $ . . . A m = A m#1 + l # A m#1 ;R N ;v 2 $ , with jA m j =1
In order to illustrate this deÖnition, let us consider the three-alternative case. We then have

v A = & v 3 ;v 2 '
, with v 3 = (1;9;0), 0 % 9 % 1, where 1, 9 and 0 are the scores of the alternatives ranked Örst, second and third, respectively, in individual preference relations in the Örst step, and v 2 = (1; 0). More generally, given a proÖle of individual preferences, the total score of an alternative is the sum of individual scores, over the whole set of individuals. A PR selects the alternative with the highest score. As a di §erence, an SPR Örst eliminates the alternative with the smallest score at the Örst step, and then selects the alternative with the highest score at the new step, among the remaining ones. Note that with three alternatives, the second step is simply a pairwise majority contest between the two remaining alternatives. Also note that for 9 equal to 0, 1, or 1=2 in v 3 , we obtain Hareís Procedure (HP ), Coombs Procedure (CP ) and Iterative Borda Procedure (IBP ), respectively. [START_REF] Smith | Aggregation of Preferences with Variable Electorate[END_REF] studies iterative -as distinguished from sequential -positional rules and he shows that they violate YRA. Under iterative positional rules, at each step the scoring vector changes only with respect to the number of alternatives, while for sequential positional voting rules, there is no Öxed relation between two scoring vectors of di §erent steps: for example, we may use plurality at step 1 and antiplurality at step 2, etc. Iterative positional rules are thus special cases of sequential positional rules. In the next section, we shall show that all SPRs violate YRA provided that jAj!3 and n ! 15.

An illustration of the violation of YRA by SPRs, is given in Example 1.

Example 1 Suppose the SPR is the Hare procedure, that is

v A = & v 3 ;v 2 '
, with v 3 = (1; 0; 0) and v 2 = (1; 0). Let A = fx; y; zg, T 1 ;T 2 ' N , and assume that R T1 2 L T1 and R T2 2 L T2 are as below: Then the scores are as follows Separate groups

1 rst Step 8 > > > > < > > > > : T 1 T 2 Sc(A;x;R T 1 ;v 3 )=16 Sc(A;x;R T 2 ;v 3 )=15 Sc(A;y;R T 1 ;v 3 )=25 Sc(A;y;R T 2 ;v 3 )=10 Sc(A;z;R T 1 ;v 3 )=15 Sc(A;z;R T 2 ;v 3 )=20
Loser: z Loser: y

2 nd Step 8 < : Sc(A#fzg;x;R T 1 ;v 2 )=31 Sc(A#fyg;x;R T 2 ;v 2 )=25 Sc(A#fzg;y;R T 1 ;v 2 )=25 Sc(A#fyg;z;R T 2 ;v 2 )=20
Winner:

x Winner:

x

The whole electorate:

N = T 1 [ T 2 1 rst Step 8 > > < > > : Sc(A;x;R N ;v 3 )=31 Sc(A;y;R N ;v 3 )=35 Sc(A;z;R N ;v 3 )=35
Loser:

x

2 nd Step 8 < : Sc(A#fxg;y;R N ;v 2 )=35 Sc(A#fxg;z;R N ;v 2 )=66
Winner: z

It appears that the Hare procedure violates YRA:

f (R T1 )=x, f (R T2 )= x but f (R N )=z.
In the next section, we will use the notions introduced above in order to prove our result on the violation of YRA by SPRs.

General results

The aim of this section is to identify SPRs that violate YRA and conditions on the number n of the voters, at which the phenomenon is susceptible to arise. We Örst consider situations at which such violations can occur for all possible scoring vectors -and it appears that this is the case roughly as soon as the number of voters is equal to 13 or at least 15 -and then, we discuss the case of electorates with less than 13 voters.

Proposition 1 In three-candidate elections with 13 voters or at least 15 voters, all SPRs violate Young reinforcement axiom.

Proof Let f be an SPR associated with the collection of scoring vectors V A = f(1;9;0) ; (1; 0)g.

Case 1: Suppose that 1 ! 9>0, and n is odd. By assumption on n, there exists a non negative integer p such that n = 13 + 2p. Consider the following proÖles: R T1 R T2 2+p 01 + p 000 xxyy z z yzxz x y zyzx y x

xxyyz z yzxzxy zyzxyx

At the Örst step and proÖle R T1 , z collects 0 point and is ruled out of the process. For the second step, x defeats y with 2+p favorable votes against 1+p for y. That is f (R T1 )=x.

At the Örst step and R T2 , x, y and z obtain respectively 3+49 points, 3+39 points and 4+39 points. Since 9>0, y gets the smallest number of points and is then ruled out of the process. At the second step, x defeats z with 6 favorable votes against 4 for z. That is f (R T2 )=x.

At the Örst step and R N , x, y and z gather respectively 5+p +(p + 5) 9 points, 4+p +(p + 5) 9 points and 4+39 points. Since 9>0 and p ! 0, z gets the smallest number of points and is ruled out of the process. At the second step, y defeats x with 8+p favorable votes against 5+p for x. That is f (R N )=y.

Case 2: Suppose that 1 ! 9>0 and n is even. By assumption on n, there exists a non negative integer p such that n = 16 + 2p. Consider the following proÖles: R T1 R T2 2+p 01 + p 000 xxyy z z yzxz x y zyzx y x

xxyyz z yzxzxy zyzxyx

Using very similar arguments as in the previous case one can easily check that f

# R T1 $ = x, f # R T2 $ = x and f # R N $ = y.
Case 3: Suppose that 9 =0and n is odd. Then there exists a non negative integer p such that n = 13 + 2p. Consider the following proÖles: R T1 R T2 3+p 02 + p 000 xxyy z z yzxz x y zyzx y x

xxyyz z yzxzxy zyzxyx

At the Örst step and R T1 , z gets 0 point and is ruled out of the process. At the second step, x defeats y with 3+p favorable votes against 2+p for y. That is f (R T1 )=x.

At the Örst step and R T2 , x, y and z get respectively 3 points, 2 points and 3 points. Therefore y gets the smallest amount of points and is then ruled out of the process. At the second step, x defeats z with 5 favorable votes against 3 for z. That is f (R T2 )=x.

At the Örst step and R N , x, y and z gather respectively 6+p points, 4+p points and 3 points. Since p ! 0, z gets the smallest number of points and is ruled out of the process. At the second step, y defeats x with 7+p favorable votes against 6+p for x. That is f (R N )=y.

Case 4: Suppose that 9 =0and n is even. By assumption on n, there exists a non negative integer p such that n =16+2p. Consider the following proÖles: R T1 R T2 3+p 02 + p 000 xxyy z z yzxz x y zyzx y x

xxyyz z yzxzxy zyzxyx

Again, as in the previous case one can easily check that f

# R T1 $ = x, f # R T2 $ = x and f # R N $ = y.
It the appears that in all cases considered above, all SPRs violate Young reinforcement axiom; and this concludes the proof.

As a direct consequence of the preceding proposition, the following holds:

Proposition 2 With 13 voters or at least 15 voters and at least three candidates, all SPRs violate the Young reinforcement axiom.

Proof Suppose A = fx; y; z; a 4 ;a 5 ; :::; a m g and let f be an SPR associated with a collection of scoring vectors

V A = & v m ; :::; v 3 ;v 2 '
, with v 2 = (1; 0) and v 3 = (1;9;0). Consider the four distinct cases in the proof of Proposition 1 and the corresponding proÖles R T1 and R T2 . Now, construct two new proÖles R T1 and R T2 from R T1 and R T2 respectively, in such a way that alternative a 4 appears at the fourth position in each individual preference relation, a 5 appears at the Öfth position, and so on. Each candidate in fx; y; zg gets the same number of points as presented in the proof of Proposition 1. Moreover all candidates in fx; y; zg record more points than every candidate in fa 4 ;a 5 ; :::; a m g until candidates in fa 4 ;a 5 ; :::; a m g are all ruled out of the process. It follows that the two last steps are exactly the same as with R T1 , R T2 and R N respectively. That is f (R T1 )=x, f (R T2 )=x and f (R N )=y. Consequently, f violates the Young reinforcement axiom.

The two propositions above show that all SPRs violate YRA when the number of voters is equal to 13 or is at least 15, provided that the number of alternatives is at least three. Note that there is no It remains to consider situations with only two alternatives, and electorates with no more than 12 voters (and also n = 14). The discussion below provides answers to these questions.

First, for all n 2 N; and jAj =2, it straightforwardly appears that there exists no SPR which violates YRA.

Second, with three alternatives, a computer based complete enumeration program (available from the authors upon simple request) reveals that there is no n for which all SPRs violate YRA. However, for some n and some 9, our computation program exhibits cases of violations of YRA. It is obvious that this is also true for four or more alternatives.

Fig. 1 below summarizes these results. It represents, for each SPR, the frequency of violation of YRA, with respect to both n (n is between 1 and 14) and the scoring vector, with 0 % 9 % 1. SpeciÖcally, it shows that the best rules, according to YRA, are clearly the usual ones: Hareís Procedure, Coombs Procedure and Iterative Borda Procedure; they minimize the frequency of violation of YRA, among all SPRs.

Fig. 1 Violation of YRA for small values of n for three alternatives, 0 % 9 % 1 and n =1; :::; 14.

These last results also suggest that one of our main goals in this paper is to evaluate how frequent SPRs violate YRA. In the next section, we focus on the special case of three alternatives.

Violations of reinforcement

In this section we present a typology of the various preference proÖles at which the violation of the YRA arises. As mentioned above, we only study the three-alternative case. With four or more alternatives, computations are much more involved.

Given A = fa 1 ;a 2 ;a 3 g, there are exactly six linear orders on A, labeled below:

R 1 : a 1 a 2 a 3 ;R 2 : a 1 a 3 a 2 ;R 3 : a 2 a 1 a 3 ;R 4 : a 2 a 3 a 1 ;R 5 : a 3 a 1 a 2 ;R 6 : a 3 a 2 a 1
We Örst introduce some further notations and deÖnitions. A situation s N is a preference proÖle obtained from a proÖle R N by rewriting it as s N = (n 1 ;n 2 ;n 3 ;n 4 ;n 5 ;n 6 ); where for each k = f1; :::; 6g ;n k is the number of individuals in N with preference relation R k . In other words, a situation is a 6+tuple of natural integers such that 5 6 k=1 n k = n.

In order to take the two disjoint groups T 1 and T 2 into consideration, with j 2f 1; 2g, t Tj =( t j 1 ; :::; t j 6 ) denotes a preference proÖle deduced from a proÖle R Tj where t j k is the number of individuals in T j with preference relation R k , and 5 6 k=1 t j k = jT j j. Note that since T 1 [ T 2 = N , t 1 k + t 2 k = n k . We now specify the mechanism we use in this paper in order to break ties among alternatives in b B (see DeÖnition 4), that is the lexicographic order; in other words, ties are broken in favor of the alternatives with the smallest index, which can be written as follows:

a h beats a k , , h<k) Sc # B; a h ;R T ;v $ ! Sc # A; a k ;R T ;v $ k<h) Sc # B; a h ;R T ;v $ >Sc # A; a k ;R T ;v $
where T ' N .

Notice that one can imagine many other ways to break ties: a random mechanism, a chairman whose vote breaks ties, etc. The mechanism chosen here -the lexicographic order -has at least two advantages: it avoids introducing chance in the determination of the outcome, and it is compatible with the anonymity of the rules since it preserves the equalitity of treatment of all voters. Further, as the number of voters rises, impact of such a mechanism becomes marginal.

Note that when the number of alternatives is reduced to three, under an SPR, there are at most two steps in order to determine the winning alternative. In situations with only one single step -that is, with an outcome selected after a unique step -the winning alternative is called the Örst step winner. We then have the proposition below.

Proposition 3 Let A = fa 1 ;a 2 ;a 3 g, B ' A, T a constituency, R T 2 L T and consider an SPR associated with V A = f(1;9;0) ; (1; 0)g. Then for all h 2f1; 2; 3g, alternative a h is the Örst step winner if

Sc # B; a h ;R T ;v 3 $ > (1 + 9) jT j 2 with (1 + 9) jT j = 3 X i=1 Sc # B; a i ;R T ;v 3 $ Proof Since 9 2 [0; 1],
we distinguish two cases: (i) First suppose 9 =1 . Then, a h cannot be selected at the Örst step. Since the sum of the scores of the three alternatives is equal to (1+9) jT j =2jT j, it follows that in order to be selected at the Örst step, a h should have a score strictly higher than 2jT j 2 = jT j, which is impossible.

(ii) Now suppose 9 2 [0; 1[. If Sc # B; a h ;R T ;v 3 $ > (1+4)jT j 2
, then clearly the score of a h must be strictly higher than the score of each of the two other alternatives in the Örst step. Furthermore, if

Sc # B; a h ;R T ;v 3 $ > (1+4)jT j 2
, then a h is a Condorcet winner, i.e. it beats all the other alternatives in the second step. Without loss of generality, assume h =3 : Sc # B; a 3 ;R T ;v 3 $ > (1+4)jT j

2

; then, we must have 9t 2 + 9t 4 + t 5 + t 6 > (1+4)jT j 2 , which can be rewritten

9t 2 + 9t 4 + t 5 + t 6 > (1+4) 2 (t 1 + t 2 + t 3 + t 4 + t 5 + t 6 ), or (1#4) 2 t 5 + (1#4) 2 t 6 > (1+4) 2 t 1 + (1#4) 2 t 2 + (1+4) 2 t 3 + (1#4) 2 t 4 , which gives t 5 + t 6 > (1+4) (1#4) t 1 + t 2 + (1+4)
(1#4) t 3 + t 4 , and Önally t 4 + t 5 + t 6 > (1+4) (1#4) t 1 + t 2 + (1+4) (1#4) t 3 +2t 4 . Besides, since 9 2 [0; 1[, then (1+4) (1#4) 2 [1; +1[, hence the last inequality above implies t 4 + t 5 + t 6 >t 1 + t 2 + t 3 +2 t 4 . And from the fact that t 4 ! 0, it follows that t 4 + t 5 + t 6 >t 1 + t 2 + t 3 , which means that a 3 beats a 1 in a pairwise contest. The same reasoning applies for a 3 and a 2 . Thus, for any alternative a h ,i fSc

# B; a h ;R T ;v 3 $ > (1+4)jT j 2
, then the alternatives di §erent from a h cannot be selected, even if we proceed with a second step between the two alternatives with the highest scores. And this completes the proof.

If a h wins with two steps, the inequalities susceptible to occur in each step are as follows:

First step 8 > > > > < > > > > : Sc # B; a h ;R T ;v 3 $ % (1+4)jT j 2 (C) Sc # B; a h ;R T ;v 3 $ ! Sc # A; a k ;R T ;v 3 $ if h<k (D) Sc # B; a h ;R T ;v 3 $ >Sc # A; a k ;R T ;v 3 $ if h>k (E) Sc # B; a j ;R T ;v 3 $ ! Sc # A; a k ;R T ;v 3 $ if j<k (F) Sc # B; a j ;R T ;v 3 $ >Sc # A; a k ;R T ;v 3 $ if j>k (") Second step , Sc # B; a h ;R T ;v 2 $ ! Sc # A; a j ;R T ;v 2 $ if h<j (H) Sc # B; a h ;R T ;v 2 $ >Sc # A; a j ;R T ;v 2 $ if h>j (I)
Below, we use sets of the Greek letters above to denote situations where any combination of inequalities above occurs; for example fC; D; Eg means that inequalities (C), (D) and (E) occur simultaneously.

DeÖnition 7 Let A = fa 1 ;a 2 ;a 3 g, B ' A, T a constituency, R T 2 L T and consider an SPR associated with V A = f(1;9;0) ; (1; 0)g. i) For all distinct h; k; j 2f1; 2; 3g, alternative a h is selected at the second step and is called the ìsecond step winnerî if any one of the following cases occurs: We then distinguish six potential possibilities, for the violation of YRA to occur: P 1 : An alternative is ìthe Örst step winnerî in both constituencies; P 2 : An alternative is ìthe second step winnerî in both constituencies, with the same ìchallengerî; P 3 : An alternative is ìthe Örst step winnerî in one constituency and ìthe second step winnerî in the other one. The ìchallengerî of the second constituency is selected in the union of the two constituencies; P 4 : An alternative is ìthe Örst step winnerî in one constituency and ìthe second step winnerî in the other one. The ìloserî of the second constituency is selected in the union of the two constituencies, with the ìchallengerî of the second constituency as the challenger of the whole electorate; P 5 : An alternative is ìthe Örst step winnerî in one constituency and ìthe second step winnerî in the other one. The ìloserî of the second constituency is selected in the union of the two constituencies, with the ìthe Örst step winnerî as the ìchallengerî of the whole electorate; P 6 : An alternative is ìthe second step winnerî in both constituencies with two di §erent ìchallengersî.

We then show that some of the possibilities above can actually lead to the violation of YRA, while others cannot. We begin with cases at which YRA is not violated.

Proposition 4 Let A = fa 1 ;a 2 ;a 3 g, B ' A, T j ' N , j =1; 2. R T1 2 L T1 and consider an SPR associated with V A = f(1;9;0) ; (1; 0)g, 0 % 9 % 1. Then, the SPR does not violate YRA if any one of the following possibilities occurs: P 1 , P 2 , P 3 , P 4 .

Proof Without loss of generality, suppose a 1 is the winner in both T 1 and T 2 . The proof for alternatives a 2 and a 3 is very similar and is then omitted. We successively consider possibilities P 1 , P 2 , P 3 and P 4 .

ñ P 1 : a 1 is the Örst step winner in both constituencies. From the proof of Proposition 3 (i), case 9 =1is not possible, then 0 % 9<1, and in particular, the two following inequalities are true:

t 1 1 + t 1 2 + 9t 1 3 + 9t 1 5 > (1+4)jT1j 2
(1) and t 2 1 + t 2 2 + 9t 2 3 + 9t 2 5 > (1+4)jT2j 2

(2). Since t 1 k + t 2 k = n k , if we add (1) and (2),w eh a v en 1 + n 2 + 9n 3 + 9n 5 > (1+4)n 2 . It follows that a 1 is the Örst step winner in . Hence, violation of YRA is not susceptible to occur. ñ P 2 : a 1 is the second step winner with a 2 as the challenger in both constituencies. Then, the following inequalities are true:

8 > > > > > > > < > > > > > > > : T 1
Step 1:

8 > > > > > < > > > > > : t 1 1 + t 1 2 + 9t 1 3 + 9t 1 5 % (1+4)jT1j 2 9t 1 1 + t 1 3 + t 1 4 + 9t 1 6 % (1+4)jT1j 2 9t 1 2 + 9t 1 4 + t 1 5 + t 1 6 % (1+4)jT1j 2 t 1 1 + t 1 2 + t 1 3 + t 1 5 ! 9t 1 2 + 9t 1 4 + t 1 5 + t 1 6 9t 1 1 + t 1 3 + t 1 4 + 9t 1 6 ! 9t 1 2 + 9t 1 4 + t 1 5 + t 1 6
Step 2:

& t 1 1 + t 1 2 + t 1 5 ! t 1 3 + t 1 4 + t 1 6 (3) 8 > > > > > > > < > > > > > > > : T 2
Step 1:

8 > > > > > < > > > > > : t 2 1 + t 2 2 + 9t 2 3 + 9t 2 5 % (1+4)jT1j 2 9t 2 1 + t 2 3 + t 2 4 + 9t 2 6 % (1+4)jT2j 2 9t 2 2 + 9t 2 4 + t 2 5 + t 2 6 % (1+4)jT3j 2 t 2 1 + t 2 2 + 9t 2 3 + 9t 2 5 ! 9t 2 2 + 9t 2 4 + t 2 5 + t 2 6 9t 2 1 + t 2 3 + t 2 4 + 9t 2 6 ! 9t 2 2 + 9t 2 4 + t 2 5 + t 2 6
Step 2:

& t 2 1 + t 2 2 + t 2 5 ! t 2 3 + t 2 4 + t 2 6 (4)
Then, when we add (3) and (4),

8 > > > > > > > < > > > > > > > : N Step 1: 8 > > > > > < > > > > > : 9n 1 + n 3 + n 4 + 9n 6 % (1+4)n 2 9n 1 + n 3 + n 4 + 9n 6 % (1+4)n 2 9n 2 + 9n 4 + n 5 + n 6 % (1+4)n 2 n 1 + n 2 + 9n 3 + 9n 5 ! 9n 2 + 9n 4 + n 5 + n 6 9n 1 + n 3 + n 4 + 9n 6 ! 9n 2 + 9n 4 + n 5 + n 6 Step 2: & n 1 + n 2 + n 5 ! n 3 + n 4 + n 6
Hence a 1 is the second step winner in N . And it follows that YRA cannot be violated. ñ P 3 : a 1 is the Örst step winner in T 1 and the second step winner with a 2 as the challenger in T 2 . Clearly, 0 % 9<1 and in particular,

t 1 1 + t 1 2 + 9t 1 3 + 9t 1 5 > (1+4)jT1j 2
(5) and t 2 1 + t 2 2 + t 2 5 ! jT2j 2 (6). We show that a 2 cannot be elected in N . Suppose on the contrary that a 2 is selected in N : in step 2, we then have n 3 + n 4 + n 6 > n 2 , which can be rewritten

t 2 3 + t 2 4 + t 2 6 | {z } (a) + t 1 3 + t 1 4 + t 1 6 | {z } (b)
> n 2 . Now, consider terms (a) and (b) separately. We begin with (a): from (6), t 2 3 + t 2 4 + t 2 6 < jT2j 2 . Now consider (b). We show that

t 1 3 + t 1 4 + t 1 6 < jT1j 2 : from (5), t 1 1 + t 1 2 + 9t 1 3 + 9t 1 5 > (1+4)jT1j 2 , which is equivalent to t 1 1 + t 1 2 + 9t 1 3 + 9t 1 5 > (1+4) 2 # t 1 1 + t 1 2 + t 1 3 + t 1 4 + t 1 5 + t 1 6 $
, and can be rewritten

(1#4) 2 t 1 1 + (1#4) 2 t 1 2 > (1#4) 2 t 1 3 + (1+4) 2 t 1 4 + (1#4) 2 t 1 5 + (1+4) 2 t 1 6 , or t 1 1 + t 1 2 >t 1 3 + (1+4) (1#4) t 1 4 + t 1 5 + (1+4) (1#4) t 1 6 , that is t 1 1 + t 1 2 + t 1 5 >t 1 3 + (1+4) (1#4) t 1 4 +2 t 1 5 + (1+4) (1#4) t 1 6 . Since 0 % 9<1, 1 % (1+4) (1#4) % +1; hence t 1 3 + (1+4) (1#4) t 1 4 +2t 1 5 + (1+4) (1#4) t 1 6 ! t 1 3 + t 1 4 + t 1 6
, from which it follows that t 1 1 + t 1 2 + t 1 5 >t 1 3 + t 1 4 + t 1 6 , and Önally t 1 3 + t 1 4 + t 1 6 < jT1j 2 . Consequently, from (a) and (b), we must have:

t 2 3 + t 2 4 + t 2 6 | {z } (a) +t 1 3 + t 1 4 + t 1 6 | {z } (b) < jT1j 2 + jT2j 2 , that is t 2 3 + t 2 4 + t 2 6 | {z } (a) +t 1 3 + t 1 4 + t 1 6 | {z } (b) < n 2
, which is equivalent to n 3 + n 4 + n 6 < n 2 , a contradiction. We then conclude that a 2 cannot be selected in N . In other words, violation of YRA is not susceptible to occur. ñ P 4 : a 1 is the Örst step winner in T 1 and the second step winner with a 2 as the challenger in T 2 . We will show that a 1 cannot be the loser in N , that is a 3 cannot be selected in N with a 2 as the challenger: in particular,

, t 1 1 + t 1 2 + 9t 1 3 + 9t 1 5 > 9t 1 1 + t 1 3 + t 1 4 + 9t 1 6 t 1 1 + t 1 2 + 9t 1 3 + 9t 1 5 > 9t 1 2 + 9t 1 4 + t 1 5 + t 1 6 (7) and t 2 1 +t 2 2 +9t 2 3 +9t 2 5 ! 9t 2 2 +9t 2 4 +t 2 5 +t 2 6 (8).
Then, in N , we have: n 1 +n 2 + 9n 3 + 9n 5 >9n 2 + 9n 4 + n 5 + n 6 , which means that Sc(A; a 1 ;R N ;v 3 ) > Sc(A; a 3 ;R N ;v 3 ). Then a 1 cannot be eliminated at the Örst step. Again, it follows that there is no violation of YRA.

Proposition 5 Let A = fa 1 ;a 2 ;a 3 g, B ' A, T j ' N , j =1; 2. R T1 2 L T1 and consider an SPR associated with V A = f(1;9;0) ; (1; 0)g, 0 % 9 % 1.

Then, the SPR is susceptible to violate YRA if any one of the following possibilities occurs: P 5 and P 6 .

Proof In order to prove our result we present an example for each possibility. Let T 1 and T 2 be two constituencies.

ñ P 5 : Suppose a 1 is the Örst step winner in T 1 and is the second step winner in T 2 with a 2 as the challenger; we show that a 3 can be selected in N . Assume the following situations t T1 = (0; 30; 0; 0; 20; 0) and t T2 = (20; 0; 0; 20; 20; 0) associated with R T1 and R T2 , respectively; then, s N = (20; 30; 0; 20; 40; 0) and the scores are as follows:

Separate groups

1 rst Step 8 > > > > < > > > > : T 1 T 2 Sc(A;a1;R T 1 ;v 3 )=30+204S c (A;a1;R T 2 ;v 3 )=20+204 Sc(A;a2;R T 1 ;v 3 )=0+04S c (A;a2;R T 2 ;v 3 )=20+204 Sc(A;a3;R T 1 ;v 3 )=20+304S c (A;a3;R T 2 ;v 3 )=20+204
Losers: a 2 ;a 3 Loser: a 3

2 nd Step 8 < : Sc(A#fa3g;a1;R T 2 ;v 2 )=40 Sc(A#fa3g;a2;R T 2 ;v 2 )=20
Winner: a 1 Winner: a 1

The whole electorate:

N = T 1 [ T 2 1 rst Step 8 > > < > > : Sc(A;a1;R N ;v 3 )=50+404 Sc(A;a2;R N ;v 3 )=20+204 Sc(A;a3;R N ;v 3 )=40+504 Loser: a 2 2 nd Step 8 < : Sc(A#fa2g;a1;R N ;v 2 )=50 Sc(A#fa2g;a3;R N ;v 2 )=60
Winner: a 3

It follows that YRA is violated. ñ P 6 : a 1 is the second step winner in constituencies with a 2 (and a 3 ) as the challenger in T 1 (and T 2 , respectively): We show that a 2 or a 3 can win in N . Assume the following situations t T1 = (18; 0; 2; 18; 17; 1) and t T2 = (2; 16; 15; 2; 4; 18) associated with R T1 and R T2 , respectively; then, (20; 16; 17; 20; 21; 19) and the scores are as follows:

s N =
Separate groups

1 rst Step 8 > > > > < > > > > : T 1 T 2 Sc(A;a1;R T 1 ;v 3 )=18+194S c (A;a1;R T 2 ;v 3 )=18+194 Sc(A;a2;R T 1 ;v 3 )=20+194S c (A;a2;R T 2 ;v 3 )=17+204 Sc(A;a3;R T 1 ;v 3 )=18+184S c (A;a3;R T 2 ;v 3 )=22+184
Loser: a 3 Loser: a 2 2 nd Step 8 < :

Sc(A#fa3g;a1;R T 1 ;v 2 )=35 Sc(A#fa2g;a1;R T 2 ;v 2 )=33 Sc(A#fa3g;a2;R T 1 ;v 2 )=21 Sc(A#fa2g;a3;R T 2 ;v 2 )=24
Winner: a 1 Winner: a 1

The whole electorate:

N = T 1 [ T 2 1 rst Step 8 > > < > > : Sc(A;a1;R N ;v 3 )=36+384 Sc(A;a2;R N ;v 3 )=37+394 Sc(A;a3;R N ;v 3 )=40+364 Loser: a 1 2 nd Step 8 < : Sc(A#fa1g;a2;R N ;v 2 )=57 Sc(A#fa1g;a3;R N ;v 2 )=56
Winner: a 2

It follows that YRA is violated. And this completes the proof.

Retrospectively, it is noticeable that cases of violations of YRA are closely linked to the fact that the winning alternative in all constituencies is selected with a di §erent challenger. Intuitively, this can be interpreted as a political situation where the two Houses are not ìmirror imagesî of each other.

The results provided in this section will subsequently be used to write systems of inequalities from which we will compute frequencies of violation of 5 The likelihood of non reinforceable proÖles Computer simulations are very often used in social choice in order to determine the frequencies of paradoxes, e. g. [START_REF] Nitzan | The vulnerability of point voting schemes to preference variation and strategic manipulation[END_REF] or [START_REF] Kelly | Almost all social choice rules are highly manipulable, but a few arenít[END_REF] among others. The Monte Carlo simulations method, used below, is very useful when a problem cannot be solved analytically. Its principle is simple: (i) Örstly, we choose a probabilistic hypothesis, that is the way probabilities are assigned to events; (ii) secondly, for any chosen values of t 1 and t 2 (the sizes of constituencies), we draw a given number of voting situations according to the selected hypothesis; voting situations are drawn independently, and (iii) thirdly, for each voting situation, we examine whether the situation induces the violation of the YRA according to the chosen scoring rule. Thus, for each value of (t 1 ;t 2 ), the estimated probability of violation of YRA is equal to the ratio number of occurrences of violations of YRA total number of drawn (possibly anonymous) proÖles where an anonymous proÖle is distinguished from a proÖle in the usual sense by the fact that in the latter the names of the voters matter and in the Örst one they do not. To illustrate, with three voters and three alternatives, conÖgurations of preferences (xyz; xzy; yzx) and (xyz; yzx; xzy) are distinct proÖles in the usual sense in which individuals 2 and 3 have swapped their preferences, but correspond to the same anonymous proÖle, where one is interested only in the number of individuals with some preference order, and not in the names of these individuals.

In our study, the total number of drawn situations is 100,000 (a greater number does not signiÖcantly improve the results). To be precise, note that the values we obtain are not exact probabilities, but rather estimates. For example, a value equal to 0 doesnít necessarily mean that the phenomenon cannot at all occur, but in fact, simply that its probability of occurrence is very small.

The assumption of independence between the two constituencies, while to some extent arbitrary, is however somewhat in accordance with the actual situations of western democracies.

Two traditional probabilistic hypotheses are taken into account: Impartial Culture (IC) and Impartial Anonymous Culture (IAC). Both are based on an equal probability assumption, but not exactly in the same way. According to IC, all preference proÖles are equally likely: for each voter, every preference order has an equal chance to be drawn ( 1 6 in the three-alternative case); and under IAC all anonymous preference proÖles are equiprobable: for every possible preference order, the number of voters reporting that order is drawn, in such a way that the total number of voters be equal to n. More precisely, for each n and each linear order we draw randomly the number n 1 of individuals with the Örst linear order. And then, again randomly, we draw the number n 2 of individuals with the second linear order between 0 and n + n 1 , and so on. For a detailed discussion of these hypotheses and some others, see Regenwetter et al. (2006).

Then in this section, for each probabilistic hypothesis and each rule under study, two series of results are provided: (a), Örst, for iterative Borda, Coombs and Hare rules, we present tables of values and graphs corresponding to some values of the total number of voters t 1 + t 2 , that is 25, 50, 75 and 100, and (b) for some speciÖc values of 9, we then focus on three special cases of bicameral congresses: the French Parliament, the US Congress and the German Parliament.

In the tables, the notation xE-y is used as an equivalent for x , 10 #y , where x and y are real numbers.

After the presentation of our results with Monte Carlo simulations, we consider cases where the number of individuals is inÖnitely large, under the IAC hypothesis. Within this framework, we use an analytical method -as distinguished from computer simulations -and provide relations giving the proportion of anonymous proÖles vulnerable to the violation of YRA.

Impartial Culture

We shall Örst consider the case where n 2f25; 50; 75; 100g and t (the size of any of the two constituencies) changes from 1 to at most n 2 (Tables 1 and2). Note that cases where t ! n 2 are symmetrical to those where t % n 2 . For example, for n = 25, the results are the same with t t = 24. For all rules under study, it appears that: (i) for any given value of n, and for all values of t, frequencies of violation of YRA are relatively small (generally smaller than 10 #2 ), (ii) for any Öxed n, those frequencies slightly rise when the value of t rises (áuctuations are due to even or odd values of t) as illustrated in Fig. 2, 3, 4 and 5 below, (iii) for any given value of t, frequencies decrease when n rises, (iv) it follows that frequencies are higher as both constituencies tend to have the same size, (v) Önally, although all frequencies are rather small, they are clearly smaller for IBP than for the two other rules, for which they are relatively close. Now, one natural question is about the limit frequencies as n gets larger. The table and the Ögure below provide an answer for houses with equal size. Simulations show that (i) values of frequencies are still smaller for IBP than for the two other rules, and (ii) they are relatively stable for HP and IBP while they remain erratic for CP.

Table 3 and Fig. 6 Frequencies of violation of YRA for the three rules, under IC, with t = n 2 and n = f100; 250; 500; 750; 1000g. We now focus on three special cases of bicameral congresses. The French Parliament is composed of two constituencies: the ìUpper Houseî, which is the French Senate (SÈnat) with 331 members, and the ìLower Houseî, the French National Assembly (AssemblÈe Nationale), which has 577 members. For the German Parliament, the two constituencies are the ìFederal Councilî, named the Bundesrat, the size of which is 68 members, and the ìFederal Dietî or Bundestag with 598 members. And Önally, the US Congress is divided into the Senate on the one hand, the number of members of which is 100, and the House of Representatives on the other hand, with 435 members.

For Table 4 and Fig. 7 below, the size n of the Congress is Öxed (n = 908 for the French Parliament, and so on for the other countries); only 9 changes from 0 to 1, and for 9 equal to 0, 1,o r1=2, we respectively obtain Hareís Procedure, Coombs Procedure and Iterative Borda Procedure. As already noted above, frequencies are relatively greater with rather equal size constituencies, although they are small for all values of 9. This is shown in Figure 7 with the German curve which is lower than the US and French curves. Besides, it must also be noticed that the three curves U which means that for all of the three Congresses appears as the best procedure among the three classical ones (HP, IBP and CP ) and moreover, IBP is one of the best rules, among all iterative positional rules.

Impartial Anonymous Culture

As for IC above, we consider n = f25; 50; 75; 100g and let t change from 1 to n 2 . Again, the Monte Carlo technique leads to the results provided in Tables 5 and6 Graphically these tables are illustrated as follows: As a di §erence with the IC hypothesis, observation of Fig. 8, 9 and 10 does not lead to a clear conclusion. However it seems that frequencies are higher as both constituencies tend to have the same size. Moreover, like Fig. 5 above, Fig. 11 shows that CP is more sensitive to violations of YRA than the two other rules. However, while under IC the performance of IBP and CP were very close, under IAC, there seems to be no clear di §erence between IBP and HP. This shows that, to some extent, the choice of the probabilistic model can have an ináuence on the results: IAC is known to assume some homogeneity within voters preferences, while IC assumes more clearly divided views among voters. Besides, it also appears that occurrences of paradoxes are higher under IC than under IAC (see Gehrlein 2006, chap. 5).

We now turn, in Table 7 and in Fig. 12, to the results obtained under IAC for the three special cases of bicameral congresses. Once again, Coombs procedure is more sensitive to the violation of YRA than Borda and Hare procedures; but, it is not clear from our results in Table 7 that one of these last two rules -Borda or Hare -would minimize the probability of violation of YRA. More importantly, the paradox is clearly unlikely to appear in the Parliaments under study. Now, suppose the size of any one of the two constituencies is equal to Cn, where C is a proportion of the total number of voters, provided that Cn is an integer. Thus, a value of C equal to 0:1 means that the number of individuals in the given constituency represents 10% of the total number of individuals in N , which implies that the 90% other individuals are members of the other constituency. Here, we focus on large electorates -and indeed on inÖnite electorates -and then we assume that C can take any value in the interval ]0; 1 2 ]. Further, as a di §erence with computer simulations from which the results provided above have been obtained, we here provide exact values of frequencies of violation of YRA. These results are obtained by use of the technique of [START_REF] Fishburn | Condorcetís paradox and anonymous preference proÖles[END_REF]. Let F (9; C; n) be the likelihood that an SPR F with the vector score V A = f(1;9;0) ; (1; 0)g violates YRA,f o rn voters. We provide closed-form formulae giving this likelihood when n = 1 and 9 2f 0; 1g. We have not been able to obtain results for 9 = 1 2 . We then can state the following results:

Proposition 6 Let HP be the voting rule. Then F (0;C;1)= 8 > > > > > > > > > > > > > > > > < > > > > > > > > > > > > > > > > :

$ 6 ( !189 517 056$+138 088 404$ 2 +12 For some values of C, we have the following results in Table 8:

Table 8 (IAC 4) Violations of YRA for Hare and Coombs with large electorates and 0 % 9 % 1.

Graphically, in Fig. 13 we have: As seen above, it appears that: (i) in general, cases of violations are rather rare, (ii) frequencies get smaller as the electorate gets larger, (iii) greatest frequencies are reached for electorates with constituencies with equal size, and Önally (iv) Coombs procedure is more sensitive to the violation of YRA than Hareís procedure.

Concluding discussion

This paper was devoted to the study of the sensitivity of sequential positional rules to Youngís reinforcement axiom. We obtain two types of results. First, we give precision about the conÖgurations of preferences and the number of voters at which violations of the axiom occur, and second, we provide frequencies of these occurrences. Most of these latter results are based on computer simulations -under two probabilistic classical hypotheses -but some other are obtained with the analytical Fishburn-Gehrlein technique. Illustrations are given for the special cases of Hareís Procedure, Coombs Procedure and Iterative Borda Procedure, and also for three actual Parliaments (France, Germany and the United States).

Our results show that: (i) in general, cases of violations of YRA are rather rare, (ii) although all frequencies are small, they are smaller for HP and IBP than CP, and (iii) frequencies decrease for electorates with constituencies with clearly uneven sizes.

Besides, although all these rules violate YRA, the frequencies of violation are very rare. Then, since the violation of this axiom is not so frequent, one should not be unduly worried about its theoretical possibility.

However, we can make some advice in order to avoid this paradox with three alternatives. First, for small constituencies, indi §erently take HP, or CP, or IBP, among all sequential positional rules. And second for large electorates, use IBP rather than HP or CP.

It must also be noted that in the special case where the winner in the Örst constituency is the most preferred alternative of all voters in the second constituency, the violation of Youngís reinforcement axiom coincides with the no-show paradox (see for example [START_REF] Lepelley | Scoring run-o § paradoxes for variable electorates[END_REF]. However, our results noticeably di §er from those of Lepelley and Merlin, mainly because, as underlined above, our simulations are based on an assumption of independence between the two constituencies, which is not the case in their study.

Finally, it is worth noting that it may be interesting to consider some di §erent contexts: more than three alternatives, more general procedures containing all SPRs (for example by possibly eliminating more than one alternative at each step), or Condorcet consistent procedures.

Fig. 2

 2 Fig. 2 Frequencies of violation Fig. 3 Frequencies of violation of of YRA under IC for Hare YRA under IC for Iterative Borda (HP ) and n 2f25; 50; 75; 100g .( IBP ) and n 2f25; 50; 75; 100g .

Fig. 4

 4 Fig. 4 Frequencies of violation Fig. 5 A comparison of the three of YRA under IC for Coombs rules, under IC, for n = 100. (CP ) and n 2f25; 50; 75; 100g .

Fig. 7

 7 Fig. 7 Comparison of the three Congresses under IC and under di §erent voting rules (0 % 9 % 1).

  below: Table 5 (IAC 1) Frequencies of violation of for Hare, Iterative Borda and Coombs, with n = 25 and n = 50.

Fig. 8

 8 Fig. 8 Frequencies of violation Fig. 9 Frequencies of violation of of YRA under IAC for Hare (HP ) YRA under IAC for Iterative Borda and n 2f25; 50; 75; 100g .( IBP ) and n 2f25; 50; 75; 100g .

Fig. 10

 10 Fig. 10 Frequencies of violation Fig. 11 A comparison of the three of YRA under IAC for Coombs rules, under IAC, for n = 100. (CP ) and n 2f25; 50; 75; 100g .

Fig. 12

 12 Fig.12 Comparison of the three Congresses under IAC and under di §erent voting rules (0 % 9 % 1).

Fig. 13

 13 Fig. 13 Frequencies of violation of YRA under IAC for Hare and Coombs with large electorates and 0 % 9 % 1.

  Alternative a h is called the , ìchallengerî if it is beaten at the second step, ìthe loserî if it is beaten at the Örst step.

	Case 1	Case 2	Case 3	Case 4
	fC; D; F; Hgf C; D; F; Igf C; D; "; Hgf C; D; "; Ig
	Case 5	Case 6	Case 7	Case 8
	fC; E; F; Hgf C; E; F; Igf C; E; "; Hgf C; E; "; Ig
	ii)

Table 1 (

 1 IC 1) Frequencies of violation of YRA for Hare, Iterative Borda and Coombs, with n = 25 and n = 50.

Table 2 (

 2 IC 1) Frequencies of violation of YRA for Hare, Iterative Borda and Coombs, with n =75and n = 100.

Table 4 (

 4 IC 3) Theoretical frequencies of violation of YRA in some Western democracies under di §erent voting rules.

Table 6 (

 6 IAC 2) Frequencies of violation of YRA for Hare, Iterative Borda and Coombs, with n =75and n = 100.

Table 7 (

 7 IAC 3) Theoretical frequencies of violation of YRA in some Western democracies under di §erent voting rules.

  570 496$ 3 !34 659 659$ 4 +140 554 848 )

		277 389 791 723 520
			if 82[0; 1 4 ]
		( 340 850$+2555 640$ 2 213 024 000$ 7 !2232 960$ 3 !1905 300$ 8 !83 435 520$ 4 +587 252 736$ 5 +235 324 320$ 6 + !19 191 680$ 9 !325 477$ 10 +15 503 )
		4160 846 875 852 800
			if 82[ 1 4 ; 1 3 ]
		( !1139 510$+16 691 760$ 2 2009 688 240$ 7 !6224 488 920$ 8 +6610 436 380$ 9 !117 557 520$ 3 +430 742 760$ 4 !2546 990 749$ 10 +24 715 !658 409 976$ 5 +1379 212 800$ 6 + )
		4160 846 875 852 800
			if 82[ 1 3 ; 1 2 ]
	Proposition 7 Let CP be the voting rule. Then F (1;C;1)= 8 $ 6 ( !189 517 056$+138 088 404$ 2 +12 570 496$ 3 !34 659 659$ 4 +140 554 848 ) 277 389 791 723 520 > > > > > > > > > <	if 82[0; 1 3 ]
	> > > > > > > > > :	#694 840!23 006 250$+314 101 800$ 2 158 559 912 000$ 6 !114 436 013 400$ 7 +13 807 845 900$ 8 +22 578 700 950$ 9 !2374 579 800$ 3 +11 071 468 800$ 4 !33 160 212 540$ 5 + !8375 899 073$ 10 81 266 540 544 000 000 if 82[ 1 3 ; 1 $ 2 ]

Acknowledgements. We are grateful to two anonymous referees for very valuable comments and suggestions.


Sébastien Courtin
email: sebastien.courtin@u-cergy.fr

Boniface Mbih
email: boni-face.mbih@unicaen.fr

Issofa Moyouwou
email: imoyouwou2@yahoo.fr

Are Condorcet procedures so bad according to the reinforcement axiom ?

Keywords: Condorcet procedures, Reinforcement axiom, Likelihood, Impartial culture, Impartial anonymous culture JEL Classification D71

A Condorcet social choice procedure elects the candidate that beats every other candidate under simple majority when such a candidate exists. The reinforcement axiom roughly states that given two groups of individuals, if these two groups select the same alternative, then this alternative must also be selected by their union. Condorcet social choice procedures are known to violate this axiom. Our goal in this paper is to put this important voting theory result into perspective. We then proceed by evaluating how frequently this phenomenon is susceptible to occur.

Introduction

Most of the literature on voting theory is concerned with the theoretical debate between Condorcet consistent social choice methods on the one hand, and positional (or scoring) systems -voting à la Borda -on the other hand. There are several arguments in favor of either type of procedures, discussed in an abundant literature (see [START_REF] Nurmi | Comparing voting systems[END_REF][START_REF] Nurmi | Voting paradoxes and how to deal with them[END_REF], and Saari 2006 among others).

One of the strongest arguments in favor of Borda's approach is the reinforcement axiom (RA). The notion of reinforcement was first introduced by [START_REF] Smith | Aggregation of preferences with variable electorate[END_REF], under the name of separability. The intuition is as follows: if two different blocks of individuals rank an alternative at least as high as some other alternative, then so does the combination of the blocks with regard to these two alternatives.

In the same way -and indeed at the same period - [START_REF] Young | An axiomatization of Borda's rule[END_REF]Young ( , 1975a) ) proposes a variant of the notion of separability, which he calls consistency. Following this idea, if two disjoint groups of individuals separately select two non-disjoint subsets of alternatives, then the union of these groups should exactly select the intersection of the two subsets. In fact, the difference between the two contributions lies in the type of social choice mechanisms studied: Young is concerned with social choice correspondences (selecting possibly more than a single alternative), while Smith is interested in aggregation functions (selecting a social ranking of alternatives from preferences reported by individual voters).

And finally, in the special context of social choice functions (correspondences selecting a unique outcome), which we consider in this paper, [START_REF] Moulin | Axioms of cooperative decision making[END_REF] introduces the phrase reinforcement axiom (RA).

In order to illustrate this phenomenon, consider the following situation, concerning two french districts. For several years now, there has been a project of reunification of two northwestern districts, the "Basse-Normandie" (almost 1 500 000 inhabitants with Caen as the capital) and the "Haute-Normandie" (almost 1 800 000 inhabitants and Rouen as the capital). Almost all the inhabitants of these two districts agree with this project, but they disagree on the new capital. They will have to choose between three cities, Caen, Rouen and Le Havre (which is another big city in Haute-Normandie). Given the importance of this decision, we assume that the chosen project must be firstly accepted by each of the two districts separately; and secondly, if both districts choose the same city, then there is a further step, in which all inhabitants of both districts will be asked to cast their votes. 1Now, after the first step of the process, assume that the inhabitants in Basse-Normandie and in Haute-Normandie have chosen Caen. Further, assume that with the same individual preferences Le Havre is the winner of the second step. Then Caen wins in each of the two districts separetely, but not in their union. This is typically an example of violation of the reinforcement axiom.

The main results concerning the violation of RA are summarized in Young's theorem (1975a). According to this theorem, all simple positional voting procedures satisfy RA, while on the contrary there is no Condorcet social choice procedure satisfying that axiom.

There is no doubt about the importance of this proposition, since it provides a clear axiomatic boundary between Borda's and Condorcet's approaches of social choice mechanisms. In this paper, we examine the violation of RA under the most widely studied Condorcet procedures (see [START_REF] Fishburn | Condorcet social choice functions[END_REF] or Rasch 1995, among others). Our goal is to examine precise conditions at which violations of RA are susceptible to arise under these rules. We give, among other statements, conditions on the minimum number of voters at which the phenomenon arises.

Besides, this normative contribution is completed by a measure of the quantitative significance of the violation of RA. Indeed, if the likelihood of such violation of RA is negligible, then one need not be unduly worried about the existence of the possibility as such. In fact, although these rules violate RA, this should not be a serious ground for rejecting them or to discriminate in favor of Borda's approach (see [START_REF] Courtin | The reinforcement axiom under sequential positional rules[END_REF] for a study on the violation of RA under sequential positional procedures).

The remainder of this work is organized as follows: Section 2 is a presentation of Condorcet social choice functions with definitions and some preliminary results about these rules. Section 3 provides some general results on the behavior of Condorcet's rules vis-à-vis the reinforcement axiom. Then, Section 4 studies the conditions at which a profile may violate the axiom, and provides the corresponding frequencies. Finally, Section 5 concludes the paper.

Condorcet social choice functions

In this section, we begin with formal definitions of Condorcet rules and we investigate some relationships between them. i2N of individual preference relations, one for each individual. The set of all profiles on N will be denoted by L N . Similarly profiles on T 1 and T 2 will be denoted R T 1 and R T 2 , respectively. Definition 1. Let N be the nonempty set of voters. Then i) A social choice function (SCF ) is a mapping that associates a single alternative with any profile over a nonempty subset of N .

Notations and definitions

ii) A social choice correspondence (SCC ) is a mapping that associates a nonempty subset of alternatives with any profile over a nonempty subset of N .

Note that F R N and C R N , given an SCF F and an SCC C, are the images determined respectively by F and C for a set of alternatives A and a profile R N on that set.

Given a profile R N and alternatives x, y 2 A, we denote by n(x, y, R N ) the number of individuals in R N who prefer x to y (or simply n(x, y) when there is no ambiguity). Hence n(x, y)+n(y, x) = n. Then the simple majority relation M on A determined by profile R N is defined by

xM y iff n(x, y) > n(y, x).
There is a tie between x and y under simple majority if n(x, y) = n(y, x). Definition 2. An SCF F is a Condorcet SCF if and only if it satisfies the Condorcet Principle, that is for all profile R N and all x 2 A, xM y for all y 2 A implies F R N = x.

We now define the rules under study in this paper. In order to do this, we follow Fishburn's 1977 terminology 2 . For each SCC (C j ), we obtain an SCF (F j ) by breaking ties in favor of the alternative coming first in the lexicographic order. For example, if two alternatives x and y are tied, then F R N = x.

C 1 . Black's function: C 1 R N = {x if yM x f or no y 2 A}. Otherwise C 1 R N = x if b(x, R N ) b(y, R N ) for all y 2 A , where b(x, R N ) = P y2A n(x, y) is the Borda score of x given R N .
According to this function, if no candidate beats or ties every other candidate on the basis of simple majority, then the SCF should be determined by the Borda scores.

C 2 . Copeland's function: C 2 R N = x 2 A : c(x, R N ) c(y, R N )f or all y 2 A , where c(x, R N ) = |y 2 A : xM y| |y 2 A : yM x| is the Copeland score of x,
that is the number of alternatives beaten by x minus the number of alternatives that beat x.

C 3 . Dodgson's function: C 3 R N = x 2 A : d(x, R N )  d(y, R N )f or all y 2 A , where d(x, R N )
is the number of binary preference reversals needed in R N to make x tie or beat every other candidate in A on the basis of simple majority. A reversal occurs when xR i y is changed into yR i x. In other words d(x, R N ) is a measure of how little R N needs to be changed to make x tie or beat every other candidate.

C 4 . Young's function: C 4 R N = x 2 A : e(x, R N ) e(y, R N )f or all y 2 A , where e(x, R N )
is the number of voters in the largest sublist from R N for which x ties or beats every other candidate in A on the basis of simple majority with respect to this sublist. If there is no such sublist then e(x, R N ) = 0. This means that E(x, R N ) = n e(x, R N ) gives the fewest number of voters that must be deleted from R N for x to tie or beat every other candidate.

In contrast to Dodgson's function, Young's function deletes voters rather than reversing preferences to obtain a majority winner.

C 5 . Condorcet's function : C 5 R N = x 2 A : f (x, R N ) f (y, R N )f or all y 2 A , where f (x, R N ) = min y2A\x n(x, y) . That is f (x, R N )
measures the worst that x does against any other candidate.

Condorcet's function selects the alternative with the largest minimum support in all pairwise comparisons. It sometimes goes under the names of minimax rule (Young 1975b) or maximin rule (see for example [START_REF] Nurmi | Voting Procedures Under Uncertainty[END_REF]. Schwartz's function elects the alternative that is beaten by no other alternative according to the majority election, directly or indirectly. C 8 . Fishburn's function: C 8 R N = {x 2 A : yF x f or no y 2 A}, where xF y iff for every a 2 A, aM x ) aM y; and there is a w 2 A such that wM y and not (wM x).

C 6 . Kemeny's function: Given R N with A = {a 1 , a 2 , ..., a m }, let L ⇤ be a subset of linear or- ders R k on A that maximize S(R k ) = P i6 =j n(a i , a j )K(a i , a j ), where K(a i , a j ) = 1 if a i R k a j and K(a i , a j ) = 0 otherwise. Then C 6 R N = {x 2 A : there is an R k 2 L ⇤ such that xR k y f or all y 2 A \ {x}}. That is C 6 R N = x 2 A : k(x, R N ) k(y, R N ) f or all y 2 A where k(x, R N ) = max {S(R k ) :
Fishburn's function is based on the notion that if every candidate that beats x also beats y under simple majority, and if there exists some candidate that beats y but not x, then x is "better than" y under simple majority comparisons.

We now introduce another Condorcet SCF, not defined by [START_REF] Fishburn | Condorcet social choice functions[END_REF], the Amendment's function (see [START_REF] Farquharson | Theory of voting[END_REF]) denoted C 9 .

C 9 . Amendment function: With A = {a 1 , a 2 , ..., a m }, the outcome of C 9 is the alternative

C 9 R N defined as follows: i) Let C 9 R N 1 = a 1 .
ii) For any 2  j  m, the winner C 9 R N j of the majority pairwise contest between

C 9 R N j 1 and a j is defined by C 9 R N j = ( a j if a j MC 9 R N j 1 C 9 R N j 1 otherwise . iii) C 9 R N = C 9 R N m
, the winner of the last majority contest against a m . The intuition of this procedure is as follows: any motion C 9 R N j 1 is saved as long as a strict majority does not reject it in favor of some other motion a j given the agenda A = {a 1 , a 2 , ..., a m }.

The special case of three alternatives

Some additional notations will be necessary to present some results in the three-alternative case.

With A = {a 1 , a 2 , a 3 }, the six possible linear orders on A are: R 1 : a 1 a 2 a 3 , R 2 : a 1 a 3 a 2 , R 3 : a 2 a 1 a 3 , R 4 : a 2 a 3 a 1 , R 5 : a 3 a 1 a 2 , R 6 : a 3 a 2 a 1 . And in the sequel, n k will be the total number of individuals in N with preference relation R k . Further, in the three-alternative case, a Condorcet cycle arises when every alternative is beaten (on the basis of simple majority) by only one other alternative. Two possible cycles can arise:

1. Condorcet cycle 1 : a 2 Ma 1 , a 1 Ma 3 and a 3 Ma 2 2. Condorcet cycle 2 : a 1 Ma 2 , a 2 Ma 3 and a 3 Ma 1 .

We shall distinguish two cases : (i) when there is a Condorcet cycle, and (ii) when there is none.

Condorcet Cycle

We first show how to compute the score of each alternative a j , a k , a l 2 {a 1 , a 2 , a 3 }, when there is a Condorcet cycle, according to Dodgson's, Young's, Condorcet's and Kemeny's procedures.

We begin with Dodgson.

, R N ) = n(a j , a k , R N ) + d(a j , a k ) n 2 . Hence d(a j , a k ) n 2 n(a j , a k , R N ) (1)
Since d(a j , a k ) is the minimum number of reversals needed to have a

j Ma k in R N , then n(a j , a k , R N )+ d(a j , a k ) 1 < n 2
, which can be rewritten as 2n(a j , a k , R N ) + 2d(a j , a k ) 2 < n, and since every term of this inequality is an integer, this can also be rewritten as 2n(a j , a k , R N )+2d(a j , a k ) 1  n, which gives

d(a j , a k )  n + 1 2 n(a j , a k , R N ) (2)
It then follows from ( 1) and ( 2)

that n 2 n(a j , a k , R N )  d(a j , a k )  n+1 2 n(a j , a k , R N ), and finally, d(a j , a k ) = ⌅ n+1 2
⇧ n(a j , a k ), which completes the proof. Proposition 2. If there exists a Condorcet cycle at some profile R N , then the Dodgson score of an alternative a j is given by d(a

j , R N ) = d(a j , a k ) = ⌅ n+1 2 ⇧ n(a j , a k )
, where a k is the unique alternative such that a k Ma j .

Proof. Consider some profile R N and some alternative a j and suppose there is a Condorcet cycle; then a j is beaten by only one other alternative, say a k . Since a j is beaten only by a k , the number of binary preference reversals needed in R N to make a j tie or beat every other candidate in A on the basis of simple majority, d(a j , R N ), is equal to the number of binary preference reversals needed in R N to make a j tie or beat a k , that is

d(a j , a k ) = ⌅ n+1 2
⇧ n(a j , a k ). As a consequence, we have the following Dodgson scores for the two Condorcet cycles, with

A = {a 1 , a 2 , a 3 }: Cycle 1 : a 2 Ma 1 , a 1 Ma 3 and a 3 Ma 2 d(a 1 , R N ) = ⌅ n+1 2 ⇧ n(a 1 , a 2 ) = ⌅ n+1 2 ⇧ (n 1 + n 2 + n 5 ) d(a 2 , R N ) = ⌅ n+1 2 ⇧ n(a 2 , a 3 ) = ⌅ n+1 2 ⇧ (n 1 + n 3 + n 4 ) d(a 3 , R N ) = ⌅ n+1 2 ⇧ n(a 3 , a 1 ) = ⌅ n+1 2 ⇧ (n 4 + n 5 + n 6 )
and Cycle 2 :

a 1 Ma 2 , a 2 Ma 3 and a 3 Ma 1 d(a 1 , R N ) = ⌅ n+1 2 ⇧ n(a 1 , a 3 ) = ⌅ n+1 2 ⇧ (n 1 + n 2 + n 3 ) d(a 2 , R N ) = ⌅ n+1 2 ⇧ n(a 2 , a 1 ) = ⌅ n+1 2 ⇧ (n 3 + n 4 + n 6 ) d(a 3 , R N ) = ⌅ n+1 2 ⇧ n(a 3 , a 2 ) = ⌅ n+1 2 ⇧ (n 2 + n 5 + n 6 )
Young's procedure Proposition 3. If for each alternative a j 2 A there exists only one a

k 2 A such that a k Ma j in some profile R N , then E(a j , R N ) = n 2n(a j , a k ).
Proof. Let A = {a j , a k , a l } and consider a profile R N such that a k Ma j in R N and not (a k Ma j ) for some profile R N 0 obtained from R N by removing some voters that strictly prefer a k to a j . Suppose that R N 0 is obtained using a minimal number of cancellations E(a

j , R N ). Let us prove that E(a j , R N ) = n 2n(a j , a k , R N ). A) We first prove that E(a j , R N ) n 2(a j , a k , R N ). By hypothesis on R N and R N 0 , it follows that n(a j , a k , R N ) < n 2 , n(a j , a k , R N 0 ) n 0 2 and n(a j , a k , R N 0 ) = n(a j , a k , R N ) with n 0 = |N 0 |. Therefore n 0 2  n(a j , a k , R N ) < n 2 . Since E(a j , R N ) = n n 0 , n E(a j ,R N ) 2  n(a j , a k , R N ) < n 2 Therefore E(a j , R N ) n 2n(a j , a k , R N ) > 0 (1) B) We now prove that E(a j , R N )  n 2n(a j , a k , R N ). Note that n(a k , a j , R N ) + n(a j , a k , R N ) = n.
Then relation (1) can be rewritten as follows:

E(a j , R N ) n(a k , a j , R N ) n(a j , a k , R N ) > 0 Consider q = n(a k , a j , R N ) n(a j , a k , R N ). Two cases arise. a) q  n(a k a l a j , R N ) + n(a l a k a j , R N ) where n(a u a v a w , R N )
is the total number of voters with preference relation of type a u a v a w . In that case by deleting exactly q voters in R N with preference relation a k a l a j or a l a k a j , we have in

R N 0 : n(a j , a k , R N 0 ) n(a k , a j , R N 0 ) = n(a j , a k , R N ) (n(a k , a j , R N ) q) = 0 Then n(a j , a k , R N 0 ) = n(a k , a j , R N 0 ) (2)
. Moreover deleted voters all prefer a l to a j . Then,

n(a j , a l , R N 0 ) = n(a j , a l , R N ) and n(a l , a j , R N 0 ) < n(a l , a j , R N ). Since not (a l Ma j ) holds, then n(a j , a l , R N ) n(a l , a j , R N ) Therefore n(a j , a l , R N 0 ) n(a l , a j , R N ) > n(a l , a j , R N 0 ) (3) Hence from (2) and (3) a j ties a k and defeats a l in R N 0 . Thus E(a j , R N )  q = n 2n(a j , a k , R N ) b) q > n(a k a l a j , R N ) + n(a l a k a j , R N ) . Then let q = n(a k a l a j , R N ) + n(a l a k a j , R N ) + r.
In that case we delete all the voters in R N with preference relation a k a l a j or a l a k a j and exactly r voters with preference relation a 5) and ( 6) a j ties a k and defeats a l in R N 0 . Therefore 1) and ( 7) it follows that

k a j a l . At R N 0 , n(a j , a k , R N 0 ) n(a k , a j , R N 0 ) = 0 and n(a j , a k , R N 0 ) = n(a k , a j , R N 0 ) (5). Moreover, we have n(a j , a l , R N 0 ) = n(a j a k a l , R N 0 ) + n(a j a l a k , R N 0 ) + n(a k a j a l , R N 0 ) = n(a j a k a l , R N ) + n(a j a l a k , R N ) + n(a k a j a l , R N ) r = n(a j a k a l , R N ) + n(a j a l a k , R N ) + n(a k a j a l , R N ) q + n(a k a l a j , R N ) + n(a l a k a j , R N ) = 2n(a j a k a l , R N ) + 2n(a j a l a k , R N ) + n(a l a j a k , R N ). and n(a l , a j , R N 0 ) = n(a l a k a j , R N 0 ) + n(a l a j a k , R N 0 ) + n(a k a l a j , R N 0 )= n(a l a j a k , R N ) Then n(a j , a l , R N 0 ) n(a l , a j , R N 0 ) = 2n(a j a k a l , R N ) + 2n(a j a l a k , R N ) 0 (6) Hence from (
E(a j , R N )  q = n 2n(a j , a k , R N ) (7) From (
E(a j , R N ) = n 2n(a j , a k , R N ).
Proposition 4. If there exists a Condorcet cycle at some profile R N , then the Young's score of an alternative a j is given by e(a j , R N ) = 2n(a j , a k ) where a k is the unique alternative such that a k Ma j .

Proof. Consider some profile R N and some alternative a j and suppose there is a Condorcet cycle; then a j is beaten by only one other alternative, say a k . And by definitition E(a j , R N ) is the fewest number of voters that must be deleted from R N in order for a j to tie or beat every other candidate. By Proposition 3, E(a

j , R N ) = n 2n(a j , a k ). And since E(a j , R N ) =n e(a j , R N ), then e(a j , R N ) = n E(a j , R N ) = n n + 2n(a j , a k ) = 2n(a j , a k ).
As a consequence, we have the following Young's scores for the two Condorcet cycles with

A = {a 1 , a 2 , a 3 }: Cycle 1 : a 2 Ma 1 , a 1 Ma 3 and a 3 Ma 2 e(a 1 , R N ) = 2.n(a 1 , a 2 ) = 2.(n 1 + n 2 + n 5 ) e(a 2 , R N ) = 2.n(a 2 , a 3 ) = 2.(n 1 + n 3 + n 4 ) e(a 3 , R N ) = 2.n(a 3 , a 1 ) = 2.(n 4 + n 5 + n 6 ) and Cycle 2 : a 1 Ma 2 , a 2 Ma 3 and a 3 Ma 1 e(a 1 , R N ) = 2.n(a 1 , a 3 ) = 2.(n 1 + n 2 + n 3 ) e(a 2 , R N ) = 2.n(a 2 , a 1 ) = 2.(n 3 + n 4 + n 6 ) e(a 3 , R N ) = 2.n(a 3 , a 2 ) = 2.(n 2 + n 5 + n 6 )
Condorcet's procedure Proposition 5. If for each alternative a j 2 A there exists only one a k 2 A such that a k Ma j at some profile R N , then the Condorcet score of a j is given by f (a

j , R N ) = n(a j , a k ). Proof. By definition, f (a j , R N ) = min x2A\a j n(a j , x
A = {a 1 , a 2 , a 3 }: Cycle 1 : a 2 Ma 1 , a 1 Ma 3 and a 3 Ma 2 f (a 1 , R N ) = n(a 1 , a 2 ) = n 1 + n 2 + n 5 f (a 2 , R N ) = n(a 2 , a 3 ) = n 1 + n 3 + n 4 f (a 3 , R N ) = n(a 3 , a 1 ) = n 4 + n 5 + n 6
and Cycle 2 :

a 1 Ma 2 , a 2 Ma 3 and a 3 Ma 1 f (a 1 , R N ) = n(a 1 , a 3 ) = n 1 + n 2 + n 3 f (a 2 , R N ) = n(a 2 , a 1 ) = n 3 + n 4 + n 6 f (a 3 , R N ) = n(a 3 , a 2 ) = n 2 + n 5 + n 6
Kemeny's procedure We start by computing the score of each of the six linear orders, from

R 1 to R 6 . For R 1 = a 1 a 2 a 3 , K(a 1 , a 2 ) = 1, K(a 1 , a 3 ) = 1, K(a 2 , a 3 ) = 1 and K(a j , a k ) = 0 otherwise. Then S(R 1 ) = n(a 1 a 2 ) + n(a 1 a 3 ) + n(a 2 , a 3 ).
In the same way, we have :

S(R 2 ) = n(a 1 , a 2 ) + n(a 1 , a 3 ) + n(a 3 , a 2 ) S(R 3 ) = n(a 2 , a 1 ) + n(a 1 , a 3 ) + n(a 2 , a 3 ) S(R 4 ) = n(a 2 , a 1 ) + n(a 3 , a 1 ) + n(a 2 , a 3 ) S(R 5 ) = n(a 1 , a 2 ) + n(a 3 , a 1 ) + n(a 3 , a 2 ) S(R 6 ) = n(a 2 , a 1 ) + n(a 3 , a 1 ) + n(a 3 , a 2 )
We can now determine when an alternative is elected under Kemeny's rule. Alternative a 1 is chosen when individual preferences in the profile are closest to R 1 or R 2 . This means that

max {S(R 1 ), S(R 2 )} max {S(R 3 ), S(R 4 )} and max {S(R 1 ), S(R 2 )} max {S(R 5 ), S(R 6 )}. In the same way a 2 is the winner if max {S(R 3 ), S(R 4 )} > max {S(R 1 ), S(R 2 )} and max {S(R 3 ), S(R 4 )} max {S(R 5 ), S(R 6 )}. And a 3 is the winner if max {S(R 5 ), S(R 6 )} > max(S(R 1 ), S(R 2 )) and max {S(R 5 ), S(R 6 )} > max {S(R 3 ), S(R 4 )}.
As a consequence, we have the following proposition.

Proposition 7. Under Kemeny's rule, if there exists a

1) Condorcet cycle 1, then k(a 1 , R N ) = max {S(R 1 ), S(R 2 )} = S(R 2 ), k(a 2 , R N ) = max {S(R 3 ), S(R 4 )} = S(R 3 ) and k(a 3 , R N ) = max {S(R 5 ), S(R 6 )} = S(R 6 ). 2) Condorcet cycle 2, then k(a 1 , R N ) = max {S(R 1 ), S(R 2 )} = S(R 1 ), k(a 2 , R N ) = max {S(R 3 ), S(R 4 )} = S(R 4 ) and k(a 3 , R N ) = max {S(R 5 ), S(R 6 )} = S(R 5 ). Proof. With a Condorcet cycle 1, a 3 Ma 2 , which means that n(a 3 , a 2 ) > n(a 2 , a 3 ). Since S(R 1 )= n(a 1 , a 2 ) + n(a 1 a 3 ) + n(a 2 , a 3 ) and S(R 2 ) = n(a 1 , a 2 ) + n(a 1 , a 3 ) + n(a 3 , a 2 ), then S(R 2 ) > S(R 1 ) and k(a 1 , R N ) = max {S(R 1 ), S(R 2 )} = S(R 2 )
. With a Condorcet cycle 2, we have the converse,

a 2 Ma 3 , n(a 2 , a 3 ) > n(a 3 , a 2 ) and S(R 1 ) > S(R 2 ). Then k(a 1 , R N ) = max {S(R 1 ), S(R 2 )} = S(R 1 ).
The proof is the same for max {S(R 3 ), S(R 4 )} and max {S(R 5 ), S(R 6 )}.

From the previous propositions, we can deduce the following results.

Proposition 8. Suppose there exists a Cycle 1 or a Cycle 2. Then for all {x, y} ✓ {a 1 , a 2 , a 3 },

d(x, R N ) d(y, R N ) = f (x, R N ) f (y, R N ) = 1 2 ⇥ e(x, R N ) e(y, R N ) ⇤ = 1 2 ⇥ k(x, R N ) k(y, R N ) ⇤ .
Proof. Suppose a Cycle 1 occurs. Consider {x, y} ✓ {a 1 , a 2 , a 3 }. For example x = a 1 and y = a 2 . It follows that d(a

1 , R N ) = ⌅ n+1 2 ⇧ (n 1 + n 2 + n 5 ) and d(a 2 , R N ) = ⌅ n+1 2 ⇧ (n 1 + n 3 + n 4 ); f (a 1 , R N ) = (n 1 + n 2 + n 5 ) and f (a 2 , R N ) = (n 1 + n 3 + n 4 ); e(a 1 , R N ) = 2(n 1 + n 2 + n 5 ) and e(a 2 , R N ) = 2(n 1 + n 3 + n 4 ); k(a 1 , R N ) = S(R 2 ) = n(a 1 , a 2 ) + n(a 1 , a 3 ) + n(a 3 , a 2 ) = 2n 1 + 3n 2 + n 3 + 2n 5 + n 6 and k(a 2 , R N ) = S(R 3 ) = n(a 2 , a 1 ) + n(a 1 , a 3 ) + n(a 2 , a 3 ) = 2n 1 + n 2 + 3n 3 + 2n 4 + n 6 . Then d(a 1 , R N ) d(a 2 , R N ) = (n 2 +n 5 ) (n 3 +n 4 ), f (a 1 , R N ) f (a 2 , R N ) = (n 2 +n 5 ) (n 3 +n 4 ), e(a 1 , R N ) e(a 2 , R N ) = 2 [(n 2 + n 5 ) (n 3 + n 4 )] and k(a 1 , R N ) k(a 2 , R N ) = 2 [(n 2 + n 5 ) (n 3 + n 4 )]. Consequently, d(a 1 , R N ) d(a 2 , R N ) = f (a 1 , R N ) f (a 2 , R N ) = 1 2 ⇥ e(a 1 , R N ) e(a 2 , R N ) ⇤ = 1 2 ⇥ k(a 1 , R N ) k(a 2 , R N ) ⇤ = (n 2 + n 5 ) (n 3 + n 4
). The same observation can easily be shown for {a 1 , a 3 } and {a 2 , a 3 }; and even when Cycle 1 is replaced with Cycle 2.

Corollary 1. Suppose Cycle 1 or Cycle 2 occur. Then F 3 , F 4 , F 5 and F 6 select the same winner.

Proof. Straightforward.

This corollary shows that when we have a cycle, Dodgson, Young, Condorcet and Kemeny's procedures always select the same winner. We will next show that this is also true when there is no Condorcet cycle.

Other cases

Proposition 9. Suppose there is no Cycle 1 or Cycle 2 at some profile. Then F 1 , F 3 , F 4 , F 5 , F 6 , F 7 and F 8 select the same winner.

Proof. We examine all possible cases and observe that the result holds.

(i) Not (a 2 Ma 1 ) and not (a 3 Ma 1 ). Then a 1 belongs to the choice set of C j , j = 1, 3, 4, 5, 6, 7, 8. Using the tie-breaking rule, a 1 is selected by Proof. The proof is straighforward from the previous propositions. Theorem 2. With three alternatives and for all profile, Fishburn's rule and Schwartz's rule always select the same winner.

F j , j = 1,
Proof. The proof is straighforward from Proposition 9. In fact the two procedures coincide when there is no cycle. Moreover by definition, the choice set from the two procedures contains the three alternatives when there is a cycle and by the tie-breaking rule, a 1 is selected by F j , j = 7, 8. Therefore F 7 = F 8 .

It can easily be checked that with more than three alternatives, the above theorems are no longer valid.

In summary, in a three-candidate context, the nine procedures can be reorganized into five classes, three of which are singletons.

Indeed, the criterion we use to distinguish those classes is the specific way they select the outcome of the election when there is no Condorcet winner.

-Whenever there is a cycle, Copeland's function directly relies on the tie-breaking device.

-Under amendment function, a 1 and a 2 can be elected if and only if they are Condorcet winners; in all other cases, the winner is a 3 .

All other functions are equivalent when there is no cycle (see Proposition 9). However, -Black's function systematically chooses the Borda winner and, if necessary, uses the tiebreaking mechanism.

-Each of Dodgson's, Young's, Condorcet's and Kemeny's rules uses a different way to determine the outcome when there is no Condorcet winner, but all these ways are equivalent.

-Finally, Schwartz's and Fishburn's rules, though based on different majority mechanisms when there is no Condorcet winner, are trivially equivalent in the three-election case.

Reinforcement axiom

We first present the reinforcement axiom property more formally.

Definition 3. Given T 1 and T 2 , an SCF F satisfies RA if for any R T 1 2 L T 1 and R T 2 2 L T 2 and for all x 2 A, ⇥ F (R T 1 ) = x and F (R T 2 ) = x ⇤ =) F (R N ) = x.
Alternatively,

Definition 4. Given T 1 and T 2 , an SCF F violates RA if there exist R T 1 2 L T 1 , R T 2 2 L T 2 and {x, y} ✓ A, x 6 = y, such that F (R T 1 ) = x, F (R T 2 ) = x and F (R N ) = y.
Smith (1973) and Young (1975a) show that all Condorcet social procedures violate the RA. In Example 1, we present a case of violation of RA by a Condorcet rule.

Example 1. Consider the Copeland procedure, and assume that R T 1 2 L T 1 and R T 2 2 L T 2 are as below:

Profiles R T 1 R T 2 Voters 4 1 1 5 4 5
w y y w z y y z x z x w x w z x y x z x w y w z

Then the scores are as follows

Separate groups

Copeland scores

T 1 T 2 c(w, R T 1 ) = 3 c(w, R T 2 ) = 1 c(x, R T 1 ) = 1 c(x, R T 2 ) = 1 c(y, R T 1 ) = 1 c(y, R T 2 ) = 1 c(z, R T 1 ) = 3 c(z, R T 2 ) = 1 Winner : w Winner : w
The whole electorate

Copeland scores

N = T 1 [ T 2 c(w, R T 1 ) = 1 c(x, R T 1 ) = 2 c(y, R T 1 ) = 3 c(z, R T 1 ) = 2
Winner : y

This shows that the Copeland procedure violates RA:

F (R T 1 ) = w, F (R T 2 ) = w, but F (R N ) = y.
Next, we identify conditions on the number of voters, under which the violation of RA is susceptible to occur for all the Condorcet procedures introduced above -and it appears that this is the case roughly as soon as the number of voters is equal to 7. This is reminiscent of Young's (1975a) result on all Condorcet procedures.

Proposition 10. With at least 13 voters and at least three candidates, all Condorcet procedures violate the reinforcement axiom.

Proof. See [START_REF] Moulin | Axioms of cooperative decision making[END_REF], Theorem 9.2, page 237. The above result covers all Condorcet voting procedures. However, for the procedures under study in this paper, we obtain slightly different versions of Proposition 10.

Proposition 11. In three-candidate elections with 7 voters, all Condorcet procedures F j with j = 1, 2, ..., 9 violate the reinforcement axiom.

Proof. Let F j be a Condorcet procedure with j = 1, 2, ..., 8; Case 1: Suppose that n is odd. By assumption on n, there exists a non negative integer p such that n = 7 + 2p. Consider the following profiles:

R T 1 R T 2 1 + p 0 1 0 0 p 1 1 1 0 0 2
x x y y z z x x y y z z y z x z x y y z x z x y z y z x y x z y z x y x In profile R T 1 , 2 + p individuals prefer x to z, versus p individuals who prefer z to x. Since p 0, by definition xM z . Likewise yM z since 2 + p prefer y to z. And x is tied with y since 1 + p individuals prefer x to y and 1 + p prefer y to x. By definition, for each C j with j = 1, 2, ..., 8, C j (R T 1 ) = {x, y}. By the lexicographic order, F j (R T 1 ) = x, for j = 1, 2, ..., 8. In R T 2 , with 3 favorable votes versus 2, x is preferred to z, y to x and z to y, then xM z, zM y and yM x. Then we have a Condorcet cycle 2. By definition of

C 1 , b(x, R T 2 ) = 2 + 3 = 5, b(y, R T 2 ) = 3 + 2 = 5 and b(z, R T 2 ) = 2 + 3 = 5, then C 1 (R T 2 ) = {x, y, z} and F 1 (R T 2 ) = x. We have also c(x, R T 2 ) = 1 1 = 0, c(y, R T 2 ) = 1 1 = 0 and c(z, R T 2 ) = 1 1 = 0, then C 2 (R T 2 ) = {x, y, z} and F 2 (R T 2 ) = x. From proposition 5, we know that f (x, R T 2 ) = n(x, y) = 2 , f (y, R T 2 ) = n(y, z) = 2 and f (z, R T 2 ) = n(z, x) = 2, then C 5 (R T 2 ) = {x, y, z} and F 5 (R T 2 ) = x. By theorem 1, F j (R T 2 ) = x for j = 3, 4, 5, 6. And finally, C 7 (R T 2 ) = C 8 (R T 2 ) = {x,
y, z}, since there are clearly no Schwartz and no Fishburn relations over {x, y, z}. Then,

F 7 (R T 2 ) = F 8 (R T 2 ) = x.
We can then conclude that F j (R T 2 ) = x, for j = 1, 2, ..., 8. In R N , y is a Condorcet winner, since yM x and yM z. Indeed, 4 + p versus 3 + p individuals prefer y to x and 4 + p versus 3 + p individuals prefer y to z. That is F j (R N ) = y for j = 1, 2, ..., 8. Case 2: Suppose that n is even. By assumption on n, there exists a non negative integer p such that n = 8 + 2p. Consider the following profiles:

R T 1 R T 2 1 + p 0 1 0 0 p 0 2 2 0 0 2
x x y y z z x x y y z z y z x z x y y z x z x y z y y x y x z y y x y x

Using very similar arguments as in the previous case one can easily check that

F j R T 1 = x, F j R T 2 =
x and F j R N = y for j = 1, 2, ..., 8. It then appears that in all cases considered above, all Condorcet procedures violate Young reinforcement axiom; and this concludes the proof for j = 1, 2, ..., 8. For j = 9, with the same reasoning as above, we reach the same conclusion, from the two following situations:

R T 1 R T 2 p 0 0 0 2 + p 0 0 2 2 0 0 1
x x y y z z x x y y z z y z x z x y y z x z x y z y z x y x z y z x y x if n is odd, and

R T 1 R T 2 p 0 0 0 1 + p 0 2 1 1 0 0 3
x x y y z z x x y y z z y z x z x y y z x z x y z y z x y x z y z x y x if n is even. And this ends the proof.

As a direct consequence of the preceding proposition, the following holds: Proposition 12. With at least seven voters and at least three candidates, all Condorcet procedure F j with j = 1, 2, ..., 9 violate the reinforcement axiom.

Proof. Suppose A = {x, y, z, a 4 , a 5 , ..., a m } and let F j be a Condorcet procedure. Consider the distinct cases in the proof of Proposition 11 and the corresponding profiles R T 1 and R T 2 . Now, construct two new profiles R T 1 and R T 2 from R T 1 and R T 2 respectively, in such a way that alternative a 4 appears at the fourth position in each individual preference relation, a 5 appears at the fifth position, and so on. Each candidate in {x, y, z} is then prefered by a majority of individuals to all candidates in {a 4 , a 5 , ..., a m }. It follows that the winners are exactly the same as with R T 1 , R T 2 and R N respectively. That is F (R T 1 ) = x, F (R T 2 ) = x and F (R N ) = y. Consequently, F j violates the reinforcement axiom.

The two propositions show that all Condorcet procedures introduced above violate RA when the number of voters is equal to or greater than seven, provided that the number of alternatives is at least three. In other words for all the procedures studied in this paper, we can always find a profile at which RA is violated, with at least seven voters and at least three alternatives. Below seven voters, we can find some pathological profiles, but not for all rules. Indeed, the reader can easily check that the minimum number of voters at which the phenomenon arise is three for Copeland, five for Fishburn, Schwartz and amendment. For the other rules, there is no case below seven. Note also, that it straightforwardly appears that with two alternatives there exists no Condorcet procedure which violates RA.

We will now turn to the evaluation on how frequent Condorcet procedures violate RA. In the next section, we focus on the special case of three alternatives.

The likelihood of reinforcement violation

Our main purpose in this section is to compare the nine Condorcet rules described above in terms of their susceptibility to violate RA. We only study the three-alternative case in this paper, since with four or more alternatives, computations are much more involved.

In order to do that we evaluate the proportion of preference profiles at which the violation of RA arises. We first have a typology of the various profiles at which this violation occurs. We can notice that if voting is restricted to profiles in which at least one Condorcet winner exists, then selecting the Condorcet winners satisfies RA. Indeed, if a majority in each of the two constituencies prefers a to x, then a majority in the whole constituency prefers a to x. Thus, the violation of RA is due to majority cycles. More precisely, given a voting rule, situations in which the voting rule violates RA are completely described by a set of linear inequalities (technicalities are available from the authors upon simple request). And from this typology, by the use of computerised evaluation processes, we obtain various frequencies measuring the susceptibility of this rule to violate RA.

The computer evaluation process consists in simulations using the traditional Monte Carlo method. Computer simulations are very often used in social choice in order to determine the frequencies of paradoxes, e.g., [START_REF] Nitzan | The vulnerability of point voting schemes to preference variation and strategic manipulation[END_REF] or [START_REF] Kelly | Almost all social choice rules are highly manipulable, but a few aren't[END_REF] among others. The Monte Carlo simulations method is useful when a problem cannot be solved analytically.

Its principle is as follows : (i) we choose a probabilistic hypothesis, that is the way probabilities are assigned to events; (ii) we randomly generate under each probability approach a set of 100,000 profiles (a greater number does not significantly improve the results); and (iii) for each voting rule under consideration, we then count the total number of pathological profiles among the list of drawn profiles. The estimation of the likelihood that the rule violates RA is then derived as the following ratio:

Number of occurences of pathological profiles randomly generated

Total number of profiles randomly generated

Note that the values we obtain are not exact probabilities, but rather estimates. For example, a value equal to 0 doesn't necessarily mean that the phenomenon cannot at all occur, but in fact, simply means that its probability of occurrence is very small. Two traditional probabilistic hypotheses are taken into account: impartial culture (IC) and impartial anonymous culture (IAC). Both are based on an equal probability assumption, but not exactly in the same way. Under the IAC probabilistic model, all anonymous preference profiles are equally likely to be observed: for every possible preference order, the number of voters reporting that order is drawn, in such a way that the total number of voters be equal to n. More precisely, for each n and each linear order we draw randomly the number n 1 of individuals with the first linear order. And then, again randomly, we draw the number n 2 of individuals with the second linear order between 0 and n n 1 , and so on. Under the IC probabilistic model and given any voter, the six rankings are equally likely to be the preference relation of this voter. In other words, for each voter, every preference order has an equal chance to be drawn ( 16 in the three-alternative case). For a more detailed discussion on these probabilistic models, see [START_REF] Reggenwetter | Behavioral social choice[END_REF], or [START_REF] Berg | Note sur le calcul de la probabilité des paradoxes du vote[END_REF], or [START_REF] Lepelley | On Ehrhart polynomials and probability calculations in voting theory[END_REF].

In this section, for each probabilistic hypothesis and each rule under study, we present tables of values and graphs corresponding to some values of the total number of voters.

Beside the results with Monte Carlo simulations, we also provide analytical results under the IAC hypothesis when the number of individuals is infinitely large. Within this framework, we use an analytical method -as distinguished from computer simulations -and provide exact relations giving the proportion of anonymous profiles vulnerable to the violation of RA. These results are obtained by the use of the technique of [START_REF] Fishburn | Condorcet's paradox and anonymous preference profiles[END_REF].

Impartial anonymous culture

Simulations results under IAC

We consider the case where n 2 {25, 50, 75, 100} and t (the size of any of the two constituencies) changes from 1 to at most n 2 (Tables 1 and2) 3 . Note that cases where t n 2 are symmetrical to those where t  n 2 . For example, for n = 25, the results are the same with t = 1 and t = 24.

Table 1 (IAC 1) Frequencies of violation of RA for Condorcet procedures with n = 25 and n = 50

t n= 25 n = 50 F 1 F 2 F 3 , F 4 , F 5 , F 6 F 7 ,F 8 F 9 F 1 F 2 F 3 , F 4 , F 5 , F 6 F 7 ,F 8 F 9 1
5,54E-03 0,00E+00 0,00E+00 0,00E+00 0,00E+00

F 1 F 2 F 3 , F 4 , F 5 , F 6 F 7 ,F 8 F 9 F 1 F 2 F 3 , F 4 , F 5 , F 6 F 7 ,F 8 F 9 1
1,07E-02 0,00E+00 0,00E+00 0,00E+00 0,00E+00 The corresponding figures are below. These results illustrate the following facts: (i) for all rules under study, it appears that for any given value of n, and for all values of t, frequencies of violation of RA are relatively small (generally smaller than 1%); (ii) although all frequencies are rather small, they are clearly smaller for F 3 F 4 F 5 F 6 than for F 2 , F 7 F 8 and F 9 ; and for these three families of rules the results are relatively close, as illustrated in Fig. 6; (iii) it must also be noticed that F 1 appears as the worst procedure, especially when the size of one of the two constituencies is relatively greater than the size of the other one; (iv) except for F 1 , for any fixed value of n, frequencies slightly increase when t rises (fluctuations are due to even or odd values of t) as illustrated in Figs. 2,3,4,5; (v) for any given value of t, frequencies for F 3 F 4 F 5 F 6 decrease when n rises, whereas this is not clear for F 2 , F 7 F 8 and F 9 ; (vi) it follows that frequencies are higher as both constituencies tend to have the same size for all the rules with the exception of F 1 ; (vii) a contrario, frequencies for F 1 are relatively smaller with equal size constituencies. Indeed as observed in Fig. 1, for any fixed n, frequencies decrease when the value of t rises, and for any given value of t, frequencies increase when the value of n rises.

To summarize, we have shown that for F j , j = 2, 3, 4, 5, 6, 7, 8, 9, the frequency of occurrence is maximal when both group tend to have the same size, whereas for F 1 we have the opposite phenomenon. Now, one natural question is about the limit frequencies as n gets larger.

Exact limiting results

We now focus on large electorates-and indeed on infinite electorates, that is on cases where the total number of voters in both constituencies tends to infinity. Let RA (F j , 1) be the vulnerability of the voting rule F j as the total number of voters tends to infinity and RA (F j , ↵, 1) be the vulnerability of the voting rule F j as the total number of voters tends to infinity assuming that the proportion of voters in the first constituency is exactly ↵, with ↵ 2 ⇤ 0, 1 2 ⇤ . We provide closed-form formulae giving this likelihood for Copeland, Schwartz, Fishburn and the Amendment rules. For the other rules similar results require very involved computations. 4We then can state the following results.

Proposition 13. Let the Copeland rule be the voting rule. Then

i) RA (F 2 , 1) = 31 6144 ⇡ 2.528 ⇥ 10 3 ii) RA (F 2 , ↵, 1) = 1 258048 ↵ (28083↵ 4 70160↵ 3 + 68490↵ 2 30960↵ + 5460) (↵ 1) 5 if 0  ↵  1 3 = ✓ 30966↵ 10 126670↵ 9 + 226755↵ 8 231480↵ 7 + 147630↵ 6 61236↵ 5 + 17010↵ 4 3240↵ 3 + 405↵ 2 30↵ + 1 ◆ 258048↵ 5 (↵ 1) 5 if 1 3  ↵  1 2
After computations it appears that RA (F 2 , ↵, 1) = RA (F 7 , ↵, 1) = RA (F 8 , ↵, 1) = RA (F 9 , ↵, 1) and RA (F 2 , 1) = RA (F 7 , 1) = RA (F 8 , 1) = RA (F 9 , 1). This result is consistent with the previous simulation observations.

For some values of ↵, we have the following results in Table 3. Graphically, we have the following results in Fig. 7. As seen above, it appears that: (i) in general, cases of violations are rather rare, (ii) highest frequencies are reached for electorates with equal size constituencies.

Impartial culture

As for IAC above, we consider n = {25, 50, 75, 100} and let t change from 1 to n 2 . Again, the Monte Carlo technique leads to the results provided in Tables 4 and5 below:

F 1 F 2 F 3 , F 4 , F 5 , F 6 F 7 ,F 8 F 9 F 1 F 2 F 3 , F 4 , F 5 , F 6 F 7 ,F 8 F 9 1
3,76E-03 0,00E+00 0,00E+00 0,00E+00 0,00E+00

F 1 F 2 F 3 , F 4 , F 5 , F 6 F 7 ,F 8 F 9 F 1 F 2 F 3 , F 4 , F 5 , F 6 F 7 ,F 8 F 9 1
9,00E-03 0,00E+00 0,00E+00 0,00E+00 0,00E+00 The corresponding figures are below. Once again, it can be observed from Figs. 9, 10, 11 and 12 that frequencies of F j , j = 2, 3, 4, 5, 6, 7, 8, 9 are higher as both constituencies tend to have the same size. Moreover, like Fig. 1 above, Fig. 8 shows that F 1 is more sensitive to the violation of RA with unequal size constituencies. It also appears that the performance of F 2 , F 7 F 8 and F 9 are again very close, whereas frequencies of occurence are the smallest for F 3 F 4 F 5 F 6 and the highest for F 1 .

Besides, it appears that the occurrences of paradoxes are higher under IC than under IAC, which is very common in voting theory (see Gehrlein 2006, chap. 5).

Conclusion

The goal of this paper was to compare nine Condorcet procedures through the study of their sensitivity to RA. In order to do that, we were first concerned with the number of voters at which violations of the axiom occur. It appears that this is the case roughly as soon as the number of voters is equal to 7. Secondly, we provide frequencies of these occurrences. Most of these latter results are based on computer simulations-under two probabilistic classical hypotheses-but some other are obtained with the analytical Fishburn-Gehrlein technique. Results are given for the nine Condorcet procedures. Our results show how these frequencies change according to changes in the values of the total number of voters, and the size of any one of the two constituencies.

Further, it appears that for all these rules: (i) in general, cases of violations of RA are rather rare, (ii) although all frequencies are small, they are smaller for Dodgson's rule, Young's rule, Condorcet's rule and Kemeny's rule, and (iii) frequencies decrease for electorates with clearly uneven constituencies, with the exception of Black's rule. The main question motivating this study was whether one should be unduly worried about the theoretical possibility of violation of RA. The negative answer to this question is the main message of our analysis. Indeed, although all these rules violate RA, the frequencies of violation are very rare, as it is the case with sequential positional rules (see [START_REF] Courtin | The reinforcement axiom under sequential positional rules[END_REF].

Finally, we have observed that in the special case of three alternatives, Dodgson's rule, Young's rule, Condorcet's rule and Kemeny's rule always select the same winner whatever the profile. The same conclusion holds for Schwartz's rule and Fishburn's rule.

3 .

 3 Consider a finite set N of n individuals or voters, with n 2 and a finite set A of m alternatives, with m Suppose two disjoint groups (or constituencies) T 1 and T 2 of individuals, with T 1 [ T 2 = N. Let 2 A be the set of nonempty subsets of A. Assume that the preference relation R i of individual i, i 2 N , is a complete, antisymmetric and transitive binary relation (or simply a linear order) on A and let L = {R k : 1  k  m!} be the set of all linear orders on A. A profile on A is an n-tuple R N = (R i )

Fig. 1 Fig. 2 Fig. 3 F 3 F 4 F 5 F 6 Fig. 4 F 7 F 8 Fig. 5 Fig. 6 A

 123345647856 Fig. 1 Frequencies of violation of RA under IAC for F 1 and n 2 {25, 50, 75, 100}

Fig. 7 F 2 F 7 F 8 F 9

 72789 Fig. 7 Frequencies of violation of RA under IAC for F 2 F 7 F 8 F 9 with large electorate

Fig. 8 Fig. 9 Fig. 10 F 3 F 4 F 5 F 6 Fig. 11 F 7 F 8 Fig. 12 Fig. 13 A

 8910345611781213 Fig. 8 Frequencies of violation of RA under IC for F 1 and n 2 {25, 50, 75, 100}

.

  Schwartz's function: C 7 R N = {x 2 A : ySx f or no y 2 A}, where xSy iff there are x 1 = x, x 2 , ..., x k = y in A such that x 1 Mx 2 , x 2 Mx 3 , ..., x k 1 Mx k , and there are no z 1 = y, z 2 , ..., z j = x in A such that z 1 Mz 2 , z 2 Mz 3 , ..., z C 2 by Copeland (1951); C 3 by Dodgson (1876); C 4 and C 5 by Young (1975b); C 6 by Kemeny (1959) and Levenglick (1975); C 7 by Schwartz (1972); and C 8 by Fishburn (1970).

	R Kemeny's function can be viewed as a measure of the closeness of individual preferences in k 2 L and xR k y f or all y 2 A} (or simply k(x) when there is no ambiguity).
	some profile with the linear order R	k derived from that profile as indicated above.
	C 7

j 1 Mz j . 2 C 1 was introduced by Black (1958);

  ) . Suppose there exists only one alternative, say a Proposition 6. If there exists a Condorcet cycle at some profile R N , then the Condorcet score of an alternative a

	j , a k ) < n(a since there is only one alternative such that xM a j , this means that n(a k , a j ), which means that n(a j , a k ) < n 2 . And j , a l ) n(a l , a j ) n 2 . Then,
	n(a

k , such that a k Ma j , in R N . Then n(a j , a k ) < n(a j , a l ). Hence, f (a j , R N ) = min x2A\a j n(a j , x) = n(a j , a k ). j is given by f (a j , R N ) = n(a j , a k )

where a k is the unique alternative such that a k Ma j .

Proof. Straightforward. As a consequence, we have the following Condorcet scores for the two Condorcet cycles with

  3, 4, 5, 6, 7, 8. (ii) a 2 Ma 1 Two subcases are distinguished. If not (a 3 Ma 2 ) occurs, then a 2 but not a 1 belongs to the choice set of C j , j = 1, 3, 4, 5, 6, 7, 8. Therefore a 2 is selected by F j , j = 1, 3, 4, 5, 6, 7, 8. Otherwise a 3 Ma 2 . Then a 2 Ma 1 and a 3 Ma 2 occur simultaneously. Since there is no cycle, not (a 1 Ma 3 ) stands. Therefore a 3 is the unique candidate in the choice set of C j , j = 1, 3, 4, 5, 6, 7, 8. Then F j , j = 1, 3, 4, 5, 6, 7, 8 selects a 3 . (iii) a 3 Ma 1 and not (a 2 Ma 1 ) Three cases are in consideration. If a 3 Ma 2 , then a 3 is the unique Condorcet winner and is selected by F j , j = 1, 3, 4, 5, 6, 7, 8. If a 2 Ma 3 , then not (a 1 Ma 2 ) occurs since there is no cycle. Moreover since not (a 2 Ma 1 ) holds, a 2 ties a 1 . Therefore a 2 is the unique candidate in the choice set of C If a 1 ties a 2 , then a 2 but not a 1 is in the choice set of C

j , j = 1, 3, 4, 5, 6, 7, 8. Thus F j , j = 1, 3, 4, 5, 6, 7, 8 all select a 2 .

Otherwise a 2 ties a 3 . Since not (a 2 Ma 1 ) holds, there are two possibilities: either a 1 ties a 2 , or

a 1 Ma 2 .

If a 1 Ma 2 , then a 3 is the unique outcome in the choice set of C j , j = 1, 3, 4, 5, 6, 7, 8 and is then selected by F j , j = 1, 3, 4, 5, 6, 7, 8. j , j = 1, 3, 4, 5, 6, 7, 8 and is then selected by F j , j = 1, 3, 4, 5, 6, 7, 8.

The theorem below follows from Corollary 1 and Proposition 9.

Theorem 1. With three alternatives and for all profile, Dodgson's rule, Young's rule, Condorcet's rule and Kemeny's rule always select the same winner.

Table 2 (

 2 IAC 2) Frequencies of violation of RA for Condorcet procedures with n = 75 and n = 100

	t	n= 75

Table 3 (

 3 IAC 3) Frequencies of violation of RA for large electorates

	↵	F 2 F 7 F 8 F 9	↵	F 2 F 7 F 8 F 9	↵	F 2 F 7 F 8 F 9	↵	F 2 F 7 F 8 F 9
	0, 025	5,20E-04	0, 15	2,80E-03	0, 275	4,40E-03	0, 4	5,13E-03
	0, 05	1,02E-03	0, 175	3,18E-03	0, 3	4,63E-03	0, 425	5,24E-03
	0, 075	1,50E-03	0, 2	3,53E-03	0, 325	4,83E-03	0, 45	5,33E-03
	0, 1	1,96-03	0, 225	3,85E-03	0, 35	4,83E-03	0, 475	5,39E-03
	0, 125	2,39E-03	0, 25	4,14E-03	0, 375	5,00E-03	0, 5	5,42E-03

Table 4 (

 4 IC 1) Frequencies of violation of RA for Condorcet procedures with n = 25 and n = 50

	t	n= 25

Table 5 (

 5 IC 2) Frequencies of violation of RA for Condorcet procedure with n = 75 and n = 100

	t	n= 75

This process is usual in France, especially for a constitutional revision.

Note that in the tables, the notation xE-y is used as an equivalent for x ⇥ 10 y , where x and y are real numbers.

The main difficulty encountered is the evaluation of an 11-dimensional volume over a parameterized polytope.