Majority properties of positional social preference correspondences

Abstract

We characterize the positional social preference correspondences (spc) satisfying the qualified majority property for any given majority threshold. We also characterize the positional spcs satisfying the minimal majority property. We next evaluate the probability that the Borda, the plurality and the antiplurality spcs fulfil the two aforementioned properties under the Impartial and Anonymous Culture assumption in the presence of three and four alternatives for various sizes of the society. Our results show that the Borda spc is the positional spc which better behaves in relation with the qualified majority principle and the minimal majority principle.

Appendices

Appendix 1: Three alternatives and qualified majority: exact probabilities

Assume that the set of alternatives is $A=\{a,b,c\}$ and the set of individuals is $N=\{1,\ldots ,n\}$. For the sake of simplicity, for every $x,y\in A$, we write in what follows xy instead of (x, y). Let ${\mathcal {G}}(A)$ be the set of asymmetric relations over A. Thus, we have that

$$\begin{aligned} {\mathcal {G}}(A)=\left\{ \begin{array}{ll} \varnothing , \{ab\}, \{ba\}, \{ac\}, \{ca\}, \{bc\}, \{cb\}, \{ab,ac\}, \{ab,ca\}, \{ab,bc\}, \\ \{ab,cb\}, \{ba,ac\}, \{ba,ca\}, \{ba,bc\}, \{ba,cb\}, \{ac,bc\}, \{ac,cb\}, \\ \{ca,bc\}, \{ca,cb\}, \{ab,ac,bc\}, \{ab,ac,cb\}, \{ab,ca,bc\}, \{ab,ca,cb\}, \\ \{ba,ac,bc\}, \{ba,ac,cb\}, \{ba,ca,bc\}, \{ba,ca,cb\} \end{array} \right\} . \end{aligned}$$

Let E be the correspondence from ${\mathcal {G}}(A)$ to ${\mathcal {L}}(A)$ defined, for every $R\in {\mathcal {G}}(A)$, by

$$\begin{aligned} E(R)=\{q\in {\mathcal {L}}(A): R\subseteq q\}. \end{aligned}$$

Let

$$\begin{aligned} \Gamma =\Big \{(R,S)\in {\mathcal {G}}(A)^2:E(R)\subseteq E(S)\Big \}, \end{aligned}$$

and consider the following subsets of ${\mathcal {G}}(A)^2$:

$$\begin{aligned} \begin{array}{l} \Gamma _1=\left\{ (\varnothing ,\varnothing )\right\} \\ \Gamma _2=\{\{ab\}\}\times \{\varnothing , \{ab\}\}\\ \Gamma _3=\{\{ba\}\}\times \{\varnothing , \{ba\}\}\\ \Gamma _4=\{\{ac\}\}\times \{\varnothing , \{ac\}\}\\ \Gamma _5=\{\{ca\}\}\times \{\varnothing , \{ca\}\}\\ \Gamma _6=\{\{bc\}\}\times \{\varnothing , \{bc\}\}\\ \Gamma _7=\{\{cb\}\}\times \{\varnothing , \{cb\}\}\\ \Gamma _8=\{\{ab,ac\}\}\times \{\varnothing , \{ab\}, \{ac\}, \{ab,ac\}\}\\ \Gamma _{9}=\{\{ba,bc\}\}\times \{\varnothing , \{ba\}, \{bc\}, \{ba,bc\}\}\\ \Gamma _{10}=\{\{ca,cb\}\}\times \{\varnothing , \{ca\}, \{cb\}, \{ca,cb\}\}\\ \Gamma _{11}=\{\{ba,ca\}\}\times \{\varnothing , \{ba\}, \{ca\}, \{ba,ca\}\}\\ \Gamma _{12}=\{\{ab,cb\}\}\times \{\varnothing , \{ab\}, \{cb\}, \{ab,cb\}\} \\ \Gamma _{13}=\{\{ac,bc\}\}\times \{\varnothing , \{ac\}, \{bc\}, \{ac,bc\}\}\\ \Gamma _{14}=\{\{ab,bc\},\{ab,ac,bc\}\}\times \{\varnothing , \{ab\}, \{ac\}, \{bc\},\{ab,ac\},\{ab,bc\},\{ac,bc\},\{ab,ac,bc\}\} \\ \Gamma _{15}=\{\{ac,cb\},\{ab,ac,cb\}\}\times \{\varnothing , \{ab\}, \{ac\}, \{cb\},\{ab,ac\},\{ab,cb\},\{ac,cb\},\{ab,ac,cb\}\}\\ \Gamma _{16}=\{\{ca,bc\},\{ba,ca,bc\}\}\times \{\varnothing , \{ba\}, \{ca\}, \{bc\},\{ba,ca\},\{ba,bc\},\{ca,bc\},\{ba,ca,bc\}\}\\ \Gamma _{17}=\{\{ba,cb\},\{ba,ca,cb\}\}\times \{\varnothing , \{ba\}, \{ca\}, \{cb\},\{ba,ca\},\{ba,cb\},\{ca,cb\},\{ba,ca,cb\}\}\\ \Gamma _{18}=\{\{ab,ca\},\{ab,ca,cb\}\}\times \{\varnothing , \{ab\}, \{ca\}, \{cb\},\{ab,ca\},\{ab,cb\},\{ca,cb\},\{ab,ca,cb\}\}\\ \Gamma _{19}=\{\{ba,ac\},\{ba,ac,bc\}\}\times \{\varnothing , \{ba\}, \{ac\}, \{bc\},\{ba,ac\},\{ba,bc\},\{ac,bc\},\{ba,ac,bc\}\}\\ \Gamma _{20}=\{\{ba,ac,cb\}\}\times {\mathcal {G}}(N)\\ \Gamma _{21}=\{\{ab,ca,bc\}\}\times {\mathcal {G}}(N).\\ \end{array} \end{aligned}$$

A cumbersome computation shows that $\{\Gamma _k\}_{k=1}^{21}$ is a partition of $\Gamma$.

Let ${\mathcal {L}}(A)=\{q_1,q_2,q_3,q_4,q_5,q_6\}$, where, using standard notation, we have that

$$\begin{aligned} q_1=abc,\; q_2=acb,\; q_3=bac,\; q_4=bca,\; q_5=cab,\; q_6=cba. \end{aligned}$$

Let

$$\begin{aligned} {\mathcal {N}}=\left\{ {\tilde{n}}=(n_1,n_2,n_3,n_4,n_5,n_6)\in {\mathbb {N}}_0^6: \sum _{j=1}^6 n_j=n\right\} , \end{aligned}$$

and let $S:{\mathcal {P}}\rightarrow {\mathcal {N}}$ be the surjective function defined, for every $p\in {\mathcal {P}}$, by

$$\begin{aligned} S(p)=\Big (|\{i\in N: p(i)=q_1\}|,\ldots ,|\{i\in N: p(i)=q_6\}|\Big ). \end{aligned}$$

Given $\lambda \in [0,1]$, let $w(\lambda )=(1,\lambda ,0)\in {\mathfrak {W}}$ and $F_{w(\lambda )}$ be the positional spc associated with the scoring vector $w(\lambda )$. Consider now $\nu \in {\mathbb {N}}\cap (\frac{n}{2},n]$. By definition of the set $\Gamma$, for every $p\in {\mathcal {P}}$, $F_{w(\lambda )}(p)\subseteq M_{\nu }(p) \text{ if and only if } (R_{w(\lambda )}(p),\Sigma _{\nu} (p))\in \Gamma$, where $R_{w(\lambda )}(p)$ and $\Sigma _{\nu }(p)$ are defined in (4) and (1). Given now ${\tilde{n}}\in {\mathcal {N}}$, it is immediate to check that, for every $p,p'\in S^{-1}({\tilde{n}})$, $R_{w(\lambda )}(p)=R_{w(\lambda )}(p')$ and $\Sigma _{\nu} (p)=\Sigma _{\nu} (p')$ (and then $F_{w(\lambda )}(p)=F_{w(\lambda )}(p')$ and $M_{\nu }(p)=M_{\nu }(p')$). Thus, for every ${\tilde{n}}\in {\mathcal {N}}$, we can set

$$\begin{aligned} R_{w(\lambda )}({\tilde{n}})=R_{w(\lambda )}(p),\quad \Sigma _{\nu} ({\tilde{n}})=\Sigma _{\nu} (p), \quad F_{w(\lambda )}({\tilde{n}})=F_{w(\lambda )}(p),\quad M_{\nu} ({\tilde{n}})=M_{\nu} (p), \end{aligned}$$

where p is any element of $S^{-1}({\tilde{n}})$. Such a definition is consistent since, as mentioned before, it does not depend on the element of $S^{-1}({\tilde{n}})$ chosen.

Our purpose is to compute the number

$$\begin{aligned} \pi =\frac{\left| \left\{ {\tilde{n}}\in {\mathcal {N}}: (R_{w(\lambda )}({\tilde{n}}),\Sigma _{\nu} ({\tilde{n}}))\in \Gamma \right\} \right| }{|{\mathcal {N}}|}, \end{aligned}$$

which corresponds to the probability to have a voting situation ${\tilde{n}}$ such that $F_{w(\lambda )}({\tilde{n}})\subseteq M_{\nu }({\tilde{n}})$ provided that each voting situation has the same probability to be picked as assumed by IAC.

Since $\{\Gamma _k\}_{k=1}^{21}$ is a partition of $\Gamma$, we have that

$$\begin{aligned} \pi =\sum _{k=1}^{21}\pi _k, \end{aligned}$$

where

$$\begin{aligned} \pi _k=\frac{\left| \left\{ {\tilde{n}}\in {\mathcal {N}}: (R_{w(\lambda )}({\tilde{n}}),\Sigma _{\nu} ({\tilde{n}}))\in \Gamma _k\right\} \right| }{|{\mathcal {N}}|}, \end{aligned}$$

Let us consider now $\psi \in {\mathrm{Sym}}(A)$, where ${\mathrm{Sym}}(A)$ is the set of permutations of the set A. For every $q\in {\mathcal {L}}(A)$, let $\psi q\in {\mathcal {L}}(A)$ be such that, for every $x,y\in A$, $\psi (x)\succ _{\psi q}\psi (y)$ if and only if $x\succ _q y$. Thus, for instance, if $q=acb$ and $\psi =\begin{pmatrix}a&{}b&{}c\\ b&{}a&{}c\end{pmatrix}$, then $\psi q= bca$. Any $\psi \in {\mathrm{Sym}}(A)$ induces a bijection $B_{\psi}$ from ${\mathcal {N}}$ to ${\mathcal {N}}$ defined, for every ${\tilde{n}}\in {\mathcal {N}}$, by

$$\begin{aligned} B_{\psi} (n_1,n_2,n_3,n_4,n_5,n_6)=(n'_1,n'_2,n'_3,n'_4,n'_5,n'_6), \end{aligned}$$

where, for every $j\in \{1,\ldots ,6\}$, $n'_j=n_{k}$ where $\psi q_{k}=q_j$. For instance, if $\psi =\begin{pmatrix}a&{}b&{}c\\ b&{}a&{}c\end{pmatrix}$, then

$$\begin{aligned} B_{\psi} (n_1,n_2,n_3,n_4,n_5,n_6)=(n_3,n_4,n_1,n_2,n_6,n_5). \end{aligned}$$

Assume now that applying $\psi \in {\mathrm{Sym}}(A)$ to each component of all the elements of $A^2$ involved in the definition of a certain $\Gamma _k$ we obtain the set $\Gamma _{k'}$ for a suitable $k'$. It can be shown that

$$\begin{aligned} B_{\psi} \left( \left\{ {\tilde{n}}\in {\mathcal {N}}: (R_{w(\lambda )}({\tilde{n}}),\Sigma _{\nu} ({\tilde{n}}))\in \Gamma _k\right\} \right) =\left\{ {\tilde{n}}\in {\mathcal {N}}: (R_{w(\lambda )}({\tilde{n}}),\Sigma _{\nu} ({\tilde{n}}))\in \Gamma _{k'}\right\} , \end{aligned}$$

so that

$$\begin{aligned} \left| \left\{ {\tilde{n}}\in {\mathcal {N}}: (R_{w(\lambda )}({\tilde{n}}),\Sigma _{\nu} ({\tilde{n}}))\in \Gamma _k\right\} \right| = \left| \left\{ {\tilde{n}}\in {\mathcal {N}}: (R_{w(\lambda )}({\tilde{n}}),\Sigma _{\nu} ({\tilde{n}}))\in \Gamma _{k'}\right\} \right| \end{aligned}$$

and then $\pi _k=\pi _{k'}$. That happens, for instance, considering $\psi =\begin{pmatrix}a&{}b&{}c\\ b&{}a&{}c\end{pmatrix}$, $\Gamma _2$ and $\Gamma _3$.

Using that argument, we can deduce that $\pi _2=\pi _3=\pi _4=\pi _5=\pi _6=\pi _7$, $\pi _8=\pi _{9}=\pi _{10}$, $\pi _{11}=\pi _{12}=\pi _{13}$, $\pi _{14}=\pi _{15}=\pi _{16}=\pi _{17}=\pi _{18}=\pi _{19}$ and $\pi _{20}=\pi _{21}$. Moreover, it is simple to show that $R_{w(\lambda )}({\tilde{n}})$ cannot be a singleton, so that $\pi _2=0$, and it cannot be a cycle, so that $\pi _{20}=0$. As a consequence,

$$\begin{aligned} \pi =\pi _1+3\pi _8+3 \pi _{11}+6 \pi _{14}. \end{aligned}$$

Then we need to compute $\pi _1$, $\pi _{8}$, $\pi _{11}$ and $\pi _{14}$.

Fix now ${\tilde{n}}=(n_1,\ldots ,n_6)\in {\mathcal {N}}$. We have that $(R_{w(\lambda )}({\tilde{n}}),\Sigma _{\nu} ({\tilde{n}}))\in \Gamma _1$ if and only if $(n_1,\ldots ,n_6)$ is solution to the system

$$\begin{aligned} \left\{ \begin{array}{ll} n_1+n_2+\lambda n_3+\lambda n_5=n_3+n_4+\lambda n_1+\lambda n_6\\ n_1+n_2+\lambda n_3+\lambda n_5=n_5+n_6+\lambda n_2+\lambda n_4\\ n_1+n_2+n_5< \nu \\ n_3+n_4+n_6< \nu \\ n_1+n_2+n_3< \nu \\ n_4+n_5+n_6< \nu \\ n_1+n_3+n_4< \nu \\ n_2+n_5+n_6< \nu \\ n_j \ge 0\quad \forall j\in \{1,\ldots 6\}\\ \sum _{j=1}^{j=6}n_{j}=n \end{array} \right. \end{aligned}$$

(7)

$(R_{w(\lambda )}({\tilde{n}}),\Sigma _{\nu} ({\tilde{n}}))\in \Gamma _{8}$ if and only if $(n_1,\ldots ,n_6)$ is solution to the system

$$\begin{aligned} \left\{ \begin{array}{ll} n_1+n_2+\lambda n_3+\lambda n_5>n_3+n_4+\lambda n_1+\lambda n_6\\ n_1+n_2+\lambda n_3+\lambda n_5>n_5+n_6+\lambda n_2+\lambda n_4\\ n_3+n_4+\lambda n_1+\lambda n_6=n_5+n_6+\lambda n_2+\lambda n_4\\ n_3+n_4+n_6<\nu \\ n_4+n_5+n_6<\nu \\ n_1+n_3+n_4<\nu \\ n_2+n_5+n_6<\nu \\ n_j \ge 0 \quad \forall j\in \{1,\ldots 6\}\\ \sum _{j=1}^{j=6}n_{j}=n \end{array} \right. \end{aligned}$$

(8)

$(R_{w(\lambda )}({\tilde{n}}),\Sigma _{\nu} ({\tilde{n}}))\in \Gamma _{11}$ if and only if $(n_1,\ldots ,n_6)$ is solution to the system

$$\begin{aligned} \left\{ \begin{array}{ll} n_3+n_4+\lambda n_1+\lambda n_6>n_1+n_2+\lambda n_3+\lambda n_5\\ n_5+n_6+\lambda n_2+\lambda n_4>n_1+n_2+\lambda n_3+\lambda n_5\\ n_3+n_4+\lambda n_1+\lambda n_6=n_5+n_6+\lambda n_2+\lambda n_4\\ n_1+n_2+n_5<\nu \\ n_1+n_2+n_3<\nu \\ n_1+n_3+n_4<\nu \\ n_2+n_5+n_6<\nu \\ n_j \ge 0 \quad \forall j\in \{1,\ldots 6\}\\ \sum _{j=1}^{j=6}n_{j}=n \end{array} \right. \end{aligned}$$

(9)

$(R_{w(\lambda )}({\tilde{n}}),\Sigma _{\nu} ({\tilde{n}}))\in \Gamma _{14}$ if and only if $(n_1,\ldots ,n_6)$ is solution to the system

$$\begin{aligned} \left\{ \begin{array}{ll} n_1+n_2+\lambda n_3+\lambda n_5>n_3+n_4+\lambda n_1+\lambda n_6\\ n_3+n_4+\lambda n_1+\lambda n_6>n_5+n_6+\lambda n_2+\lambda n_4\\ n_3+n_4+n_6<\nu \\ n_2+n_5+n_6<\nu \\ n_4+n_5+n_6<\nu \\ n_j \ge 0 \quad \forall j\in \{1,\ldots 6\}\\ \sum _{j=1}^{j=6}n_{j}=n. \end{array} \right. \end{aligned}$$

(10)

From the analysis of the systems (7)–(10), we can compute the exact value of $\pi _1$, $\pi _{8}$, $\pi _{11}$, and $\pi _{14}$ by means of Ehrhart polynomials. That allows to get the exact value of $\pi$. More exactly, we use the parametrized Barvinok’s algorithm (Barvinok, 1994; Barvinok & Pommersheim, 1999; Verdoolaege et al., 2004) in order to solve those systems.¹⁰ We provide in what follows the pseudo-polynomials for the Borda spc. The number $\left[ \frac{19}{4}, \frac{41}{8} \right] _n$, for instance, in the first validity domain of the pseudo-polynomial of system (7) is a 2-periodic coefficient, meaning that such a coefficient depends on the parity of the parameter n: the coefficient is equal to $\frac{19}{4}$ for even n and to $\frac{41}{8}$ for odd n. The pseudo-polynomials corresponding to the other spcs are available upon request.

• 1. System (7) for the Borda spc

  • Validity domain 1: $2n -3\nu + 3 \ge 0$ and $- n + 2\nu -2 \ge 0$

    $$\begin{aligned}&-\frac{7}{8} n^3 + \left( 4 \nu -\frac{31}{8} \right) n^2 + \left( -6 \nu ^2 + 12 \nu - \left[ \frac{19}{4}, \frac{41}{8} \right] _n \right) n + 3 \nu ^3 -9 \nu ^2 + 8 \nu - \left[ 1, \frac{17}{8} \right] _n. \end{aligned}$$

  • Validity domain 2: $-2n + 3\nu -4 \ge 0$

    $$\begin{aligned}&\frac{1}{72} n^3 + \frac{1}{8} n^2 + \left[ \frac{7}{12} , \frac{5}{24} \right] _n n + \left[ 1, -\frac{25}{72} , \frac{11}{9}, -\frac{1}{8} , \frac{7}{9} , \frac{7}{72} \right] _n. \end{aligned}$$

• 2. Systems (8) and (9) for the Borda spc

  • Validity domain 1: $- n + 2\nu -2 \ge 0$ and $n -2\nu + 3 \ge 0$

    $$\begin{aligned}&-\frac{5}{64} n^4 + \left( \frac{1}{3} \nu - \frac{11}{24} \right) n^3 + \left( -\frac{1}{2} \nu ^2 + \frac{3}{2} \nu - \left[ \frac{13}{16}, \frac{27}{32} \right] _n \right) n^2\\&+ \left( \frac{1}{3} \nu ^3 -\frac{3}{2} \nu ^2 + \frac{11}{6} \nu - \left[ \frac{5}{12}, \frac{13}{24} \right] _n \right) n \\&+ \left( -\frac{1}{12} \nu ^4 + \frac{1}{2} \nu ^3 -\frac{11}{12} \nu ^2 + \frac{1}{2} \nu + \left[ 0, -\frac{5}{64} \right] _n \right). \end{aligned}$$

  • Validity domain 2: $2n -3\nu + 3 \ge 0$ and $- n + 2\nu -4 \ge 0$

    $$\begin{aligned}&\frac{29}{192} n^4 + \left( -\frac{4}{3} \nu + \frac{11}{12} \right) n^3 + \left( 4 \nu ^2 -6 \nu + \left[ \frac{85}{48}, \frac{167}{96} \right] _n \right) n^2 + \left( -5 \nu ^3 + 12 \nu ^2 -\frac{23}{3} \nu + \frac{13}{12} \right) n \\ &+ \frac{9}{4} \nu ^4 -\frac{15}{2} \nu ^3 + \frac{31}{4} \nu ^2 -\frac{5}{2} \nu + \left[ 0, \frac{7}{64} \right] _n. \end{aligned}$$

  • Validity domain 3: $-2n + 3\nu -4 \ge 0$

    $$\begin{aligned}&\frac{5}{1728} n^4 + \frac{1}{36} n^3 + \left[ \frac{5}{48}, \frac{7}{96} \right] _n n^2 + \left[ \frac{1}{12}, \frac{1}{108}, \frac{17}{108} \right] _n n + \left[ 0, -\frac{65}{576}, 0, \frac{7}{64}, -\frac{2}{9}, \frac{7}{64} \right] _n. \end{aligned}$$

• 3. System (10) for the Borda spc

  • Validity domain 1: $n =2$ and $\nu =2$

    $$2.$$

  • Validity domain 2: $n -2\nu + 2 \ge 0$ and $2n -3\nu + 1 \ge 0$ and $- n + 2\nu -1 \ge 0$

    $$\begin{aligned}&-\frac{193}{4320} n^5 + \left( \frac{11}{36} \nu -\frac{437}{1728} \right) n^4 + \left( -\frac{5}{6} \nu ^2 + \frac{25}{18} \nu -\frac{707}{1296} \right) n^3 \\&+ \left( \frac{7}{6} \nu ^3 -\frac{11}{4} \nu ^2 + \frac{43}{18} \nu - \left[ \frac{73}{144}, \frac{137}{288} \right] _n \right) n^2 \\&+ \left( \frac{ -5}{6} \nu ^4 + \frac{5}{2} \nu ^3 -\frac{37}{12} \nu ^2 + \frac{7}{4} \nu + \left[ -\frac{23}{180}, \frac{41}{1440}, -\frac{29}{540}, \frac{41}{1440}, -\frac{23}{180}, \frac{443}{4320} \right] _n \right) n \\&+\frac{29}{120} \nu ^5 -\frac{7}{8} \nu ^4 + \frac{11}{8} \nu ^3 -\frac{9}{8} \nu ^2 + \left[ \frac{23}{60}, \frac{23}{60}, \frac{29}{180 }\right] _n \nu + \left[ 0, \frac{1505}{5184}, -\frac{2}{81}, \frac{17}{64}, \frac{2}{81}, \frac{1249}{5184} \right] _n. \end{aligned}$$

  • Validity domain 3: $2n -3\nu + 2 = 0$ and $\nu -4 \ge 0$

    $$\begin{aligned}&\frac{27}{2560} \nu ^5 + \frac{57}{1024} \nu ^4 + \frac{13}{128} \nu ^3 + \left[ \frac{7}{64}, \frac{19}{256}, \frac{5}{128}, \frac{37}{256} \right] _{\nu} \nu ^2\\&+ \left[ -\frac{3}{40}, \frac{37}{640} \right] _{\nu} \nu + \left[ 0, \frac{7}{64}, -\frac{3}{64}, \frac{5}{32} \right] _{\nu}. \end{aligned}$$

  • Validity domain 4: $-2n + 3\nu -3 \ge 0$

    $$\begin{aligned}&\frac{1}{720} n^5 + \frac{31}{1728} n^4 + \frac{19}{216} n^3 + \left[ \frac{3}{16}, \frac{7}{32} \right] _n n^2 + \left[ \frac{1}{5}, \frac{727}{2160}, \frac{17}{135}, \frac{21}{80}, \frac{37}{135}, \frac{407}{2160} \right] _n n \\&+ \left[ 0, \frac{583}{1728}, -\frac{1}{27}, \frac{5}{64}, \frac{7}{27}, \frac{71}{1728} \right] _n. \end{aligned}$$

  • Validity domain 5: $2n -3\nu + 1 \ge 0$ and $- n + 2\nu -3 \ge 0$

    $$\begin{aligned}&\frac{259}{2160} n^5 + \left( -\frac{10}{9} \nu + \frac{1183}{1728} \right) n^4 + \left( 4 \nu ^2 -\frac{46}{9} \nu + \frac{713}{648} \right) n^3 \\&+ \left( -7 \nu ^3 + 14 \nu ^2 -\frac{113}{18} \nu + \left[ \frac{35}{144}, \frac{79}{288} \right] _n \right) n^2 \\&+ \left( 6 \nu ^4 -\frac{33}{2} \nu ^3 + 12 \nu ^2 - \nu - \left[ \frac{31}{90}, \frac{203}{720}, \frac{73}{270}, \frac{203}{720}, \frac{31}{90}, \frac{449}{2160} \right] _n \right) n \\&-\frac{81}{40} \nu ^5 + \frac{57}{8} \nu ^4 -\frac{175}{24} \nu ^3 + \frac{11}{8} \nu ^2 + \left[ \frac{49}{60}, \frac{49}{60}, \frac{107}{180} \right] _n \nu + \left[ 0, \frac{533}{5184}, -\frac{2}{81}, \frac{5}{64}, \frac{2}{81}, \frac{277}{5184} \right] _n. \end{aligned}$$

For n individuals and three alternatives, it is well known that the number $|{\mathcal {N}}|$ is given by the fifth-degree polynomial:

$$\begin{aligned} \frac{1}{120}n^{5}+ \frac{1}{8}n^{4}+\frac{17}{24}n^{3}+\frac{15}{8}n^{2}+\frac{137}{60}n+1. \end{aligned}$$

Taking into account the various validity domains, we have all the ingredients needed to calculate our probabilities. Note that the complexity of the pseudo-polynomials above makes the general mathematical representation for $\pi$ not beautiful to present. This is because the number of validity domains and the length of the periodic numbers are high. Fortunately, when n is fixed, compact expressions can be found. Let us provide for instance the corresponding representation of $\pi$ for the Borda spc with $n=51$:¹¹

$$\begin{aligned} \pi = {\left\{ \begin{array}{ll} -\frac{27}{8488480}\nu ^5 +\frac{841}{1697696}\nu ^4 -\frac{463627}{15279264}\nu ^3+\frac{4628821}{5093088}\nu ^2 -\frac{24006551}{1818960}\nu +\frac{10130609}{136422}, &{} \text{for } 26\le \nu \le 34\\ 1, &{} \text{for } 35\le \nu \le 51. \end{array}\right. } \end{aligned}$$

Note finally that if we assume large electorates, that is when the total number of voters tends to infinity, replace $\nu$ by rn in the results, and only consider the term of higher degree in n, we can find the probability that $F_w\subseteq M_{\nu}$ for $m=3$ and $n\rightarrow \infty$ as a function of $r=\frac{\nu }{n}$ for the three considered spcs. For instance, using the Ehrhart polynomials previously presented, we find the following function for the Borda spc:

$$\begin{aligned} \pi = {\left\{ \begin{array}{ll} \frac{259}{3}-800r+2880r^2-5040r^3+4320r^4-1458r^5, &{} \text{for } \frac{1}{2}< r<\frac{2}{3} \\ 1, &{} \text{for } \frac{2}{3}\le r\le 1. \end{array}\right. } \end{aligned}$$

The corresponding function for the plurality and the antiplurality spcs is given as follows:

$$\begin{aligned} \pi = {\left\{ \begin{array}{ll} -\frac{19}{3}+50r-180r^2+360r^3-360r^4+\frac{279}{2}r^5, &{} \text{for } \frac{1}{2}< r<\frac{2}{3} \\ 18r^5-90r^4+180r^3-180r^2+90r-17, &{} \text{for } \frac{2}{3}\le r\le 1. \end{array}\right. } \end{aligned}$$

Those functions are plotted on Fig. 3.

Appendix 2

Table 1 The exact probability that $F_w\subseteq M_{\nu}$ for $m=3$ and $n=50/51$

Table 2 The exact probability that $F_w\subseteq M_{\nu}$ for $m=3$ and $n=100/101$

Table 3 The exact probability that $F_w\subseteq M_{\nu}$ for $m=3$ and $n=1000/1001$

Table 4 The exact probability that $F_w\subseteq M_{\nu}$ for $m=3$ and $n\rightarrow \infty$ as a function of $r=\frac{\nu }{n}$

Table 5 The simulated probability that $F_w\subseteq M_{\nu}$ for $m=4$ and $n=50/51$

Table 6 The simulated probability that $F_w\subseteq M_{\nu}$ for $m=4$ and $n=100/101$

Table 7 The simulated probability that $F_w\subseteq M_{\nu}$ for $m=4$ and $n=1000/1001$

Table 8 The simulated probability that $F_w\subseteq M_{\nu}$ for $m=4$ and $n=10^6$ as a function of $r=\frac{\nu }{n}$

Table 9 The simulated probability that $F_w\subseteq M$ for $m=3$ and $m=4$