The likelihood of a Condorcet winner in the logrolling setting

Abstract

The purpose of this note is to compute the probability of logrolling for three different probabilistic cultures. The primary finding is that the restriction of preferences to be in accord with the condition of separable preferences creates enough additional structure among voters’ preference rankings to create an increase in the likelihood that a Condorcet winner will exist with both IC and IAC-based scenarios. However this increase is limited and the probabilities remain close to the conventional probabilities of having a Condorcet winner when preferences are unrestricted.

Appendix: moving from uniform to Dirichlet: how to deal with the curse of dimensionality

Let $S_{m}$ be the m-dimensional unit simplex i.e. the set of vectors p in ${{\mathbb {R}}}^{m}$ such that:

$$\begin{aligned} p_{i}\ge 0 \text{ for } \text{ all } i=1,2,\ldots ,m \text{ and } \sum _{i=1}^{m}p_{i}=1 \end{aligned}$$

The uniform distribution on $S_{m}$ is a special case of a Dirichlet distribution. The Dirichlet distribution of order m with parameters $\alpha _{1},\ldots ,\alpha _{m}>0$ has a probability density function with respect to Lebesgue measure defined by :

$$\begin{aligned} f(x;\alpha )=\frac{1}{B(\alpha )}\prod \limits _{k=1}^{m}x_{k}^{\alpha _{k}-1} \ \text{ where } \alpha \equiv \left( \alpha _{1},\ldots ,\alpha _{m}\right) \end{aligned}$$

with:¹³

$$\begin{aligned} B(\alpha )=\frac{\prod \nolimits _{k=1}^{m}\Gamma \left( \alpha _{k}\right) }{ \Gamma \left( \sum _{k=1}^{m}\alpha _{k}\right) } \end{aligned}$$

When $\alpha \equiv \left( 1,\ldots ,1\right) ,$ we obtain the uniform distribution. We know a lot of things about this parametric family of distributions. For our purpose, we only need the property called aggregation property. It asserts that if p follows a Dirichlet distribution of order m with parameters $\alpha$, then if we sum the coordinates i and j while leaving the others the same, the new vector $\left( p_{1},\ldots ,p_{i}+p_{j},\ldots ,p_{m}\right)$ follows a Dirichlet distribution of order $m-1$ with parameters $\left( \alpha _{1},\ldots ,\alpha _{i}+\alpha _{j},\ldots ,\alpha _{m}\right)$.

This result is used with the observation that a number of common subsets of the 96 $p_i$ terms appear in the seven constraints that are required to have (1, 1, 1) as the Condorcet winner. These subsets consist of groups of 2, 4 and 12 $p_i$ terms that are combined to create 32 variables denoted as $q_i$ for $1 \le i \le 32$.

$$\begin{aligned} q_{1}= & {} p_{1}+p_{2}+p_{3}+p_{4}+p_{5}+p_{6}+p_{7}+p_{8}+p_{9}+p_{10}+p_{11}+p_{12},\\ q_{2}= & {} p_{19}+p_{20}, \ \ q_{3}=p_{15}+p_{16}, \ \ q_{4}=p_{21}+p_{23}, \ \ q_{5}=p_{22}+p_{24},\\ q_{6}= & {} p_{13}+p_{14}+p_{17}+p_{18}, \ \ q_{7}=p_{25}+p_{26}, \ \ q_{8}=p_{35}+p_{36},\\ q_{9}= & {} p_{29}+p_{31}, \ \ q_{10}=p_{30}+p_{32}, \ \ q_{11}=p_{27}+p_{28}+p_{33}+p_{34},\\ q_{12}= & {} p_{37}+p_{39}, \ \ q_{13}=p_{38}+p_{40}, \ \ q_{14}=p_{45}+p_{46}, \ \ q_{15}=p_{41}+p_{42},\\ q_{16}= & {} p_{43}+p_{44}+p_{47}+p_{48}, \ \ q_{17}=p_{49}+p_{51}, \ \ q_{18}=p_{50}+p_{52},\\ q_{19}= & {} p_{53}+p_{54}, \ \ q_{20}=p_{57}+p_{58}, \ \ q_{21}=p_{55}+p_{56}+p_{59}+p_{60},\\ q_{22}= & {} p_{61}+p_{62}, \ \ q_{23}=p_{65}+p_{67}, \ \ q_{24}=p_{66}+p_{68}, \ \ q_{25}=p_{71}+p_{72},\\ q_{26}= & {} p_{63}+p_{64}+p_{69}+p_{70}, \ \ q_{27}=p_{75}+p_{76}, \ \ q_{28}=p_{79}+p_{80},\\ q_{29}= & {} p_{81}+p_{83}, \ \ q_{30}=p_{82}+p_{84}, \ \ q_{31}=p_{73}+p_{74}+p_{77}+p_{78},\\ q_{32}= & {} p_{85}+p_{86}+p_{87}+p_{88}+p_{89}+p_{90}+p_{91}+p_{92}+p_{93}+p_{94}+p_{95}+p_{96}. \end{aligned}$$

From the aggregation property, since the vector p follows a Dirichlet distribution of order 96 with parameter $\left( 1,1,1,\ldots ,1\right)$, we deduce that the vector q follows a Dirichlet distribution of order 32 with parameters:

$$\begin{aligned} \alpha&=(12,2,2,2,2,4,2,2,2,2,4,2,2,2,2,4,2,2,2,2,\\&\quad 4,2,2,2,2,4,2,2,2,2,4,12) \end{aligned}$$

So, we have moved from 96 dimensions to 32 dimensions but at the cost of moving from a simple Dirichlet (the uniform) to a more sophisticated one. We rewrite the seven inequalities to make (1, 1, 1) the Condorcet winner in terms of the 32 variables as:

$$\begin{aligned}&q_{1} +q_{7}+q_{8}+q_{9}+q_{10}+q_{11} +q_{12}+q_{13}+q_{14}\\&\quad +q_{15}+q_{16} +q_{27}+q_{28}+q_{29}+q_{30}+q_{31}> \frac{1}{2};\\&q_{1} +q_{2}+q_{3}+q_{4}+q_{5}+q_{6} +q_{12}+q_{13}+q_{14}\\&\quad +q_{15}+q_{16} +q_{22}+q_{23}+q_{24}+q_{25}+q_{26}> \frac{1}{2};\\&q_{1}+q_{2}+q_{3}+q_{4}+q_{5}+q_{6} +q_{7}+q_{8}+q_{9}\\&\quad +q_{10}+q_{11} +q_{17}+q_{18}+q_{19}+q_{20}+q_{21}> \frac{1}{2};\\&q_{1} +q_{2}+q_{6} +q_{8}+q_{11} +q_{12}+q_{13}+q_{14}+q_{15}\\&\quad +q_{16}+q_{22}+q_{23}+q_{24} +q_{27}+q_{29}+q_{30}> \frac{1}{2};\\&q_{1} +q_{3}+q_{6} +q_{7}+q_{8}+q_{9}+q_{10}+q_{11} +q_{14}\\&\quad +q_{16} +q_{17}+q_{18}+q_{19} +q_{28}+q_{29}+q_{30}> \frac{1}{2};\\&q_{1} +q_{2}+q_{3}+q_{4}+q_{5}+q_{6} +q_{7}+q_{11}+q_{15}\\&\quad +q_{16} +q_{17}+q_{18}+q_{20}+q_{23}+q_{24}+q_{25}> \frac{1}{2};\\&q_{1} +q_{2}+q_{3}+q_{5}+q_{6}+q_{7}+q_{8}+q_{10}+q_{11}\\&\quad +q_{13}+q_{14}+q_{15}+q_{16}+q_{17} +q_{23}+q_{29} > \frac{1}{2}. \end{aligned}$$

Let $H^{D}$ denote the polytope associated to these inequalities.

The probability of being in $H^{D}$ when q is drawn over $S_{32}$ according to the Dirichlet (12, 2, 2, 2, 2, 4, 2, 2, 2, 2, 4, 2, 2, 2, 2, 4, 2, 2, 2, 2, 4, 2, 2, 2, 2, 4, 2, 2, 2, 2, 4, 12) is the same as the probability of being in H when p is drawn uniformly over $S_{96}$.