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% Shirley Jowell, of Morgan Kaufmann Publishers, and Peter F.
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\title{How Likely Are Two Voting Rules Different?
%Likelihood of Incompatibility: Positional and Edge-Order-Based Rules
}


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\begin{document}

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\maketitle

\begin{abstract}
    We characterize the maximum likelihood that two voting rule outcomes are different and that the winner of one voting rule is the loser of another (implying that they are {\em drastically different}) on positional scoring rules, Condorcet, Copeland, and ranked pairs, under any fixed number of alternatives. The most famous problem in this scope is strong Borda's paradox, in which the winner of the plurality rule is the Condorcet loser. Under mild assumptions, we show that the maximum likelihood that different rules are drastically different is $\Theta(1)$ except for a few special cases, demonstrating the difference between these rules. We also prove that two scoring rules with linear independent scoring vectors have different winners with probability $\Theta(1)$, no matter how similar they are. Our analysis adopts the {\em smoothed social choice framework}~\citep{xia2020smoothed} and can be applied to a variety of statistical models including the standard impartial culture (IC).
\end{abstract}

\section{Introduction}
% I update my version of intro here. 
{\input{intro_new}}

% The old version from Ziqi is at below. 

% Variety of electoral systems are used all over the world to elect a single winner to a unique position, such as prime minister, president etc., or multiple winners, such as members of a football team or the parliment. Many voting rules have been designed in order to make the election efficient, fair and desirable. However, the outcome of voting rules can be strange and counterintuitive. These events are called voting paradoxes, and they have been categorized to several types in Nurmi's book \cite{nurmi1999voting}. The founder fathers of this theory are Borda and Condorcet, who presented first formally mathematically analyses of such voting paradoxes. We begin with an example of Borda's Paradox. 

% \begin{example}[Borda's Paradox]\label{exmp:borda}
%     We consider a 3-alternatives election and let $\{A,B,C\}$ be the set of alternatives, where each voter has some complete preference ranking on the alternatives.
    
%     \begin{table}[htbp]
%     \centering
% \begin{tabular}{l|llll}
% \hline
%     & 1 voters & 7 voters & 7 voters & 6 voters \\ \hline
% 1st & A        & A        & B        & C        \\
% 2nd & B        & C        & C        & B        \\
% 3rd & C        & B        & A        & A        \\ \hline
% \end{tabular}
% \end{table}

% In the plurality rule, only top-ranked alternative is counted in each vote. The plurality winner is the alternative with maximum votes which think it is top-ranked. Hence, $A$ is the plurality winner. However, we observes that the majority, which consists of $13$ voters , ranked $B$ higher than $A$. The similar case holds for $C$. From this angle, $A$ is the worst choice, since the majority believes both $B$ and $C$ is a better choice than $A$.

% \end{example}

% If we consider two voting rules \emph{Plurality} and \emph{Pairwise Comparison}, 
% example \ref{exmp:borda} illustrates the fact that the winner of plurality rule can be the loser of the pairwise comparison. Generally speaking, for any two different rules, there are possibility that the best choice of one rule can be the worst choice of the other. On the one hand, it is inevitable that different voting rules generates different results, otherwise these two rules are essentially the same. On the other hand, it is a controversial topic, since there were criticism on the result of an election (example of criticism on US election). 
% Also, \cite{diss2012borda} summerized the studies had been conducted to determine if either form of Borda's paradox has ever been observed in actual elections and the answer is `Yes'. Due to the mathematical impossibility of avoiding these events, we concerns about the following probiem: 
% \[
% \textbf{How Likely are Paradoxes occur under realistic model?}
% \]

% There are a number of theoretical studies \cite{tataru1997relationship,cervone2005scoring,gehrlein2010probability,diss2012borda}
% calculating the likelihood of several kinds of Borda Paradox in a 3-alternatives election under some IC or IAC \cite{eugeciouglu2013impartial}. However, the assumption of IC has been widely criticized of being unrealistic \cite{nurmi1999voting}. Also, it is hard to generalize their work, i.e. representing the likelihood when considering an $m$-alternatives election system.



\subsection{Our Contributions} 

We analyze the asymptotic max-likelihood that the outcomes between voting rules are different (named {\em different winner}, \EI{}) and that the winner under one voting rule is the loser under another rule (named {\em drastically different}, \DI{}), including the strong Borda's paradox. Our analysis across positional scoring rules, Copeland, ranked pairs, and the Condorcet winner/loser. We adopt the smoothed social choice framework in~\citet{xia2020smoothed}, in which the votes are randomly generated from a distribution chosen by an adversary from a given set. 

\begin{table}[]
\centering
\begin{tabular}{@{}llll@{}}
\toprule
Rule 1 & Rule 2 & Likelihood &  \\ \midrule
Borda & \multirow{2}{*}{Condorcet} & 0 & \multirow{2}{*}{\begin{tabular}[c]{@{}l@{}}Thm.~\ref{thm:SgBorda}\\ Coro.~\ref{coro:reverse1}\end{tabular}} \\ \cmidrule(r){1-1}
Other PSR &  & $\Theta (1)$ &  \\ \midrule
Any PSR & Copeland & $\Theta (1)$ & \begin{tabular}[c]{@{}l@{}}Thm.~\ref{thm:CdandPosi}\\ Coro.~\ref{coro:reverse_Cpd}\end{tabular} \\ \midrule
\multirow{2}{*}{\begin{tabular}[c]{@{}l@{}}Borda,\\ $m = 3$\end{tabular}} & \multirow{3}{*}{Ranked pairs} & 0, $n$ is odd & \multirow{3}{*}{\begin{tabular}[c]{@{}l@{}}Thm.~\ref{thm:psr_rp}\\ Coro.~\ref{coro:reverse_rp}\end{tabular}} \\
 &  & $\Theta (n^{-\frac32})$, n is even &  \\ \cmidrule(r){1-1}
Otherwise &  & $\Theta (1)$ &  \\ \midrule
\multicolumn{2}{l}{Linear independent PSR} & \EI{}: $\Theta (1)$ & Thm.\ref{thm:diffposi} \\ \bottomrule
\end{tabular}
\caption{Our results. All but the last row is for the drastically different problem, while the last row is for the different winner problem. PSR is the abbreviation of the positional scoring rule, and $m$ is the number of alternatives. \label{tbl:result}}
\end{table}

Our main results are two-fold. We show that, under natural assumptions, there is $\Theta (1)$ probability for two scoring rules whose scoring vectors are linearly independent to have different outcomes, no matter how similar they are (Theorem~\ref{thm:diffposi}).
We also characterize the likelihood of the drastically different problem between a wide range of voting rules, as shown in Table~\ref{tbl:result}. 
Our results demonstrate that, under natural assumptions, these voting rules are indeed significantly different from each other except for a few very special cases. Specifically, Theorem~\ref{thm:SgBorda} shows that strong Borda's paradox occurs with $\Theta(1)$ probability in positional scoring rules except for Borda's rule. 

% For drastically incompatible, we consider the generalized version of strong Borda's paradox and show that the likelihood of a winner in any integer positional scoring rule (except for Borda's rule) being a Condorcet loser is $\Theta (1)$, while a Borda's rule winner can never be a Condorcet loser. On the other hand, all positional scoring rules, including Borda's rule, have a likelihood $\Theta (1)$ to be drastically incompatible with Coepland. For essentially incompatible, we show that for any two integer positional scoring rules whose scoring vectors are linearly independent, the likelihood that their outcomes are different is $\Theta(1)$. \qishen{Should we add a Table?}



Our work extends previous studies in three aspects. Firstly, our likelihood results cover a wider range of voting rules. Secondly, the semi-random analysis provides a more general and practical characterization than the standard IC. With the semi-random framework, our results can be applied for analysis of the likelihood under a variety of statistical models including IC. Finally, our results hold for any fixed number of alternatives, while most previous work focused on three or four alternatives. 

 % Therefore, we address the open questions in~\citet{gehrlein2004probability} and ~\citet{WILSON200probability} of exploring more realistic models on the likelihood of voting paradoxes.

% Technically, we convert voting profiles into vectors and \DI{}/\EI{} events into polyhedra, and use the geometric properties of the polyhedra to bound the likelihood that the vectors are located in them. This allows our results to hold for any fixed number of alternatives. 

% We utilized the \emph{smoothed social choice} framework \cite{xia2020smoothed}, which is more realistic and general than common i.i.d. distrubutions, such as IC and IAC. We considered a broader range of voting rules under an arbitrarily fixed number of alternatives system ,and their mutually incompatibility, i.e. the likelihood of different rules generates different winners under realistic models. 

% We first consider the incompatibility of positional rules, and we have
% \begin{theorem}[Incompatibility of positional rules, Informal]
%     For fixed $m\geq 3$ and any pair of essentially different integer positional rules,
%     the probability of their winners are different is $\Theta(1)$ under IC.
% \end{theorem}

% We also consider the incompatibility between positional rules and several Graph-order-based rules, such as Copeland and Maxmin. 


\subsection{Related Works and Discussions}

There is a large literature characterizing the likelihood of Borda's paradox. Under settings of three alternatives,  \citet{tataru1997relationship} and~\citet{cervone2005scoring} characterize the likelihood of strong Borda's paradox under IC and IAC respectively, and \citet{diss2012borda} study the likelihood of the strict Borda's Paradox under IC and IAC. \citet{gehrlein2010probability} examine the impact that degrees of mutual coherence among voters' preferences will have on the likelihood of the paradox.
\citet{diss2018another} characterize conditions for a profile to show Borda's paradox or not in weighted scoring rules. There are also many empirical studies of Borda's paradox in real-world elections, including the 2016 Republican presidential primaries~\citep{kk2018trump} and the 2017 Iran 
presidential election~\citep{feizi2020iran}. The reader is referred to the work by~\citet{diss2012borda} for additional empirical literature. 

Studies have also been conducted on paradoxes that are closely related to Borda's paradox.  
\citet{diss2018note} study the likelihood of a similar class of paradoxes named {\em absolute majority winner/loser paradox}, in which an alternative ranked top by more than half of the voters is not elected as the winner. 
\citet{Diss2015true} study the likelihood that a Condorcet winner is elected by a scoring rule (named {\em Condorcet efficiency}) under a modified version of IAC. 
\citet{Brandt2022minimal} leverage ILP techniques to identify the minimum number of voters and alternatives for multiple voting paradoxes (including Borda's paradoxes and absolute majority winner/loser paradox) to occur under a wide range of voting rules. 
\citet{bruns2019computations} identify the likelihood of strong Borda's paradox and Condorcet efficiency of plurality with runoff under a four-alternative setting and IAC.

Smoothed analysis~\citep{spielman2004simplex,spielman2009smoothed} refers to a type of worst-average analysis in which an adversary (the ``worst'' component) first selects an instance, and then random noises (the ``average'' component) are added. Smoothed analysis has been widely applied to analyze the practical performance of algorithms, circumventing the hard worst-case. For example, \citet{spielman2004simplex} use smoothed complexity analysis to explain why simplex algorithms take polynomial time most of the time despite their hardness in the worst case. Readers are referred to~\cite{spielman2009smoothed} for more applications. 

Recent studies introduce smoothed analysis into computational social choice~\citep{baumeister2020towards,xia2020smoothed} to study the likelihood of axiom satisfactions~\citep{xia2021semi,xia2023semi}, manipulations and paradoxes~\citep{xia2020smoothed,liu2022semi,xia2023impact}, and ties~\citep{xia2021likely}. More specifically, \citet{baumeister2020towards} first proposes the blue-sky idea of incorporating smoothed analysis into social choice, and \citet{xia2020smoothed} first implements the idea by proposing the formal smoothed social choice framework and characterizing the smoothed likelihood of famous Condorcet's paradox and ANR impossibility. \citet{xia2021kemeny,xia2022beyond} study the smoothed complexity of the winner determination for the NP-hard Kemeny, Dogson, and Young rule.

% Semi-random analysis~\citep{blum1990tools,blum1995color} is an analysis technique in which the process of generating instances contains both average components and adversary components. This technique has been widely applied to characterize 
% \noindent\textbf{Likelihood of Paradoxes under IC or IAC.}
% A number of empirical studies have been conducted to represent the likelihood of such paradoxes occurring. Tataru et al\cite{tataru1997relationship} gives a representation for the conditional probability of a Strong Borda Paradox observed in an 3-alternatives election under \emph{Impartial Culture (IC)}. Here IC assumes for $k$ alternatives each of the $k!$ rankings has the same probability $1/k!$ of being any voter's preference ranking. Similarly, the \emph{Impartial Anonymous Culture (IAC)} \cite{eugeciouglu2013impartial} assumes a uniform distrubution on each pattern of preference profile. Cervone et al \cite{cervone2005scoring} calculated the same event under the assumption of IAC. Gehrlein et al \cite{gehrlein2010probability} examined the impact that degrees of mutual coherence among voters' preferences will have on the likelihood of this paradox being observed. Monstapha et al \cite{diss2012borda} represented the likelihood of Strict Borda Paradox under IC and IAC. 

% \noindent\textbf{Smoothed analysis}







\section{Preliminary}
%\subsection{Basic Setting}

For any integer $q\in\mathbb{N}$, let $[q]=\{1,2,\cdots,q\}$. 
The set of alternatives is defined as $\mathcal{A}:=[m]$, and the set of all linear orders over $\mathcal{A}$ is defined as $\mathcal{L}(\mathcal{A})$. Let $n$ be the number of agents. Each agent uses a linear order $R\in\mathcal{L}(\mathcal{A})$ to represent his/her preference, called a \emph{vote}. The vector of $n$ agents' votes is a \emph{(preference) profile}, denoted by $P$. We also say $n$-profile, where $n$ implies the number of voters. The set of $n$-profiles for all $n\in \mathbb{N}$ is denoted by $\mathcal{L}(\mathcal{A})^*:=\cup_{n=1}^\infty\mathcal{L}(\mathcal{A})^n$. 

An \emph{(irresolute) voting rule} $r:\mathcal{L}(\mathcal{A})^*\to(2^\mathcal{A}\setminus \{\emptyset\})$ maps every profile to its corresponding set of winners in $\mathcal{A}$. 
% We say a voting rule is \emph{resolute} if it always chooses a unique winner. 
% A \emph{generalized voting rule} $r^*:\mathcal{L}(\mathcal{A})^*\to 2^\mathcal{A}$, where a winner does not exist for some profiles.
For a voting rule $r$, let $\rlos$ be the mapping from a profile to its corresponding set of losers (see the definition of each voting rule).
Let $\Hist(P)\in\mathbb{Z}^{m!}_{\geq 0}$ denote the \emph{histogram} of $P$, whose $R-$element is the multiplicity of linear order $R\in\mathcal{L}(\mathcal{A})$ occurs in profile $P$. 

An \emph{integer positional scoring rule} is characterized by an integer scoring vector $\vec{s}=(s_1,\cdots,s_m)\in\mathbb {Z}^m$ where $s_1\geq s_2\geq\cdots\geq s_m$ and $s_1>s_m=0$\ziqi{I add an restriction that $s_m=0$, for the convenience of ``linear independency''.}. Each alternative receives $s_i$ score from each vote if he/she is ranked at $i$-th position and the alternatives with the maximum overall score win. The loser is defined by the set of alternatives with the minimum score.  For example, \emph{plurality} is characterized by $(1,0,\cdots,0)$ and \emph{Borda's count} is characterized by $(m-1, m-2,\cdots,1,0)$. 

For a score vector $\vec s=(s_1,\cdots,s_m)$, its \emph{normalized score vector} is defined by 
\[\vec s':=\frac{1}{s_1-s_m}[(s_1,s_2,\cdots,s_m)-(s_m,s_m,\cdots,s_m)].
        \] 
It is easy to see $s'_1=1$ and $s'_m=0$. Since the score of each ranking is scaled and shifted simultaneously, the winner and the loser of the positional rule characterized by $\vec s'$ are the same as that of $\vec s$.

Given a profile $P$, let $P[a\succ b]$ be the number of votes in $P$ which rank $a$ higher than $b$. The \emph{weighted majority graph} of $P$ is a weighted directed graph, where each vertex represents an alternative, and the weight of an edge $a\to b$ is $w_P(a,b)=P[a\succ b]-P[b\succ a]$. The {\em unweighted majority graph} $\UMG(P)$ is the unweighted directed graph obtained from  $\WMG(P)$ by keeping the edges with strictly positive weights.  We say a voting rule $r$ is \emph{weighted-majority-graph-based (WMG-based)}, if $r(P_1)=r(P_2)$ whenever $WMG(P_1)=WMG(P_2)$. 
\begin{itemize}
    \item \textbf{Condorcet winner/loser.} Given profile $P$, an alternative $a$ is a Condorcet winner if $w_P(a,b)>0$ for any $b\neq a$. In other words, $a$ wins all pairwise comparisons. Similarly, $a$ is a Condorcet loser if and only if $w_P(a,b)<0$ for any alternatives $b\neq a$. We will use $r_C(P)$ and $\rlos_{C}(P)$ to denote the set of Condorcet winners and losers in profile $P$ when they exist, respectively. 
    \item \textbf{Copeland.} 
    The Copeland ($\mathrm{Cd}_\alpha$) rule with parameter $0\leq\alpha\leq 1$, assigns $1$ point to alternative $a$ every time he/she wins a pairwise comparison and $\alpha$ points for both alternatives in each tie. Then the winners of $\mathrm{Copeland}_\alpha$ are the set of all alternatives with the highest total score. The losers are the alternatives with the lowest total score.
    \item \textbf{Ranked Pairs.} The ranked pairs rule ($\RP$) sorts every pair of alternatives $(a,b)$ according to $w_P(a, b)$ in the descending order. Then it creates an acyclic graph $G''$ by adding each pair as an edge (from the winner to the loser in the pairwise comparison) in the sorted order unless it creates a cycle. Then the winner is the source node in $G'$ and the loser is the sink in $G'$.
    %\item \textbf{Maxmin.} The Maximin rule chooses all alternatives $a$ with the highest $\min_{b\in\mathcal{A}}w_P(a,b)$, i.e. the lowest weight among all the outgoing edges of the alternative in the WMG. Intuitively, the Maximin rule selects the alternatives that are beaten by other alternatives at the lightest extent in the pairwise comparison. 
\end{itemize}

\subsection{Semi-random likelihood of different winners} 

In this paper, we consider two levels of difference between voting rules. We say two voting rules have {\em different winners} (\EI{}) if their winners are different and {\em drastically different} (\DI{}) if a winner of the first voting rule is a loser of the second rule.
\begin{definition}
    Two voting rules $r_1$ and $r_2$ have different winners on profile $P$, denoted by $\EI{}(r_1, r_2, P) = 1$, if $r_1(P) \neq r_2(P)$. $r_1$ and $r_2$ are drastically different, denoted by $\DI{}(r_1, r_2, P) = 1$,  on $P$ if $ r_1(P)\cap \rlos_2(P) \neq \emptyset$. 
\end{definition}
Note that $\DI{}$ is not symmetric since we care about $r_1$'s winner and $r_2$'s loser. When $r_1$ is plurality and $r_2$ is Condorcet, \DI{} is exactly strong Borda's paradox. 

We now recall the statistic model applied in the semi-random analysis model. A {\em single-agent preference model}~\citep{xia2020smoothed} $\mathcal{M} = (\Theta, \mathcal{L}(\mathcal{A}), \Pi)$ is characterized by a parameter space $\Theta$, the sample space $\mathcal{L}(\mathcal{A})$, and a set of distributions $\Pi$ parameterized by $\Theta$. The distribution $\Pi$ is closed if it's a closed set of $\mathbb{R}^{m!}$ and is strictly positive if there exists a constant $\varepsilon > 0$ such that the probability of any linear order in $\mathcal{L}(\mathcal{A})$ under any distribution in $\Pi$ is at least $\varepsilon$. 
In this paper, we assume that $\Pi$ is closed and strictly positive. 

\begin{example}
\label{ex:mallows}
Here we take a single-agent Mallows' model as an example~\citep[Example 2 in the appendix]{xia2020smoothed}.
For any $0\leq\underline{\varphi}<\overline{\varphi}\leq 1$, we let $\mathcal{M}_{Ma}^{[\underline{\varphi},\overline{\varphi}]}$ denote the Mallows model whose parameter space  $\Theta = \mathcal{L}(\mathcal{A})\times[\underline{\varphi},\overline{\varphi}]$. For each $(R, \varphi) \in \Theta$, $R$ is the central ranking, and $\varphi$ is the parameter of dispersion. For distribution $\pi_{(R, \varphi)}$, any ranking $R'$ is generated with probability $\varphi^{KT(R,R')}/Z_\varphi$, where $KT(R,R')$ is the Kendall Tau distance between $R$ and $R'$, defined by the number of pairwise disagreements between $R$ and $R'$, and $Z_\varphi=\sum_{R'\in\mathcal{L}(A)}\varphi^{KT(R,R')}$ is the normalization constant. The distribution set $\Pi$ in single-agent Mallows' model is closed and strictly positive.
\end{example}

\begin{definition}[Max semi-random likelihood of different winner and drastically different problem]

Given a single-agent preference model $\mathcal{M} = (\Theta, \mathcal{L}(\mathcal{A}), \Pi)$, and two voting rules $r_1,r_2$, the max-adversary aims to maximize the likelihood for \EI{} (\DI{}, respectively) by choosing distribution $\pi \in \Pi$ for each agent, whose votes are generated independent. Let $\vec \pi$ be the vector of distributions for all agents.  The max semi-random likelihood of \EI{} and \DI{} are defined as follows.
\begin{align*}
\widetilde{\EI}^{\max}_{r_1,r_2}:=&\ \sup_{\vec{\pi}\in\Pi^n}\mathrm{Pr}_{P\sim \vec \pi}(\EI(r_1,r_2,P)=1),\\
\widetilde{\DI}^{\max}_{r_1,r_2}:=&\ \sup_{\vec{\pi}\in\Pi^n}\mathrm{Pr}_{P\sim \vec \pi}(\DI(r_1,r_2,P)=1).
    \end{align*}
\end{definition}


% Let $\Pi$ be a set of distributions over $\mathcal{L}(\mathcal{A})$, and $r_1,r_2$ are two voting rules. 

% We say two voting rules are \emph{essentially incompatible} under profile $P$ if 
% $r_1(P)\neq r_2(P)$.
% Given profile $P$, we say the ordered pair $(r_1,r_2)$ is \emph{drastically incompatible}, if
% there exists an alternative $a$ such that 
% $a\in r_1(P)\cap \rlos_2(P)$. 
% Then, a \emph{distribution adversary} tries to maximize the probability of such incompatibility to be $1$ by choosing a distribution for each agent, whose vote is generated independently. To be exact, we are interested in 
% \[\widetilde{\EI}^{\max}_{r_1,r_2}:=\sup_{\vec{\pi}\in\Pi^n}\mathrm{Pr}_{P\sim \vec \pi}(\EI(r_1,r_2,P)=1),
% \]
% and 
% \[\widetilde{\DI}^{\max}_{r_1,r_2}:=\sup_{\vec{\pi}\in\Pi^n}\mathrm{Pr}_{P\sim \vec \pi}(\DI(r_1,r_2,P)=1),
% \]
% where $\EI(r_1,r_2,P)=1$ if and only if $r_1$ and $r_2$ are essentially incompatible under profile $P$, and the similar one holds for $\DI$.




\section{PMV-in-Polyhedron Problem}

We adopt the techniques in~\citet{xia2021likely} to characterize the likelihood of different winners. 

% Here PMV is the abbreviation of Poission multivariate variables and we first formally states its definition.

\begin{definition}[Poisson multivariate variables (PMVs)~\citep{xia2021likely}]
    Given any $q,n\in\mathbb{N}$ and any vector $\vec \pi=(\pi_1,\cdots,\pi_n)$ of $n$ distributions over $[q]$, a $(n,q)$-PMV, denoted by $\vec{X}_{\vec \pi}$, is the histogram of $n$ independent random variables $Y_1,Y_2,\cdots,Y_n$, of which $Y_i$ follows distribution $\pi_i$.  
    
    % hat corresponds to $\vec \pi$. To be exact, let $\vec \pi=(\pi_1,\cdots,\pi_n)$ and $Y_1,Y_2,\cdots,Y_n$ be n independent random variables over $[q]$, where $Y_j$ is distributed as $\pi_j$ for $j\in[n]$. Then $\Xpi$ is a $q$-dim vector whose $i$-th component is the multiplicity of $Y_j$ take value $i$.
\end{definition}

Intuitively, $\vec \pi$ is the distribution of $n$ votes, and the PMV $\Xpi$ is the random vector that counts each linear order $R\in\mathcal{L}(\mathcal{A})$. Sampling the $\vec\pi$ generates a profile $P$, and $\Hist(P)$ is the corresponding sample of $\Xpi$. 

A {\em polyhedron} $\mathcal{H}\subseteq\mathbb{R}^q$ can be characterized by a matrix $A$ and a vector $\vec b$, i.e. $\mathcal{H}:=\{\vec x\in\mathbb{R}^q:A\vec x\leq \vec b\}$. 
The characteristic cone of $\mathcal{H}$ is $\mathcal{H}_{\leq 0}:=\{\vec x\in\mathbb{R}^q:A\vec x\leq \vec 0\}$. The dimension of polyhedron $\mathcal{H}$, denoted by $\dim(\mathcal{H})$, is the dimension of the smallest affine subspace of $\mathcal{H}$. A polyhedron $\mathcal{H}\subseteq \mathbb{R}^q$ is full dimensional if $\dim(\mathcal{H})=q$.

\begin{definition}[PMV in polyhedron problem~\citep{xia2021likely}]
    Given a $q\in \mathbb{N}$, a polyhderon $\mathcal{H}\subseteq\mathbb{R}^q$, and a set $\Pi$ of distributions over $[q]$, we problem we interested in is $\sup_{\pi \in \Pi^n} \Pr[\vec{X}_{\vec \pi} \in \mathcal{H}]$. 
\end{definition}

\paragraph{Methodology} For each different winner and drastically different problem, we represent the event by a polyhedron $\mathcal{H}$ such that the event occurs in a profile $P$ if and only if $\Hist(P)\in\mathcal{H}$. This converts the \EI{} or \DI{} problem into a PMV-in-polyhedron problem. Then we apply the main technical theorem in~\citet{xia2021likely} to characterize the max semi-random likelihood of the problem. 

Let $\mathcal{H}_n := \{\vec x \in \mathcal{H} \cap \mathbb{R}^{q}_{\ge 0} : \vec x \cdot \vec 1 = n\}$, and $\mathcal{H}_n^{\mathbb{Z}} :=\mathcal{H}_n \cap \mathbb{Z}^{q}$. And let $CH(\Pi)$ be the convex hull of set $\Pi$. The theorem is as follows.


% Given $q\in\mathbb{N}$, let $\Pi$ be the set of distrubutions on $[q]$. Each $\pi\in\Pi$ is a $q$-dim vector in $\mathbb {R}^q$. We say $\Pi$ is strict positive by $\varepsilon>0$, if for all $i\in[q]$, $\pi(i)\geq \varepsilon$. We say $\Pi$ is closed if it is closed as a subset of $\mathbb{R}^q$. We use $CH(\Pi)$ to denote the convex hull of $\Pi\subseteq \mathbb{R}^q$.





\begin{theorem}[Smooth Likelihood of PMV-in-polyhedron~\citep{xia2021likely}]\label{thm:pmv1}
    Given any $q\in N$, any closed and strictly positive $\Pi$ over $[q]$, and any polyhedron $\mathcal{H}$ characterized by an integer matrix $A$, for any $n\in \mathbb{N}$, $\sup_{\vec{\pi}\in \Pi^n}\mathrm{Pr}(\vec{X}_{\vec{\pi}}\in \mathcal{H}) =$
    
    \[\begin{cases}
        0 & \ if\mathcal{H}_n^\mathbb{Z}=\emptyset \\
        \exp(-\Theta(n)) & \begin{aligned}
             if\mathcal{H}_n^\mathbb{Z}\neq\emptyset\ and\ \mathcal{H}_{\leq 0}\cap CH(\Pi)=\emptyset
        \end{aligned} \\
        \Theta\left(\sqrt{n}^{\dim(\mathcal{H}_{\leq 0})-q}\right) & otherwise
    \end{cases}
        \]

\end{theorem}
\section{Results}

\subsection{Strong Borda's Paradox}

% Here we give a formal definition of the Strong Borda Paradox when the number of alternatives is $m$. Given profile $P$, we rank all alternatives by their plurality score. A \emph{Strong Borda Paradox} happens when a Condorcet loser ranked the first place of the plurality score. 

% Roughly speaking, Borda's Paradox is the incompatibility between \emph{plurality rule} and the \emph{Condorcet rule}. Notice that the Condorcet rule is a generalized voting rule, since its winner or loser does not always exist.
% We first consider the the likelihood of general \emph{posituional rule} and the \emph{Condorcet rule} results in different winners, and the Borda's Paradox is one of the special cases, since \emph{plurality rule} is the scroing rule characterized by $(1,0,\cdots,0)$.

% Let $r_{\vec s}$ denotes the positional rule characterized by $\vec s$.
%  We use $r_C$ to denote the \emph{Condorcet rule}, which selects a Condorcet winner when exists.

Our first result characterizes the likelihood that a winner of a positional scoring rule is a Condorcet loser, which is a generalized case of strong Borda's paradox. Let $r_{\vec s}$ denote the positional rule characterized by $\vec s$, and $\pi_{uni}$ is the uniform distribution on $\mathcal{L}(\mathcal{A})$.

\begin{theorem}\label{thm:SgBorda}
    Suppose the set of distributions $\Pi$ be strictly positive, closed and $\pi_{uni}\in CH(\Pi)$ then
    \[\widetilde{\DI}^{\max}_{r_{\vec s},r_C}(n)=\begin{cases}
        0 & \text{if}\ r_{\vec s}\ \text{is the Borda's rule} \\
        \Theta(1) & otherwise 
    \end{cases}
    \]
\end{theorem}

Theorem~\ref{thm:SgBorda} enables us the analyze the likelihood of (drastically) different winners under a wider range of statistical models, including IC and single-agent Mallows' (Example~\ref{ex:mallows}), bringing a deeper implication to the votings in practice. As shown in the following example, the theorem works even when $\Pi$ does not include uniform distribution once its convex hull includes it. 

\begin{example}\label{Mallows}
   
    We consider a single-agent Mallows' model $\mathcal{M}_{Ma}^{[\underline{\varphi},\overline{\varphi}]}$ mentioned in Example~\ref{ex:mallows}. 
    Note that the  $\pi_{uni}\notin \Pi$ when $\overline{\varphi} < 1$ because for any $\pi_{(R, \varphi)}$, the likelihood of generating any ranking $R'$ other than the central ranking $R$ is strictly lower than that of $R$ itself. However, for a fixed $\underline{\varphi}\le \varphi \le \overline{\varphi}$, let $\pi_0 = \frac{1}{m!} \sum_{R\in \mathcal{L}(\mathcal{A})} \pi_{(R, \varphi)}$ be the average of all $\pi_{(R, \varphi)}$ for all ranking $R$. Then due the symmetricity, $\pi_0 = \pi_{uni}$. Therefore, We have $\pi_{uni} \in CH(\Pi)$. And by applying Theorem~\ref{thm:SgBorda}, we show that under single-agent Mallows' model, and for any $r_{\vec s}$ not being a Borda's rule, the max semi-random likelihood that the winner of $r_{\vec s}$ is a Condorcet loser is $\Theta (1)$. 
   
\end{example}






\noindent\textbf{Proof Sketch of Theorem 
\ref{thm:SgBorda}.} 
The Borda's rule's case comes from the following theorem. 
\begin{theorem}[\citep{fishburn1976borda}]\label{thm:bordaisspecial}
        For every $m\geq 3$ and any scoring vector $\vec s$, there exists a profile $P$ such that a winner of $r_{\vec s}$ is a Condorcet loser if and only if $r_{\vec s}$ is NOT the Borda's rule.
    \end{theorem}

Hence, it suffices to show the case of other positional scoring rules. The following lemma serves as an outline of the proofs in this paper. In every proof, we first specify the event $X$ and the sub-event $Y$. Then we show that $Y$ satisfies the conditions in the lemma. Finally, we apply Theorem~\ref{thm:pmv1} and show that the likelihood of $Y$ (consequently $X$) is $\Theta (1)$. 

\begin{lemma}\label{lem:Theta1}
    Let $X$ be the event we are interested in and 
    \[\Xmax:=\sup_{\vec\pi \in\Pi^n}\mathrm{Pr}_{P\sim\vec \pi}(X\ \text{happens}),\]
    where $\Pi$ is closed and strictly positive. 
    We say an event $Y$ is a subset of $X$, if $Y\subseteq X$ in the sample space. If we can find a subset $Y\subseteq X$, such that 
    \begin{itemize}
        \item $Y$ can be characterized by a polyhedron $\mathcal{H}=\{A\vec x\leq \vec b\}$, where $A$ is an integer matrix and $\vec b$.
        \item $\dim(\mathcal{H}_{\le 0}) = m!$.
        \item $\mathcal{H}_{n}^{\mathbb{Z}} \neq \emptyset$ .
        \item $\mathcal{H}_{\le 0}\cap CH(\Pi)\neq\emptyset$.
    \end{itemize}
    Then we can apply Theorem~\ref{thm:pmv1} and show that $\Xmax=\Theta(1)$.
\end{lemma}

To prove Lemma~\ref{lem:Theta1}, it suffice to show $\Ymax=\Theta(1)$, since $Y\subseteq X$ implies $\Xmax(n)\geq\Ymax(n)$. The first condition of $Y$ allows us to convert $Y$'s likelihood into a PMV-in-polyhedron problem. The rest conditions guarantee this probability falls into the $\Theta\left(\sqrt{n}^{\dim(\mathcal{H}_{\leq 0})-q}\right)$ case. $\dim(\mathcal{H}_{\le 0}) = m!$ implies the probability is $\Theta (1)$.  The proof of Lemma~\ref{lem:Theta1} is in Appendix~\ref{apx:lemma}. 

\begin{proof}[Proof of Theorem \ref{thm:SgBorda} ] We give the proof when $r_{\vec s}$ is not Borda's rule. 
    % As it follows from the definition, $(r_{\vec s},r_C)$ are drastically different under profile $P$, if and only if there exists an alternative $a$ such that $a\in r_{\vec{s}}(P)\cap r_{C}^{-1}(P)$. Notice that the Condorcet loser is unique if exists, and the winner set of $r_{\vec s}$ is non-empty.
    % Here we omit the subscript $P\sim\vec \pi$ for the simplicity of notations, then
    % \[
    %     \mathrm{Pr}(\DI(r_{\vec s}, r_C,P)=1)=\sum_{a\in[m]}\mathrm{Pr}(a\in r_{\vec s}(P)\cap r_C^{-1}(P)) 
    % \]\qishen{Here say is sufficiently to show Theta 1 on a fixed a.}

    % The Borda count case is directly implied by the following theorem.
    % \begin{theorem}[\citep{fishburn1976borda}]\label{thm:bordaisspecial}
    %     For every $m\geq 3$ and any scoring vector $\vec s$, there exists a profile $P$ such that a winner of$r_{\vec s}$ is a Condorcet loser if and only if $\vec s$ is NOT the Borda count. 
    % \end{theorem}
    
    % Hence, $r_{\vec s}(P)\cap r_C^{-1}(P)$ is empty for any profile $P$ when $\vec s$ is a Borda count\ziqi{BORDA COUNT HERE!}, and clearly $\widetilde{\DI}^{\max}_{r_{\vec s},r_C}(n)=0$.

    % We then start the proof when $\vec {s}$ is not the Borda Count\ziqi{BORDA COUNT HERE!}.
    
    
    \noindent\textbf{Step 1:Define the Sub-Event and the Polyhedron $\mathcal{H}^{a}$.} 

    We define the sub-event $Y$ as follows. We first fix an arbitrary alternative $a$. Then $Y$ is the event that given profile $P$, $a$ is the unique winner in $r_{\vec s}$ and the Condorcet loser. Now we construct the polyhedron representing $Y$ via the following vectors. 
    
    \begin{definition}[Score difference vector~\citep{xia2021likely}]
        For any scoring vector $\vec s=(s_1,\cdots,s_m)$ and any pair of different alternatives $x,y$, let $Score_{x,y}^{\vec{s}} $ denote the $m!$-dimensional vector indexed by rankings 
    in $\mathcal{L}(\mathcal{A})$, the $R$-element of $Score_{x,y}^{\vec s}$ is $s_{j_1}-s_{j_2}$, where $j_1$ and $j_2$ are the ranks of $x$ and $y$ in $R$, respectively.
    \end{definition}
    \begin{definition}[Pairwise difference vector~\citep{xia2020smoothed}]
        For any pair of different alternatives $x,y$, $Pair_{x,y}$ denotes the $m!$-dimensional vector indexed by $R\in \mathcal{L}(\mathcal{A})$, whose $R$-element is $1$ if $x$ ranks higher than $y$ in $R$ and otherwise $-1$.
    \end{definition}
    
     %Let $x_{\mathcal{A}}=(x_R: R\in\mathcal{L}(A))$ denote the $m!-$dimensional vector where each component represents the multiplicity of a linear order $R$ in a profile. 
     For a profile $P$,  $Score_{x,y}^{\vec s}\cdot \Hist(P)$ is the score difference between $x$ and $y$ under $\vec s$, and $Pair_{x,y}\cdot \Hist(P)>0$ if and only if $x$ wins the pairwise comparison with $y$. 
     
     We define $\mathcal{H}^{a}$ as follows. 
     Let $A^{a,\vec s}$ be the matrix whose rows are $\{Score_{x,a}^{\vec s}:x\neq a, x\in[m]\}$. 
     Let $C^{a}$ be the matrix whose rows are $\{Pair_{a,x}:n\neq a, x\in[m]\}$. Let $A^a:=\begin{pmatrix}
         A^{a,\vec s} \\
         C^a
     \end{pmatrix}$ and $\vec b=-\vec 1$. 
     $\mathcal{H}^a$ is the polyhedron characterized by $A$ and $\vec b$.

    We then show that $\mathcal{H}^a$ is the polyhedron representing $Y$. First, $A^{a,\vec s}\cdot\Hist(P)\leq -\vec 1$ implies the score of $a$ strictly higher than that of any other alternative $x$. Hence, $a$ is the unique winner of $r_{\vec s}$. Then, $C^a\cdot\Hist(P)\leq -\vec 1$ implies $a$ is the Condorcet loser. Therefore, $\mathcal{H}^a$ is the polyhedron representing $Y$.
    
    \noindent\textbf{Step 2: Prove the conditions on $\mathcal{H}^a$.}
    %\begin{claim}\label{lem:proha}
    %The polyhedron $\mathcal{H}^a$ satisfies the following properties.
    %    \begin{enumerate}
    %        \item For any profile $P$, $\Hist(P)\in\mathcal{H}^a$ if and only if $a\in r_{\vec s}(P)\cap %\rlos_C(P)$, where $a$ is the unique winner in $r_{\vec s}$.
    %        \item $\dim(\mathcal{H}_{\leq 0})=m!$.
    %        \item $\mathcal{H}_n^{\mathbb {Z}}\neq \emptyset$ for sufficiently large $n$.
    %    \end{enumerate}
    %\end{claim}
    %\noindent\textbf{Proof of Claim \ref{lem:proha}.} It is easy to verify that $A^{a,\vec s}\cdot\Hist(P)\leq -\vec 1$ implies the score of $a$ strictly higher than that of any other alternative $x$. Hence, $a$ is the unique winner of $r_{\vec s}$. Also, $C^a\cdot\Hist(P)\leq -\vec 1$ implies $a$ is the Condorcet loser. Then 1 holds. 

    %We then show $\dim(\mathcal{H}^a_{\leq 0})=m!$, where $\mathcal{H}_{\leq 0}^a=\{A\vec x\leq \vec 0\}$. 
    %We consider the polyhedron $\Hastar:=\{A\vec x\leq -\vec 1\}$. Clearly $\Hastar\subseteq \mathcal{H}^a_{\leq 0}$.
    
    \begin{claim}
    For $\mathcal{H}^a$, the conditions in Lemma~\ref{lem:Theta1} hold: 
       (1)  $\mathcal{H}_{\le 0} \cap CH(\Pi)\neq \emptyset$,
       (2) $\dim (\mathcal{H}_{\le 0}) = (m!)$, and 
       (3)$\mathcal{H}_{n}^{\mathbb{Z}} \neq \emptyset$. 
    \end{claim}
    
    (1) is guaranteed by the assumption that $\pi_{uni} \in CH(\Pi)$, and that $\pi_{uni}\in\mathcal{H}_{\leq 0}$ since  $Score_{x,y}\cdot \vec \pi_{uni}=0$ and $Pair_{x,y}\cdot\vec\pi_{uni}=0$ for every pair of alternatives $x,y$. 

    %We then present a theorem, which showd that $\mathcal{H}^{\mathbb{Z}}_n$ is not empty for sufficiently large $n$ for a special class of full-dimensional polyhedra.

    (2) comes from applying the following lemma given that $\mathcal{H}^a$ is not empty is guaranteed by Theorem~\ref{thm:bordaisspecial}. 
    \begin{lemma}\label{lem:<-1}
    Any non-empty polyhedron $\mathcal{H}=\{\vec x\in\mathbb{R}^q:A\vec x\leq \vec b\}$ where $\vec b<\vec 0$ satisfies $\dim(\mathcal{H}_{\leq 0}) = q$.

\end{lemma}

For any $\vec x_0\in\mathcal{H}$ and any $\lambda > 1$,  $\vec b<\vec 0$ implies $A\cdot \lambda\vec{x}_0=\lambda A\vec {x}_0\leq \lambda \vec b< \vec 0$. Then we can show that $\lambda\vec{x}_0$ is an inner point of $\mathcal{H}_{\leq 0}$. The existence of an inner point implies $\dim(\mathcal{H}_{\leq 0}) = q$.  

Finally, (3) comes from applying another lemma as follows.

    \begin{lemma}\label{thm:intpoint_fulldim}
    Let $\mathcal{H}$ be the polyhedron characterized by $A,\vec b$, where $\vec b\leq \vec 0$.
    Suppose $\mathcal{H}$ is full-dimensional, then there exists $N\in\mathbb{N}$, such that $\mathcal{H}_n^{\mathbb{Z}}$
    is not empty for every $n>N$.
\end{lemma}

Since $\mathcal{H}$ is full-dimensional, we can find an open ball in $\mathcal{H}_{n}$ (the intersection of $\mathcal{H}$ and hyperplane $\vec x \cdot 1^{\top} = n$) for any $n$. Then according to the fact that $A\cdot \lambda\vec{x}_0=\lambda A\vec {x}_0\leq \lambda \vec b< \vec b$, the radius of such open ball is proportional to $n$. 
The larger $n$ is, the larger open ball we can find. Therefore, for all sufficiently large $n$, we can find an open ball in $\mathcal{H}_{n}$ that contains an integer point. 


The full proof of Lemma~\ref{lem:<-1} and~\ref{thm:intpoint_fulldim} is in Appendix~\ref{apx:lemma}.

    \noindent\textbf{Step 3: Apply Lemma \ref{lem:Theta1}.} 

    By assumption, $\Pi$ is a closed and strictly positive set of distributions over $[m!]$. And we have proved that sub-event $Y$ is characterized by$\mathcal{H}^a$ satisfying all the conditions. %Finally, $\dim(\mathcal{H}^a_{\leq 0})=m!$ and $\mathcal{H}^{\mathbb{Z}}_n\neq \emptyset$ for sufficiently large $n$. Hence, 
    %\[\begin{aligned}
    %    \mathrm{Pr}_{P\sim\pi}(a\in r_{\vec s}(P)\cap r_C^{-1}(P))&=\mathrm{Pr}(\vec X_{\vec \pi}\in\mathcal{H}^a) \\
    %   & = \Theta\left(n^{\frac{\dim(\mathcal{H}^a_{\leq 0})-m!}{2}}\right)=\Theta(1),
    %\end{aligned}
    %\]
    %when $\vec s$ is not a Borda count. Also, we have 
    %\[
    %\mathrm{Pr}(\DI(r_{\vec s}, r_C,P)=1)\geq \mathrm{Pr}(a\in r_{\vec s}(P)\cap r_C^{-1}(P)) =\Theta(1).
    %\]
    %Hence, $\mathrm{Pr}(\DI(r_{\vec s}, r_C,P)=1)=\Theta(1)$. 
    Hence, by applying Lemma \ref{lem:Theta1} we have $\widetilde{\DI}^{\max}_{r_{\vec s},r_C}(n)=\Theta(1)$. 
\end{proof}

The following corollary characterizes the likelihood of the reverse case in Theorem~\ref{thm:SgBorda} when a Condorcet winner is a loser of a positional scoring rule. 

\begin{corollary}\label{coro:reverse1}
    Suppose the set of distributions $\Pi$ be strictly positive, closed and $\pi_{uni}\in CH(\Pi)$ then
    \[\widetilde{\DI}^{\max}_{r_C,r_{\vec s}}(n)=\begin{cases}
        0 & \text{if}\ r_{\vec s}\ \text{is the Borda's rule} \\
        \Theta(1) & otherwise 
    \end{cases}
    \]
\end{corollary}
\begin{proof}
    The proof follows a similar route as Theorem~\ref{thm:SgBorda}. Let $\mathcal{H}^{a}$ characterized by $A$ and $\vec b$ be the polyhedron of subevent $Y$ ($a$ is the unique winner of $r_{\vec s}$ and the Condorcet loser) in Theorem \ref{thm:SgBorda}. Then the reverse event $\hat{Y}$ ($a$ is the unique loser of $r_{\vec s}$ and the Condorcet winner) can be characterized by the polyhedron $\hat{\mathcal{H}}^a:=\{-A\vec x\leq \vec b\}$. Once we show that $\hat{\mathcal{H}}^a$ is non-empty, we can follow the reasoning in the proof for Theorem 2 and apply Lemma~\ref{lem:Theta1} to achieve the goal. 

    Now we show that $\hat{\mathcal{H}}^a$ by constructing a profile $\hat{P}$ such that $\Hist(\hat{P})$ falls in $\hat{\mathcal{H}}^a$. Suppose $P$ is a profile such that $\Hist(P) \in \mathcal{H}^a$. Then for each $R \in P$, $\hat{P}$ has one copy of each $R' \in \mathcal{L}(\mathcal{A}) \setminus \{R\}$. This means $\hat{P}$ contains $n(m! -1)$ rankings.  
    We call $\hat{P}$ the {\em reverse profile} of $P$ since it is the rank-wise compliment on $P$. Then we can verify that for any score vector $\vec {s}$ and any pair of different alternatives $a,b$, $Score_{a,b}^{\vec s}(P)=-Score_{a,b}^{\vec s}(\hat{P})$. Also, $Pair_{a,b}(P)=-Pair_{a,b}(\hat{P})$.  This implies that $Hist(\hat{P})$ is in $\hat{\mathcal{H}}^a$, which finishes the proof. 

    \end{proof}
    
%     For each profile $P$, we construct its \emph{reverse profile} $\hat P$ as follows. For each vote in $R\in P$, let $P_R$ be the profile where the $W$-element of $\Hist(P_R)$ is one for every $W\neq R$ and its $R$-element is zero. Then, $\hat{P}:=\sum_{R\in P}P_R$. Here the sum of profiles stands for the combination of each profile.
% We can easily observe that for any score vector $\vec {s}$ and any pair of different alternatives $a,b$, $Score_{a,b}^{\vec s}(P)=-Score_{a,b}^{\vec s}(\hat{P})$. Also, $Pair_{a,b}(P)=-Pair_{a,b}(\hat{P})$. Hence, the order of $Score^{\vec{s}}(P)$ from high to low is the rank of $Score^{\vec{s}}_a(P)$ from low to high. Also, for any $\WMG-$based rule, the winner in profile $P$ is the loser of $\hat{P}$, since $w_P(a,b)=-w_{\hat{P}}(a,b)$ for any $P,a,b$. 

% We follow the same route in Theorem \ref{thm:SgBorda}. The polyhedron in Theorem \ref{thm:SgBorda} is characterized by $A$ and $\vec b$, and it can be verified that the reverse event can be characterized by $-A$ and $\vec b$. Also, $\hat{\mathcal{H}}:=\{-A\vec x\leq \vec b\}$ is not empty, since there exists $P$ such that $\Hist(P)\in\mathcal{H}$ and clearly $\Hist(\hat{P})\in\hat{\mathcal{H}}$. Finally, $\vec\pi_{uni}\in CH(\Pi)\cap \hat{\mathcal{H}}_{\leq 0}$, since $Score_{a,b}^{\vec s}\cdot\Hist(P)=Pair_{a,b}^{\vec s}\cdot\Hist(P)=0$. Then we can conclude the proof by applying Lemma \ref{lem:Theta1}.
% \end{proof}


% We can see Lemma \ref{lem:<-1}, \ref{lem:ballinhyper} in the Appendix.

\begin{conjecture}\ziqi{update 20240604: Conjecture of min-likelihood, Strong Borda paradox.}
    Suppose the set of distributions $\Pi$ is strictly positive and closed.
    If for any $i\in[m]$, there exists a corresponding fractional profile $\vec \pi_i\in CH(\Pi)$ such that $i$ is not the winner of positional rule and a Condorcet loser (tie permit) simultaneously, then
    \[\widetilde{\DI}^{\min}_{r_{\vec s},r_C}(n)=\begin{cases}
        0 & \text{if}\ r_{\vec s}\ \text{is the Borda's rule} \\
        \exp(\Theta(n)) & otherwise.
    \end{cases}
    \]
    Otherwise, 
    \[\widetilde{\DI}^{\min}_{r_{\vec s},r_C}(n)=\begin{cases}
        0 & \text{if}\ r_{\vec s}\ \text{is the Borda's rule} \\
       \Theta(1) & otherwise.
    \end{cases}
    \]
\end{conjecture}


\input{psr}


\subsection{Different Winners between Positional Rules}\label{sect:diffposi}
We are interested in the fact that the different positional rules can result in different winners under the same profile. But two positional rules characterized by $\vec s$ and $2\vec s$ are essentially the same. Hence, we consider linear independent vectors.

\begin{theorem}\label{thm:diffposi}
    For fixed $m\geq 3$ and any two linear independent score vectors $\Vsi$ and $\Vsii$. Suppose the set of distributions $\Pi$ is strictly positive and closed, and $\vec \pi_{uni}\in CH(\Pi)$, then 
    \[
        \widetilde{\EI}^{\max}_{r_{\Vsi},r_{\Vsii}}(n)=\Theta(1).
    \]
\end{theorem}



\begin{proof}
    

    \noindent\textbf{Step 1: Sub-Event and Polyhedron $\mathcal{H}^{a,b}$}
    
    We fix two different alternatives $a$ and $b$. The sub-event $Y$ is ``$a$ is the winner under $r_{\Vsi}$ while $b$ is the winner under $r_{\Vsii}$ in $P$''. 

    % First, we consider the special case, when the winner of $\Vsi$ and $\Vsii$ is unique and are not the same. We omit $P\sim\vec \pi$ and $P$ for the simplicity of notations, then 
    % \[\begin{aligned}
    %     \mathrm{Pr}(r_{\Vsi}\neq r_{\Vsii})& \geq \mathrm{Pr}(r_{\Vsi}\neq r_{\Vsii},\lvert r_{\Vsi}\rvert=\lvert r_{\Vsii}\rvert=1) \\
    %     & =\sum_{a,b\in[m], a\neq b}\mathrm{Pr}(r_{\Vsi}=\{a\},r_{\Vsii}=\{b\}).
    % \end{aligned}
    % \]

    

    The polyhedron $\mathcal{H}^{a,b}$ is defined as follows. Let $A^{\Vsi,a}$ be the matrix whose row vectors are $\{Score_{k,a}^{\Vsi}:k\in[m], k\neq a\}$ and $A^{\Vsii,b}$ be the matrix 
    with row vectors $\{Score_{k,b}^{\Vsii}:k\in[m], k\neq b\}$. Let $A^{a,b}=\begin{pmatrix}
        A^{\Vsi,a} \\
        A^{\Vsii,b}
    \end{pmatrix}$ and $\vec b=-\vec{1}$. Then $\mathcal{H}^{a,b}:=\{\vec x\in\mathbb{R}^{m!}:A^{i,j}\vec x\leq \vec b\}$. Then it's not hard to verify that $\mathcal{H}^{a, b}$ represents sub-event $Y$. 
    % \begin{claim}\label{claim:event}
    %     For any integral profile $P$, $Hist(P)\in\mathcal{H}^{a,b}$ if and only if $r_{\Vsi}(P)=\{a\}$ and $r_{\Vsii}(P)=\{b\}$. 
    % \end{claim}
    
    \noindent\textbf{Step 2: Prove the conditions on $\mathcal{H}^{a, b}$.}
    
    Once we can show that $\mathcal{H}^{a, b}$ is not empty, the rest of the reasoning is similar to the proof of Theorem~\ref{thm:SgBorda}. In this step, we construct a profile $P$ such that $\Hist(P) \in \mathcal{H}^{a, b}$. 

    % \begin{definition}
    %     For a profile $P$ and an integer $k\in\mathbb{N}$, let $kP$ be the profile which consists of $k$ copies of $P$. 
    %     For two profiles $P_1$ and $P_2$, let $P_1+P_2$ be the profile which is the combination of $P_1$ and $P_2$.
    % \end{definition} 
    To construct such $P$, we separately construct two profiles $P_1$ and $P_2$, and construct $P$ by combining copies of $P_1$ and $P_2$.
        We construct $P_1$ as follows. Reindex the alternative set by $\mathcal{A}=\{a,b,c_1,c_2,\cdots,c_{m-2}\}$.
        % \[P_1^{a\succ b}:=\{(a,b,c_k, \cdots, c_{m-2}, c_1,\cdots,c_{k-1}): k\in[m-2]\}, 
        %     \]
        % \[P_1^{b\succ a}:=\{(b,a,c_k, \cdots, c_{m-2}, c_1,\cdots,c_{k-1}): k\in[m-2]\}. 
        %     \]
        Let $P_1=P_1^{a\succ b}+P_1^{b\succ a}$, such that for every $k \in [m-2]$, $P_1^{a\succ b}$ contains one copy of $[a\succ b\succ c_k\succ \cdots\succ  c_{m-2}\succ c_1\cdots\succ c_{k-1}]$, and $P_1^{b\succ a}$ contains one copy of $[b\succ a\succ c_k\succ \cdots\succ  c_{m-2}\succ c_1\cdots\succ c_{k-1}]$. Then for any positional scoring vector $\vec s$, $a$ and $b$ will have the same highest score which is strictly higher than any other alternative under $P_1$.

        We then construct $P_2$. Since $\Vsi$ and $\Vsii$ are linear independent, their normalized vectors $\vec s^{1'}\neq \vec{s}^{2'}$. 
        Then there exists $\ell\in\mathbb{N}$, such that $ s^{1'}_\ell\neq {s}^{2'}_\ell$. Without loss of generality we suppose $0\leq s^{1'}_\ell<{s}^{2'}_\ell\leq 1$.
        We consider three kinds of votes $v^1, v^2$ and $v^3$, defined by
        \[v^1_{-\ell}=[a\succ  c_1\succ c_2\succ \cdots\succ c_{m-2}\succ b],\ v^1_\ell=b,
            \]
        \[v^2_{-\ell}=[c_1\succ c_2\succ \cdots\succ c_{m-2}],\ v^2_\ell=a,
            \]
        \[v^3=[b\succ  c_1\succ c_2\succ \cdots\succ c_{m-2}\succ a].
            \]
        
        Let $P_2$ be the profile: for $i =1 ,2, 3$, $n_i$ voters choose $v^i$, where $n_i$ are positve integers and satisfy 
        \[s^{2'}_\ell(n_1-n_2)>n_1-n_3>s^{1'}_\ell(n_1-n_2).
        \]  
        Now we show that such $n_1,n_2,n_3$ always exists. We first a find a sufficiently large integer $N$ such that $(s^{2'}_\ell-s^{1'}_\ell) N>2$. 
        Since $0\leq  s^{1'}_\ell<{s}^{2'}_\ell\leq 0$, there exists a positive integer $N_0$ between $s^{1'}_\ell N$ and $s^{2'}_\ell N$. 
        Then $n_1=N+1, n_2=1, n_3=N-N_0+1$ will be a satisfiable solution. Then in profile $P_2$, the score of $a$ is higher than $b$ under $\Vsi$ and the score of $b$ is higher than that of $a$ under $\Vsii$. 

    Finally the profile $P$ is constructed by combining $2t$ copies of $P_1$ and one copy of $P_2$, where $t$ equals to $|P_2|\cdot \max(s^{1'}_1 - s^{1'}_m, s^{2'}_1 - s^{2'}_m)$, which is an upper bound of the score difference in $P_2$ between two alternative under vector $\Vsi$ or $\Vsii$. Then, the $P_1$ copies in $P$ guarantees that the winner is either $a$ or $b$. And the $P_2$ copy makes $a$ be the $\Vsi$ winner and $b$ be the $\Vsii$ winner . Hence, $\Hist(P)\in\mathcal{H}^{a,b}$, which finishes our construction. 

    % We complete the proof of Claim \ref{claim:existdwp} by showing that there exists an integer $t$ such that $\Hist(P)\in\mathcal{H}^{a,b}$ where $P=tP_1+P_2$.\qishen{I remember you change it to $t^2 P1 + T P2$?} Let $N_i$, $i\in\{1,2\}$ be the maximum score gap\qishen{what is score gap?} in profile $P$ under $\Vsi$ or $\Vsii$, respectively. Let $N_3:=\max\{N_1,N_2\}$. Then let $t=2N_3$ be satisfiable. First, any alternative other than $a,b$ cannot be the winner under $\Vsi$ or $\Vsii$ in $P$, since $a,b$ get at least $2N_3$ points higher than others in $tP_1$ and $P_2$ cannot change this result. Also, $b$ ranks higher than $a$ under $\Vsi$, since $a$ is higher than $b$ in $P_2$ and they get a tie in $tP_1$. Hence, $a$ is the unique $\Vsi$ winner. Similarly, $b$ is the unique $\Vsii$ winner. Hence, $\Hist(P)\in\mathcal{H}^{a,b}$ and we finish the proof of Claim \ref{claim:existdwp}.
    
    \noindent\textbf{Step 3: Apply Lemma \ref{lem:Theta1}.}
    Finally, we apply Lemma 1 and get 
    \[        \widetilde{\EI{}}^{\max}_{r_{\Vsi},r_{\Vsii}}(n)=\Theta(1).
    \] 

    \end{proof}
    % We consider the following three claims.
    % \begin{claim}\label{claim:existdwp}
    %     There exists profile $P$, such that $\Hist(P)\in\mathcal{H}^{a,b}$.
    % \end{claim}
    % \begin{claim}\label{claim:CHposi}
    %     $\pi_{uni}\in\mathcal{H}_{\leq 0}^{a,b}\cap CH(\Pi)$.
    % \end{claim}
    % We then verify the conditions of Lemma \ref{lem:Theta1}. 
    % Since $\mathcal{H}^{a,b}$ is characterized by $A$ and $\vec b$ where $A$ is an integer matrix and $\vec b=-\vec 1<\vec 0$. Then by Lemma \ref{thm:intpoint_fulldim}, we have $\dim{\mathcal{H}^{a,b}}=m!$.
    % Also, $r_{\Vsi}=\{a\},r_{\Vsii}=\{b\}$ is a subset of $r_{\Vsi}\neq r_{\Vsii}$. 
    % Hence, by Lemma \ref{lem:Theta1}, 
    % \[
    %     \widetilde{EI}^{\max}_{r_{\Vsi},r_{\Vsii}}(n)=\Theta(1).
    % \] 
    % We then prove these three claims, which conclude the Theorem \ref{thm:diffposi}.
    % \end{proof}
    
    % \noindent\textbf{Proof of Claim \ref{claim:event}.}
    % For each row $Score_{k,a}^{\vec s^1}\cdot \Hist(P)\leq -1$, the left-hand side represents the difference of positional scores and $LHS\leq -1$ implies the score of alternative $a$ is higher than 
    % that of alternative $k$, since $Score_{k,a}^{\vec s^1}\cdot\Hist(P)<0$ implies $Score_{k,a}^{\vec s^1}\cdot\Hist(P)\leq -1$ because they are both integer vectors.
    % Hence, $A^{\vec s^1,a}\cdot\Hist(P)\leq -\vec{1}$ if and only if $a$ is the winner when the scoring vector is $\vec s^1$. The same thing holds for $A^{\Vsii,b}\cdot\Hist(P)\leq -\vec 1$
    % Then clearly the claim holds. \qed 

 %    \noindent\textbf{Proof of Claim \ref{claim:existdwp}.} We will construct the profile $P$ and start from defining the linear combination of profiles.
 %    \begin{definition}
 %        For a profile $P$ and an integer $k\in\mathbb{N}$, let $kP$ be the profile which consists of $k$ copies of $P$. 
 %        For two profiles $P_1$ and $P_2$, let $P_1+P_2$ be the profile which is the combination of $P_1$ and $P_2$.
 %    \end{definition} 
 %    To construct such $P$, we separately construct two profiles $P_1$ and $P_2$, and $P$ is a linear combination of $P_1$ and $P_2$.
 %        We construct $P_1$ as follows. Reindex the alternative set by $\mathcal{A}=\{a,b,c_1,c_2,\cdots,c_{m-2}\}$. Here we use a permutation of $[m]$,\qishen{what is permutation of m here?} 
 %        which can be considered as an $m$-dimensional vector $v$, to represent a preference order. Let 
 %        \[P_1^{a\succ b}:=\{(a,b,c_k, \cdots, c_{m-2}, c_1,\cdots,c_{k-1}): k\in[m-2]\}, 
 %            \]
 %        \[P_1^{b\succ a}:=\{(b,a,c_k, \cdots, c_{m-2}, c_1,\cdots,c_{k-1}): k\in[m-2]\}. 
 %            \]
 %        Let $P_1=P_1^{a\succ b}+P_1^{b\succ a}$. \qishen{why $t$ here?}
 %    \begin{claim}\label{claim:tiehigh}
 %        For any positional vector $\vec s$, $a$ and $b$ have the same highest score among all alternatives in profile $P_1$, and the gap between this highest score and other ones is at least $1$.
 %    \end{claim}
 %    \begin{proof}
 %        It is easy to see the score of $a$ and $b$ are the same under any positional vectors, i.e.
 %        \[Score^{\vec s}(a)=Score^{\vec{s}}(b)=(m-2)(s_1+s_2).
 %            \]
 %        For alternative $x$ other than $a,b$, 
 %        \[Score^{\vec s}(x)=2\sum_{k=3}^m s_q, 
 %            \]
 %        Hence, $Score^{\vec s}(x)+1\leq Score^{\vec s}(a)=Score^{\vec{s}}(b)$ for any alternative $x$ other than $a$ and $b$. 
 %    \end{proof}
    
 % We then construct $P_2$. Since $\Vsi$ and $\Vsii$ are essentially different, we have $\vec s^{1'}\neq \vec{s}^{2'}$, where $\vec s^{1'},\vec{s}^{2'}$ are the corresponding 
 %        normalized positional vector. 
 %        Since $\vec s^{1'}\neq \vec{s}^{2'}$, there exists $\ell\in\mathbb{N}$, such that $ s^{1'}_\ell\neq {s}^{2'}_\ell$. Without loss of generality we suppose $0\leq s^{1'}_\ell<{s}^{2'}_\ell\leq 1$.
 %        We consider three kinds of votes $v^1, v^2$ and $v^3$, defined by
 %        \[v^1_{-\ell}=(a, c_1,c_2,\cdots,c_{m-2}), v^1_\ell=b,
 %            \]
 %        \[v^2_{-\ell}=(c_1,c_2,\cdots,c_{m-2},b), v^2_\ell=a,
 %            \]
 %        \[v^3=(b,c_1, c_2, \cdots,c_{m-2}, a)
 %            \]
        
 %        Let $P_2$ be the profile: for $i =1 ,2, 3$, $n_i$ voters choose $v^i$, where $n_i$ are positve integers and satisfy 
 %        \[s^{2'}_\ell(n_1-n_2)>n_1-n_3>s^{1'}_\ell(n_1-n_2).
 %        \]  
 %        Now we show that such $n_1,n_2,n_3$ always exists. We first a find a sufficiently large integer $N$ such that $(s^{2'}_\ell-s^{1'}_\ell) N>2$. 
 %        Since $0\leq  s^{1'}_\ell<{s}^{2'}_\ell\leq 0$, there exists a positive integer $N_0$ between $s^{1'}_\ell N$ and $s^{2'}_\ell N$. 
 %        Then $n_1=N+1, n_2=1, n_3=N-N_0+1$ will be a satisfiable solution.

 %        \begin{claim}\label{claim:i>j,j>i}
 %            In profile $P_2$, the score of $a$ is higher than $b$ under $\Vsi$ and the score of $b$ is higher than that of $a$ under $\Vsii$. \qishen{For example. this kind of calculation can be thrown into Appendix. }
 %        \end{claim}
 %    \begin{proof}
 %        We can calculate that 
 %        $Score^{\vec s^1}(a)=s_1^1 n_1+s^{1}_l n_2+s^1_m n_3$, 
 %        $Score^{\vec s^1}(b)=s_l^1n_1+s_m^1n_2+s_1^1n_3$,
 %        $Score^{\vec s^2}(a)=s_1^2 n_1+s^{2}_l n_2+s^2_m n_3$, and
 %        $Score^{\vec s^2}(b)=s_l^2n_1+s_m^2n_2+s_1^2n_3$.
            
 %        Also, since $s_\ell^{1'}=\frac{1}{s_1^1-s_m^1}(s_\ell^1-s_m^1)$, then 
 %        \[\begin{aligned}
 %            & Score^{\vec s^1}(a)-Score^{\vec s^1}(b) \\
 %            & =(s^1_1-s^1_\ell)n_1+(s^1_\ell-s^1_m)n_2+(s^1_m-s^1_1)n_3 \\
 %            & =(s^1_1-s^1_m)\left[(1-s_\ell^{1'})n_1+s_\ell^{1'}n_2-n_3\right] \\
 %            & =(s^1_1-s^1_m)\left[(n_1-n_3)-s_\ell^{1'}(n_1-n_2)\right]>0.
 %        \end{aligned}
 %            \]
 %        Similarly, we have 
 %        \[\begin{aligned}
 %            & Score^{\vec s^2}(a)-Score^{\vec s^2}(b) \\
 %            & =(s^2_1-s^2_m)\left[(n_1-n_3)-s_\ell^{2'}(n_1-n_2)\right]<0.
 %        \end{aligned}
 %            \]
 %        Hence, we have completed the proof of Claim \ref{claim:i>j,j>i}.
 %    \end{proof}
 %    We complete the proof of Claim \ref{claim:existdwp} by showing that there exists an integer $t$ such that $\Hist(P)\in\mathcal{H}^{a,b}$ where $P=tP_1+P_2$.\qishen{I remember you change it to $t^2 P1 + T P2$?} Let $N_i$, $i\in\{1,2\}$ be the maximum score gap\qishen{what is score gap?} in profile $P$ under $\Vsi$ or $\Vsii$, respectively. Let $N_3:=\max\{N_1,N_2\}$. Then let $t=2N_3$ be satisfiable. First, any alternative other than $a,b$ cannot be the winner under $\Vsi$ or $\Vsii$ in $P$, since $a,b$ get at least $2N_3$ points higher than others in $tP_1$ and $P_2$ cannot change this result. Also, $b$ ranks higher than $a$ under $\Vsi$, since $a$ is higher than $b$ in $P_2$ and they get a tie in $tP_1$. Hence, $a$ is the unique $\Vsi$ winner. Similarly, $b$ is the unique $\Vsii$ winner. Hence, $\Hist(P)\in\mathcal{H}^{a,b}$ and we finish the proof of Claim \ref{claim:existdwp}.
 %    \qed %qed of Claim 3
    
    % \noindent\textbf{Proof of Claim \ref{claim:CHposi}.} It suffices to verify $\vec\pi_{uni}\in\mathcal{H}^{a,b}_{\leq 0}$. This is clear since $Score_{x,y}^{\vec s}\cdot\vec\pi_{uni}=0$ holds for any pair of alternatives $x,y$ and any score vector $\vec s$.\qed

    \subsection{Copeland and Integer Positional Rules are Drastically Different}\label{sect:CdPosi}
    \begin{theorem}\label{thm:CdandPosi}
    Let ${\mathrm{Cd}_\alpha}$ denotes the copeland rule with parameter $\alpha$.
    %and score vector $\vec s$.
    For fixed $m\geq 3$, $\alpha\in[0,1]$, and $\vec s$ which is an integer score vector, 
    suppose the set of distributions $\Pi$ be strictly positive, closed, and $\vec \pi_{uni}\in CH(\Pi)$. Then 
    \[
        \widetilde{\DI}^{\max}_{r_{\vec s},{\mathrm{Cd}_\alpha}}(n)=\Theta(1).
    \]
\end{theorem}



\paragraph{Proof Sketch.} 
When $r_{\vec s}$ is not the Borda's rule, we claim that the Strong Borda Paradox is a special case of the desired event, hence its likelihood is $\Theta(1)$. 

When $r_{\vec s}$ is the Borda's rule, we construct sub-event $Y$ as follows. We re-index the alternatives as $\mathcal{A} = \{a, 1, \cdots, m-1\}$. Let $G$ be the unweighted majority graph such that (1) $a$ beats 1, (2) for $2\le j \le m-1$, $j$ beats $a$, and for $1\le i < j \le m-1$, $i$ beats $j$. 
We set the sub-event $Y$ to be that $a$ is the unique winner of $r_{\vec s}$ and $UMG(P) = G$.  We define $\mathcal{H}^{a,G}$ as follows.  
    Let 
    $A^{a,G}:=\begin{pmatrix}
        A^{a,\vec s^0} \\
        S^G
    \end{pmatrix}$ and $\vec b=-\vec 1$. Here $A^{a,\vec s^0}$ is the matrix with row vectors $\{Score^{\vec s^0}_{k,a}:k\in[m],k\neq a\}$. Let $S^G$ be the matrix whose row vectors are $\{Pair_{y,x}:(x,y)\in Edge(G)\}$. Then following the reasoning as in the proof of Theorem~\ref{thm:SgBorda}, it suffices to show that $\mathcal{H}^{a,G}$ is non-empty. 

    We construct a profile $P$ such that $\Hist(P) \in \mathcal{H}^{a,G}$. Let $BC(a)$ be the Borda score of alternative $a$, we have the following observation, of which the proof is in Appendix~\ref{apx:lemma}. 

     \begin{observation}
     \label{obs:borda}
     For a profile $P$, we have 
     $$\sum_{b\neq a}P[a\succ b]=BC(a) = \frac{(m-1)n}{2}+\frac{1}{2}\sum_{b\neq a}w_P(a,b).$$
    \end{observation}

    By Observation~\ref{obs:borda}, for any alternative $x$ and $y$, $BC(x)>BC(y)$ if and only if $\sum_{k\neq x}w_P(x,k)>\sum_{k\neq y}w_P(y,k)$. We then construct $P$ by constructing a weighted majority graph $WMG(P)$ as follows.  For every $1\leq i<j\leq m-1$, let $w(i,j)=2$. Let $w(a,1)=2m^2$ and $w(j,a)=2$ for $2\leq j\leq m-1$. \citet{mcgarvey1953theorem} guarantee that such profile $P$ exists.  
    
    Then we show that $P$ is the desired profile. First, $a$ is the unique $\vec s_0$ winner, since $\sum_{b\neq a}w_P(a,b)\geq 2m(m-1)>\sum_{j\neq i} w_P(i,j)$ for any $i\in[m-1]$. Also, $a\in \rlos_{\mathrm{Cd}_\alpha}$, since its $\mathrm{Cd}_\alpha$ score is one and any other agent wins at least one pairwise comparison. Hence, we can follow the reasoning as in the proof of Theorem~\ref{thm:SgBorda} and applies Lemma~\ref{lem:Theta1} to get $\widetilde{\DI{}}^{\max}_{r_{\vec s^0},\mathrm{Cd}_\alpha}(n)=\Theta(1)$.

    The full proof is in Appendix~\ref{apx:Cpd}. 

We also state the reverse version, which can be proved by similar reasoning and construct a reverse profile as in the proof of Corollary \ref{coro:reverse1}.
\begin{corollary}
\label{coro:reverse_Cpd}
    Suppose the set of distributions $\Pi$ be strictly positive, closed, and $\vec \pi_{uni}\in CH(\Pi)$. Then 
    \[
        \widetilde{\DI}^{\max}_{\mathrm{Cd}_\alpha,r_{\vec s}}(n)=\Theta(1).
    \]
\end{corollary}

\subsection{Positional Rules and Ranked Pairs are Drastically Different}\label{sect:RPPosi}
\begin{theorem}
\label{thm:psr_rp}
    Let $r_{\mathrm{RP}}$ denote the ranked pairs rule. 
    For fixed $m\geq 3$ and an integer score vector $\vec s$, suppose the set of distributions $\Pi$ be strictly positive, closed, and $\vec \pi_{uni}\in CH(\Pi)$. Then 
    \[
        \widetilde{\DI}^{\max}_{r_{\vec s},r_{\mathrm{RP}}}(n)=
        \begin{cases}
            0 & m=3,\ r_{\vec s} \text{ is Borda},\ n\ \text{ is odd} \\
            \Theta(n^{-\frac{3}{2}}) & m=3,\ r_{\vec s} \text{ is Borda},\ n\ \text{ is even}  \\
            \Theta(1)& otherwise
        \end{cases}
    \]
\end{theorem}
We also study the reverse case whose proof follows the same route in Corollary \ref{coro:reverse1}.
\begin{corollary}
\label{coro:reverse_rp}
    Suppose the set of distributions $\Pi$ be strictly positive, closed, and $\vec \pi_{uni}\in CH(\Pi)$. Then 
    \[
        \widetilde{\DI}^{\max}_{r_{\RP},r_{\vec s}}(n)=
        \begin{cases}
            0 & m=3,\ r_{\vec s} \text{ is Borda},\ n\ \text{ is odd} \\
            \Theta(n^{-\frac{3}{2}}) & m=3,\ r_{\vec s} \text{ is Borda},\ n\ \text{ is even}  \\
            \Theta(1)& otherwise
        \end{cases}
    \]
\end{corollary}

\noindent\textbf{Proof of Theorem \ref{thm:psr_rp}.} When $r_{\vec s}$ is not Borda's rule, the statement holds with the same reason on Case 1 in the proof of Theorem \ref{thm:CdandPosi}. This is because a Condorcet winner (loser) is always a Ranked Pair winner (loser). Hence, we follow the same processes and have 
    \[
        \widetilde{\DI}^{\max}_{r_{\vec s},r_{\mathrm{RP}}}(n)=\Theta(1).
    \]

    We separately consider the case when $m\geq 4$ and $m=3$ when $r_{\vec s}$ is Borda's rule.  
    
    \noindent\textbf{Case 1: $m\geq 4$.}
    When$r_{\vec s}$ is Borda's rule and $m\geq 4$, we follow a similar reasoning in Theorem~\ref{thm:CdandPosi}. And it suffices to specify a profile $P$, such that $a$ is both the unique winner of $r_{\vec s}$ and the loser (not necessarily unique) of ranked pairs and there are no ties in any pairwise comparison.

    We then construct $P$ as follows.  Reindex the set of alternatives by $\mathcal{A}=\{a,1,\cdots,m-1\}$. Consider the weighted majority graph $G$ as follows. For $i\in[m-2]$, let $w(a,i)=2m^2$, and $-w(m-1,a)=w(a,m-1)=-2m^2-2$. For $j\in[m-2]$, let $w(j,j+1)=4m^2+2m-2j$, and $w(m-1,1)=2m^2+2$. For any other undefined edges $(i,j)$, let $w(i,j)=2$ if $i<j$. We can calculate that $\sum_{i\neq a}w_P(a,i)=2m^3-6m^2-2$, and for all $j\in [m-1]$, $\sum_{k\neq j}w_{P}(j,k)\leq 4m$. Since $m\geq 4$, we have 
    \[
        2(m^3-3m^2-1)=2[(m-3)m^2-1]\geq 2m^2-2> 4m.
    \]
     The existence of the $P$ such that $WMG(P) = G$ is guaranteed by~\citep{mcgarvey1953theorem}. 
    Hence, $a$ is the unique $r_{\vec s^0}$ winner under this profile. However, if we apply the Ranked Pair rule, we will first fix $(1,2),\cdots,(m-2,m-1)$ and finally fix $(m-1,a)$. Hence, $a$ is the Ranked Pair loser. The sub-event $Y$ is ``$a$ is the Borda winner in $P$, and the weighted majority graph of $P$ is $G$''. 

    \noindent\textbf{Case 2: $m=3$.} First, let $\WMG^*$ be the set of all $\WMG(P)$.  Then
    we have 
    \[\begin{aligned}
        &\Pr[r_{\vec s^0}(P)\cap \rlos_{\mathrm{RP}}(P)\neq\emptyset] \\
        &=\sum_{G\in\WMG^*} \Pr[r_{\vec s^0}(P)\cap \rlos_{\mathrm{RP}}(P)\neq\emptyset,\WMG(P)=G] (*)
    \end{aligned}
    \]
    %where the last equality is because whether $a$ is an element of $\rlos_{RP}(P)$ can be  determined by $\WMG(P)=G$. 
    Let $G_0$ denote the directed graph whose weight of each edge is zero. Suppose $\WMG(P)\neq G_0$, then it can be verified that $w_P(a,b)+w_P(a,c)<0$ when $a$ is a Ranked Pairs loser. 
    Hence, only $\Pr[r_{\vec s^0}(P)\cap \rlos_{\mathrm{RP}}(P)\neq\emptyset,\WMG(P)=G_0]$ is not zero in $(*)$.
    Also, we notice that $\WMG(P)=G_0$ implies $r_{\vec s^0}(P)\cap \rlos_{\mathrm{RP}}(P) = \mathcal{A}$ since all alternatives are tied. Hence, 
    \[\begin{aligned}
        & \widetilde{\DI}^{\max}_{r_{\vec s^0},r_{\mathrm{RP}}}(n) \\
        & =\sup_{\vec \pi\in\Pi^n}\mathrm{Pr}_{P\sim\vec\pi}[r_{\vec s^0}(P)\cap \rlos_{\mathrm{RP}}(P)\neq\emptyset] \\
        & = \sup_{\vec \pi\in\Pi^n}\mathrm{Pr}_{P\sim\vec\pi}[r_{\vec s^0}(P)\cap \rlos_{\mathrm{RP}}(P)\neq\emptyset,WMG(P)=G_0] \\
        & = \sup_{\vec \pi\in\Pi^n}\mathrm{Pr}_{P\sim\vec\pi}[WMG(P)=G_0]
    \end{aligned} 
    \]
    
    When $n$ is odd, it is clear that $\WMG(P)\neq G_0$, since every pairwise comparison will create a unique winner. 
    We prove the case when $n$ is even by three steps. 

    \noindent\textbf{Step 1: Characterize $\WMG(P)=G_0$ by $\mathcal{H}^{0}$.}
     Let $\mathcal{H}^0:=\{A\vec x\leq \vec b\}$ where $A$ is the matrix whose six rows are $\{Pair_{i,j}: i,j\in[3], i\neq j\}$ and $\vec b=\vec 0$.

     \noindent\textbf{Step 2: Prove Properties of $\mathcal{H}^0$.} 
    \begin{claim}
        $\dim(\mathcal{H}_{\leq 0}^0)=m!-3$.
    \end{claim}
    This is because $\mathcal{H}_{\leq 0}^0=\mathcal{H}^0$, and $\mathcal{H}^0$ is defined by three linear independent equations. Therefore, the dimension of the null space of $(\mathcal{H}^0_{\leq 0}$ is 3, and $\dim(\mathcal{H}^0_{\leq 0})=m!-3$. 
    \begin{claim}
        $CH(\Pi)\cap \mathcal{H}^0_{\leq 0}\neq \emptyset$.
    \end{claim}
    This is because $\vec\pi_{uni}\in CH(\Pi)$ and $\vec\pi_{uni}\in\mathcal{H}^0_{\leq 0}$.
    \begin{claim}
        $(\mathcal{H}^0)^{\mathbb{Z}}_n\neq \emptyset$ for every even integer $n$.
    \end{claim}
    Let $P_0$ be a profile with two votes $\{[1\succ 2\succ 3],[3\succ 2\succ 1]\}$. Then $\WMG(kP_1)=G_0$ for all $k\in\mathbb{N}$, where $kP_1$ is the profile that contains $k$ copies of $P_1$. Hence, $(\mathcal{H}^0)_n^{\mathbb{N}}\neq \emptyset$ for every even $n$. 
    
     \noindent\textbf{Step 3: Apply Theorem \ref{thm:pmv1}.}
     We check all conditions of Theorem \ref{thm:pmv1}. First, $\Pi$ is closed and strictly positive, and $A$ is an integer matrix. Then, $CH(\Pi)\cap\mathcal{H}_{\leq 0}^0\neq \emptyset$ and $(\mathcal{H}^0)_n^{\mathbb{Z}}$, then it suffices to consider $\Theta(n^{-\frac{\dim({\mathcal{H}_{\leq 0}})-q}{2}})$ case. Finally, we have $\dim(\mathcal{H}^0_{\leq 0})=m!-3$. Hence, by Theorem \ref{thm:pmv1}, we have

    \[
    \begin{aligned}
        \widetilde{\DI}^{\max}_{r_{\vec s^0},r_{\mathrm{RP}}}(n)&=\sup_{\vec \pi\in\Pi^n}\mathrm{Pr}_{P\sim\vec\pi}[\WMG(P)=G_0] \\ 
        & = \Theta(n^{-\frac{m!-3-m!}{2}})=\Theta(n^{-\frac{3}{2}}),
    \end{aligned}
    \]
    when $n$ is odd and $\vec s^0$ represents the Borda's rule. \qed
    
\begin{conjecture}
    Let $r_{\mathrm{RP}}$ denote the ranked pairs rule. 
    For fixed $m\geq 4$ and an integer score vector $\vec s$, suppose the set of distributions $\Pi$ be strictly positive and closed. If all fractional profiles $P$ in $CH(\Pi)$ satisfy
    \begin{itemize}
        \item $WMG(P_i)=G_0$ holds for all $P_i\in CH(\Pi)$, 
        \item The winner of positional rule is the loser of the ranked pair, given any $P\in CH(\Pi)$,
    \end{itemize}
    then 
    \[
        \widetilde{\DI}^{\min}_{r_{\vec s},r_{\mathrm{RP}}}(n)=\Theta(1).
    \]
\end{conjecture}



\subsection{STV and Plurality have different winners.}
The single transferable vote (STV) is a voting rule selecting the winner in $m-1$ rounds. In each round, the plurality rule is used and the plurality loser is eliminated from the election. An easy observation is, that the STV winner cannot be a plurality loser, who is eliminated in the first round following the STV rule.

\begin{theorem}
    Given $m\geq 3$, 
    \[
    \widetilde{\EI}^{max}_{STV, plurality}(n)=\Theta(1).
    \]
\end{theorem}

\begin{proof}
    \noindent\textbf{Step 1: Sub-Event and Polyhedron}
    We characterize the sub-event $Y$ by a permutation of $[m-1]$ (denoted by $\sigma([m-1])$) and an alternative $i\in [m-1]$. The sub-event $Y$ is ``$\sigma(j)$ is eliminated in $j$-th round, while the plurality winner is $i$''. We further suppose $\sigma(j)$ is the only loser in $j$-th round, where it is unnecessary to break the tie. 
    
    and we consider polyhedral $\mathcal{H}^{\sigma, i}$ defined as follows. Let 
    \[
        A=\begin{pmatrix}
            A^{\sigma}_1 \\ A^\sigma_2 \\ \vdots \\A^\sigma_{m-1} \\ A^i
        \end{pmatrix}, \vec b=-\vec 1, 
    \]
    where $A^i\cdot \Hist(P)\leq -\vec 1$ characterizes ``i'' is the plurality winner and $A^\sigma_j\cdot \Hist(P)\leq -\vec 1$ characterizes ``given $\sigma(1), \cdots, \sigma(j-1)$ are eliminated in former rounds, $\sigma(j)$ is eliminated in $j$-th round''. It can be verified these events can be exactly described by linear constraints. 

    \noindent\textbf{Step 2: Prove the conditions on $\mathcal{H}^{\sigma, i}$.} We consider a special case where $\sigma_0(1,2,\cdots,m-1)=(1,2,\cdots,m-1)$ and $i_0=m-1$. It suffices to show that $\mathcal{H}^{\sigma_0, i_0}$ is not empty and the rest of the proof is similar to that of Theorem \ref{thm:SgBorda}. Consider the voting profile in Table \ref{tab:STVandPlur}. Since $m\geq 3$, $2m-3>2$ and $1$ will be the first alternative being eliminated, while the score of $m$ in the second round becomes $2m-1$, and hence $2$ will be eliminated in the second round. It can be verified that $m$ is the STV winner and $m-1$ is the plurality winner, which implies $\mathcal{H}^{\sigma_0,m-1}$ is not empty.

    \begin{table}[htbp]
    \centering
    \begin{tabular}{@{}llllll@{}}
    \toprule
    Num of votes & $2$   & $4$   & $\cdots$ & $2(m-1)$  & $2m-3$ \\ \midrule
    First place  & $1$ & $2$ & $\cdots$ & ${m-1}$ & $m$  \\
    Second place & $m$ & $m$ & $\cdots$ & $m$     & $1$  \\ \bottomrule
    \end{tabular}
    \caption{Example: STV winner but not plurality winner.}
    \label{tab:STVandPlur}
    \end{table}

     \noindent\textbf{Step 3: Apply Lemma \ref{lem:Theta1}.}
    Finally, we apply Lemma 1 and obtain 
    \[ \widetilde{\EI}^{max}_{STV, plurality}(n)=\Theta(1).       
    \]
\end{proof}


%\input{mallows.tex}




\section{Conclusion and Future Works}
We study the max semi-random likelihood of different voting rules being (drastically) different. Our results imply that these voting rules are indeed different in most cases, as there is a constant probability that the winner of one rule is the loser of the other. Also, even tiny differences between scoring rules may lead to different winners eventually. Our analysis can be applied to a wide range of statistical models including the standard IC assumption, and any fixed number of alternatives. 

While we focus on the max semi-random likelihood, similar techniques of our work can be applied to analyze the min semi-random likelihood. However, this may require a finer characterization and case discussions. Another question that remains open is the smoothed likelihood of strict Borda's paradox. We are also interested in applying semi-random analysis to more voting paradoxes, including the no-show paradox and Condorcet's paradox, for example. 


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\appendix

\section{Proof of Lemmas}
\label{apx:lemma}

\noindent\textbf{Proof of Lemma \ref{lem:Theta1}.}
    
    It suffice to show $\Ymax=\Theta(1)$, since $Y\subseteq X$ implies $\Xmax(n)\geq\Ymax(n)$. Since $Y$ is characterized by $\mathcal{H}$, then $\Ymax=\sup_{\vec \pi\in\Pi^n}\mathrm{Pr}_{P\sim\pi}(\Xpi\in\mathcal{H})$, and $\mathcal{H}$ is not empty. Since we also have $\vec b<0$, $\dim(\mathcal{H})=\dim(\mathcal{H}_{\leq 0})=m!$ by Lemma \ref{lem:<-1}. Also, $\vec b<\vec 0$ and $\dim(\mathcal{H})=m!$ implies $\mathcal{H}_n^{\mathbb{Z}}\neq\emptyset$ for suffciently large $n$ by applying Lemma \ref{thm:intpoint_fulldim}. Finally, $\mathcal{H}_{\leq 0}\cap CH(\Pi)\neq \emptyset$. Then by Theorem \ref{thm:pmv1} we have 
    \[\begin{aligned}
        \Ymax& =\sup_{\vec \pi\in\Pi^n}\mathrm{Pr}_{P\sim\pi}(\Xpi\in\mathcal{H}) \\
        &=\Theta(n^{\frac{\dim(\mathcal{H}_{\leq 0})-m!}{2}})=\Theta(1).
    \end{aligned}
    \] \qed

\begin{lembis}{lem:<-1}
    Suppose a polyhedron $\mathcal{H}=\{\vec x\in\mathbb{R}^q:A\vec x\leq \vec b\}$ where $\vec b<\vec 0$. Then $\mathcal{H}$ is either empty or full-dimensional. Furthermore, $\dim(\mathcal{H}_{\leq 0})=q$ when $\mathcal{H}\neq \emptyset$.
\end{lembis}
\begin{proof}
    Suppose $\vec x_0\in\mathcal{H}$. It suffices to show $\dim(\mathcal{H})=q$. For any $\lambda\geq 1$, we have 
    \[A\cdot \lambda\vec{x}_0=\lambda A\vec {x}_0\leq \lambda \vec b< \vec b.
        \]
    Hence, $\lambda \vec {x}_0\in \mathcal{H}$ for any $\lambda\geq 1$. 
    In Euclidean space, a polyhedron $\mathcal{H}$ contains an interior point that implies $\mathcal{H}$ is full-dimensional.
    Then it suffices to find an interior point of $\mathcal{H}$. We consider the $\mathcal{L}^\infty$ norm for vector $\vec n=(n_1,\cdots,n_q)$ and matrix $A=(a_{ij})_{L\times q}$, where 
    \[\norm*{\vec n}_\infty:=\max_{i\in[q]}\lvert n_i\rvert.
        \]
    \[\norm*{A}_\infty:=\max_{i,j}\lvert a_{ij}\rvert.
    \]
    Suppose $\vec b=(b_1,\cdots,b_L)$, let $\norm*{\vec b}_0:=\min_{k\in [L]}\lvert b_k\rvert$. We have $\norm*{\vec b}_0>0$ since $\vec b<\vec 0$. Then there exists $\lambda_1\geq 1$ such that $\lambda_1\norm*{\vec b}_0\geq 2\norm*{\vec b}_\infty$.
    For each row of $A$, denoted by $\vec {a}_i$, we have $\vec a_i\cdot \vec x_0\leq b_i\leq -\norm*{\vec b}_0$. 
    Hence, for any vector $d\in\mathbb{R}^q$ satisfy $\norm*{d}_\infty\leq \frac{\norm*{b}_\infty}{2\norm*{A}_\infty}$, we have 
    \[\begin{aligned}
        \vec{a}_i(\lambda_1\vec x_0+\vec d) &= \vec{a}_i\cdot \lambda_1\vec {x}_0+\vec a_i\cdot\vec d \\
        & \leq -\lambda_1\norm*{\vec b}_0+\norm*{\vec a_i}_\infty\cdot\norm*{\vec{d}}_\infty \\
        & \leq -\lambda_1\norm*{\vec b}_0+\norm*{A}_\infty\norm*{\vec d}_\infty \\
        & \leq -\lambda_1\norm*{\vec b}_0+\frac{1}{2}\norm*{\vec b}_\infty \\
        & \leq -\frac{3}{2}\norm*{\vec b}_\infty\leq b_i.
    \end{aligned}
        \]
    Hence, $\lambda_i \vec x_0+\vec d\in\mathcal{H}$. Since the  $\mathcal{L}^\infty$ norm is equivalent to that of $\mathcal{L}^2$, $\lambda_i\vec x_0$ is an interior point of $\mathcal{H}$. Also, since $\vec b< \vec 0$, $\mathcal{H}$ is a subset of the characteristic cone $\mathcal{H}_{\leq 0}$. Hence, $\mathcal{H}_{\leq 0}$ is also full-dimensional. 
\end{proof}

\begin{lemma}\label{lem:ballinhyper}
    Let $F_n$ denote the hyperplane 
    \[\{\vec x\in\mathbb{R}^q:\vec x\cdot \vec 1^\top=n\}.
        \]
    Let $B_{q-1}(\vec x_0,r)$ be a $q-1$ dimensional ball, which centered at $\vec x_0$ with radius $r$ and contained in $F_n$. Then there exists an integer point $\alpha_n\in B_{q-1}(\vec x_0,r)$ when $r>\sqrt{q}$. 
\end{lemma}
\noindent\textbf{Proof of Lemma \ref{lem:ballinhyper}:} Let $\vec x_0=(x_0^1,\cdots,x_0^q)$, and 
$k:=\sum_{i=1}^q \lfloor x_0^i\rfloor$. Clearly, $n-q<k\leq n$ and $k\in\mathbb{Z}$, and the equality holds if and only if 
all components of $\vec x_0$ are integers. Hence, $\alpha_0=x_0$ will be the desired integer point on $F_n$ when $k=n$. 

If not so, then $n-k\geq 1$. We define $\alpha_n$ by 
\[\begin{aligned}
    \alpha_n &:=(\lfloor x_0^1\rfloor +1, \lfloor x_0^2\rfloor +1,\cdots, \\
   & \lfloor x_0^{n-k}\rfloor +1,\lfloor x_0^{n-k+1}\rfloor, \lfloor x_0^{n-k+2}\rfloor, \cdots, \lfloor x_0^q\rfloor).
\end{aligned}
    \]

    It is easy to verify $\alpha_n\in F_n$, since $\alpha_n\cdot \vec{1}^\top=n-k+\sum_{i=1}^q \lfloor x_0^i\rfloor=n$. Also, 
    \[\mathrm{Dist}(x_0,\alpha_n)=\left(\sum_{i=1}^q (x_0^i-\lfloor x_0^i\rfloor -1)^2\right)^{1/2}< \sqrt{\sum_{i=1}^q 1^2}=\sqrt{q}.
        \]
    
    Notice that $r>\sqrt{q}$, then show the existence of the integer point $\alpha_n$ in a constructive way. \qed

\begin{lembis}{thm:intpoint_fulldim}
    Let $\mathcal{H}$ be the polyhedron characterized by $A,\vec b$, where $\vec b\leq \vec 0$.
    Suppose $\dim(\mathcal{H})=q$, then there exists $N\in\mathbb{N}$, such that 
    \[\mathcal{H}_n^{\mathbb{Z}}:=\mathcal{H}\cap \{\vec x\in\mathbb{Z}^q: \vec x\cdot 1^{\top}=n\}
        \]
    is not empty for every $n>N$.
\end{lembis}
\begin{proof}
    %First, notice that $\mathcal{H}$ is the intersection of several half spaces, which are convex sets. Any intersection of convex sets are also convex. 
    %Hence, we have shown $\mathcal{H}$ is convex.

    %We then consider the faces of $\mathcal{H}$. Denote the $i^{th}$ row of $A$ by $a^i$, we notice that
    %$a^ix=0$ is a $q-1$-dimensional hyperplane in $\mathbb{R}^q$. We denote this hyperplane by $F^i$. For any $x\in\mathbb{R}^q$, let 
    %$Dist_i(x)$ be the distance between vector $x$ and hyperplane $F^i$.

    Since $\dim(\mathcal{H})=q$, there exists $\vec x_0\in \mathrm{int}(\mathcal{H})$. In other word, there exists an open ball $B(\vec x_0,r),r>0$, 
    such that $B(\vec x_0,r)\subseteq \mathcal{H}$. We claim that $B(\lambda \vec x_0,\lambda r)\in \mathcal{H}$ for any $\lambda >1$.This is because  
    $A\vec x\leq \vec b\implies A(\lambda \vec x)\leq \lambda\vec b\leq \vec b$, since $\lambda>1$.

    Let $\lambda_n:=n/(\vec x_0\cdot \vec 1^\top)\in \mathbb{R}$, then $\lambda_n \vec x_0\in F_n$. Clearly, there exists $N$, such that 
    \[n>N\implies \lambda_nr>\sqrt q.
        \]
    Then for any $n>N$, $B_{q-1}(\lambda_n \vec x_0,\lambda_n r)\cap F_n$ is a ball in hyperplane $F_n$. Since its radius is greater than $\sqrt{q}$, by Lemma \ref{lem:ballinhyper} there exists an integer point $\alpha_n$ such that 
    $\alpha_n \in B_{q-1}(\lambda_n \vec x_0,\lambda_n r)\subseteq \mathcal{H}$ and $\alpha_n\cdot 1^\top=n$. Hence, $\mathcal{H}_n^{\mathbb{Z}}\neq \emptyset$ for any $n>N$. 
\end{proof}

\section{Proof of Theorem \ref{thm:CdandPosi}.}
\label{apx:Cpd}
    We prove this theorem through the case study of $\vec s$ is a Borda Count or not.
    
    \noindent\textbf{Case 1: $\vec s$ is not a Borda Count.}

    By Theorem \ref{thm:bordaisspecial}, there exists a profile $P$ such that the winner of positional rule under $\vec s$ is the Condorcet loser with probability $\Theta(1)$. Since a Condorcet loser always ranks the last place in $\mathrm{Cd}_\alpha$ for any $\alpha\in[0,1]$, $P$ is also a profile where the positional rule winner is the $\mathrm{Cd}_\alpha$ loser. Recall that $r_C$ is the Condorcet rule, hence, 
    \[\begin{aligned}
        \widetilde{\DI}^{\max}_{r_{\vec s}, r_{\mathrm{Cd}_\alpha}}(n) & = \sup_{\vec \pi\in\Pi^n}\sum_{a\in[m]}\mathrm{Pr}_{P\sim\vec\pi}[a\in r_{\vec s}(P)\cap \rlos_{\mathrm{Cd}_\alpha}(P)] \\ 
        & \geq \sup_{\vec \pi\in\Pi^n}\sum_{a\in[m]}\mathrm{Pr}_{P\sim\vec\pi}[a\in r_{\vec s}(P)\cap \rlos_{C}(P)] \\
        & =\widetilde{\DI}^{\max}_{r_{\vec s}, r_{C}}(n)=\Theta(1) \quad \text{by Theorem \ref{thm:SgBorda}.} 
    \end{aligned}
    \]


    \noindent\textbf{Case 2: $\vec s$ is a Borda Count.}

    \noindent\textbf{Step 1: Find the Sub-event and Characterize Polyhdron $\mathcal{H}^{a,G}$.}
    
    We use $\vec s^0$ denotes the Borda Count. 
    For a profile $P$, we consider the \emph{unweighted majority graph}, denoted by $\UMG(P)$
    where each vertex represents an alternative
    and a directed edge $(a,b)\in E(G)$ if and only if $w_P(a,b)=P[a\succ b]-P[b\succ a]>0$. We say there is a \emph{tie} between $a$ and $b$ if $w_P(a,b)=0$. A tournament graph is a simple (no loop or bidirected edges) and complete (each pair of vertex is joined by at least one directed edge) directed graph. 

    We consider the following event: $a$ is the unique winner of $r_{\vec s}$ and $\UMG(P)=G$, where $G$ is a tournament graph such that (1) $a$ beats 1, (2) for $2\le j \le m-1$, $j$ beats $a$, and for $1\le i < j \le m-1$, $i$ beats $j$.  We denote this event by $Y$.
    
    %For an alternative $a$ and a tournament graph $G$ with $m$ vertices, 
    We define $\mathcal{H}^{a,G}$ as follows.  
    Let 
    $A^{a,G}:=\begin{pmatrix}
        A^{a,\vec s^0} \\
        S^G
    \end{pmatrix}$ and $\vec b=-\vec 1$. Here $A^{a,\vec s^0}$ is the matrix with row vectors $\{Score^{\vec s^0}_{k,a}:k\in[m],k\neq a\}$. Let $S^G$ be the matrix whose row vectors are $\{Pair_{y,x}:(x,y)\in Edge(G)\}$.
    
    \begin{claim}\label{claim:a,G}
        $\Hist(P)\in\mathcal{H}^{a,G}$ if and only if $Y$ happens.
    \end{claim}
    \noindent\textbf{Proof of Claim \ref{claim:a,G}}
        As we have discussed previously, $A^{a,\vec s^0}\cdot\Hist(P)\leq -1$ implies $a$ is the unique $r_{\vec s^0}$ winner. 
       Also, $Pair_{y,x}\cdot\Hist(P)\leq -1$ if and only if $x$ wins $y$ in the pairwise comparison. Hence, the claim holds.
    
    \noindent\textbf{Step 2: Show the conditions on $Y$ and $\mathcal{H}^{a,G}$.}
    \begin{claim}
        $Y\neq \emptyset$.
    \end{claim}
    We give a construction later. 
    \begin{claim}\label{claim:existCdPosi}
        $\mathcal{H}_{\leq 0}\cap CH(\Pi)\neq \emptyset$. 
    \end{claim}
    Our assumption guarantees that $\pi_{uni} \in CH(\Pi)$.  
    And $\pi_{uni}\in\mathcal{H}_{\leq 0}$ since  $Score_{x,y}\cdot \vec \pi_{uni}=0$ and $Pair_{x,y}\cdot\vec\pi_{uni}=0$ for every pair of alternatives $x,y$.
    \begin{claim}
        $\dim(\mathcal{H}^{a,G})=m!$
    \end{claim}
    Since $\mathcal{H}^{a,G}$ can be characterized by an integer matrix $A$ and $\vec b<\vec 0$, 
    we have $\dim(\mathcal{H}^{a,G})=m!$ if $\mathcal{H}^{a,G}$ is not empty by Lemma \ref{lem:<-1}. 

    %Notice that if $\lvert E(G)\rvert=m(m-1)/2$, the set of constraint $E^G\vec x\leq \vec 0$ will be empty. Also, $r_{\mathrm{Cd}_\alpha}$ can be determined by the $\UMG$. Let $G_a$ be a $m$-vertex directed graph satisfies: 
    %\begin{itemize}
        %\item If $G_a$ is considered as an $\UMG$, $a\in \rlos_{\mathrm{Cd}_\alpha}$.
        %\item $\lvert E(G)\rvert=m(m-1)/2$.
    %\end{itemize}\qishen{Can we remove everything on ties here? In the way "it suffices to show in the subcases where there are no ties in pairwise comparisons.}

    \noindent\textbf{Step 3: Apply Lemma \ref{lem:Theta1}}    We check the conditions in Lemma \ref{lem:Theta1}. Clearly, $Y$ is a subset of $r_{\vec s^0}\cap r_{\mathrm{Cd}_\alpha}^{-1}\neq \emptyset$, which is the desired event in this theorem. Also, $\mathcal{H}^{a,G}$ can be characterized by an integer matirx $A$ and $\vec b$. Finally, $\mathcal{H}_{\leq 0}^{a,G}\cap CH(\Pi)\neq \emptyset$ and $\dim(\mathcal{H}^{a,G})=m!$. Then by Lemma \ref{lem:Theta1}, it suffices to construct such a profile $P$, such that $r_{\vec s^0}(P)=\{a\}$, $a\in \rlos_{\mathrm{Cd}_\alpha}$m and $\UMG(P)$ is a tournament graph.
    
    % \noindent\textbf{Case 1: $\vec s$ is not a Borda Count.} By Theorem \ref{thm:bordaisspecial}, there exists a profile $P$ such that the winner of positional rule under $\vec s$ is the Condorcet loser. Since a Condorcet loser always ranks the last place in $\mathrm{Cd}_\alpha$ for any $\alpha\in[0,1]$, $P$ is also a profile where the positional rule winner is the $\mathrm{Cd}_\alpha$ loser.

    %\noindent\textbf{Case 2: $\vec s$ is a Borda Count.} 
    We then 
    construct a profile $P$ such that $a\in r_{\vec s^0}\cap \rlos_{\mathrm{Cd}_\alpha}$. We start with several observations. 
    We use $BC(a)$ to represent the Borda Count of $a$. 
    \begin{observation}
        We consider a profile $P$ and the corresponding weighted majorith graph $\WMG(P)$, then for any alternative $a\in[m]$,
        \begin{enumerate}
            \item 
        \[\sum_{b\neq a}P[a\succ b]=BC(a).\]
            \item 
            \[
                BC(a)=\frac{(m-1)n}{2}+\frac{1}{2}\sum_{b\neq a}w_P(a,b).
            \]
        \end{enumerate}
        
    \end{observation}
    \begin{proof}
        We first prove the first equality. For each vote $v_i$, we consider its contribution to both left hand side (LHS) and right hand side (RHS). Suppose $a$ ranks higher than $k$ other alternative in this vote. Then clearly the contribution of $v_i$ to the RHS is $k$. Also, for every agent $b$ ranked lower than $a$, $v_i$ contributes one point to $P[a\succ b]$ in LHS, which is in total $k$ points. Hence we have equality.

        Then we prove the second equality. Recall that $w_P(a,b) = P[a\succ b] - P[b\succ a]$. By definition, we have $P[a\succ b]=(n+w_P(a,b))/2$. Then we take the sum of every $b\neq a$, and we have 
        \[
            BC(a)=\sum_{b\neq a}P[a\succ b]=\frac{(m-1)n}{2}+\frac{1}{2}\sum_{b\neq a}w_P(a,b)
        \]

    \end{proof}
    By this observation, $BC(x)>BC(y)$ if and only if $\sum_{k\neq x}w_P(x,k)>\sum_{k\neq y}w_P(y,k)$. We then construct $P$ by constructing a weighted directed graph $G$ as follows. Reindex the set of alternatives by $\mathcal{A}=\{a,1,\cdots,m-1\}$. For every $1\leq i<j\leq m-1$, let $w(i,j)=2$. Let $w(a,1)=2m^2$ and $w(j,a)=2$ for $2\leq j\leq m-1$. \citet{mcgarvey1953theorem} guarantee that there exists a profile $P$ such that $\WMG(P)=G$. 
    
    Then it suffices to show $P$ is the desired profile. First, $a$ is the unique $\vec s_0$ winner, since $\sum_{b\neq a}w_P(a,b)\geq 2m(m-1)>\sum_{j\neq i} w_P(i,j)$ for any $i\in[m-1]$. Also, $a\in r_{\mathrm{Cd}_\alpha}^{-1}$, since its $\mathrm{Cd}_\alpha$ score is one and any other agent win at least one pairwise comparison. Hence, $P$ is the desired profile and we can conclude $\widetilde{\DI}^{\max}_{r_{\vec s^0},r_{\mathrm{Cd}_\alpha}}(n)=\Theta(1)$.


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