\subsection{STV and Plurality}

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

\noindent\textbf{Proof Sketch.} Given a permutation $\sigma$ of $[m]$, the sub-event $Y$ is ``$\sigma(j)$ is the only loser and being eliminated in $j$-th round, while the plurality winner is $i$''. Given the sequence of eliminated alternatives, we can characterize $Y$ by a polyhedron $\mathcal{H}^{\sigma,i}$. We then construct a profile in $\mathcal{H}^{\sigma,i}$ and hence the conditions of Lemma~\ref{lem:Theta1} holds. Finally, we apply Lemma~\ref{lem:Theta1} and derive the conclusion.
Full proof is presented in Appendix~\ref{apx:stv_plur}.

\begin{theorem}\label{thm:stv_plur_dd}
    Given $m\geq 3$, 
    \[
        \widetilde{\DI}^{max}_{STV, Plurality}(n)=\Theta(n^{-\frac{1}{2}}).
    \]
\end{theorem}

%The event ``one of the STV winners is also one of the plurality losers" occurs only when there are at least two plurality losers, one of whom is an STV winner and survives the first round under a specific tie-breaking scheme. Consequently, $\DI$ is not an appropriate measure for capturing the difference between STV and Plurality. Hence, $\EI$ is a more suitable measure for these two voting rules.
\paragraph{Proof Sketch of Theorem~\ref{apx:thm:stv_plur_dd}.} 
We analyze the desired event by decomposing it into the union of sub-events defined by the elimination sequence and the tie-breaking sequence. Let $\mathcal{H}$ denote the corresponding polyhedron, and let $\mathcal{H}_{\leq 0}$ represent its characteristic cone. We prove that $\dim(\mathcal{H}{\leq 0}) \leq m! - 1$ in the case of a tie, which concludes our proof. The full proof is presented in Appendix~\ref{apx:thm:stv_plur_dd}.

For the converse, the event ``one of the plurality winners is the STV loser (eliminated in the first round)'' can only occur when all candidates have the same plurality score. By following a similar approach to the proof of Theorem~\ref{thm:stv_plur_dd}, we derive the following corollary.
\begin{corollary}\label{coro:stv_plur_dd}
    \[
        \widetilde{\DI}^{max}_{Plurality, STV}(n)=\Theta(n^{-\frac{m-1}{2}}).
    \]
\end{corollary}
