SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\left| \mathbb {E} [ \mathrm {d s w} (j, \boldsymbol {\sigma}) ] - \frac {\operatorname {S c o r e} (j)}{m + 1} \right| \leq n \varepsilon .
$$
Because  $\mathrm{Borda}(\sigma)$  is the alternative that maximizes  $\mathrm{Score}(j)$ , we can apply this inequality twice to get
$$
\begin{array}{l} \mathbb {E} [ \mathrm {d s w} (\operatorname {B o r d a}, \sigma) ] + n \varepsilon \geq \frac {\operatorname {S c o r e} (\operatorname {B o r d a} (\sigma))}{m + 1} \\ \geq \frac {\operatorname {S c o r e} (\operatorname {e w m r} (\sigma))}{m + 1} \\ \geq \mathbb {E} [ \mathrm {d s w} (\mathrm {e w m r}, \boldsymbol {\sigma}) ] - n \varepsilon . \\ \end{array}
$$
To relate  $n \varepsilon$  to  $\mathbb{E}[\mathrm{dsw}(\mathrm{ewmr}, \sigma)]$ , we need to lower bound the expectation. If two distributions supported on  $[0,1]$  have total variation distance of  $\varepsilon$ , then the means of the two distributions differ by at most  $\varepsilon$ . Therefore, if  $\mu$  is the expected value of  $P$ , then  $\mu \geq 0.5 - \varepsilon$ . This means Lemma F.1 implies:
$$
\mathbb {E} [ \mathrm {d s w} (\operatorname {e w m r}, \sigma) ] \geq n \mu \geq n (0. 5 - \varepsilon).
$$
Therefore, the two equations above combine to:
$$
\begin{array}{l} \mathbb {E} [ \mathrm {d s w} (\text {B o r d a}, \boldsymbol {\sigma}) ] \geq \mathbb {E} [ \mathrm {d s w} (\text {e w m r}, \boldsymbol {\sigma}) ] - 2 n \varepsilon \\ = \left(1 - \frac {2 n \varepsilon}{\mathbb {E} [ \mathrm {d s w} (\mathrm {e w m r} , \boldsymbol {\sigma}) ]}\right) \mathbb {E} [ \mathrm {d s w} (\mathrm {e w m r}, \boldsymbol {\sigma}) ] \\ \geq \left(1 - \frac {2 n \varepsilon}{\frac {n}{2} - n \varepsilon}\right) \mathbb {E} [ \mathrm {d s w (e w m r}, \pmb {\sigma}) ] \\ \geq (1 - 5 \varepsilon) \mathbb {E} [ \mathrm {d s w} (\mathrm {e w m r}, \sigma) ] \\ \end{array}
$$
The last inequality comes from taking  $\varepsilon \leq \frac{1}{10}$ .
O Plurality
Theorem O.1 For every value of  $n, m$  and every distribution  $\mathcal{D}$ , plurality is an  $\alpha$ -SWMR for
$$
\alpha = \max  \left(\frac {1}{m}, \frac {1}{n}\right)
$$
Proof. Define  $\text{Plurality}(\sigma)$  as the plurality winner. Then by definition,  $\text{Plurality}(\sigma)$  is the top-ranked alternative by at least  $\lceil \frac{n}{m} \rceil$  voters. The maximum total expected welfare for an alternative is  $n \mathbb{E}[X_{(m)}]$ , where  $\mathbb{E}[X_{(m)}]$  is the expected value of the maximum from  $m$  i.i.d. draws from  $\mathcal{D}$ . Therefore, the expected welfare of  $\text{Plurality}(\sigma)$  is lower bounded as follows:
$$
\begin{array}{l} \mathbb {E} \left[ \mathrm {d s w} (\text {P l u r a l i t y}, \boldsymbol {\sigma}) \right] \geq \lceil \frac {n}{m} \rceil \cdot \mathbb {E} \left[ X _ {(m)} \right] \\ \geq \frac {\left[ \frac {n}{m} \right]}{n} \cdot \mathbb {E} [ \mathrm {d s w} (\mathrm {e w m r}, \boldsymbol {\sigma}) ] \\ \geq \max  \left(\frac {1}{m}, \frac {1}{n}\right) \cdot \mathbb {E} [ \mathrm {d s w} (\operatorname {e w m r}, \boldsymbol {\sigma}) ]. \\ \end{array}
$$