SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
p, q = \underset {p, q} {\arg \min } \left(\max  \left(\frac {q + \alpha}{p - \alpha}, \frac {p + \alpha}{1 + q - \alpha}\right)\right)
$$
Solving this gives that the minimum is achieved at  $(p,q) = (\sqrt{0.75},0.5)$ . For this choice of  $p,q$ , there is no choice of alternative that achieves at least a
$$
\frac {0 . 5 + \alpha}{\sqrt {0 . 7 5} - \alpha} \leq \sqrt {\frac {1}{3}} + 4 \alpha \leq \sqrt {\frac {1}{3}} + \alpha_ {0}
$$
fraction of the maximum expected welfare alternative if the underlying distribution can be either  $P_{1}$  or  $P_{2}$ . Therefore, we have shown that no rule can be better than  $(\sqrt{\frac{1}{3}} + \alpha_{0})$ -EWMR for all symmetric distributions, and taking the limit as  $\alpha_{0} \to 0$  gives the desired theorem result.
J Proof of Theorem 5.7
Lemma J.1 If  $X_{(m + 1 - k)}$  is the  $(m + 1 - k)^{th}$  order statistic of  $m$  draws from a distribution supported on  $[0,1]$  with median  $\nu$ , then:
$$
\mathbb {E} [ X _ {(m + 1 - k)} ] \geq 2 ^ {- m} \cdot \nu \cdot \sum_ {\ell = k} ^ {m} \binom {m} {\ell}.
$$
Proof. Note that a random variable  $X \sim \mathcal{D}$  is greater than or equal to the median with probability at least  $\frac{1}{2}$ . Therefore the probability that the  $(m + 1 - k)^{th}$  largest out of  $m$  draws is greater than the median is at least:
$$
\begin{array}{l} \Pr \left(X _ {(m + 1 - k)} \geq \nu\right) \geq \Pr \left(\operatorname {B i n} (m, \frac {1}{2}) \geq k\right) \\ = \sum_ {\ell = k} ^ {m} \binom {m} {\ell} 2 ^ {- m} \\ = 2 ^ {- m} \sum_ {\ell = k} ^ {m} \left( \begin{array}{c} m \\ \ell \end{array} \right). \\ \end{array}
$$
Therefore, we can lower bound the expectation of  $X_{(m + 1 - k)}$  by
$$
\begin{array}{l} \mathbb {E} \left[ X _ {(m + 1 - k)} \right] \geq \nu \Pr \left(X _ {(m + 1 - k)} \geq \nu\right) \\ \geq \nu \cdot 2 ^ {- m} \sum_ {\ell = k} ^ {m} \left( \begin{array}{c} m \\ \ell \end{array} \right). \\ \end{array}
$$
Proof of Theorem 5.7. Define  $\operatorname{Score}(j)$  to be the score of  $j$  under the Binomial Voting:
$$
\operatorname {S c o r e} (j) := \sum_ {i = 1} ^ {n} \sum_ {\ell = r _ {i}} ^ {m} \left( \begin{array}{c} m \\ \ell \end{array} \right).
$$
The scaled sum of the scores of all the alternatives can then be written as
$$
\begin{array}{l} \sum_ {j = 1} ^ {m} 2 ^ {- m} \operatorname {S c o r e} (j) = n \sum_ {k = 1} ^ {m} \sum_ {\ell = k} ^ {m} \binom {m} {\ell} 2 ^ {- m} \\ = n \sum_ {k = 1} ^ {m} k \binom {m} {k} 2 ^ {- m} \\ = n \cdot E \left[ \operatorname {B i n} \left(m, \frac {1}{2}\right) \right] \\ = \frac {n m}{2}, \\ \end{array}
$$
where the second equality follows by expanding the inner sum. Let  $\mathrm{Binom}(\sigma)$  be the alternative that is chosen by the Binomial Voting. Then by the pigeonhole principle, the score of  $\mathrm{Binom}(\sigma)$  must satisfy
$$
2 ^ {- m} \cdot \operatorname {S c o r e} (\operatorname {B i n o m} (\sigma)) \geq \frac {n}{2}.
$$
We next provide a lower bound on the expected social welfare of an alternative based on the Binomial Voting, and use this to lower bound the expected social welfare of  $\operatorname{Binom}(\sigma)$ . Recall that if an alternative  $j$  is ranked  $r_1, \ldots, r_n$  by the  $n$  voters, then Boutilier et al. [2015] tells us that:
$$
\mathbb {E} \left[ \mathrm {d s w} (j, \boldsymbol {\sigma}) \right] = \sum_ {i = 1} ^ {n} E \left[ X _ {(m + 1 - r _ {1})} \right].
$$
Applying Lemma J.1, we then find that
$$
\begin{array}{l} \mathbb {E} [ \mathrm {d s w} (j, \boldsymbol {\sigma}) ] \geq \sum_ {i = 1} ^ {n} 2 ^ {- m} \nu \sum_ {\ell = r _ {i}} ^ {m} \left( \begin{array}{c} m \\ \ell \end{array} \right) \\ = \nu \cdot 2 ^ {- m} \cdot \operatorname {S c o r e} (j). \\ \end{array}
$$
For  $\operatorname{Binom}(\sigma)$  this yields
$$
\begin{array}{l} \mathbb {E} [ \mathrm {d s w} (\operatorname {B i n o m}, \sigma) ] \geq \nu \cdot 2 ^ {- m} \cdot \operatorname {S c o r e} (\operatorname {B i n o m} (\sigma)) \\ \geq \frac {n \nu}{2} \\ \geq \frac {\nu}{2} \mathbb {E} [ \mathrm {d s w} (\operatorname {e w m r}, \boldsymbol {\sigma}) ], \\ \end{array}
$$
K Generalized Binomial Voting
Lemma K.1 If  $X_{(m + 1 - k)}$  is the  $(m + 1 - k)^{th}$  order statistic of  $m$  draws from a distribution  $\mathcal{D}$  supported on  $[0,1]$ , where  $Q$  is value of the  $p^{th}$  quantile of  $\mathcal{D}$ , then
$$
\mathbb{E}[X_{(m + 1 - k)}]\geq Q\cdot \sum_{\ell = k}^{m}\binom {m}{\ell}(1 - p)^{\ell}p^{m - \ell}.
$$
Proof. As in the proof of Lemma J.1,
$$
\begin{array}{l} \Pr (X _ {(m + 1 - k)} \geq Q) \geq \Pr (\operatorname {B i n} (m, 1 - p) \geq k) \\ = \sum_ {\ell = k} ^ {m} \binom {m} {\ell} (1 - p) ^ {\ell} p ^ {m - \ell}. \\ \end{array}
$$
$$
\begin{array}{l} \mathbb {E} \left[ X _ {(m + 1 - k)} \right] \geq Q \cdot \Pr \left(X _ {(m + 1 - k)} \geq Q\right) \\ \geq Q \cdot \sum_ {\ell = k} ^ {m} \binom {m} {\ell} (1 - p) ^ {\ell} p ^ {m - \ell}. \\ \end{array}
$$
Proof of Corollary 5.9. If we define  $r_1, \ldots, r_n$  as the rankings of alternative  $j$  and
$$
\operatorname {S c o r e} (j) := \sum_ {i = 1} ^ {n} \sum_ {\ell = r _ {i}} ^ {m} \binom {m} {\ell} (1 - p) ^ {\ell} p ^ {m - \ell},
$$
then by Lemma K.1,
$$
\begin{array}{l} \mathbb {E} [ \mathrm {d s w} (j, \boldsymbol {\sigma}) ] = \sum_ {i = 1} ^ {n} \mathbb {E} \left[ X _ {(m + 1 - r _ {i})} \right] \\ \geq \sum_ {i = 1} ^ {n} Q \sum_ {\ell = r _ {i}} ^ {m} \binom {m} {\ell} (1 - p) ^ {\ell} p ^ {m - \ell} \\ = Q \cdot \operatorname {S c o r e} (j). \\ \end{array}
$$
The sum of the scores of all the alternatives can then be written as:
$$
\begin{array}{l} \sum_ {j = 1} ^ {m} \operatorname {S c o r e} (j) = n \sum_ {k = 1} ^ {m} \sum_ {\ell = k} ^ {m} \binom {m} {\ell} (1 - p) ^ {\ell} p ^ {m - \ell} \\ = n \mathbb {E} [ \operatorname {B i n} (m, 1 - p) ] \\ = n m (1 - p). \\ \end{array}
$$
The rest of the proof follows as in the proof of Theorem 5.7.
Lemma K.2 Let  $X$  be a random variable supported on  $[0,1]$ . Choose  $T$  values in  $[0,1]$ , denoted  $0 \leq Q_1 \leq \ldots \leq Q_T \leq 1$ . Define  $Q_0 = 0$  and  $Q_{t+1} = 2$ . Then
$$
\mathbb {E} [ X ] \geq \sum_ {t = 1} ^ {T} Q _ {t} \cdot \Pr (Q _ {t} \leq X <   Q _ {t + 1}).
$$
Proof. The expectation of  $X$  can be written as
$$
\begin{array}{l} \mathbb {E} [ X ] = \int_ {0} ^ {1} \Pr (X \geq x) d x \\ = \int_ {Q _ {T}} ^ {1} \Pr (X \geq x) d x + \sum_ {t = 0} ^ {T - 1} \int_ {Q _ {t}} ^ {Q _ {t + 1}} \Pr (X \geq x) d x \\ \geq \sum_ {t = 0} ^ {T - 1} \int_ {Q _ {t}} ^ {Q _ {t + 1}} \Pr (X \geq Q _ {t + 1}) d x \\ \geq \sum_ {t = 0} ^ {T - 1} \left(Q _ {t + 1} - Q _ {t}\right) \cdot \Pr (X \geq Q _ {t + 1}) \\ = \sum_ {t = 1} ^ {T} Q _ {t} \cdot \Pr (Q _ {t} \leq X <   Q _ {t + 1}). \\ \end{array}
$$
Proof of Theorem 5.11. Let  $X_{(m - k + 1)}$  be the  $(m - k + 1)^{th}$  order statistic from  $m$  i.i.d. draws from distribution  $\mathcal{D}$ . Consider the probability  $\operatorname*{Pr}(A \leq X_{(m - k + 1)} < B)$  for  $0 \leq A \leq B \leq 1$ . Define  $p_A := \operatorname*{Pr}(X < A)$  and  $p_B := \operatorname*{Pr}(X < B)$ . In order for  $A \leq X_{(m - k + 1)} < B$ , there must be at least  $k$  draws greater than or equal to  $A$ , and at most  $k - 1$  draws greater than or equal to  $B$ . The probability that there are at least  $k$  draws greater than or equal to  $A$  is
$$
\sum_ {\ell = k} ^ {m} \binom {m} {\ell} (1 - p _ {A}) ^ {\ell} p _ {A} ^ {m - \ell},
$$
and the probability of having at least  $k$  draws that are greater than or equal to  $B$  is
$$
\sum_ {\ell = k} ^ {m} \binom {m} {\ell} (1 - p _ {B}) ^ {\ell} p _ {B} ^ {m - \ell}.
$$
Now using the fact that the second event is a subset of the first event, we have that
$$
\Pr (A \leq X _ {(m - k + 1)} <   B) = \sum_ {\ell = k} ^ {m} \binom {m} {\ell} \left(1 - p _ {A}\right) ^ {\ell} p _ {A} ^ {m - \ell} - \sum_ {\ell = k} ^ {m} \binom {m} {\ell} \left(1 - p _ {B}\right) ^ {\ell} p _ {B} ^ {m - \ell}. \tag {17}
$$
Therefore the score in Generalized Binomial Voting is a lower bound for the expectation of the  $(m - k + 1)^{th}$  order statistic, as shown below. We start by applying Lemma K.2 to the order statistic  $X_{(m - k + 1)}$ . By combining this with (17) we get
$$
\begin{array}{l} \mathbb {E} \left[ X _ {(m - k + 1)} \right] \geq \sum_ {t = 1} ^ {T} Q _ {t} \cdot \Pr \left(Q _ {t} \leq X _ {(m - k + 1)} <   Q _ {t + 1}\right) \\ = \sum_ {t = 1} ^ {T} Q _ {t} \cdot \left(\sum_ {\ell = k} ^ {m} \binom {m} {\ell} \left(1 - p _ {t}\right) ^ {\ell} p _ {t} ^ {m - \ell} - \sum_ {\ell = k} ^ {m} \binom {m} {\ell} \left(1 - p _ {t + 1}\right) ^ {\ell} p _ {t + 1} ^ {m - \ell}\right). \tag {18} \\ \end{array}
$$
Suppose alternative  $j$  has rankings  $r_1, \ldots, r_n$  by the  $n$  voters, and again define  $\mathrm{Score}(j)$  as the total score of this alternative under the Generalized Binomial Voting rule. Then by linearity of expectation and the above inequality,
$$
\begin{array}{l} \mathbb {E} [ \mathrm {d s w} (j, \boldsymbol {\sigma}) ] = \sum_ {i = 1} ^ {n} \mathbb {E} [ X _ {(m + 1 - r _ {i})} ] \\ \geq \sum_ {i = 1} ^ {n} \sum_ {t = 1} ^ {T} Q _ {t} \cdot \left(\sum_ {\ell = r _ {i}} ^ {m} \binom {m} {\ell} (1 - p _ {t}) ^ {\ell} p _ {t} ^ {m - \ell} - \sum_ {\ell = r _ {i}} ^ {m} \binom {m} {\ell} (1 - p _ {t + 1}) ^ {\ell} p _ {t + 1} ^ {m - \ell}\right) \\ = \operatorname {S c o r e} (j). \tag {19} \\ \end{array}
$$
Now the last step is to find a lower bound for the score of  $\mathrm{GBinom}(\pmb{\sigma})$ , the index chosen by Generalized Binomial Voting. Note that if we add all of the scores across all of the alternatives, we have that the value can be written as the sum of expectations of the order statistics  $Z_{(1)},\ldots Z_{(m)}$  of  $m$  draws from a Multinomial distribution with  $T + 1$  outcomes, with probabilities  $\{p_{t + 1} - p_t\}_{t = 0}^T$  and values  $0,Q_1,\dots ,Q_T$  respectively, where we define  $p_0 = 0$ . This allows us to simplify the sum of scores as follows: In a manner similar to the proof of Theorem 5.7, we have
$$
\begin{array}{l} \sum_{j = 1}^{n}\operatorname {Score}(j) = n\cdot \sum_{k = 1}^{m}\sum_{t = 1}^{T}Q_{t}\left(\sum_{\ell = k}^{m}\binom {m}{\ell}(1 - p_{t})^{\ell}p_{t}^{m - \ell} - \sum_{\ell = k}^{m}\binom {m}{\ell}(1 - p_{t + 1})^{\ell}p_{t + 1}^{m - \ell}\right) \\ = n \cdot \sum_ {t = 1} ^ {T} Q _ {t} \left(\sum_ {k = 1} ^ {m} \sum_ {\ell = k} ^ {m} \binom {m} {\ell} (1 - p _ {t}) ^ {\ell} p _ {t} ^ {m - \ell} - \sum_ {k = 1} ^ {m} \sum_ {\ell = k} ^ {m} \binom {m} {\ell} (1 - p _ {t + 1}) ^ {\ell} p _ {t + 1} ^ {m - \ell}\right) \\ = n \cdot \sum_ {t = 1} ^ {T} Q _ {t} \left(\sum_ {k = 1} ^ {m} \Pr [ \operatorname {B i n} (m, 1 - p _ {t}) \geq k ] - \sum_ {k = 1} ^ {m} \Pr [ \operatorname {B i n} (m, 1 - p _ {t + 1}) \geq k ]\right) \\ = n \cdot \sum_ {t = 1} ^ {T} Q _ {t} \left(\mathbb {E} [ \operatorname {B i n} (m, 1 - p _ {t}) ] - \mathbb {E} [ \operatorname {B i n} (m, 1 - p _ {t + 1}) ]\right) \\ = n m \cdot \sum_ {t = 1} ^ {T} Q _ {t} \left(p _ {t + 1} - p _ {t}\right). \tag {20} \\ \end{array}
$$
Finally, we conclude as in the proof of Theorem 5.7 that the score of  $\mathrm{GBinom}(\sigma)$  can be bounded as
$$
\begin{array}{l} \mathbb {E} [ \mathrm {d s w} (\mathrm {G B i n o m}, \boldsymbol {\sigma}) ] \geq \operatorname {S c o r e} (\mathrm {G B i n o m} (\boldsymbol {\sigma})) \\ \geq n \sum_ {t = 1} ^ {T} Q _ {t} \left(p _ {t + 1} - p _ {t}\right) \\ \geq \sum_ {t = 1} ^ {T} Q _ {t} \left(p _ {t + 1} - p _ {t}\right) \mathbb {E} \left[ \mathrm {d s w} \left(\operatorname {e w m r}, \boldsymbol {\sigma}\right) \right], \\ \end{array}
$$
where we used (19), an averaging argument applied to (20), and that  $n \geq \mathbb{E}[\mathrm{dsw}(\mathrm{ewmr},\sigma)]$ , since  $\mathcal{D}$  takes values in  $[0,1]$ .
Proof of Corollary 5.12. To prove this result, we need to show that the scores in Generalized Binomial Voting approach the expected values of the order statistics, which are the scores in the ewmr.
Recall in the above proof, the fact that there is an inequality in the equation  $\mathbb{E}[\mathrm{dsw}(j,\pmb {\sigma})]\geq \mathrm{Score}(j)$  is because of the inequality in Lemma K.2. That inequality came from the fact that, for random
variable  $X$
$$
\begin{array}{l} \mathbb {E} [ X ] = \int_ {0} ^ {1} \Pr (X \geq x) d x \\ = \int_ {Q _ {T}} ^ {1} \Pr (X \geq x) d x + \sum_ {t = 0} ^ {T - 1} \int_ {Q _ {t}} ^ {Q _ {t + 1}} \Pr (X \geq x) d x \\ \geq \sum_ {t = 0} ^ {T - 1} \int_ {Q _ {t}} ^ {Q _ {t + 1}} \Pr (X \geq Q _ {t + 1}) d x \\ \end{array}
$$
In other words, the score in Generalized Binomial Voting can be thought of as a rectangular lower bound approximation of the integral of expectation of the order statistics of the underlying distribution. By the definition of the Riemann Integral, this rectangle approximation will exactly approach the true integral (making the inequality an equality) as  $T \to \infty$ .
L Sampling
As mentioned in Section 6, we can directly calculate an approximation to the EDMR rule by sampling from the known underlying distribution  $\mathcal{D}$ . We formalize this result below.
Consider the following sampling algorithm. Take  $T$  batches of  $nm$  i.i.d. samples each from distribution  $\mathcal{D}$ . For each batch, assign  $m$  samples to each voter  $i$ , and denote the sorted samples as  $X_{(1)}^{i}, \ldots, X_{(m)}^{i}$ . Let  $r_1^j, \ldots, r_n^j$  be the rankings of alternative  $j$  under  $\sigma$ . For each  $j$ , let
$$
\hat {\mu} _ {j} = \frac {1}{T} \sum_ {t = 1} ^ {T} \mu_ {j} ^ {t} = \frac {1}{T} \sum_ {t = 1} ^ {T} \frac {\sum_ {i = 1} ^ {n} X _ {(r _ {i} ^ {j})} ^ {i}}{\max _ {k} \sum_ {i = 1} ^ {n} X _ {(r _ {i} ^ {k})} ^ {i}}
$$
be the mean sample distortion of alternative  $j$ .
Theorem L.1 Let  $\mathcal{D}$  be bounded on  $(0,1]$  with mean  $\mu$ . For any  $\varepsilon, \delta > 0$ , take  $T = \frac{2\log\left(\frac{2m}{\delta}\right)}{\mu^2\varepsilon^2}$ . Let  $j_{\mathrm{Sampling}} = \arg \max_j \hat{\mu}_j$  be the alternative with the highest mean sample distortion, as defined above. With probability  $1 - \delta$ ,  $j_{\mathrm{Sampling}}$  is an  $\varepsilon$ -approximation of the EDMR:
$$
\mathbb {E} [ \mathrm {d d i s t} (\mathrm {S a m p l i n g}, \boldsymbol {\sigma}) ] \geq (1 - \varepsilon) \mathbb {E} [ \mathrm {d d i s t} (\mathrm {e d m r}, \boldsymbol {\sigma}) ].
$$
Proof. By linearity of expectation, clearly  $\mathbb{E}[\hat{\mu}_j] = \mathbb{E}[\mu_j^t ] = \mathbb{E}[\mathrm{ddist}(j,\sigma)]$ . Applying the standard Hoeffding's inequality to  $\mu_{j}$  gives:
$$
\begin{array}{l} P \left(| \hat {\mu} _ {j} - \mathbb {E} [ \mathrm {d d i s t} (j, \pmb {\sigma}) ] | \geq \frac {\varepsilon}{2} \cdot \mathbb {E} [ \mathrm {d d i s t} (\mathrm {e d m r}, \pmb {\sigma}) ]\right) \leq 2 \exp \left(- T \varepsilon^ {2} \mathbb {E} [ \mathrm {d d i s t} (\mathrm {e d m r}, \pmb {\sigma}) ] ^ {2} / 2\right) \\ \leq 2 \exp (- T \varepsilon^ {2} \mu^ {2} / 2) \\ = \frac {\delta}{m}. \\ \end{array}
$$
Let  $\mathcal{E}$  be the event that for all  $j\in [1:m]$
$$
\left| \hat {\mu} _ {j} - \mathbb {E} [ \mathrm {d d i s t} (j, \boldsymbol {\sigma}) ] \right| \leq \frac {\varepsilon}{2} \cdot \mathbb {E} [ \mathrm {d d i s t} (\mathrm {e d m r}, \boldsymbol {\sigma}) ].
$$
By a union bound,  $P(\mathcal{E}) \geq 1 - \delta$ . By construction,  $j_{\mathrm{Sampling}}$  satisfies  $\hat{\mu}_{j_{\mathrm{Sampling}}} \geq \hat{\mu}_{j_{\mathrm{edmr}}}$ . Under event  $\mathcal{E}$ , this implies that
$$
\mathbb {E} [ \mathrm {d d i s t} (\mathrm {S a m p l i n g}, \boldsymbol {\sigma}) ] + \frac {\varepsilon}{2} \mathbb {E} [ \mathrm {d d i s t} (\mathrm {e d m r}, \boldsymbol {\sigma}) ] \geq \mathbb {E} [ \mathrm {d d i s t} (\mathrm {e d m r}, \boldsymbol {\sigma}) ] - \frac {\varepsilon}{2} \mathbb {E} [ \mathrm {d d i s t} (\mathrm {e d m r}, \boldsymbol {\sigma}) ].
$$
This equation simplifies to
$$
\mathbb {E} [ \mathrm {d d i s t} (\mathrm {S a m p l i n g}, \boldsymbol {\sigma}) ] \geq (1 - \varepsilon) \mathbb {E} [ \mathrm {d d i s t} (\mathrm {e d m r}, \boldsymbol {\sigma}) ],
$$
Note that each of the  $T$  batches includes  $nm$  samples, so the total number of samples used in this algorithm is  $Tnm = \frac{nm\log(\frac{2m}{\delta})}{\mu^2\varepsilon^2}$ .
M Benchmarks
In this section we elaborate on alternative benchmarks for our results, as briefly mentioned in the discussion. While our results use the EDMR and EwMR as benchmarks, we can also rewrite the statements to instead compare to the actual expected values. We restate Theorem 4.2 and Theorem 5.7 below as Corollary M.1 and Corollary M.2 respectively.
Corollary M.1 Assume  $\mathcal{D}$  is supported on  $[0,1]$  and has constant (relative to  $n,m$ ) mean  $\mu$  and variance  $s^2$ . Then for every  $\varepsilon_0 > 0$ , there exists  $n_0$  such that if  $n \geq n_0$ ,  $\mathbb{E}\big[\mathrm{dist}(\mathrm{ewmr},\sigma)\big] \geq 1 - \varepsilon_0$ .
Proof. Define  $\mathcal{E}$  as in the proof of Theorem 4.2. Under event  $\mathcal{E}$ , we have that
$$
\begin{array}{l} \operatorname {d d i s t} (\operatorname {e w m r}, \boldsymbol {\sigma}) = \frac {\operatorname {s w} (\operatorname {e w m r} , \boldsymbol {\sigma})}{\max  _ {i} \operatorname {s w} (i , \boldsymbol {\sigma})} \\ \geq \frac {(1 - \varepsilon) \mathbb {E} [ \mathrm {s w} (\mathrm {e w m r} , \boldsymbol {\sigma}) ]}{(1 + \varepsilon) \mathbb {E} [ \mathrm {s w} (\mathrm {e w m r} , \boldsymbol {\sigma}) ]} \\ = \frac {1 - \varepsilon}{1 + \varepsilon} \\ \end{array}
$$
We can then write
$$
\begin{array}{l} \mathbb {E} [ \mathrm {d d i s t} (\mathrm {e w m r}, \boldsymbol {\sigma}) ] \geq \Pr (\mathcal {E}) \cdot \frac {1 - \varepsilon}{1 + \varepsilon} + \Pr (\neg \mathcal {E}) \cdot 0 \\ \geq \left(1 - 2 e ^ {- 2 \varepsilon^ {2} n \mu^ {2}}\right) \frac {1 - \varepsilon}{1 + \varepsilon} \\ \geq \left(1 - 2 e ^ {- 2 \varepsilon^ {2} n \mu^ {2}}\right) \left(1 - \frac {2 \varepsilon}{1 + \varepsilon}\right) \\ \geq 1 - 2 e ^ {- 2 \varepsilon^ {2} n \mu^ {2}} - 2 \varepsilon \\ \geq 1 - \varepsilon_ {0} \\ \end{array}
$$
where the last inequality comes from taking  $n$  suitably large and  $\varepsilon = \varepsilon_0 / 3$ .
Corollary M.2 Let  $\mathcal{D}$  be a distribution supported on  $[0,1]$  whose largest median is  $\nu$ , i.e.,  $\nu = \sup \left\{y \mid \operatorname*{Pr}_{x \sim \mathcal{D}}(x \leq y) \leq \frac{1}{2}\right\}$ . Then binomial voting achieves expected welfare of at least  $\frac{n\nu}{2}$ .
Proof. In the last expression of the proof of Theorem 5.7, we exactly have
$$
\begin{array}{l} \mathbb {E} [ \mathrm {d s w} (\operatorname {B i n o m}, \sigma) ] \geq \nu \cdot 2 ^ {- m} \cdot \operatorname {S c o r e} (\operatorname {B i n o m} (\sigma)) \\ \geq \frac {n \nu}{2} \\ \end{array}
$$
N Total Variation Distance
A desirable result would be that the EwMR for one distribution generalizes well to similar distributions. Unfortunately, this is not the case, and in fact the EwMR for one distribution can have arbitrarily bad expected welfare on an a second distribution that is arbitrarily close in total variation distance.
Theorem N.1 For every  $0 \leq \delta \leq 1$  and  $0 < \varepsilon \leq 1$  there exists  $n, m, \sigma$  and two distributions  $D_{1}, D_{2}$  both supported on  $[0,1]$  such that  $TV(D_{1}, D_{2}) \leq \varepsilon$  and the social welfare maximizing rule on alternative  $D_{1}$  is not a  $\delta$ -EWMR for  $D_{2}$ .
Proof. This follows directly from the construction in the proof of Theorem 5.2 in Appendix G.
N.1 Borda Count
Borda count has been previously shown to be the scoring rule that maximizes expected welfare for the uniform distribution (which we also prove in Lemma N.2. Despite the negative result of Theorem N.1, a somewhat surprising result is that Borda count is also a  $(1 - \varepsilon)$ -EWMR for distributions close in total variation distance to the uniform distribution. We formalize this result in Theorem N.4.
Lemma N.2 (Weber [1978]) Define Borda count as the scoring rule that assigns scores  $(1,2,\dots,m)$  to the alternatives in rank order. Then Borda count is equivalent to the expected welfare maximizing rule when the underlying distribution is Uniform[0, 1].
Proof. Let  $X_{(1)}, \ldots, X_{(m)}$  be the order statistics from  $m$  i.i.d. draws from the Unif(0,1) distribution. Then  $\mathbb{E}[X_{(k)}] = \frac{k}{m + 1}$ . For alternative  $j$ , let  $r_1, \ldots, r_n$  be the ranks of alternative  $j$  by voters  $1, \ldots, n$  and let  $\mathrm{Score}(j)$  be the Borda count score of alternative  $j$ . Then by linearity of expectation,
$$
\begin{array}{l} \mathbb {E} \left[ \mathrm {d s w} (j, \boldsymbol {\sigma}) \right] = \mathbb {E} \left[ \sum_ {i = 1} ^ {n} X _ {(m + 1 - r _ {i})} \right] \\ = \sum_ {i = 1} ^ {n} \mathbb {E} \left[ X _ {\left(m + 1 - r _ {i}\right)} \right] \\ = \frac {1}{m + 1} \sum_ {i = 1} ^ {n} m + 1 - r _ {i} \\ = \frac {1}{m + 1} \cdot \operatorname {S c o r e} (j) \\ \end{array}
$$
Therefore, maximizing the expected social welfare is equivalent to maximizing the Borda count score  $\operatorname{Score}(j)$ .
Lemma N.3 Let  $U \sim \mathrm{Unif}(0,1)$  be the uniform distribution and  $P$  be a continuous distribution supported on  $[0,1]$  with an invertible cdf such that
$$
T V (U, P) = \sup  _ {A \subseteq [ 0, 1 ]} | P (A) - U (A) | \leq \varepsilon .
$$
Define  $X_{(k)}$  as the kth order statistic out of  $m$  i.i.d. draws from  $P$ . Then
$$
\left| \mathbb {E} [ X _ {(k)} ] - \frac {k}{m + 1} \right| \leq \varepsilon .
$$
Proof. Let  $F_P$  and  $F_U$  be the cdfs of  $P$  and  $U$  respectively. Under the assumption, for all  $x \in [0,1]$ :
$$
\left| F _ {P} (x) - F _ {U} (x) \right| = \left| F _ {P} (x) - x \right| \leq \varepsilon .
$$
Because the slope of  $x$  is 1 and these distributions are supported on  $[0,1]$ , this implies that for all  $y \in [0,1]$ :
$$
\left| F _ {P} ^ {- 1} (y) - y \right| \leq \varepsilon .
$$
The probability integral transform for order statistics gives that  $X_{(k)} \sim F^{-1}(U_{(k)})$  where  $U_{(k)}$  is the  $k$ th order statistic from  $m$  i.i.d. draws from  $U$ . Using that  $\mathbb{E}[U_{(k)}] = \frac{k}{m + 1}$ , we have the following
result for the expectation of  $X_{(k)}$ .
$$
\begin{array}{l} \left| \mathbb {E} \left[ X _ {(k)} \right] - \frac {k}{m + 1} \right| = \left| \mathbb {E} \left[ F ^ {- 1} \left(U _ {(k)}\right) \right] - \frac {k}{m + 1} \right| \\ = \left| \int_ {0} ^ {1} F ^ {- 1} (y) f _ {U _ {(k)}} (y) d y - \frac {k}{m + 1} \right| \\ = \left| \int_ {0} ^ {1} \left(F ^ {- 1} (y) - y\right) f _ {U _ {(k)}} (y) d y + \mathbb {E} [ U _ {(k)} ] - \frac {k}{m + 1} \right| \\ = \left| \int_ {0} ^ {1} \left(F ^ {- 1} (y) - y\right) f _ {U _ {(k)}} (y) d y \right| \\ \leq \int_ {0} ^ {1} | F ^ {- 1} (y) - y | f _ {U _ {(k)}} (y) d y \\ \leq \int_ {0} ^ {1} \varepsilon f _ {U _ {(k)}} (y) d y \\ = \varepsilon . \\ \end{array}
$$
Theorem N.4 Let  $P$  be a continuous distribution with invertible cdf that is supported on  $[0,1]$  such that  $TV(U(0,1),P)\leq \varepsilon \leq \frac{1}{10}$ . Let Score represent the Borda score. Then the Borda count winner is a 5-approximation of the expected welfare maximizing rule.
$$
\mathbb {E} [ \mathrm {d s w} (\text {B o r d a}, \sigma) ] \geq (1 - 5 \varepsilon) \mathbb {E} [ \mathrm {d s w} (\text {e w m r}, \sigma) ].
$$
Proof. Recall that lemma N.2 shows that
$$
\begin{array}{l} \operatorname {S c o r e} (j) = \sum_ {i = 1} ^ {n} m + 1 - r _ {i} ^ {j} \\ = (m + 1) \left(n - \sum_ {i = 1} ^ {n} \frac {r _ {j} ^ {i}}{m + 1}\right). \\ \end{array}
$$
Define  $X_{(m + 1 - r_i^j)}$  to be the  $(m + 1 - r_i^j)^{th}$  order statistic. Recall that Lemma N.3 states that for all  $r_i^j$ ,
$$
\left| \mathbb {E} \left[ X _ {\left(m + 1 - r _ {i} ^ {j}\right)} \right] - \frac {m + 1 - r _ {i} ^ {j}}{m + 1} \right| \leq \varepsilon .
$$
$$
E \left[ \operatorname {d s w} (j, \boldsymbol {\sigma}) \right] = \sum_ {i = 1} ^ {n} \mathbb {E} \left[ X _ {(m + 1 - r _ {i} ^ {j})} \right].
$$
Combining these two equations gives
$$
\left| \mathbb {E} [ \mathrm {d s w} (j, \boldsymbol {\sigma}) ] - \sum_ {i = 1} ^ {n} \frac {m + 1 - r _ {i} ^ {j}}{m + 1} \right| \leq n \varepsilon ,
$$