SlowGuess/NeurIPS_2023_OCR huggingface dataset:
where the expectation in the numerator is taken only over randomness in the voting rule  $f$ . In this paper we choose to only consider the deterministic rules because linearity of expectation implies that any randomized voting rule  $f$  is weakly worse than a deterministic rule. Therefore, although the number of randomized rules is infinite, considering the finite number of deterministic rules is sufficient for finding an EDMR. Note that the same argument holds for EwMR as well. This result is formalized in the following lemma.
Lemma A.1 For any randomized voting rule, there exists a deterministic rule that has expected distortion as least as large as that of the randomized rule. The same result holds for expected welfare.
Proof. We will start by proving the result for expected distortion. The expected distortion of a randomized voting rule  $f$  can be written as follows, where the inner expectation is over the randomness in the voting rule  $f$  and the outer expectation is over the random utilities  $\mathbf{u} \triangleright \sigma$ . Note that the randomness in  $f$  is over the  $m$  alternatives that  $f$  can select.
$$
\begin{array}{l} \underset {\mathbf {u} \triangleright \boldsymbol {\sigma}} {\mathbb {E}} \left[ \frac {\mathbb {E} [ \operatorname {s w} (f (\boldsymbol {\sigma}) , \mathbf {u}) ]}{\operatorname* {m a x} _ {k \in A} \operatorname {s w} (k , \mathbf {u})} \right] = \underset {\mathbf {u} \triangleright \boldsymbol {\sigma}} {\mathbb {E}} \left[ \frac {\sum_ {j = 1} ^ {m} \Pr (f (\boldsymbol {\sigma}) = j) \cdot \operatorname {s w} (j , \mathbf {u})}{\operatorname* {m a x} _ {k \in A} \operatorname {s w} (k , \mathbf {u})} \right] \\ = \underset {\mathbf {u} \triangleright \boldsymbol {\sigma}} {\mathbb {E}} \left[ \sum_ {j = 1} ^ {m} \Pr (f (\boldsymbol {\sigma}) = j) \cdot \frac {\operatorname {s w} (j , \mathbf {u})}{\max  _ {k \in A} \operatorname {s w} (k , \mathbf {u})} \right] \\ = \sum_ {j = 1} ^ {m} \Pr (f (\boldsymbol {\sigma}) = j) \cdot \underset {\mathbf {u} \triangleright \boldsymbol {\sigma}} {\mathbb {E}} \left[ \frac {\operatorname {s w} (j , \mathbf {u})}{\max  _ {k \in A} \operatorname {s w} (k , \mathbf {u})} \right] \\ \leq \max  _ {j} \left(\mathbb {E} _ {\mathbf {u} \triangleright \sigma} \left[ \frac {\operatorname {s w} (j , \mathbf {u})}{\operatorname* {m a x} _ {k \in A} \operatorname {s w} (k , \mathbf {u})} \right]\right). \\ \end{array}
$$
This completes the proof for expected distortion, as the final expression is a deterministic rule.
For expected welfare, the result follows in a similar manner:
$$
\begin{array}{l} \underset {\mathbf {u} \triangleright \boldsymbol {\sigma}} {\mathbb {E}} \left[ \mathbb {E} \left[ \operatorname {s w} (f (\boldsymbol {\sigma}), \mathbf {u}) \right] \right] = \underset {\mathbf {u} \triangleright \boldsymbol {\sigma}} {\mathbb {E}} \left[ \sum_ {j = 1} ^ {m} \Pr (f (\boldsymbol {\sigma}) = j) \cdot \operatorname {s w} (j, \mathbf {u}) \right] \\ = \sum_ {j = 1} ^ {m} \Pr (f (\boldsymbol {\sigma}) = j) \cdot \underset {\mathbf {u} \triangleright \boldsymbol {\sigma}} {\mathbb {E}} [ \operatorname {s w} (j, \mathbf {u}) ] \\ \leq \max  _ {j} \left(\mathop {\mathbb {E}} _ {\mathbf {u} \triangleright \sigma} [ \operatorname {s w} (j, \mathbf {u}) ]\right). \\ \end{array}
$$
B Infinite Expected Inverse Distributional Distortion
As mentioned in Section 2, we chose to define distributional distortion with the ratio in the opposite direction as worst-case distortion. Another possible definition would have been to define the inverse distributional distortion as follows, which is exactly the inverse of distributional distortion. Note that we have defined inverse distributional distortion using the more general definition for randomized voting rules introduced above, as in (3).
Definition B.1 We define the inverse distributional distortion of a rule  $f$  under preference profile  $\sigma$  as
$$
\operatorname {i n v d d i s t} (f, \boldsymbol {\sigma}) = \frac {\max  _ {j \in A} \operatorname {s w} (j , \mathbf {u})}{\mathbb {E} [ \operatorname {s w} (f (\boldsymbol {\sigma}), \mathbf {u}) ]}. \tag {4}
$$
where the expectation is over the the randomness of the voting rule.
As before, we can consider the expectation of the random variable  $\mathrm{invddist}(f,\sigma)$  with respect to a random  $\mathbf{u} \triangleright \sigma$ . Unfortunately, even for relatively simple distributions supported on [0, 1], the expectation of  $\mathrm{invddist}(f,\sigma)$  can be infinite. This makes comparing rules more difficult when using expected inverse distortion as the metric.
Lemma B.2 Let  $m = n = 2$  and  $\sigma$  be a symmetric preference profile. If  $\mathcal{D} \sim \beta\left(\frac{1}{4}, 1\right)$ , then no deterministic rule  $f$  has finite  $\mathbb{E}[\mathrm{invddist}(f, \sigma)]$ .
Proof. Let  $f(x) = 0.25x^{-0.75}$  be the pdf of  $\mathcal{D}$ . Note that the joint pdf, denoted  $f_{X_{(1)}X_{(2)}}(x,y)$ , of the minimum  $X_{(1)}$  and maximum  $X_{(2)}$  of two i.i.d. draws from  $\mathcal{D}$  is:
$$
f _ {X _ {(1)} X _ {(2)}} (x, y) = 2 f (x) f (y).
$$
Because  $n = m = 2$ , each of the two voters is drawing two utilities from  $\mathcal{D}$ , for a total of 4 i.i.d. draws from  $\mathcal{D}$ . Denote the utilities of voter 1 as  $(w, x)$  and the utilities of voter 2 as  $(y, z)$  respectively, and consider the case when  $x \geq w$  and  $y \geq z$ , with corresponding ranking  $\sigma$ . The preference profile is symmetric, and therefore the expected inverse distortions of the two alternatives are equal. We can assume WLOG that alternative 1 has welfare of  $y + w$  (ranked second by voter 1 and first by voter 2) and alternative 2 has welfare of  $x + z$  (ranked first by voter 1 and second by voter 2). Using the joint pdf for the two sets of two draws, this allows us to calculate the expected inverse distortion of alternative 1 as follows:
$$
\begin{array}{l} \mathbb {E} \left[ \operatorname {i n v d d i s t} (1, \sigma) \right] = \int_ {0} ^ {1} \int_ {0} ^ {1} \int_ {0} ^ {x} \int_ {0} ^ {y} \frac {\max  (x + z , y + w)}{y + w} 4 f (w) f (z) f (x) f (y) d z d w d y d x \\ \geq 4 \int_ {0} ^ {1} \int_ {0} ^ {1} \int_ {0} ^ {x} \int_ {0} ^ {y} \frac {x + z}{y + w} f (w) f (z) f (x) f (y) d z d w d y d x \\ = \frac {1}{6 4} \int_ {0} ^ {1} \int_ {0} ^ {1} \int_ {0} ^ {x} \int_ {0} ^ {y} \frac {x + z}{y + w} \cdot \frac {1}{(x y z w) ^ {0 . 7 5}} d z d w d y d x \\ \longrightarrow \infty . \\ \end{array}
$$
Therefore no deterministically chosen alternative has finite  $\mathbb{E}[\mathrm{invddist}(f,\sigma)]$ .
C Randomization and Inverse Distributional Distortion
In Appendix A, we showed that a deterministic rule always achieves the highest expected distortion. For expected inverse distortion as defined in Appendix B, this is no longer the case. In fact, there exist distributions  $\mathcal{D}$  supported on  $[0,1]$  such that the best deterministic rule does arbitrarily worse than the best randomized rule.
Theorem C.1 For all  $\varepsilon > 0$ , there exists a distribution  $\mathcal{D}$  and preference profile  $\sigma$  such that the deterministic rule that maximizes expected inverse distortion has expected inverse distortion less than  $\varepsilon$  fraction of that of the best random rule, i.e:
$$
\max  _ {j} \mathbb {E} [ \operatorname {i n v d d i s t} (j, \boldsymbol {\sigma}) ] \leq \varepsilon \cdot \sup  _ {f} \mathbb {E} [ \operatorname {i n v d d i s t} (f, \boldsymbol {\sigma}) ]
$$
Proof. To prove the desired result, we will show that there exist  $n, m, \sigma$ , and  $\mathcal{D}$  such that there is a random rule that performs at least  $1 / \varepsilon$  times better than the best deterministic rule. Let  $n = m = 100$ . Define  $\mathcal{D}$  as follows for  $d \sim \mathcal{D}$ .
$$
d = \left\{ \begin{array}{l l} \frac {\varepsilon}{2 n m} & \mathrm {w . p .} 1 - \frac {1}{n m} \\ 1 & \mathrm {w . p .} \frac {1}{n m} \end{array} \right.
$$
Consider the uniform preference profile  $\sigma$ , where every alternative is ranked  $i$ th by exactly 1 voter for all  $i \in [1:n]$ . First, note that the rule that chooses an alternative uniformly at random achieves expected inverse distortion at most  $m$ , as uniformly randomly selecting an alternative will choose the alternative with the maximum social welfare with probability  $1/m$ . Therefore, the best randomized rule must have expected inverse distortion of at most  $m$ .
Define  $N_{j}$  to be the total number of voters with utility 1 for alternative  $j$ . Define  $N_{tot} = \sum_{j=1}^{m} N_{j}$ . By construction of  $\mathcal{D}$ ,
$$
\Pr (N _ {t o t} = 0) = \left(1 - \frac {1}{n m}\right) ^ {n m} \leq 0. 4
$$
$$
\Pr (N _ {j} = 0) = \left(1 - \frac {1}{n m}\right) ^ {n} \geq 0. 9
$$
Now consider the deterministic voting rule that selects alternative  $j$ . By a union bound, the probability that  $N_{tot} > 0$  and  $N_{j} = 0$  is at least
$$
\Pr \left[ N _ {t o t} > 0 \cup N _ {j} = 0 \right] \geq 1 - 0. 4 - 0. 1 = 0. 5
$$
If  $N_{tot} > 0$  and  $N_{j} = 0$ , then alternative  $j$  has total social welfare of  $\frac{\varepsilon}{2m}$  and there is at least one alternative with social welfare greater than 1. Therefore, we can write
$$
\mathbb {E} \left[ \operatorname {i n v d d i s t} (j, \sigma) \right] \geq 0. 5 * \frac {1}{\frac {\varepsilon}{2 m}} = \frac {m}{\varepsilon}
$$
Because the preference profile  $\sigma$  is symmetric across voters, the above inequality is true for all  $j$ . Therefore, the best deterministic rule gives expected inverse distortion that is a factor of  $\varepsilon$  worse than that of the rule which selects an alternative uniformly at random.
The above result is especially disturbing if we allow ourselves to relax the independent and identically distributed assumption. For example, suppose that among the  $nm$  pairs of voters and alternatives, there is a single voter that has value 1 for a single alternative, and every other voter/alternative pair has utility  $\frac{\varepsilon}{2nm}$ . By symmetry, the best deterministic rule is to choose an arbitrary alternative. This rule has expected inverse distortion at least
$$
\frac {m - 1}{m} \left(\frac {1}{\frac {\varepsilon}{2 m}}\right) \geq \frac {m}{\varepsilon}.
$$
In contrast, the randomized rule that chooses one of the  $m$  alternatives uniformly at random has expected inverse distortion of at most  $m$ . Notice that the joint distribution of the social welfare of the chosen alternative and the maximum social welfare across all alternatives is the same for both rules. Therefore, one would expect them to have the same expected inverse distortion value. However, the random rule can have arbitrarily better inverse distortion as shown above. In contrast, these two rules will have the same expected distortion. This is another reason why we chose to define distributional distortion with the ratio in the opposite direction as worst-case distortion.
D Proof of Theorem 3.1: Omitted Lemma
Lemma D.1 If  $x \geq y \geq 0$ , then for all  $Z_{1}, Z_{2} \geq 0$ , the following inequality holds:
$$
\frac {Z _ {1} + x}{\max  \left(Z _ {1} + x , Z _ {2} + y\right)} \geq \frac {Z _ {1} + y}{\max  \left(Z _ {1} + y , Z _ {2} + x\right)}.
$$
Proof. Consider the following two cases.
Case 1: If  $Z_{1} + x \geq Z_{2} + y$ , then
$$
\frac {Z _ {1} + x}{\max  \left(Z _ {1} + x , Z _ {2} + y\right)} = 1 \geq \frac {Z _ {1} + y}{\max  \left(Z _ {1} + y , Z _ {2} + x\right)}
$$
Case 2: If  $Z_{1} + x < Z_{2} + y$ , then since  $x \geq y$  we have that
$$
\frac {Z _ {1} + x}{\max  \left(Z _ {1} + x , Z _ {2} + y\right)} = \frac {Z _ {1} + x}{Z _ {2} + y} \geq \frac {Z _ {1} + y}{Z _ {2} + x} = \frac {Z _ {1} + y}{\max  \left(Z _ {1} + y , Z _ {2} + x\right)}.
$$
Lemma E.1 For  $a \geq 0$  and integer  $b \geq 2$  such that  $ab \leq 1/5$ ,
$$
1 - a b \leq (1 - a) ^ {b} \leq 1 - a b + (a b) ^ {2}
$$
Proof. Expanding the binomial gives:
$$
(1 - a) ^ {b} = 1 - a b + \sum_ {k = 2} ^ {b} \binom {b} {k} (- a) ^ {k}
$$
Examining the last terms, we have that
$$
\begin{array}{l} \left| \sum_ {k = 2} ^ {b} \binom {b} {k} (- a) ^ {k} \right| \leq \frac {a ^ {2} b (b - 1)}{2} + \left| \sum_ {k = 3} ^ {b} \binom {b} {k} (- a) ^ {k} \right| \\ \leq \frac {a ^ {2} b ^ {2}}{2} + \sum_ {k = 3} ^ {b} \binom {b} {k} (- a) ^ {k} \\ \leq \frac {a ^ {2} b ^ {2}}{2} + \sum_ {k = 3} ^ {b} (a b) ^ {k} \\ \leq \frac {a ^ {2} b ^ {2}}{2} + \frac {a ^ {3} b ^ {3}}{1 - a b} \\ \leq a ^ {2} b ^ {2}. \\ \end{array}
$$
$$
\begin{array}{l} \sum_ {k = 2} ^ {b} \binom {b} {k} (- a) ^ {k} = \frac {a ^ {2} b (b - 1)}{2} + \sum_ {k = 3} ^ {b} \binom {b} {k} (- a) ^ {k} \\ = \frac {a ^ {2} b ^ {2}}{2} - \frac {a ^ {2} b}{2} + \sum_ {k = 3} ^ {b} \binom {b} {k} (- a) ^ {k} \\ \geq \frac {a ^ {2} b ^ {2}}{2} - \frac {a ^ {2} b}{2} - \sum_ {k = 3} ^ {b} (a b) ^ {k} \\ = \frac {a ^ {2} b ^ {2}}{2} - \frac {a ^ {3} b ^ {3}}{1 - a b} - \frac {a ^ {2} b}{2} \\ \geq 0. \\ \end{array}
$$
Proof of Theorem 4.1. Let  $m \geq 5$ . Choose any  $n > 1$  such that  $\frac{\sqrt{n}}{\log(n)} \geq m \geq 2\log(n) + 2$ . Define  $\mathcal{D}$  as follows: for  $u_{ij} \sim \mathcal{D}$ ,
$$
u _ {i j} = \left\{ \begin{array}{l l} n ^ {4} m & \text {w . p .} \frac {1}{n ^ {2} m} \\ n & \text {w . p .} p := e ^ {- \log (m) / m} - \frac {1}{n ^ {2} m} \\ 1 & \text {w . p .} q := 1 - e ^ {- \log (m) / m}. \end{array} \right.
$$
Draw  $m$  samples from  $\mathcal{D}$  and denote them  $X_{(1)},\ldots ,X_{(m)}$ , in increasing order. The expectation of the largest sample is:
$$
\begin{array}{l} \mathbb {E} [ X _ {(m)} ] = \operatorname * {P r} (X _ {(m)} = 1) + \operatorname * {P r} (X _ {(m)} = n) \cdot n + \operatorname * {P r} (X _ {(m)} = n ^ {4} m) \cdot n ^ {4} m \\ = q ^ {m} + \left(\left(1 - \frac {1}{n ^ {2} m}\right) ^ {m} - q ^ {m}\right) n + \left(1 - \left(1 - \frac {1}{n ^ {2} m}\right) ^ {m}\right) n ^ {4} m \\ \geq q ^ {m} + \left(\left(1 - \frac {1}{n ^ {2} m}\right) ^ {m} - q ^ {m}\right) n + \left(1 - \left(1 - \frac {1}{n ^ {2}} + \frac {1}{n ^ {4}}\right)\right) n ^ {4} m \\ = q ^ {m} + \left(\left(1 - \frac {1}{n ^ {2} m}\right) ^ {m} - q ^ {m}\right) n + n ^ {2} m - m \\ \geq n ^ {2} m - m, \\ \end{array}
$$
where we used Lemma E.1 and the fact that  $1 - 1 / n^{2}m \geq q$ .
For every  $k < m$ , we have that
$$
\begin{array}{l} \mathbb {E} \left[ X _ {(k)} \right] \leq \mathbb {E} \left[ X _ {(m - 1)} \right] \\ = \Pr (X _ {(m - 1)} = 1) + \Pr (X _ {(m - 1)} = n) \cdot n + \Pr (X _ {(m - 1)} = n ^ {4} m) \cdot n ^ {4} m \\ \leq n \Pr (X _ {(m - 1)} \neq n ^ {4} m) + \left(1 - \left(1 - \frac {1}{n ^ {2} m}\right) ^ {m} - m \left(1 - \frac {1}{n ^ {2} m}\right) ^ {m - 1} \left(\frac {1}{n ^ {2} m}\right)\right) n ^ {4} m \\ \leq n + \left(\frac {1}{n ^ {2}} - \left(1 - \frac {m - 1}{n ^ {2} m}\right) \left(\frac {1}{n ^ {2}}\right)\right) n ^ {4} m \\ \leq n + 2 m, \\ \end{array}
$$
where again we used Lemma E.1.
Combining these two inequalities, we have that  $\mathbb{E}[X_{(m)}] > n\mathbb{E}[X_{(k)}]$  for every  $k$ . Therefore, one voter ranking a chosen alternative first has higher expected utility than all  $n$  voters ranking a chosen alternative second. This implies that if there is a unique alternative that is ranked first by the most voters, then that alternative is the unique expected welfare maximizing alternative.
We consider the following preference profile. Alternative 1 is ranked first by  $\frac{n}{m-1} + 1$  voters and ranked  $m^{th}$  by  $n - \left( \frac{n}{m-1} + 1 \right) = n \cdot \frac{m-2}{m-1} - 1$  voters. Alternative 2 is ranked second by all  $n$  voters. Finally, all other alternatives are ranked first by at most  $\frac{n}{m-1}$  voters.
In this construction, alternative 1 has the most first place ranks and therefore is the unique expected welfare maximizing alternative. We now show that
$$
\mathbb {E} [ \mathrm {d d i s t} (1, \boldsymbol {\sigma}) ] \leq O \left(\frac {\mathbb {E} [ \mathrm {d d i s t} (2 , \boldsymbol {\sigma}) ]}{m}\right).
$$
Let  $\mathcal{E}_1$  be the event that no voter has utility  $n^4 m$  for any alternative. By Lemma E.1, we have that
$$
\Pr (\mathcal {E} _ {1}) = \left(1 - \frac {1}{n ^ {2} m}\right) ^ {n m} \geq 1 - \frac {1}{n}.
$$
Let  $\mathcal{E}_2$  be the event that every voter has utility at least  $n$  for alternative 2. Under event  $\mathcal{E}_2$ , the total welfare of alternative 2 is at least  $\mathrm{dsw}(2, \sigma) \geq n^2$ . We have that
$$
\begin{array}{l} \Pr (\mathcal {E} _ {2}) = \Pr (X _ {(m - 1)} \geq n) ^ {n} \\ \geq \Pr (X _ {(m - 1)} = n) ^ {n} \\ = \left(1 - q ^ {m} - m q ^ {m - 1} (1 - q)\right) ^ {n} \\ \geq \left(1 - q ^ {m} - m q ^ {m - 1}\right) ^ {n} \\ \geq \left(1 - \left(\frac {\log (m)}{m}\right) ^ {m} - m \left(\frac {\log (m)}{m}\right) ^ {m - 1}\right) ^ {n} \\ \geq \left(1 - \left(\frac {\log (m)}{m}\right) ^ {m} - \left(\frac {(m) ^ {1 / (m - 1)} \log (m)}{m}\right) ^ {m - 1}\right) ^ {n} \\ \geq \left(1 - \left(\frac {1}{2}\right) ^ {m} - \left(\frac {1}{2}\right) ^ {m - 1}\right) ^ {n} \\ \geq \left(1 - \frac {1}{n ^ {2}}\right) ^ {n} \\ \geq 1 - \frac {1}{n}, \\ \end{array}
$$
where we simplified using that  $q \leq \frac{\log(m)}{m}$ ,  $m \geq 5$ , and  $m \geq 2\log(n) + 2$ .
We finally want to establish a tighter upper bound on the typical utility of alternative 1 than event  $\mathcal{E}_1$  alone provides. Let  $N_{1}$  be the number of voters, among the  $n\cdot \frac{m - 2}{m - 1} -1$  voters that ranked alternative 1 in last place, who have utility strictly greater than 1 for alternative 1. Then let  $\mathcal{E}_3$  be the event that:
$$
N _ {1} \leq \frac {1}{m} (n \cdot \frac {m - 2}{m - 1} - 1) + \log (n) \sqrt {n}.
$$
We will show  $\mathcal{E}_3$  is also likely. To calculate  $\operatorname*{Pr}(\mathcal{E}_3)$ , note that by definition of the probability  $p$ ,
$$
\Pr (X _ {(1)} > 1) = \left(p + \frac {1}{n ^ {2} m}\right) ^ {m} = \left(m ^ {- 1 / m}\right) ^ {m} = \frac {1}{m}.
$$
Therefore,  $N_{1}\sim Bin(\frac{1}{m},n\cdot \frac{m - 2}{m - 1} -1)$  . Applying Hoeffding's inequality to this random variable gives:
$$
\begin{array}{l} \Pr (\mathcal {E} _ {3}) = 1 - \Pr \left(N _ {1} \geq \frac {1}{m} (n \cdot \frac {m - 2}{m - 1} - 1) + \log (n) \sqrt {n}\right) \\ = 1 - \Pr \left(N _ {1} - \mathbb {E} [ N _ {1} ] \geq \log (n) \sqrt {n}\right) \\ \geq 1 - \exp \left(- 2 \left(\log (n) \sqrt {n}\right) ^ {2} / \left(n \cdot \frac {m - 2}{m - 1} - 1\right)\right) \\ \geq 1 - \exp (- 2 (\log (n) \sqrt {n}) ^ {2} / n) \\ = 1 - \exp (- 2 \log^ {2} (n)) \\ \geq 1 - \frac {1}{n}. \\ \end{array}
$$
Define  $\mathcal{E} \coloneqq \mathcal{E}_1 \cap \mathcal{E}_2 \cap \mathcal{E}_3$ . By inclusion/exclusion and a union bound,
$$
\begin{array}{l} \Pr (\mathcal {E}) = 1 - \Pr (\neg \mathcal {E} _ {1} \cup \neg \mathcal {E} _ {2} \cup \neg \mathcal {E} _ {3}) \\ \geq 1 - \Pr (\neg \mathcal {E} _ {1}) - \Pr (\neg \mathcal {E} _ {2}) - \Pr (\neg \mathcal {E} _ {3}) \\ \geq 1 - \frac {3}{n}. \\ \end{array}
$$
We consider the social welfare of alternatives 1 and 2 under event  $\mathcal{E}$  to obtain bounds on their respective expected distributional distortions.
Event  $\mathcal{E}_3$  implies that the total number of voters that have utility greater than 1 for alternative 1 (including the  $\frac{n}{m - 1} + 1$  voters who rank alternative 1 first) is at most:
$$
\frac {1}{m} \left(n \cdot \frac {m - 2}{m - 1} - 1\right) + \log (n) \sqrt {n} + \frac {n}{m - 1} + 1.
$$
Under event  $\mathcal{E}_1$ , no voter has utility greater than  $n$  for any alternative. Therefore, every voter has utility either 1 or  $n$  for alternative 1. Since we have an upper bound on the number of voters that have utility  $n$  for alternative 1, and trivially at most  $n$  voters have utility 1 for alternative 1, we upper bound the total social welfare for alternative 1 by:
$$
\begin{array}{l} \operatorname {d s w} (1, \sigma) \leq n \cdot \left(\frac {1}{m} \left(n \cdot \frac {m - 2}{m - 1} - 1\right) + \log (n) \sqrt {n} + \frac {n}{m - 1} + 1\right) + 1 \cdot n \\ \leq \frac {n ^ {2}}{m} + \log (n) n ^ {3 / 2} + \frac {2 n ^ {2}}{m} + n \\ \leq \frac {4 n ^ {2}}{m} + \log (n) n ^ {3 / 2}. \\ \end{array}
$$
Combining this with the fact that  $\mathrm{dsw}(2,\sigma) \geq n^2$  under event  $\mathcal{E}_2$ , we get that the distributional distortion under event  $\mathcal{E}$  is at most the following, using that  $m \leq \frac{\sqrt{n}}{\log(n)}$ :
$$
\frac {\frac {4 n ^ {2}}{m} + \log (n) n ^ {3 / 2}}{n ^ {2}} = \frac {4}{m} + \frac {\log (n)}{\sqrt {n}} \leq \frac {5}{m}.
$$
Since  $\operatorname{Pr}(\mathcal{E}) \geq 1 - \frac{3}{n}$ , the distributional distortion of alternative 1 is upper bounded as follows:
$$
\mathbb {E} [ \mathrm {d d i s t} (1, \boldsymbol {\sigma}) ] \leq \left(1 - \frac {3}{n}\right) \cdot \frac {5}{m} + \frac {3}{n} \cdot 1 \leq \frac {8}{m}.
$$
Under event  $\mathcal{E}$ , alternative 2 has total social welfare of  $n^2$  and the max social welfare of any alternative is  $n^2$ . This implies that alternative 2 has distributional distortion 1. The expected distributional distortion of alternative 2 is thus lower bounded as follows:
$$
\mathbb {E} [ \mathrm {d d i s t} (2, \sigma) ] \geq \left(1 - \frac {3}{n}\right) \cdot 1 + \frac {3}{n} \cdot 0 \geq \frac {1}{2}.
$$
$$
\begin{array}{l} \mathbb {E} [ \mathrm {d d i s t} (\operatorname {e w m r}, \sigma) ] = \mathbb {E} [ \mathrm {d d i s t} (1, \sigma) ] \\ \leq \frac {1 6}{m} \mathbb {E} [ \mathrm {d d i s t} (2, \sigma) ] \\ \leq \frac {1 6}{m} \mathbb {E} [ \mathrm {d d i s t} (\operatorname {e d m r}, \boldsymbol {\sigma}) ]. \\ \end{array}
$$
Note that while this distribution is not supported on  $[0, 1]$ , simply scaling the values of the distribution by  $\frac{1}{n^4m}$  will make the distribution supported on  $[0, 1]$ . Furthermore, expected distortion is invariant to scaling the distribution, and therefore the result still holds.
F Proof of Theorem 4.2
Lemma F.1 Suppose the distribution  $\mathcal{D}$  has mean  $\mu$  and is supported on  $(0,1]$ . Then the expected distributional distortion of the alternative selected by the expected welfare maximizing rule is at least  $\mu$ . Furthermore, the expected welfare of the expected welfare maximizing alternative is at least  $n\mu$ .
Proof. First, note that we can add the expected distortion for all  $m$  players and apply linearity of expectation to get that
$$
\sum_ {j = 1} ^ {m} \mathbb {E} [ \mathrm {d s w} (j, \boldsymbol {\sigma}) ] = \sum_ {j = 1} ^ {m} \sum_ {i = 1} ^ {n} \mathbb {E} [ u _ {i j} ] = n m \mu .
$$
Furthermore, the maximum expected social welfare alternative must have expected social welfare at least as large as the average expected social welfare across all alternatives. This implies that there must exist some  $j$  such that
$$
\mathbb {E} [ \mathrm {d s w} (j, \boldsymbol {\sigma}) ] \geq n \mu .
$$
Therefore, the alternative selected by the expected welfare maximizing rule must have expected social welfare of at least  $n\mu$ . Therefore, using the fact that no alternative can have social welfare of greater than  $n$ ,
$$
\begin{array}{l} \mathbb {E} [ \mathrm {d d i s t} (\mathrm {e w m r}, \boldsymbol {\sigma}) ] = \mathbb {E} \left[ \frac {\mathrm {d s w} (\mathrm {e w m r} , \boldsymbol {\sigma})}{\max  _ {j} \mathrm {d s w} (j , \boldsymbol {\sigma})} \right] \\ \geq \mathbb {E} \left[ \frac {\operatorname {d s w} (\operatorname {e w m r} , \boldsymbol {\sigma})}{n} \right] \\ = \frac {\mathbb {E} [ \mathrm {d s w (e w m r} , \boldsymbol {\sigma}) ]}{n} \\ \geq \frac {n \mu}{n} \\ = \mu . \\ \end{array}
$$
The below is a purely technical lemma.
Lemma F.2 If  $X_{1},\ldots ,X_{m}$  are i.i.d. draws from a continuously differentiable distribution with cdf  $F$  Suppose that  $\frac{dF(x)}{dx} >0$  for all  $x$ . Then for every  $s_\varepsilon >0$  there exists  $M$  such that for every  $m > M$  and for every  $j$ ,
$$
\operatorname {V a r} \left(X _ {(j)}\right) \leq s _ {\varepsilon} ^ {2}.
$$
Proof. Let  $F$  be the cdf of  $\mathcal{D}$  and let  $F_{m}$  be the empirical cdf from  $m$  samples. Let  $|F - F_{m}| = \sup_{x} |F(x) - F_{m}(x)|$ . The iterated logarithm law of Smirnov (see Smirnov [1944], Chung [1949]) states that: