How the Republic of Venice chose its Doge: lot-based elections and supermajority rule

Abstract

We study a family of voting rules inspired by the peculiar protocol used for over 500 years by the Republic of Venice to elect its Doge. Lot-based indirect elections have two main features: a pool of delegates is chosen by lot out of a general assembly, and then they vote in a single winner election with qualified majority. Under the assumption that the assembly is divided into two factions, we characterise the win probability of the minority and show that these features promote a more equitable allocation of political representation, striking a balance between protecting the minority and giving proper recognition to the majority. We then consider this family of voting procedures from a constitutional perspective: we analyse how the electoral result varies with the college size and the winning threshold in order to understand how these two parameters can be tuned when drawing up electoral law. We find that minorities are better off with larger majority thresholds. The role of the college size, on the other hand, is ambiguous: a smaller college size offers more protection to sparse minorities; for more sizeable ones, it depends instead on the qualified majority required for the election.

Appendix

1.1 Proof of Proposition 1

Proof

First, consider $x\le m$, which implies $p_x> 0$. Then either $y>m,$ in which case $p_y=0$ and thus $p_x>p_y=0$, or else $y\le m$ and $p_y\ne 0.$ In the latter case the definition of the binomial coefficient and $y=c-x$ imply:

$$\begin{aligned} \left( {\begin{array}{c}m\\ x\end{array}}\right) =\frac{m!}{x!(m-x)!} = \left( {\begin{array}{c}m\\ y\end{array}}\right) \frac{y!(m-y)!}{x!(m-x)!}=\left( {\begin{array}{c}m\\ y\end{array}}\right) \frac{(c-x)!(m-c+x)!}{x!(m-x)!}. \end{aligned}$$

(2)

By an analogous reasoning we obtain:

$$\begin{aligned} \begin{aligned} \left( {\begin{array}{c}n-m\\ c-x\end{array}}\right) =&\,\left( {\begin{array}{c}n-m\\ c-y\end{array}}\right) \frac{(c-y)!(n-m-c+y)!}{(c-x)!(n-m-c-x)!}\\ =&\,\left( {\begin{array}{c}n-m\\ c-y\end{array}}\right) \frac{x!(n-m-x)!}{(c-x)!(n-m-x-c)!}.\end{aligned}\end{aligned}$$

(3)

We now use (2) and (3) to express $p_x$ as a function of $p_y$:

$$\begin{aligned} \begin{aligned} p_x=&\,\frac{\displaystyle \left( {\begin{array}{c}m\\ x\end{array}}\right) \left( {\begin{array}{c}n-m\\ c-x\end{array}}\right) }{\displaystyle \left( {\begin{array}{c}n\\ c\end{array}}\right) }\\ =&\,\frac{\displaystyle \left( {\begin{array}{c}m\\ y\end{array}}\right) \left( {\begin{array}{c}n-m\\ c-y\end{array}}\right) }{\displaystyle \left( {\begin{array}{c}n\\ c\end{array}}\right) } \frac{(m-x-c)!}{(m-x)!}\frac{(n-m-x)!}{(n-m-x-c)!}\\ =&\,p_y \frac{(m-x-c)!}{(m-x)!}\frac{(n-m-x)!}{(n-m-x-c)!}\\ =&\,p_y \frac{(n-m-x)}{m-x}\cdot \frac{(n-m-x-1)}{(m-x-1)}\dots \frac{(n-m-x-c+1)}{(m-x-c+1)} >p_y, \end{aligned} \end{aligned}$$

where the last inequality follows from the fact that, being $m<n/2$, each fraction in the last line is greater than one.

Next, consider the case $x> m.$ Since $x<y,$ we have $y>m$ and, therefore, $p_x=p_y=0.$ This completes the proof. $\square$

1.2 Proof of Proposition 2

Proof

Suppose $w=0$. Given the definition of w in (1), it must be $p_x=0$ for any $x\ge c-t+1$. But with the hypergeometric distribution $p_x=0$ if and only if $x>m$. It follows that $m<c-t+1,$ a contradiction. $\square$

1.3 Proof of Proposition 3

Proof

First we prove that $w\le \mu .$ Start with Eq. (1):

$$\begin{aligned} w=\sum _{x=c-t+1}^{t-1}p_x\frac{x}{c}+ \sum _{x=t}^{c}p_x. \end{aligned}$$

Add and subtract both $\sum _{x=0}^{c-t}p_x\frac{x}{c}$ and $\sum _{x=t}^{c}p_x\frac{x}{c}$ to get

$$\begin{aligned} w=\sum _{x=0}^{c}p_x\frac{x}{c}-\sum _{x=0}^{c-t}p_x\frac{x}{c} -\sum _{x=t}^{c}p_x\frac{x}{c}+ \sum _{x=t}^{c}p_x. \end{aligned}$$

For the hypergeometric distribution the expected value $\sum _{x=0}^{c}p_x x$ is equal to $\dfrac{m}{n}c$. Inserting this value in the equation above and rearranging the terms yields

$$\begin{aligned} w=\frac{m}{n} -\sum _{x=0}^{c-t}p_x\frac{x}{c} +\sum _{x=t}^{c}p_x \left( 1-\frac{x}{c}\right) \end{aligned}$$

We can simplify this expression and switch to a new summation index in the last term, yielding:

$$\begin{aligned} \begin{aligned} w=&\,\frac{m}{n} -\sum _{x=0}^{c-t}p_x\frac{x}{c} +\sum _{x=0}^{c-t}p_{c-x} \left( 1-\frac{c-x}{c}\right) \\ =&\,\frac{m}{n} -\sum _{x=0}^{c-t}(p_x-p_{c-x}) \frac{x}{c}.\\\end{aligned}\end{aligned}$$

(4)

We now use Proposition 1 to sign the last term of this equality. By definition of supermajority threshold, $t >\dfrac{c}{2}$. Then, for any value of $x\in \{0, 1,\dots , c-t\}$, we know that $x<c-x$. Since $x+(c-x)=c$ the proposition applies and, thus, each term $(p_x-p_{c-x})$ in the sum is non-negative. This implies $w\le \mu$.

To complete the proof consider $t=c$. Then neither faction can be decisive and Eq. (4) reduces to

$$\begin{aligned} w=\frac{m}{n}-\sum _{x=0}^{0}(p_x-p_{c-x})\frac{x}{c}=\frac{m}{n}=\mu . \end{aligned}$$

$\square$

1.4 Proof of Proposition 4

Proof

Fix n, m and c. Abusing notation, we drop these three parameters and write w[t] for the mwp given a value of the supermajority threshold $t<c$. It is enough to show that when we raise the threshold from t to $t+1$, the difference $\Delta _t w=w[t+1]-w[t]$ is positive.

Using the definition of mwp given in Eq. (1) we write

$$\begin{aligned} w[t]=\sum _{x=c-t+1}^{t-1}p_x\frac{x}{c}+ \sum _{x=t}^{c}p_x. \end{aligned}$$

Similarly, the winning probability for a threshold $t+1$ is

$$\begin{aligned} w[t+1]=\sum _{x=c-t}^{t}p_x\frac{x}{c}+ \sum _{x=t+1}^{c}p_x. \end{aligned}$$

Notice that all the terms in w[t] and $w[t+1]$ are the same except for those indexed by $c-t$ and t. Then the difference $\Delta _t w$ reduces to:

$$\begin{aligned} \Delta _t w=p_{c-t}\frac{c-t}{c}+p_t\frac{t}{c}-p_t=\frac{c-t}{c}(p_{c-t}-p_t).\end{aligned}$$

(5)

To determine the sign of $\Delta _t w$ notice that $c-t\ge 0$ by definition of supermajority threshold. This implies that $\Delta _t w$ has the same sign as $(p_{c-t}-p_t)$. Since the pair of probabilities $p_{c-t}$ and $p_t$ satisfies the conditions of Proposition 1 and we are computing w for a minority faction, we know that $(p_{c-t}-p_t)\ge 0$ and thus conclude that $\Delta _t w\ge 0.$

To complete the proof consider the unanimity rule. When $t=c$ the mwp is:

$$\begin{aligned} w=\sum _{x=1}^{c-1}p_x\frac{x}{c}+ p_c= \frac{1}{c}\left(\sum _{x=1}^{c}p_x\,x-p_c c\right)+ p_c=\frac{1}{c}\sum _{x=1}^{c}p_x\,x=\frac{m}{n} \end{aligned}$$

where the last equality follows from the fact that the expected value $\sum _{x=1}^{c}p_x\,x$ is equal to $\dfrac{mc}{n}$ for a hypergeometric random variable. $\square$