The Fairness of Redistricting Ghost

TL;DR

This paper analyzes Redistricting Ghost, an abstract redistricting protocol where two parties alternately assign voters to districts without an independent arbiter. In a nongeometric setting, the minority can guarantee at least $p-1$ districts, where $p = \text{round}(j n / v)$, while the majority can employ a cracking strategy to cap the minority at $p-1$ when the minority share is small; the authors provide precise bounds and strategies for both players. They show a concrete lower bound $n \ge 2q(m+1)$ guaranteeing $q$ districts for the minority, and a complementary lower bound $n < f(q)$ (with $f(q) = 2q \left( 1 - \frac{q}{j+q} \right)(m+1) - 1$) preventing such wins under a cracking majority. A key result is the existence of a gap between these bounds, indicating ranges where proportional fairness or gerrymandering-like outcomes are not fully characterized, with conjectures about the dependence on minority size and first-mover effects. The work motivates study of fairness under abstract protocols and highlights the need to extend analyses to geometric and graph-based models for more realistic redistricting scenarios.

Abstract

We explore the fairness of a redistricting game introduced by Mixon and Villar, which provides a two-party protocol for dividing a state into electoral districts, without the participation of an independent authority. We analyze the game in an abstract setting that ignores the geographic distribution of voters and assumes that voter preferences are fixed and known. We show that the minority player can always win at least $p-1$ districts, where $p$ is proportional to the percentage of minority voters. We give an upper bound on the number of districts won by the minority based on a "cracking" strategy for the majority.

Figures (4)

• Figure 1: A simple example of Redistricting Ghost. An interactive version of the game can be found at http://ballsandbins.com. Each row represents a district; so here $j = 4$. The size of a district is $2m+1$, $m=5$ here. We draw bricks as red squares and apples as green circles; a brick played by $A$ has a green outline, and an apple played by $B$ has a red outline. The number inside each brick or apple represents the turn at which it was played ($B$ plays odd turns and $A$ plays even). For visual clarity, we place bricks into districts from left to right and apples from right to left, and we sort the rows by the number of bricks they contain, and within that by the number of empty spaces. Here, we see the state of the game after the last brick has been played; at every subsequent turn an apple will be played to an empty position, and the order in which they are played will have no effect on the outcome. So $n$, the number of bricks, is $19$, and $q$, the number of districts won by $B$, is $2$.
• Figure 2: An example of a game in which the majority player $A$ uses the "cracking" strategy in Algorithm \ref{['alg:majorityNew']}, at the point at which the bricks run out. The main idea of of the algorithm is that $A$ fills in as many columns as she can with bricks. Player $B$ may do anything. In this example, $B$ concentrates on winning one district at a time. The lower bound (Theorem \ref{['thm:lower']}) says that if the number of bricks $n < 28$, then $B$ cannot win three districts. Here $n = 33$, and $B$ just barely wins three districts. With $q=3, j=7, m=6$, we prove that $A$ can always fill $c=2$ columns, but in this example $A$ actually manages to fill $3$ columns.
• Figure 3: Another example game in which $A$ plays the "cracking" strategy, at the point at which the bricks run out. In this game $B$ fills in a row with apples so as to block $A$ from adding the final brick to columns 3 and 4, and then switches to winning districts one by one. Using the notation of Theorem \ref{['thm:lower']}, we have $j=7, m=6, q=3, c=2$, and $n=37$. We notice that $B$ needs $n=36$ to win three districts, while he only needed $33$ in Figure \ref{['fig:majorityExample1']}.
• Figure 4: The bounds illustrated for the case $j=10$. The $x$ axis is the number of bricks (voters of the minority party $B$) as a function of $m$, the size of a bare majority in a district. The $y$ axis is the number of districts that $B$ can win. The grey plot shows $p$, the number of districts $B$ wins if they are distributed proportionally. The red plot shows $n=\frac{2jq}{j+q}m$; Theorem \ref{['thm:lower']} says that when $n$ is below this value, $B$ cannot win $q$ districts. The blue line shows $n=2qm$; Theorem \ref{['thm:upperBound']} says that when $n$ is above this value, $B$ can always win $q$ districts.

Theorems & Definitions (9)

• Theorem 1
• Lemma 2
• Lemma 3
• Corollary 4
• Lemma 6
• Lemma 7
• Theorem 8
• Theorem 9
• Lemma 12