Can You Gerrymander Like a Dummy?¶
Fiddler¶
Fiddler Nation has an upcoming election between its two major political parties: the Suns and the Moons. You are drawing a new electoral map, and you have been tasked with gerrymandering on behalf of your party, the Moons. Fortunately, you know exactly how every citizen will vote in the election.
Fiddler Nation has seven districts, each with an equal population. For simplicity, assume that the Moons win any district in which they receive at least half of the vote.
What is the smallest total fraction of Fiddler Nation’s population needed to vote Moon such that it’s possible for the Moons to win a majority of the seven districts?
Solution¶
To win a district, the minimal number of votes necessary is $\dfrac{1}{14}$ of Fiddler Nation's population.
Answer¶
$$4\times\dfrac{1}{14} = \boxed{ \dfrac{2}{7} \approx 28.57\%}$$
Extra Credit¶
In the next election cycle you’re back at it, gerrymandering on behalf of the Moon party once more. This time, you can draw a map that results in any set of probabilities P1, P2, P3, P4, P5, P6, and P7 that you’ll win each of the corresponding districts—subject to the constraint that the average of the probabilities is 1/2. In other words, P1 + P2 + P3 + P4 + P5 + P6 + P7 = 7/2.
The bosses of the Moon party want to go big. More specifically, they want to maximize the party’s chances of winning at least five of the seven districts. However, you are concerned that such a greedy attempt at gerrymandering may backfire, a phenomenon that has become known as dummymandering. When you draw a map that maximizes your chances of winning at least five districts, the probability the Moon party blows it and winds up losing a majority of the seven districts is LA.
Your personal preference is to simply maximize the party’s chances of winning at least four of the seven districts—a simple majority of them. When you draw a map that maximizes these chances, the probability the Moon party loses a majority of the seven districts is LB.
How much greater than LB is LA?
From the AM-GM inequality, the maximum product of $n$ elements with a fixed average is when the elements are equal.
Thus, $P_1 = P_2 = P_3 = P_4 = P_5 = \dfrac{7}{10}$.
The probability of winning all five districts is $\left(\dfrac{7}{10}\right)^5 = 0.16807$.
However, one can still win the majority of the seven districts by losing any one of the five and winning the other four. This probability is $5 * \left(\dfrac{7}{10}\right)^4\left(\dfrac{3}{10}\right) = 0.36015$.
$L_A = 1- (0.16807+0.36015) = \boxed{0.47178}$
LB¶
Similarly for LB, $P_1 = P_2 = P_3 = P_4 = \dfrac{7}{8}$.
The probability of winning all four districts is $\left(\dfrac{7}{8}\right)^4 \approx 0.58618$.
All four districts are critical to get a majority of seven districts.
$L_B = 1 - 0.58618 = \boxed{0.41382}$
Answer¶
$$0.47178 - 0.41382 \approx \boxed{0.05796}$$
The exact answer is $\dfrac{741909}{12800000} = 0.057961640625$