From Peter Mowrey comes an elegant electoral enigma:
Riddler Township is having its quadrennial presidential election. Each of the town’s 10 “shires” is allotted a certain number of electoral votes: two, plus one additional vote for every 10 citizens (rounded to the nearest 10).
The names and populations of the 10 shires are summarized in the table below.
| Shire | Population | Electoral votes |
|---|---|---|
| Oneshire | 11 | 3 |
| Twoshire | 21 | 4 |
| Threeshire | 31 | 5 |
| Fourshire | 41 | 6 |
| Fiveshire | 51 | 7 |
| Sixshire | 61 | 8 |
| Sevenshire | 71 | 9 |
| Eightshire | 81 | 10 |
| Nineshire | 91 | 11 |
| Tenshire | 101 | 12 |
As you may know, under this sort of electoral system, it is quite possible for a presidential candidate to lose the popular vote and still win the election.
If there are two candidates running for president of Riddler Township, and every single citizen votes for one or the other, then what is the lowest percentage of the popular vote that a candidate can get while still winning the election?
For Shire n-shire in the Riddler Township, the population can be expressed as :
$$P(n) = 10n + 1$$And the total population is : $$\sum_{n=1}^{10} (10n+1) = \Big(10\sum_{n=1}^{10}n\Big) + 10 = 560 $$
The number of electoral votes can be expressed as : $$V(n) = n + 2$$
And the total number of votes is :
$$\sum_{n=1}^{10} (n+2) = \Big(\sum_{n=1}^{10}n\Big) + 20 = 75$$The majority cutoff for electoral votes is $38$. The sum of the three largest shires is too small $(10 + 11 + 12 = 33)$ and the sum of the seven smallest shires is too large $(3 + 4 + 5 + 6 + 7 + 8 + 9 = 42)$. In order to achieve exactly $38$, either four, five, or six shires is necessary.
The cutoff for majority of the population for each n-shire is :
$$M(n) = 5n + 1$$In order to minimize the percentage of the popular vote, assume exactly the cutoff majority voting population, which means the weight of each constituent for n-shire is :
$$V(n) = \frac {V(n)}{M(n)} = \frac{n + 2}{5n + 1} = \frac{n + \frac{1}{5}}{5n + 1} + \frac{\frac{9}{5}}{5n + 1} = \frac{1}{5} + \frac{9}{25n+5}$$It follows that smaller shire constituents have more weight than that of larger shires. Making a coalition of six shires,
$$s_1, s_2, s_3, s_4, s_5, s_6$$the respective sum of their votes can be expressed as:
$$CoalitionVotes = 38 = \sum_{k=1}^6 V(s_k) = \sum_{k=1}^6 (s_k + 2) = 12 + \sum_{k=1}^6 s_k$$$$\sum_{k=1}^6 s_k = 26$$The population of that coalition can be expressed as :
$$\sum_{k=1}^6 M(s_k) = \sum_{k=1}^6 (5s_k + 1) = 6 + 5\sum_{k=1}^6 s_k = 136$$The following coalitions satisfy both of these conditions.
[3, 4, 5, 6, 8, 12] [3, 4, 5, 7, 8, 11] [3, 4, 6, 7, 8, 10] [3, 4, 5, 6, 9, 11] [3, 4, 5, 7, 9, 10] [3, 5, 6, 7, 8, 9]
Each of these coalitions has $38$ electoral votes and only $136$ out of the $560$ constituents, $≈24.29\%$, voting for the winner.
Note that for a coalition of five shires,
$$CoalitionPopulation = 5 + 5(38 - 10) = 145$$
And for a coalition of four shires, $$CoalitionPopulation = 4 + 5(38 - 8) = 154$$