Not-So-Fair Democracy

Let us begin by assuming the worst-case win scenario in a riding, that each elected candidate in our party received only one more vote than their adversary. If each riding has a population of $m$ people, then the fraction of the riding who voted for this candidate would be

$\frac{m+1}{2m}=\frac{1}{2}+\frac{1}{2m}$

As $m\to\infty$ , this expression approaches $\frac{1}{2}$ .

We repeat this process with the ridings: there are $k$ seats, which elect one more member of our party than any other. The amount of seats in Parliament that our party has would be

$\frac{k+1}{2k}=\frac{1}{2}+\frac{1}{2k}$

Once again, as $k\to\infty$ , the expression approaches $\frac{1}{2}$ .

The candidates received $\frac{1}{2}$ of the votes in their riding: the majority party received $\frac{1}{2}$ of the seats in Parliament. The minimal proportion of the popular vote a majoritarian party could receive is then $(\frac{1}{2})(\frac{1}{2})=\boxed{\frac{1}{4}}$ , or $25\%$ .

Of course, keep in mind that this can only happen in an infinitely-sized population, divided into infinitely many geographic regions.

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Bonus ( This is a straightforward generalization of the calculations above ): For each number $n$ of candidates, there is the possibility of a tie, where each elected party candidate receives $\frac{1}{n}$ of the votes. Suppose that there is a tiebreaker of one vote. For a riding population $m$ , the number of votes the elected candidate receives is then

$\frac{m+1}{nm}=\frac{1}{n}+\frac{1}{nm}$

As $m\to\infty$ , this expression approaches $\frac{1}{n}$ .

We repeat the process with parliamentary seats: in order for a party to receive majoritarian status, they require one more seat than any other party. In other terms, for $k$ parties, they need

$\frac{k+1}{nk}=\frac{1}{n}+\frac{1}{nk}$

seats in order to attain majoritarian status. Once again, as $m\to\infty$ , this expression approaches $\frac{1}{n}$ . There being $\frac{1}{n}$ votes for each elected party candidate, and $\frac{1}{n}$ seats owned by the party, the minimal proportion of popular vote required is

$(\frac{1}{n})(\frac{1}{n})=\frac{1}{n^2}$

For $n=2, 3, 4...$ , the corresponding minimal popular vote proportions are $\frac{1}{4}, \frac{1}{9}, \frac{1}{16}...$ .

Imagine the case with 100 people in each of the 100 ridings. If 51 of the ridings have 51 supporters for party $A$ , and all other ridings have 100 supporters for party $B$ . This means that party $A$ would win, even as a minority. Instead of 100 people, let's generalize for $n$ people and ridings.

The minimum percentage for $n$ people and ridings is $\frac{50+\frac{100}{n}}{2}+\frac{100}{n}$ . The 50 represents half of the vote that the party needs to tie, and the $\frac{100}{n}$ represents the tiny sliver party $A$ needs to win. There are 2 of these fractions due to the ridings having to have that same sliver to get the majority.

Now, taking the limit,

$\displaystyle\lim_{n\rightarrow \infty} \frac{50+\frac{100}{n}}{2}+\frac{100}{n}=\frac{50}{2}=25$ $\beta_{\lceil \mid \rceil}$

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To generalize, each riding needs $\frac{100}{p}$ % for party $A$ to win if there are $p$ parties. Therefore, we can generalize to the expression: $\displaystyle\lim_{n\rightarrow \infty} \frac{\frac{100}{p}+\frac{100}{n}}{p}+\frac{100}{n}=\frac{100}{p^2}$

Therefore, the proportion needed to win the election is $\frac{1}{p^2}$ .