As you may have seen in FiveThirtyEight’s reporting, there’s an election coming up. Inspired, Vikrant Kulkarni has an electoral enigma for you:
On Nov. 3, the residents of Riddler City will elect a mayor from among three candidates. The winner will be the candidate who receives an outright majority (i.e., more than 50 percent of the vote). But if no one achieves this outright majority, there will be a runoff election among the top two candidates.
If the voting shares of each candidate are uniformly distributed between 0 percent and 100 percent (subject to the constraint that they add up to 100 percent, of course), then what is the probability of a runoff?
Extra credit: Suppose there are $N$ candidates instead of three. What is the probability of a runoff?
It is easier to complementary count, that is find the probability of outright majority occurences and subtract from $1$, in order to determine runoff occurences.
A more generalized problem is to look at a line of length $1$ with $N-1$ points. The segements between the points represent the fraction of the vote one of the $N$ candidates receives. The points should be randomly spread between $\frac{1}{2}$ and $1$ to ensure a majority occurence.
Let the first point be at location $x$. The remaining $N-2$ points will be randomly spread between $x$ and $1$. This can be represented as:
$$\int_{\frac{1}{2}}^{1} (1-x)^{N-2} \, dx$$Note that any of the $N$ candidates can recieve the majority. The probability of a runoff is therefore:
$$1 - N \int_{\frac{1}{2}}^{1} (1-x)^{N-2} \, dx$$
$$= 1 - N \bigg( \Big[ \frac{-(1-x)^{N-1}}{N-1} \Big]_\frac{1}{2}^1 \bigg)$$
$$= 1 - N \bigg( \Big[0 + \frac{1}{2^{N-1} \cdot (N-1)} \Big] \bigg)$$
$$= 1 - \frac{N}{2^{N-1} \cdot (N-1)}$$
For $N = 3$ the probability of a runoff is:
$$1 - \frac{3}{4 \cdot 2} = \frac{5}{8}$$