The main disadvantage of the plurality method is that it does not allow each voter to express her full preference order: the top-ranked candidate receives 1 "point" and all other candidates receive 0. In the positional method, the second-ranked candidate receives a fractional amount of votes. We have a parameter s that is between 0 and 1, inclusive, and each voter's three candidates receives 1, s, and 0 points, in order of preference.
When s = 0, the positional method reduces to the plurality method that we have already discussed.
When s = 1, the voter's first and second place candidates both receive an equal vote, and the third place receives zero. In effect, a vote has been placed against the third-place candidate. This method is called antiplurality.
When s = 0.5, the positional method reduces to the Borda count, a method developed by Jean-Charles de Borda in 1770. An equivalent way to perform the Borda count is to assign 2 points for each first-place vote, 1 point for each second-place vote, and 0 points for third-place votes.
In the interactive mathlet below, experiment with changing the value of s. Can you construct profiles where you can change the winner of the election, not by adding or removing voters, but by simply changing the value of s?
Consider another voter profile:
|6||A > B > C|
|4||A > C > B|
|0||B > A > C|
|8||B > C > A|
|3||C > A > B|
|4||C > B > A|
Who should win this election? We will analyze this election using the positional method, but we will not choose a value of s right away.
|A > B > C||6||6s||0|
|A > C > B||4||0||4s|
|B > A > C||0||0||0|
|B > C > A||0||8||8s|
|C > A > B||3s||0||3|
|C > B > A||0||4s||4|
|Totals||10 + 3s||8 + 10s||7 + 12s|
Performing the calculations this way allows us to easily change the value of s and determine if the winner of the election changes. In this example, for different values of s, any of the three candidates could win this election:
So which candidate should win? In this case there is no easy answer.