Weighted positional method

A weighted positional method is a preferential voting method that assigns $$a_i$$ points to the candidate a voter ranks in ith place. It then sums the scores and the candidate with the greatest score wins.

These methods can be characterized by the $$a$$ vector. For instance, the Borda count, the First Past the Post electoral system, and Antiplurality are all weighted positional methods, and their vectors are:


 * $$a_\mathrm{Borda} = (m-1, m-2, \dots, 0)$$
 * $$a_\mathrm{Plurality} = (1, 0, \dots, 0)$$
 * $$a_\mathrm{Antiplurality} = (1, \dots, 1, 0)$$

where $$m$$ is the number of candidates.

Criterion compliances
Every weighted positional method passes the participation criterion, the consistency criterion, and mono-add-top; and is summable with order k=1.

Every weighted positional method except for First past the post fails the later-no-harm criterion and the majority criterion. Since First past the post fails the Condorcet criterion, and Condorcet implies majority, every weighted positional method fails Condorcet.

The Borda count is the only weighted positional method that never ranks the Condorcet winner last. It follows that the only Condorcet-compliant runoff method that eliminates one loser at a time and is based on a weighted positional method is Baldwin (Borda-elimination).

Every resolvable weighted positional method fails clone independence: Plurality fails to vote-splitting, and every other method can be made to fail the majority criterion even for clones, hence turning a majority loser into a winner.

Majority criterion
Consider an election with three candidates. The method's $$a$$ vector can be normalized to one of $$a = (1, \alpha, 0),\, a = (0, \alpha, 1)$$ or $$a = (0, 0, 0)$$.

In the latter two cases, the method trivially fails unanimity and thus also majority. So normalize $$a$$ so that $$a = (1,\alpha,0)$$ with $$\alpha \ne 0$$ since the method is not First past the post.

If $$\alpha > 0$$, construct the following election:


 * x: A>B>C
 * y: B>C>A

with $$x = \frac{1}{\min(1,\alpha)},\, y = x-1$$.

A and B will be tied even though A has a majority of the first preferences, thus constituting a violation of the majority criterion.

On the other hand, if $$\alpha < 0$$, construct the following election:


 * x: A>C>B
 * y: B>A>C

with $$x = \frac{\alpha - 1}{\alpha},\, y = x-1$$. Again A and B will be tied even though A has a majority of the first preferences.

With more work, the examples can be generalized to any number of candidates greater then 3 by assuming every voter ranks all the other candidates in the same order.

Generalizations
It is possible to view Approval voting and Score voting as a more general weighted positional method, where each voter has some freedom in what $$a$$ vector to choose. For Approval, the voter's $$a$$ vector has value 1 for every approved candidate and 0 otherwise - i.e. 1 down to the voter's approval cutoff and then 0 below - while for Score voting, the voter directly specifies $$a$$.

As a result, Score voting fails every criterion that does not involve removing or adding candidates, and that least one weighted positional method fails. The voters could just happen to rate the candidates the same way a weighted positional method would score them, and then the failure example for that method would also apply to Score.

Some methods

 * Nauru island adopted a positional method with harmonic weights where the kth-ranked candidate gets a score of 1/k. It can be represented by the $$a$$ vector: $$a_\mathrm{Nauru} = (1, 1/2, ..., 1/k, ..., 1/m)$$. This is also sometimes called the Dowdall method.
 * Borda Count
 * As shown above, many systems can be represented as weighted positional methods including Plurality, Antiplurality, Vote For and Against, and Dabagh's "vote and a half" method where a voter assigns 1 point to their favorite and half a point to their second favorite.
 * The Eurovision Song Contest uses a method where each country's panel of judges rates the top 10 songs. The first pick gets 12 points, the second 10, the third 8, and the rest get $$m - i$$ (i.e. 7, 6, ..., 1) with all the rest of the songs getting 0. This can be represented by the vector $$a_\mathrm{Eurovision} = (12, 10, 8, 7, ..., max(m - i, 0), ..., 0)$$