# Weighted positional method

(Redirected from Positional voting system) Wikipedia has an article on:

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).

### 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.