[EM] Nanson

[Stephane Rouillon]

This amazes me.
You are not the first to tell me Borda and Condorcet are "equivalent".
It could be the case in term of determining the winner when there is a
Condorcet winner.
However, Borda is not cloneproof and I always believed Condorcet methods were.

So, does a susceptibility to cloning affects only second or latest rankings
when ordering candidates after the election
or what is my mistake?

Steph

Simon Gazeley a écrit :

> Dear List
>
> For interest.
>
> Simon
>
>
> NANSON'S PROPOSITION
>
> Nanson proved that if a series of Borda counts is conducted on the same
> set of votes, eliminating each time the candidates whose Borda scores
> are below the average, the one candidate who is left at the end is the
> Condorcet winner if there is one.
>
> BORDA COUNTING
>
> I have seen various specifications for a Borda count, many of which are
> mutually inconsistent. The specification I use is as follows:
>
> a. Voters rank the candidates in order of preference, starting at 1; the
> second favourite is 2, the third favourite is 3 and so on. Equal
> preferences can be catered for and there is no need to rank all the
> candidates.
>
> b. Each ballot-paper has a value of r(r-1)/2 points, where r is the
> number of candidates.
>
> c. At the count, each ballot-paper is examined and each candidate is
> awarded the number of points that corresponds with the number of
> non-eliminated candidates ranked below him/her on that paper. If the
> voter has awarded equal preferences to two or more candidates, those
> candidates are awarded the arithmetic mean of the points they would have
> had if the preferences had been discrete. Any candidates left unranked
> on a ballot-paper are treated as equal preferences and the remaining
> points are distributed equally among them.
>
> d. A candidate's "Borda score" is the total number of points awarded to
> him/her in the count.
>
> NANSON'S PROPOSITION PROVED
>
> Suppose that v votes have been cast in an election in which r candidates
> are standing. In a Borda count involving only two of those candidates a
> total of v points are scored of which, barring ties, one of the
> candidates has more than v/2, the other has fewer. The Condorcet winner
> by definition scores more than v/2 points in such pairwise contests with
> each of the r-1 other candidates, making a total of more than v(r-1)/2
> points. The total number of points scored in a Borda count is vr(r-1)/2
> and the arithmetic mean is v(r-1)/2, less than the Condorcet winner's
> total: therefore, if candidates who have less than the arithmetic mean
> of points are eliminated, the Condorcet winner, if there is one, will be
> among those who survive.
>
>
>
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