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. > > > > ---- > Election-methods mailing list - see http://electorama.com/em for list info ---- Election-methods mailing list - see http://electorama.com/em for list info
