[EM] Re: Hand counting election methods
Gervase Lam
gervase at group.force9.co.uk
Wed Nov 19 16:07:02 PST 2003
> From: Gervase Lam <gervase at group.force9.co.uk>
> Subject: Re: [EM] Re: Hand counting election methods
> Date: Wed, 19 Nov 2003 01:19:20 +0000
I was mostly using intuition in this particular post, which was about hand
counting a Condorcet vote. I was mostly focussing on whether the
counting would obtain all of the required data in order to make up the
results matrix. And the counting did.
As a result, I didn't think about how the hand counting method would be
translated into a pairwise matrix. Therefore, I want to make some
comments on this.
> Have (N - 1) teams. Each team is assigned to a candidate A, B, C, D,
> etc...
>
> For each ballot, a member of team A marks down the candidates who beat
> candidate A and the candidates who lost to candidate A. The ballots are
> then passed to team B, the members of which do the same thing except for
> candidate B. And so on.
What is happening here is that team A is filling in the row and column for
candidate A in the pairwise matrix. Team B is doing likewise for
candidate B and so on.
As a result, what you would get is a results matrix with each cell/element
in the matrix getting two counts. For example, one of the counts would be
from team A counting the number who ranked A above B. The other count
would come from team B, who would be counting the number who ranked B
below A.
It can therefore be seen that one of the counts is redundant. That is
unless you want a second count to check the first count.
Therefore, all you want to do is get each team to fill in only its row of
a matrix. There is no need for the team to bother filling in its column.
The simple way to do this is to have team A, for example, only mark down
the candidates who beat A. Similarly, team B only mark down the
candidates who beat B and so on.
This halves the amount of work and time needed to do the count. This
excludes overheads, of course.
I mentioned that there should be (N - 1) teams. This is wrong. For the
elements in the results matrix to be all consistently filled in, there
needs to be N teams.
If the original method were followed, all of the elements in the matrix
would contain two counts except for the team "missed out". The candidate
that would have been assigned to that team would only have one count in
each of its matrix elements.
Thanks,
Gervase.
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