# [EM] Re: Hand counting election methods

Gervase Lam gervase at group.force9.co.uk
Tue Nov 18 17:15:12 PST 2003

```> Date: Mon, 17 Nov 2003 20:45:01 -0800
> From: Bart Ingles <bartman at netgate.net>
> Subject: Re: [EM] Re: Hand counting election methods

> I would start by separating out ballots containing bullet votes.  You
> would only need to go through these once, and the gap between first and
> second place in these ballots might be greater then the total number of
> multicandidate ballots.
>
> If not, sort out all ballots containing exactly two votes into NxN
> piles, ordered into rows and columns according to the positions of the
> first and second approval in list order.

You could extend this Approval counting method to three votes by having N
piles of (N - 1) x (N - 1) grids.  After that, it is probably best to
recount on a per candidate basis.

> Someone on another list once suggested perforated ballots, which you
> could separate into separate piles & treat the same as plurality.

I think this is probably the best option.

Your gridding idea has made me start thinking of extending my original
Condorcet hand counting method.  Instead of just looking at the top ranked
candidates (i.e. beats all candidates) on each ballot, why not look at the
bottom ranked candidate as well.  All the ballots can therefore be
arranged in a NxN grid.

However, what you do next, I don't know.  Also, it can't handle tied top
and bottom ranks.

Then I began thinking about how similar Condorcet is to pairwise
"approval."  Basically, each pairwise contest is really an "approval"
vote, with the voters being allowed rank candidates right on a cut-off
point in the case of tied rankings.  Also, in contrast with Borda, the
distance between the rankings of two candidates does not matter.

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.

Any candidates who are not marked down have obviously been given the same
ranking as the candidate assigned to the team.  Therefore, this is ideal
for handling tied rankings.

Each team could even be divided into two halves.  One half marks down the
candidates who beat their assigned candidate, while the other half marks
down the candidates who lost to their assigned candidate.

I think it is best to exclude the team who would have been assigned to the
expected winning candidate.  The idea being that this would give an
approximately balanced work load between the teams.

You could have N teams instead of (N - 1).  The Nth team would end up
being a count checking team to see if the ballots have been counted up
correctly or not.

However, I don't think this is necessary.  In fact, I think it could cause
confusion.  I think it is better to have a tolerance for each pairwise
contest and re-count if that tolerance is not met.

This method is very similar to counting the Approval ballots in a per
candidate basis.  The only added complexity is that you have to look at
who has been ranked higher and lower for each candidate.

Thanks,
Gervase.

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