# [EM] More easily hand-counting three-slot Condorcet

Kevin Venzke stepjak at yahoo.fr
Fri Nov 5 22:54:51 PST 2004

```Jobst,

Thanks for taking the time to look at this idea.

--- Jobst Heitzig <heitzig-j at web.de> a écrit :
> But the vector
>   (0,1,1,0,0,1,0,0,0)
> is linearly independent from
>   (1,1,1,1,1,1,1,1,1), [...]

Interesting, this is precisely what I ended up doing to convince
myself that the magnitudes couldn't be pulled out.  I'm surprised
to see it reproduced so readily.

> > I believe all rules are then equivalent to "Elect the least-
> > approved candidate who pairwise beats every candidate with
> > greater approval."
>
> That's true, but it would erroneously count those having A in slot 2 and
> B in slot 1 towards the strength of defeat A>B.

Hmm.  I note that if A is going to be elected despite being defeated
pairwise by B, this must mean that A has greater approval than B, and
that A's wins will be locked first.

(The scenario I have in mind is:
Approval ranking: A>B>C.  Pairwise: B>A>C>B.  B wins.
If B's approval is simply reduced somehow, with no other changes:
Approval ranking: A>C>B.  Now A wins.)

I think my suggestion is very similar to yours in spirit.  They both
involve compressing the top two ranks.

> In accordance with my
> recent "grand compromise" proposal, I suggest to use instead the number
> of voters who have A in slot 1 or 2 and B in slot 3, which corresponds
> to V13+V23.
>
> Although also this number cannot be determined from the numbers you
> suggest to tally, it is not too expensive to make it possible by also
> counting all voters who have both A and B in slot 1 or both A and B in
> slot 2. This corresponds to counting V11+V22 also.

This would require a second matrix, wouldn't it?  That would seem to
defeat the purpose somewhat!

Also, if you consider my second example:

49 B>>GN
24 G>>BN
27 N>G>B

I notice that the N>G win has a strength of 0, which the N>G>B voters
can adjust in a straightforward manner (by downranking G).  I can't
put my finger on it, but it seems to me that the best placement of G
is less uncertain using my suggested measure of defeat strength.
(That is, I think the desire to register a G>B vote would far outweigh
the concern that if G beats N, the N>G>B vote strengthens that
victory.  If I'm right about that, then I see it as a positive thing,
because we can get the most information from the N>G>B voters.)

I'm not yet able to picture clearly how your measure of defeat strength
would change everything, though.

Kevin Venzke

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