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

Jobst Heitzig heitzig-j at web.de
Sat Nov 6 04:23:18 PST 2004

```Dear Kevin!

you wrote:
> 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.

This is what happens in your version, right. And at first it seems to be
intuitive to require that the winner should be more approved of than any
candidate beating her/him. But let's look at your own example:

> (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.)

Isn't that very strange? We have a cycle A>C>B>A and the one with the
*second* largest approval (B) should win? If you like to put more weight
on approval, it would only seem natural to me to require that in a
3-cycle, the most approved one should win. To achieve this, you could
define defeat strength as the negative of the approval of the defeated
option instead of the positive of the approval of the defeating option,
which is likewise easy to determine from your matrix. Equivalently and
even more simply, you can use the number of voters not approving of the
defeated option. Formally:
strength of defeat A>B := - number of voters having B in slot 1 or 2
or
strength of defeat A>B := number of voters having B in slot 3
With this definition, cycles are broken at the defeat against the most
approved candidate.

However, like your original definition, also this defeat strength is
somewhat counterintuitive for two reasons: First, the strength of a
defeat depends only on one of the two candidates. Second, it counts some
voters prefering the defeated candidate over the defeating one.

The latter fault seems very bad to me when using approval of the
defeating option since it means that when trying to convince people to
go voting, you will have to admit that when they put A in slot 1 and B
in slot 2, they can actually strengthen the reverse defeat B>A! When
using the above variant instead, the effect is not quite so important
since it then only means that when you place both A and B in slot 3 and
actually prefer A over B, you might strengthen the reverse defeat B>A by

So, in situations in which it seems really important to have a counting
procedure as easy as you demand, I suggest to use the above variant

Now for your 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).

With my original proposal, we have strengths N>G:0, G>B:51, B>N:49 here,
so that G wins. As in every deterministic method, the N voters can elect
N instead. That is a problem you can only solve by using a method which
chooses non-deterministically in case of cycles, a point I stressed
several times in the past. However, simply downranking G from N>G>B to
N>>GB (which is nothing else than truncation) does not work here, since
it makes B a Condorcet winner, which is even worse for the N voters!

By the way, note that my original definition of defeat strength
corresponds to weighted pairwise with slots 1 and 2 getting a rating of
100 and slot 3 getting a rating of 0, so we can expect it to have the
same good anti-strategy properties!

With my the alternative definition of defeat strength, we get N>G:-51,
G>B:-49, B>N:-27 (or N>G:49, G>B:51, B>N:73), which would also elect G.

Yours, Jobst

```