# False votes in Condorcet ties ?

Steve Eppley seppley at alumni.caltech.edu
Sun Nov 3 22:33:25 PST 1996

```DEMOREP1 wrote:
>If false, then does it matter?

In what sense do you think there's something false?  The people
who voted A1B2 have voted that they prefer B more than C.

(You might want to use a more compact syntax, like in Donald's
vote-sums: AB could mean the same as A1B2.)

Why would you think their preferences for B over C should be
discounted?  If A weren't running, these voters would have voted B1
and you'd certainly count them as preferences for B over C then.

If first choices are privileged in the tally, then similar
candidates will fragment the votes if they run.  This is likely
to defeat the center candidate, since wing candidates will try
to position themselves just slightly left or right of the center
candidate and this will peel away most of the center's first choice

Suppose candidate A prefers B more than C.  If the method doesn't
count some voters' preferences for B over C if A runs, then A will
face the spoiler dilemma when deciding whether to run.  By defeating
the candidate nearer the center, this can elect the candidate at the
candidates from running, if you're trying to change from a two-party
system to a multi-party system.

>Are such same votes (such as A1B2) on the other two lines true?

They all look "true" to me.  They should be counted as preferences
for A over B, A over C, and B over C.

>If the fewest votes against is to be the tie breaker (i.e. the
>lowest total amount on the left)

That may not be quite accurate, depending on your meaning.  The only
pairings in which "votes against" can count against a candidate are
the pairings lost by that candidate.

By total amount, I hope you don't mean to total the votes against
a candidate in all that candidate's lost pairings.  A candidate's
Condorcet score is not such a total; it's the maximum of the votes
against him in his lost pairings.  (Lower scores are better, like
in golf.)

For example, in the 46/20/34 RML "sincere" scenario, L loses the
"L vs R" pairing with 46 "against" and L loses the "L vs M" pairing
with 66 against.  L's score is 66 (the maximum), not 112 (the total).

>, then I will say again it will be the strategy for the candidate
>with the plurality of first choice votes (as shown in the polls)
>to especially try to maximize the votes against the other two
>candidates so that such plurality candidate would win.

I don't see why you mention plurality of first choices in this
paragraph, since first choices are irrelevant in Condorcet.

In practice, what tactics do you mean?  For a candidate X to
increase the "votes against" another candidate Y can mean one of
two things:
1. getting more voters to rank X > Y, or
2. getting more voters to rank Z (and others) > Y.
(Of course, none of these "votes against Y" hurt Y if they're
in pairings won by Y.)

The first is the same as what happens no matter what method is used.
The second is double-edged, so X has to be careful: by getting more
voters to rank Z>Y, X may be helping to elect Z.

If there's a pairing involving X where a majority of the voters
prefers the pairing-opponent more than X, X *can't win* unless
*every* other candidate loses some pairing by an even greater
majority.

>With 4 or more candidates in a tie, the number of first choice