# Lowest Number of First Choice votes tie breaker

Steve Eppley seppley at alumni.caltech.edu
Thu Oct 17 16:19:54 PDT 1996

```Demorep1 wrote:
>As a tie breaker I suggest that the candidate (in the tie group of
>candidates) with the lowest number of first choice votes should be
>the initial loser.
>
>I note that if the fewest-votes-against-in-worst-defeat (FVA)
>method is used as a tie breaker then some of the voters by voting
>sincerely may defeat their first choice.

I thought we'd discussed this enough, but maybe you're saying
something I'm misinterpreting.

The game-playing and insincere voting would be worse using Lowest
Number of First Choice Votes.  Have another look at the 46/20/34
examples, using this tie-breaker:

46: R>M>L            46: R
20: M                20: M
34: L>M>R            34: L>M>R

M beats all in the voting on the left.

In the voting on the right, M wins if the tie-breaker is Condorcet.
But R wins by the tie-breaker you're suggesting (if my understanding
of the rest of the steps is correct).

This means the 46 R>M>L voters can win by voting R instead, defeating
a beats-all winner.

The scenario you constructed where the A>C>B voters could elect A
by changing a few ballots to A>B>C depended on the voters having a
level of knowledge that is implausibly detailed.  There's a high risk
they'd elect B if they're wrong, since the numbers are so close.

The most important thing, seems to me, is to elect the candidate who
would beat all, if there is one, according to the voters' sincere
preferences.  So an example where there's a beats-all winner, but
another candidate would win if some voters rationally misrepresent
themselves, is a more compelling "bad example" of a method.

---Steve     (Steve Eppley    seppley at alumni.caltech.edu)

```