# [EM] Single-candidate DMTBR idea

Kristofer Munsterhjelm km_elmet at t-online.de
Fri Mar 11 04:42:51 PST 2022

```On 3/11/22 6:04 AM, Kevin Venzke wrote:

> This is interesting though. The "obvious" way to expand IFPP to many candidates
> is to eliminate candidates with a below-average vote count. But it seems like
> the 1/3 rule was the important thing, as it's what enforces that always either
> one or two candidates are eligible to win, and these candidates can't be harmed
>
> Tricky, to reduce a scenario to this state without breaking mono-raise.

Actually, now that I think about it, I think the rule I provided
combined with Condorcet *is* fpA-fpC (disregarding ties for now).

Suppose we have an ABCA cycle without pairwise ties. Since there are
three candidates, at least one of them must have more than 1/3 fpp,
unless there's an exact tie for first preferences. And in that case, the
election is a perfect tie too, because, since we have an ABCA cycle, the
ballots must be

x: A>B>C
x: B>C>A
x: C>A>B.

If there's only one candidate with more than 1/3 fpp, then that
candidate wins (both with the given rule and in fpA-fpC). So suppose the
candidates with more than 1/3 fpp are A and B (without loss of
generality; you can always go along the cycle to make it that way).

Then the rule given will elect A since he beats B pairwise. For fpA-fpC,
A's score is fpA-fpC and B's score is fpB - fpA. So let's try to
maximize B's score to produce an example where fpA-fpC and the rule
don't agree.

The margin between these is fpB - fpA - (fpA - fpC) = fpB - 2 fpA + fpC.
So to maximize the margin in favor of B, we should minimize A's number
of first preferences, and the distribution of first preferences between
B and C doesn't matter. So let A's fpp be 1/3 + epsilon. And for the
sake of convenience, set fpC = 0, so that B's first preference count is
2/3 - eps.

The margin in favor of B is then 2/3 - eps - 2/3 - 2 eps. For epsilon=0,
this is 0, but for any positive epsilon, the margin is negative and A
wins. Which was what was wanted.

(It's most likely impossible for an ABCA cycle to exist if B has 2/3 of
the first preferences, so the argument is generous to B.)

-km
```