[EM] Single-candidate DMTBR idea

[Kristofer Munsterhjelm]

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
> by getting more votes.
> 
> 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