[EM] Single-candidate DMTBR idea

Kevin Venzke stepjak at yahoo.fr
Thu Mar 10 21:04:27 PST 2022

Hi Kristofer,

> On 3/11/22 12:33 AM, Kristofer Munsterhjelm wrote:
> > If there are two candidates with more than 1/3 fpp, elect the one that
> > pairwise beats the other. If there is only one candidate with more than
> > 1/3 fpp, elect that candidate. And if there are none, do something else
> > that doesn't violate monotonicity given the rules above.
> I was thinking, perhaps this isn't monotone after all. The usual
> nonmonotonicity problem (e.g. in IRV) has the pattern that you're in an
> ABCA cycle, then some BAC voters rank A higher so that B drops below C
> in first preferences, and then A is defeated by C in the second round.
> But suppose A and B both have > 1/3 first preferences. Then upranking A
> on a BAC ballot has no effect if B stays above 1/3 first preferences,
> and if it pushes B below the threshold, then A wins outright as the sole
> candidate with more than 1/3 fpp. Upranking A can't give C more first
> preferences, and three candidates can't all have more than 1/3 of the
> first preferences.
> So is monotonicity preserved after all? If so, that's a nice trick!
> Does it also apply to the generalization where you just take the two
> candidates with the most first preferences? I'm not sure.

In the three-candidate case, electing the pairwise winner between the top two
candidates is basically IRV.

Without the 1/3 limit it could happen that the FPW gets more votes and changes
who the second place candidate is. He might not beat the new one.

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.


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