Re: Some ways to extend social rankings to scores

Hi Kristofer,

Might the inverse of the keep value under a Warren or Meek STV count suit
as well?

Regards,

On Wed, 11 Feb 2026, 12:03 am Kristofer Munsterhjelm via Election-Methods, <
[email protected]> wrote:

> Suppose we want to make a method return not just who won (and the order
> of finish), but how well each candidate did - how close to each other
> the candidates were - by also returning a score for each.
>
> (See the end of the post for 2009 Burlington results :-)
>
> This is easy in FPTP: just count the number of first preferences and
> divide by the number of voters.
>
> But suppose that we'd like to have less of a spoiler effect than FPTP
> *and* return scores.
>
> Ideally, we'd like the scores to not be affected by what other
> candidates are running. But that's impossible, at least for a
> majoritarian method.
>
> Let's say that we have three candidates: A, B, and C, and it's a
> Condorcet order: A beats B and C, B beats C, and C is the Condorcet
> loser. Say furthermore that B's win over C is 75-25, and A's win over B
> is 60-40. Then the straightforward scores if only two of them were
> present would be (as pretty much every method, including FPTP, would
> tell you):
>
> In A vs B:
> A: 60%
> B: 40%
>
> In B vs C:
> B: 75%
> C: 25%
>
> The very strictest IIA interpretation would have these scores not change
> when the third candidate is introduced. (That's what cardinal methods
> with absolute interpersonal comparability do.) Since there's a Condorcet
> order, the *ranking* of the other candidates don't change when we
> introduce a third, e.g.
>
> A>B becomes A>B>C after adding C,
> B>C becomes A>B>C after adding A.
>
> But if the scores were to stay the same, then B's score would have to be
> 40% and 75% at once. That's clearly impossible.
>
> So majoritarian methods' scores, if they're numbers on a scale, must to
> some degree be relative.
>
> I've found two ways to more or less consistently normalize the scores to
> the number of candidates. One is to keep the top two scorers' score the
> same, and the other is to make the scores sum to 100%. (For lack of a
> better term, I'd call the first "minmax-style" because that's what
> minmax does.)
>
> I've also found two ways to calculate these scores - one that's
> appropriate for LIIA methods, and another that should work on a much
> broader range of methods.
>
> So let's do the calculation types first:
>
> The LIIA style is this: Suppose that candidates are ordered x_1 > x_2 >
> x_3 > ... > x_n, and the pairwise victory of x_k against x_(k+1) is
> d(x_k, x_(k+1)). Let the score of candidate x_k be s_k. Then set
>
> s_(k+1)/s_k = d(x_k, x_(k+1))/d(x_(k+1), x_k)
>
> for k = 1..n-1.
>
> This is a set of n-1 equations with n unknowns: the normalization method
> fixes the last unknown.
>
> For the A>B>C example above, we'd have:
>         s_A/s_B = 60/40
>         s_B/s_C = 75/25.
>
> The nice thing about this approach is that the relative scores of
> adjacent candidates stay the same as long as the social ranking/ordering
> stays the same; in particular when losers or winners drop out and the
> method passes LIIA, the relative scores of the other candidates stay the
> same.
>
> The plump style that's applicable to more methods is this: For each
> non-winning candidate x_k, let P_k be the number of plump/bullet votes
> for x_k that have to be added to the election to make x_k the winner.
> Let P_none be the number of such votes that have to be added for a new
> candidate (that currently has no support) to win.
>
> Then set up a linear scale that maps 0% to P_none, and set x_1/x_2 =
> d(x_1, x_2)/d(x_2, x_1).[1]
>
> This again has one more unknown than equations (the rate of change of
> the linear scale, or equivalently, the "virtual" negative value P_1 that
> should be assigned to a winner, since the winner needs no plump votes to
> win).
>
> And again, the normalization choice determines that unknown.
>
>
> Now for the normalization approaches:
>
> Minmax style is simply this: Let the winner x_1's score be d(x_1,
> x_2)/d(x_2, x_1), so that the top two's scores stays the same no matter
> how many losers are removed from the election (assuming LIIA).
>
> For the A>B>C example and LIIA-style relative scores, that gives:
> s_A: 0.6
> s_A/s_B = 60/40 = 0.4
> s_B/s_C = 75/25 = 2/15 = 0.1333...
>
> so
>         A: 60%
>         B: 40%
>         C: 13.3%
>
> Sum-to-100% is just what it says. For the A>B>C example:
>
> s_A/s_B = 60/40
> s_B/s_C = 75/25
> s_A + s_B + s_C = 100%
>
> which gives
>
> s_A = 9/17 = 52.9%
> s_B = 6/17 = 35.3%
> s_C = 2/17 = 11.8%
>
>
> Finally, here are the different scores for the Burlington election with
> RP(margins) as the base method for the plump calculations:
>
> (Minmax style)
>
> Candidate     LIIA-relative %      Plump-based %
>
> Montroll      53.91%               53.91%
> Kiss          46.09%               46.09%
> Wright        43.42%               43.53%
> Smith         41.43%               38.67%
> Simpson        5.36%                7.78%
> Write-in       0.26%                1.15%
>
> (Sums-to-100)
>
> Candidate     LIIA-relative %      Plump-based %
>
> Montroll      28.38%               28.21%
> Kiss          24.20%               24.11%
> Wright        22.80%               22.78%
> Smith         21.75%               20.23%
> Simpson        2.81%                4.07%
> Write-in       0.14%                0.60%
>
> Minmax-style is more like "approval ratings" while sums-to-100 is more
> like "how big a share of the total". I'd be inclined to say sums-to-100
> is more natural given that ranks are relative anyway, but what do you
> think?
>
> -km
>
> [1] Alternatively make the algorithm accept negative ballot counts and
> see how many negative plump votes have to be added until adding more
> makes the winner lose, and fix the scale so that zero plumpers added
> would give 50%. "Negative plumping" like this is possible with ranked
> pairs, and the values and two scale points determine the system, thus
> making normalization unnecessary. It gives results similar to
> minmax-style, but slightly different (e.g. Montroll gets 54.44% instead
> of 53.91%)
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