Some ways to extend social rankings to scores
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%)
Score is not the same as STAR, but they have something in common. I'll, admit I don't think about Score Voting much outside of STAR.
Just to keep the problem tractable, I would leave it to single-winner and 3 significant candidates: the voter's favorite candidate, the 2nd-choice (or lesser evil) candidate, and the candidate that the voter hates. The interaction (and number of permutations) with 4 or more candidates is just too many to get a good grip all of the contingencies.
There really is no reason for a voter to mark their STAR ballot any differently than 5-1-0 unless they anticipate the very same Center Squeeze effect that happened to Burlington 2009 or Alaska in August 2022. In every other case, there is no incentive, whatsoever, to ever score your 2nd-choice or lessor evil candidate any higher than 1.
The only time a voter would ever want to score their lesser evil higher is if they actually want their lesser evil to defeat their favorite candidate in getting to the STAR automatic runoff. This would be because the voter understands that their favorite candidate will not be able to defeat the candidate they hate in the final runoff. In every other scenario, there is no incentive whatsoever to score one's lesser evil (or 2nd-choice) higher than 1, just enough to beat the candidate you hate in case your favorite cannot make it into the runoff.
I had a little argument with Mark Frohnmayer (the big STAR advocate/promoter, sorta like the Rob Richie of STAR) about this.
On 02/10/2026 8:02 AM EST Kristofer Munsterhjelm via Election-Methods election-methods@lists.electorama.com 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 :-)
A couple of years ago, I did an analysis of STAR (where everyone votes 5-1-0) applied to Burlington 2009. (I didn't worry about Smith or Simpson, while I will say Smith was not insignificant, he was eliminated before the big three that demonstrated the Center Squeeze. Simpson and Write_in are insignificant, I wouldn't bother with any other keystrokes regarding Simpson or Write-In.) Turns out it makes the same mistake as IRV (when people score 5-1-0).
https://www.reddit.com/r/EndFPTP/comments/1bbq6hu/heres_a_good_hypothetical_for_how_star_fails/
Sorry Kristofer, I'm just reacting. I haven't read through your post, yet. It looks complicated, but I'll get through it.
--
r b-j . _ . _ . _ . _ rbj@audioimagination.com
"Imagination is more important than knowledge."
.
.
.
On 2026-02-10 19:22, robert bristow-johnson via Election-Methods wrote:
On 02/10/2026 8:02 AM EST Kristofer Munsterhjelm via Election-Methods election-methods@lists.electorama.com 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 :-)
A couple of years ago, I did an analysis of STAR (where everyone
votes 5-1-0) applied to Burlington 2009. (I didn't worry about Smith or
Simpson, while I will say Smith was not insignificant, he was
eliminated before the big three that demonstrated the Center Squeeze.
Simpson and Write_in are insignificant, I wouldn't bother with any
other keystrokes regarding Simpson or Write-In.) Turns out it makes the
same mistake as IRV (when people score 5-1-0).
https://www.reddit.com/r/EndFPTP/comments/1bbq6hu/heres_a_good_hypothetical_for_how_star_fails/
Sorry Kristofer, I'm just reacting. I haven't read through your post, yet. It looks complicated, but I'll get through it.
I was talking about scores as outputs, not inputs :-)
I.e. a Condorcet method that doesn't just say "Montroll is in first,
Kiss is in second", but whether Montroll's win was comfortable or close
(and similarly, if Wright was close to passing Kiss, Simpson close to
Wright, etc.)
The voters would still use plain old ranked ballots.
The method would just return more detailed information -- like how FPTP
shows the strength of victory on a scale (as percent of first
preferences gained), even though FPTP isn't a cardinal method.
As for STAR strategy, it makes sense that STAR would behave that way if
everybody does 5-1-0, because that's more or less reproducing the logic
of IRV/TTR (first preferences count a lot, other preferences don't). I
suspect that STAR strategy is less unbalanced, though; that some times,
5-4-0 is good, other times 5-1-0 is good, and you don't get skewered the
same way you can in Approval if you get it wrong.
-km
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, <
election-methods@lists.electorama.com> 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%)
Election-Methods mailing list - see https://electorama.com/em for list
info
On 2026-02-11 05:22, Etjon Basha wrote:
Hi Kristofer,
Might the inverse of the keep value under a Warren or Meek STV count
suit as well?
That's an interesting thought; they might, but I'd have to think more
about how to represent scores for multiwinner, what they would mean, and
how keep values would tally with it.
For instance, I imagine using keep values would be better at showing a
distinction between the winners than between losers (who have no surplus
to redistribute as they don't reach the quota).
The combinatorial Condorcet view (CPO-STV, Schulze STV) would be to give
each assembly outcome a different score, but that's very hard to
interpret.
In any case, straightforward scores (one per candidate) would always
lose some context because, unlike a single-winner method where closer to
some favored point in space is better (median voter, or some wing
position for center squeeze methods), what one candidate's optimum
location is depends on what other candidates were elected.
-km
One of the advantages of things like approval voting and FPTP, along with score voting and even Borda Count is that you can simply give a list of scores for all candidates, and give them as a percentage of the maximum.
I quite like the idea of doing this for Condorcet methods as well, but it's really only ever going to be an academic exercise. I can't imagine them being published as part of any official results.
Also, methods that don't pass LIIA presumably would still sometimes return an order as well as just the winner. Would it not be the case that ordering by plump score would sometimes contradict the order given by the method?
Also if the total is set to be 100, then some candidates will inevitably get a lower percentage than the percentage of ballots where they are top ranked. This will likely not make much sense to people.
Toby
On Tuesday, 10 February 2026 at 13:03:00 GMT, Kristofer Munsterhjelm via Election-Methods election-methods@lists.electorama.com 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%)
Election-Methods mailing list - see https://electorama.com/em for list info
On 02/10/2026 2:07 PM EST Kristofer Munsterhjelm km-elmet@munsterhjelm.no wrote:
I was talking about scores as outputs, not inputs :-)
My point is, with STAR, that if your input is A>B>C, unless you anticipate the Center Squeeze effect (something that happens less than 0.5% of the time in RCV elections in the U.S.), there is no reason whatsoever to mark your STAR ballot any differently than A:5, B:1, C:0 .
The voters would still use plain old ranked ballots.
So this is some Borda-ish scheme that turns ranked ballots into scores.
The method would just return more detailed information -- like how FPTP
shows the strength of victory on a scale (as percent of first
preferences gained), even though FPTP isn't a cardinal method.
As for STAR strategy, it makes sense that STAR would behave that way if
everybody does 5-1-0, because that's more or less reproducing the logic
of IRV/TTR (first preferences count a lot, other preferences don't). I
suspect that STAR strategy is less unbalanced, though; that some times,
5-4-0 is good, other times 5-1-0 is good, and you don't get skewered the
same way you can in Approval if you get it wrong.
The only time (assuming 3 candidates, or 3 significant candidates) that you ever want to mark your STAR ballot 5-4-0 is when you actually want your 2nd-choice (or lesser evil) candidate to actually beat your favorite candidate to get into the STAR final runoff. This would be because you fear your favorite cannot beat the candidate you hate in the runoff but your lesser evil candidate can beat your greater evil candidate.
That's my spin on it.
--
r b-j . _ . _ . _ . _ rbj@audioimagination.com
"Imagination is more important than knowledge."
.
.
.
Hi Kristofer,
Sorry, I should have added that the inverse of the keep value will most
likely only make sense if a very particular application of STV where
instead of electing n winners from k candidates we "elect" K winners (i.e.,
we just redistribute votes among all candidates such as to leave each with
exactly the same vote count). Only in this scenario does the keep value
truly measure support.
Regards,
On Thu, 12 Feb 2026, 12:55 am Kristofer Munsterhjelm, <
km-elmet@munsterhjelm.no> wrote:
On 2026-02-11 05:22, Etjon Basha wrote:
Hi Kristofer,
Might the inverse of the keep value under a Warren or Meek STV count
suit as well?
That's an interesting thought; they might, but I'd have to think more
about how to represent scores for multiwinner, what they would mean, and
how keep values would tally with it.
For instance, I imagine using keep values would be better at showing a
distinction between the winners than between losers (who have no surplus
to redistribute as they don't reach the quota).
The combinatorial Condorcet view (CPO-STV, Schulze STV) would be to give
each assembly outcome a different score, but that's very hard to
interpret.
In any case, straightforward scores (one per candidate) would always
lose some context because, unlike a single-winner method where closer to
some favored point in space is better (median voter, or some wing
position for center squeeze methods), what one candidate's optimum
location is depends on what other candidates were elected.
-km
The other thing with the LIIA version of the scores is that it can make wins look much bigger than they are. If you have the classic A>B>C>A cycle with each victory a 2:1 ratio, then you could add a few ballots to break the tie and make A the winner, but the ratios would be largely the same (we could be talking about thousands of voters and a handful of extra ballots). So the score ratios would be 4:2:1. In other words out of 100:
A: 57B: 29C: 14
I don't think this really captures the true nature of the election that took place.
Toby
On Wednesday, 11 February 2026 at 15:31:06 GMT, Toby Pereira tdp201b@yahoo.co.uk wrote:
One of the advantages of things like approval voting and FPTP, along with score voting and even Borda Count is that you can simply give a list of scores for all candidates, and give them as a percentage of the maximum.
I quite like the idea of doing this for Condorcet methods as well, but it's really only ever going to be an academic exercise. I can't imagine them being published as part of any official results.
Also, methods that don't pass LIIA presumably would still sometimes return an order as well as just the winner. Would it not be the case that ordering by plump score would sometimes contradict the order given by the method?
Also if the total is set to be 100, then some candidates will inevitably get a lower percentage than the percentage of ballots where they are top ranked. This will likely not make much sense to people.
Toby
On Tuesday, 10 February 2026 at 13:03:00 GMT, Kristofer Munsterhjelm via Election-Methods election-methods@lists.electorama.com 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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