J

Someone asked me how I thought the following factions would vote in this system and how it would respond:
20 = A:5, B:2, C:2, D:0
20 = A:2, B:5, C:2, D:0
20 = A:2, B:2, C:5, D:0
40 = A:0, B:0, C:0, D:5
I begin my analysis of the situation by asking how fear may play into the voters' considerations. The D-lovers are in a plurality, so the fears of the A-, B-, and C-lovers, I predict, center on D. So I'm going to have them vote their first line toward doing their best to eliminate D.
A-lovers:
1. K(D) P(A B C).
2. K(A) P(A).
B-lovers:
1. K(D) P(A B C).
2. K(B) P(B).
C-lovers:
1. K(D) P(A B C).
2. K(C) P(C).
D-lovers:
1. K(D) P(D).
Round the first:
A 60, B 60, C 60, D 40.
Eliminate D.
Second round:
A 20, B 20, C 20.
It's a tie. Usually ties are decided randomly with all voting systems, and if that is done in this case, A, B, or C should win. Let's say it is A.
Voter satisfaction or disappointment by linear and sum-of-squared-errors methods:
My questioner used a scale of 0-5, but I'm going to normalize it to 0-1.
Linear method:
For the linear method, I will treat satisfaction rather than disappointment.
A-lovers: 20 people completely satisfied, so sum is 20.
B-lovers: Their true score for A is 2/5, which is .4, and they number 20 people, so their total satisfaction is 8.
C-lovers: similarly, 8.
D-lovers: 0.
Total over all voters: 36.
Sum-of-squared-errors method:
I probably have less understanding of statistics than most people ought to have. My education is weak in that area. So, I could be off the deep end by thinking of this method of evaluating overall electoral outcomes. But because I have heard of least-squares curvefitting, and know that squared errors figure into variance, I have a general idea that errors are usually squared. And I have a notion that compassion and fairness maybe should consider that mild pain distributed among many is better than sharp pain visited on a few. And it occurs to me that from a voter's viewpoint, a disappointing outcome is kind of an error caused by poor reasoning or lack of morals on the part of the other voters with whose positions and values the voter disagrees. So that gives me a rationale to treat voter disappointment like a measure of error. So for these naive reasons, I think to use this method. Here I treat disappointment rather than satisfaction.
A-lovers: no disappointment, so error is zero and squared error is zero.
B- and C-lovers: Satisfaction of each individual was .4 so I calculate disappointment by subtracting from 1 and arrive at .6. Squared is .36. Multiply by 40 people, get 14.4.
D-lovers: total disappointment, 1 for each person. Square is 1. Total for 40 people is 40.
Sum of squared errors for election 40 + 14.4 = 54.4.
Now I will repeat the election but using fine-grained Score Voting. Coarse-grained is inappropriate for this small a count of voters.
Score does not provide a strategic incentive to invert rankings. But because of fear, the A-, B- and C-lovers will exaggerate their support for their compromise candidates. Each of them knows that overall support for their top candidate is only 20/100 = .2. My tentative model of exaggeration of support for compromise candidates says they should score their compromises at 1 - .2 = .8.
A-lovers: A 1, B .8, C .8, D 0.
B-lovers: A .8, B 1, C .8, D 0.
C-lovers: A .8, B .8, C 1, D 0.
D-lovers: A 0, B 0, C 0, D 1.
Multiply by numerosity of each faction:
A B C D
20 16 16 00
16 20 16 00
16 16 20 00
00 00 00 40
totals:
52 52 52 40
It is the same outcome as with my system, a tie among A, B, and C for the win.