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AB>AC>AD>BC>BD>CD

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When I first envisioned this project, I had no idea that it would be so easy to completely ruin the elections.  Even starting with the very first and arguably simplest matrix, AB>AC>AD>BC>BD>CD, I’m finding it possible for any candidate to win.

To reiterate the notation as used here, the matrix specification above says that there are four candidates (A, B, C, and D), that A beats B more often than A beats C, and so on.  I also need to introduce a notation for the ballots themselves; for example, “acbd” will refer to a ballot cast by a person who prefers A over C, C over B, and B over D.  The preferences of individual ballots are forced to be transitive; that is, if an individual prefers A over B and B over C, then they cannot prefer C over A.  With that notation, we can express the matrix specification as a set of inequalities in terms of the twenty-four possible complete ballots:

AC > CA:  abcd + adbc + adcb + dbac + acbd + dabc + acdb + bacd + abdc + bdac + badc + dacb > cbad + dcab + bcda + dcba + bcad + cabd + cadb + cdab + bdca + cdba + dbca + cbda
AB > BA:  abcd + adbc + adcb + acbd + dabc + acdb + abdc + cabd + cadb + cdab + dcab + dacb > cbad + bacd + bcda + dbac + dcba + cbda + bcad + bdac + badc + bdca + cdba + dbca
AD > DA:  abcd + cbad + adbc + adcb + acbd + acdb + abdc + cabd + cadb + badc + bacd + bcad > dcab + bcda + dbac + dabc + cbda + cdba + dcba + bdac + cdab + bdca + dacb + dbca
BC > CB:  abcd + adbc + bcda + dbac + dabc + bcad + bacd + abdc + bdac + badc + bdca + dbca > cbad + dcab + adcb + acbd + dcba + acdb + cabd + dacb + cadb + cdab + cdba + cbda
CD > DC:  abcd + cbad + bacd + bcda + acbd + acdb + bcad + cabd + cadb + cdab + cdba + cbda > adbc + adcb + dbac + dabc + bdca + bdac + dcba + abdc + badc + dcab + dacb + dbca
BD > DB:  abcd + cbad + bacd + bcda + acbd + cbda + bcad + abdc + cabd + bdac + badc + bdca > adbc + adcb + dbac + dabc + acdb + cdba + dcab + dcba + cadb + cdab + dacb + dbca
AB > AC:  cadb + cdab + dcab + cabd > bacd + badc + bdac + dbac
AC > AD:  dabc + bdac + dacb + dbac > cbad + cadb + bcad + cabd
AD > BC:  cbad + adcb + acbd + acdb + cabd + cadb > bcda + dbac + dabc + bdca + bdac + dbca
BC > BD:  dabc + adbc + dbca + dbac > cbad + cbda + cabd + acbd
BD > CD:  bdca + bdac + badc + abdc > acdb + cdab + cdba + cadb

The first six inequalities are implicit in the fact that the second term is not listed in the matrix; the last five are explicit in the specification.  Note that the last five have been shortened by removing identical terms from each side of the inequality; for example, “abcd” satisfies both AB and AC.  We could expand the list to include the transitive inequalities, like AB > AD, but that doesn’t buy us anything.

This is solved fairly easily, but has multiple solutions.  An infinite number of them.  In fact, if we assume that the election will be determined by a plurality vote, and each voter will vote only for his or her top preference, we can add a few more inequalities to specify a winner:

A > C: abcd + adbc + adcb + acbd + acdb + abdc > cbad + cdba + cabd + cadb + cdab + cbda
A > B: abcd + adbc + adcb + acbd + acdb + abdc > bacd + bcda + bcad + bdac + badc + bdca
A > D: abcd + adbc + adcb + acbd + acdb + abdc > dcab + dbac + dabc + dacb + dcba + dbca

To give one example for each candidate, feel free to check that each of the following give the same ordering for the pairwise matrix:

A B C D
abcd 13 10 10 11
abdc 11 11 11 13
acbd 13 13 10 16
acdb 10 10 10 10
adbc 13 12 13 10
adcb 10 7 6 5
bacd 10 15 10 10
badc 10 10 10 10
bcad 10 10 10 10
bcda 10 10 10 10
bdac 10 10 12 10
bdca 10 10 8 5
cabd 11 12 20 11
cadb 10 10 11 10
cbad 8 7 5 10
cbda 7 6 7 6
cdab 10 10 10 10
cdba 9 9 8 7
dabc 12 14 15 19
dacb 10 10 10 11
dbac 9 7 10 9
dbca 7 7 5 7
dcab 10 11 10 10
dcba 7 9 9 10

At first glance, they’re not so very different.  From having started with a default of 10, no ballot goes low in one and high in another.  That’s probably not always true, but it does highlight a potential trend to investigate.

The really upsetting part about this, however, is that it doesn’t seem fair.  If only two candidates were running, A would win against any other candidate, and D would likewise lose.  How, then, can A not win when all four run?

The good news is that of the voting methods I’ve implemented, only plurality fails quite so miserably here.  The pair-wise methods, of course, have absolutely no disagreement.  Even instant-runoff always elects A for these for cases.  IRV, Borda count, and the Bucklin method can show a skewed total ordering, but they’re really only meant for choosing a single winner.

Then again, perhaps I can break even those three.  Borda should be convenient enough to provide inequalities for, but Bucklin and IRV are more iterative in nature.  I’ll have to think about this some more…

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Written by eswald

19 Jul 2011 at 9:49 pm

Posted in Politics

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