Donkey Vote

[VotingPatterns]

You're stuck in the polling booth. You have before you a list of candidates. The voting system in force is one of preferential voting where all the boxes need a number entered (your first choice being indicated with a "1", your second choice indicated with a "2", and so on). You really don't care who you vote for.

Whoever is listed first on the page gets vote #1, and whoever is listed second gets vote #2, and so on.

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In Australia (one of the few countries to use preferential voting), "donkey voting" isn't all that common. The order of candidates on the ballot is randomised prior to printing, so no party can claim to be disadvantaged, anyway.

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On a whim one evening before dinner, I took up this problem. Took a few minutes to write down the design, including honoring ties and skipped numbers, and leaving open the possibility of "lasts" (so you can vote " ", "2", "1", "2", "8", " " in a six-candidate instant-runoff-vote election, number of winners irrelevant).

Key is to count pairwise preferences across all ballots. For each ballot, fill in the holes in the numbering and normalize. The above example is equivalent to 4 2 1 2 6 4, which produces the preferences AE BA BE BF CA CB CD CE CF DA DE DF FE.

• If one candidate is preferred over each other candidate by those who express a preference, that person fills one seat. Eliminate from consideration the least-preferred candidate(s) -- those preferred in the fewest pairings; drop all tied for lowest unless that prevents you from filling the seat(s). Tie-break by dropping the least-preferred among just those tied. When no other means is available to break a tie, use whatever method is already in place in law (eg the tied candidates dice for it).

• If this were the sole ballot to elect two seats, we'd notice that C is preferred over all others, and takes one seat; E is preferred over no others and is eliminated. Among the remainder (A B D F), we have to fill one seat, and note that both A and F are preferred over no others and eliminate them. This leaves B and D tied to fill one seat; existing law would cover how to resolve it (perhaps they guess separate numbers between 1 and 6, C rolls a die, and the closer guess wins the seat).

• You can thus treat it like a regular "vote for one", and it will show as a preference for that over all others; you can vote for "anyone but him" and it will show as a preference for anyone but him; you can leave it blank and it will show as "no preference stated". Very straightforward. With slight modification in the underlying data structure (track + 0 - for each pair of candidates), undercounts and partial counts can be detected by inspection.

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''For 1 seat elections this is very similar to DeCondorcet. Where there are multiple seats up, you will lose proportionality. For discussion of a possible solution that keeps proportionality, see http://www.electoral-reform.org.uk/sep/publications/votingmatters/nov00a.htm#Ordered --JoeOtten''