Auros (auros) wrote,

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Geekout! Election Methods edition...

I have a new favorite election method. It's called Definite Majority Choice, and it works like this:

Voters are given a ballot on which they provide both rankings and approvals. This can be done either on a paper ballot, where you assign some candidates numbers or letter grades (equal rankings are permitted), and then there's a specification of the lowest approved level (either given by you, or pre-defined -- e.g. if the scale is letter grades, A+ to F-, then D+ or below is disapproved, C- or above is approved); or it can be done with something more sophisticated, like a GUI on which you drag candidates onto a grid (equal-ranked candidates at the same level, left-to-right, and the vertical dimension represents unequal ranks), and drag a literal "bar" to the level of approval. In text, rank/approval ballots can be written like this: "A{BC}D|EF". The pipe | represents the approval bar, the {braces} represent equal-ranked candidates. Any candidates not explicitly ranked are considered to be disapproved, and ranked equal to each other below the last explicitly-ranked candidate.

An alternate way of expressing the same vote is: "A>B=C>D>>E>F". I personally find this method annoying, since a) it puts a symbol between EVERY pair of candidates, when the left-to-right reading order is quite adequate for the "default" concept of precedence, and b) the "much-greater-than" sign, ">>", gets lost among the other symbols.

OK, so now we have ballots. How do we tell who wins?

Well, first, let's define a "pairwise contest": you can evaluate the pairwise contest between any two selected candidates, say, A and B. There are really two contests, one on approval, and one on rankings. For the approval contest, we just count the number of ballots that approve each one. For the ranking contest, we check whether there are more ballots that rank A over B, or vice versa.

A candidate is considered "unacceptable" if he suffers a double-defeat -- losing on both approval AND rankings -- in ANY pairwise contest. A candidate who suffers no such double-defeat is "acceptable".

Usually there is only one such "acceptable" candidate. But it turns out that even if there is more than one, we can prove that among the set, one of them will necessarily defeat all others in the pairwise ranking contests. Why are we guaranteed this? The only way to avoid having a single champion in the ranking contests is to have a "cycle", where A beats B, B beats C, and C beats A. Here's an example of a cycle, with just three voters:

1 person votes ABC
1 person votes BCA
1 person votes CAB

But, if we have such a cycle, one of them will have the highest Approval score in the cycle (or if they're tied, we can settle the tie with a coin flip, or rock-paper-scissors, or whatever). The one they defeat on rankings will then be double-defeated, and thus not part of the "acceptable" set! QED.

What I like about this is that, unlike other Condorcet methods, it settles cycles in a way that avoids ever actually talking about them. "Double defeats" are pretty easy to understand as a way of eliminating contenders; selecting the guy who beats everybody else, from the final set of contenders, is also pretty easy to understand. This elegantly resolves the biggest problem in pitching a Condorcet system.

I still like the "compromise exception" I came up with, ages back. The exception works like this:

We're faced with a situation where the straight Approval winner (call him A) is different from the Condorcet-ish winner (call him C) selected by the method (e.g. DMC, or any other method combining rank and approval info). Look at the pairwise contest between A and C. In that contest, the maximum possible score for C would've been if every voter said C>A. What percent did C actually get? This is some value higher than 50%. Call this the "compromise threshhold".

Split the electorate into two groups -- those who rank C over A (C>A), and those who don't (A>C, or A=C). If, within both groups, A's approval score (percent of the group who approved A) is higher than the compromise threshhold, then A wins. Else, C wins.

Here's how it works in practice. We have a situation like this, with Left-wing, Right-wing, and Centrist candidates:

15 L|CR
36 LC|R
35 RC|L
14 R|CL

We see that L is the Condorcet winner (and the winner under DMC, "Smith//Approval", or a variety of other Condorcet-related methods). But C is the Approval winner.

In the pairwise contest L vs C, L receives 51/100 = 51% of the votes.

In the block voting L>C, C receives 36/51 = 71% approval > 51%.
Thus, a compromise has been offered by the block that won on straight rankings.

In the block voting C>L, C receives 35/49 = 71% approval > 51%.
Thus, the compromise is desirable to the block that opposed the rank-based winner.

So, C wins.

This idea of looking for the situation where "both sides" have agreed to compromise appeals to me -- I really think that there are cases where "everybody's second choice" has a better mandate, and will probably govern more effectively. The extreme case is something like:

5000001 LC|R
4999999 RC|L

But I'm aware that trying to deal with this adds a lot of complexity, and would make the method harder to pitch to voters. Given how rarely we'd expect it to be invoked, I'm fine with sacrificing it for the sake of actually doing something about our crummy electoral system.

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