With a little bit of tweaking, I have the new version of Modern Ballots easily embedding into iFrames. That might end up being useful to someone that wants one on their website.

Vote page

Results page

Just include something like <iframe style="width: 100%; height: 400px;" src="https://modernballots.com/public/zombies/vote/"></iframe>

• Consider a scheduled movie night where 30 people vote on 10 different films to decide what they’ll watch.
• Presume for a moment that all movies can be placed on an axis with comedy at one end and tragedy at the other.
• Assume that each voter’s preference falls somewhere on that axis.
• Lastly, assume that the collective preference is bimodal, with some clustering around an 80/20 comedy/tragedy mix, the others a 20/80 mix.

The election occurs, the results come in and a compromise 50/50 film is declared the winner. Well, by some peculiarity of preferences, this just doesn’t sit well with the comedy-lovers, so come movie night, most of them don’t show up.

Should the votes of those that didn’t even show up count? If we reuse the ballots of those that show up, we’re likely to reach a far more tragedy-heavy winner. Further, if this movie night is held monthly, do we favourably weigh the votes of those that didn’t show up?

I’ve been trying to find a way to combine both geographic and proportional representation. First, consider allowing MPs to vote with varying strengths (e.g. if an MP represents 45k people, they vote in parliament with a strength of 45k). The only trouble with this approach is that it isn’t proportional to the popular vote.

Instead, consider something akin to a federal MMP. If your chosen candidate doesn’t win, your vote is transferred to the party leader of your choice: a two-part ballot. In our current system, you aren’t represented unless you voted for the winning candidate of your riding.

To estimate the outcome of this new system, let’s assume that vast majority voting for a candidate support their party too. Each geographic seat would be the same as it is now. For example, Vancouver Centre’s Hedy Fry still represents the riding, but only at a power of 18k. The remaining 41k ballots are transferred to the appropriate parties.

Nationally, 2.17m ballots would be transferred to the Liberals from voters in ridings where they didn’t win, 2.13m to the NDP, 1.44m to the Conservatives, 818k to the Bloc, and 539k to the Green Party. The end result has the geographic representatives holding 7.42m geographically proportional votes and the party representatives holding 7.10m, resulting in a party proportional parliament.

• Elections result data
• Script used to calculate these results

Gerrymandering in this system would be moot, although I gather voting fraud might rise. There’s also the question of who would hold the party seats. The party leaders seems like the most direct answer, but I’m not sure how I feel about any one representative wielding upwards to 15% of the total voting power in parliament.

I wanted to visualize patterns of data related to the most recent election. For those of you familiar with Gapminder.org and Hans Rosling’s TED presentations, the following Google Doc will seem quite familiar.

With the Canadian 2011 election at a close, I thought I’d take a minute to estimate the Condorcet winners per riding. The data’s available in preliminary form through Elections Canada. Let’s look at Scarborough Centre as an example for how FPTP voting fails to elect a representative candidate

Using the CBC vote compass, we can approximate how these voters might have ranked each candidate if given the option for a more expressive ballot.

Given these preferences (which I admit are a wild estimate; people vote tactically concealing their true preferences), the Condorcet winner would be the Liberal Party’s candidate. Indeed, the national breakdown would have produced results like so (full estimate attached):

For the curious, the code used to generate the above is attached.

• Elections Canada preliminary data
• Sample source code
• Python vote core

An example scenario

Let’s say you run a summer camp. At the start of the summer, 100 campers arrive and learn that you offer 5 unique courses. These courses can’t have more than 25 campers each. The campers all submit ballots expressing their preferences for the courses. How do we maximize course allocation satisfaction?

The general case

• V voters that each submit a ballot of preferences
• G groups, each with a maximum capacity (e.g. G₁^cap = 25)
• Maximize satisfaction (open to interpretation)

Algorithm sketches

A horrible algorithm

for group-x in groups:
    while group-x has remaining capacity:
        allocate any unallocated voter to group-x

Clearly this isn’t going to cut it. Preferences aren’t being taken into account at all. We’re just dumping voters into groups.

Linear optimization

convert each ballot into an allocation of points where
    all group preferences are in [0, 1]
    sum of group preferences = 1
system satisfaction = sum of voter satisfaction for all voters
linear optimization to maximize system satisfaction

As an illustration of the principle, consider an example with voter V₁ that prefers G₁ and voter V₂ that prefers G₂. Assigning the voters to their preferred groups optimizes the system satisfaction at 2, whereas switching them around produces a system satisfaction of 0. Okay, this is a step forward.

Possible problem case:

5 groups, each with a capacity of 1
voter 1:  0.4  0.2  0.2  0.2  0.0
voter 2:  0.7  0.2  0.0  0.0  0.1
voter 3:  0.0  0.7  0.2  0.0  0.1
voter 4:  0.0  0.0  0.7  0.2  0.1
voter 5:  0.2  0.0  0.0  0.7  0.1
system optimizes with voter 1 in group 5
voter 1 allocated to their least preferred group

This algorithm provides a measure of preference debt that we can use in the iterative versions of this problem. Also, I’m not sure how much of an issue tactical misrepresentation is here.

Additional problems

Consider the additional problem where voters vote once per day for new sets of courses. This can vary on fixed or variable set of voters, on a fixed or variable set of groups. Consider further the constraint where groups have both a minimum and a maximum capacity. Or perhaps where C has preferences for G and G has preferences for C.

I don’t quite have the solutions to any of this just yet, but it’s an interesting problem space. Someone has to have studied this already. Pointers in the right direction?

Consider an electorate divided solely on some linear issue X. Assume the breakdown is as follows:

Assume also that each position has only two representatives. Pretend you’re a voting system and imagine the ballots these people submit to you. How do you aggregate these votes to best represent the electorate? I’d argue the following is fair:

As an aside, note that the result at 6 winners isn’t really proportional to the popular vote, but that we were constrained by the number of running candidates. Had More X another candidate to run, More/More/More/Same/Less/Less would have been a fairer result.

More to the point though, notice the oscillation between the 1 and 2 winner scenarios. Although the electorate is best represented with a single Same X winner, More/Less better represents the electorate than Same/Same. While this holds true when the number of winners is known and fixed, our notion of fair changes when the number of winners is variable.

Consider the scenario where you have one representative, but occasionally need a second. Since the first representative is fixed, you choose your second proportional to the electorate. We’re looking at generating an ordered list of candidates.

Note the difference here between a fixed two-winner election (where we chose More/Less) and the proportional ranking (which starts off Same/More). In this different problem space, a fair outcome produces different results.

Lastly, consider the scenario where you only need one winner, but you need a runner up in case the first winner is unable to take the position.

Where in the first example we provided More/Less as a fairer alternative to Same/Same, the problem space for this third example would dictate Same/Same over More/Less. Again, different results for the same notion of fair.

These three scenarios have produced significantly different results, illustrating fixed-winner allocation, proportional ranking and non-proportional ranking. Hopefully this highlights the need to determine in advance the context in which you’re bringing things to a vote.

The one question I have in all of this is how to distinguish between these two problems:

• One winner every couple of years (e.g. federal elections) is likely best suited with fixed-winner allocation
• One winner every hour (e.g. a board game marathon) is likely best suited with proportional ranking

Is the salient difference here that preferences can change over time? That the voters themselves change over time? Or is it that you can opt to take a break during the next board game if you don’t like it, but you can’t quite opt out of being a citizen (at least, not with as much ease). I’m really not sure here.

When you’re in a group trying to reach a collective decision by voting, chances are it sounds a little like this.

The problem here is that the majority might not want X, which only got 35% of the vote.

The Alternative

Next time, try the following. It takes a little longer, but the results will be much fairer.

Bam. Done.

As an aside, you could also infer this information from a set of collected ballots.

The Edge Cases

Disclaimer: Edge cases are exceedingly rare in large elections. What follows is simply for completeness.

In the case where two or more candidates never lose a pairing, that’s a tie. Randomly choose between these candidates.

In the case that all candidates lose a pairing, that’s a Condorcet paradox. The chances of this occurring are asymptotically rare as the number of voters increases, but here’s how to handle it if it comes up.

The first image shows the tallies. The second shows the cycles that prevented us from finding a simple winner. All subsequent steps are simply removing the weakest edge and any loose candidates.

For more information, see Wikipedia’s criteria analysis. To see it in action, try this mock election out.