I just noticed this and it seemed worth writing down. It may be a well known result, but 30 seconds of Googling doesn’t turn up everything obvious.

It turns out, that under a certain (highly unrealistic) model of voter behaviour, if you use Approval Voting then you get the Borda Count winner.

In Approval Voting you get to cast a vote for as many candidates as you like, and then the candidate with the most total votes wins.

The question is: How many votes do you cast? Do you cast a vote for only your top candidate (in a close race you probably should)? Do you vote for everyone except the candidate you think is literally Hitler?

I’m actually not sure what the generic tactical answer to this is. For the sake of this post it doesn’t matter, because I’m going to be looking at the following model: What happens if everyone just picks the answer at random? i.e. you have \(N\) candidates, they rank the candidates in order of preference, and then they pick a number c between \(1\) and \(N – 1\) and vote for their top c candidates.

And the answer is that what happens is that you elect the Borda winner.

The Borda Count works as follows:

Everyone ranks every candidate. For each voter, a candidate is given a score: The top candidate gets N, the second candidate gets N – 1, and so on. The candidate with the highest total score wins.

The reason why approval voting results in the Borda winner is quite straightforward: The expected number of votes a voter gives a candidate is just an increasing linear function of the Borda score they would give that candidate.

If you rank a candidate in position t, then you vote for them unless you choose \(c < t\), which happens with probability \(\frac{t – 1}{N – 1}\). So the expected number of votes is \(0\) if this happens and \(1\) otherwise, i.e. \(1 – \frac{t – 1}{N – 1} = \frac{N – t – 2}{N – 1} = \frac{1}{N – 1} B – \frac{1}{N – 1}\), where \(B\) is the Borda score.

Because the the Borda winer is just determined by adding up scores, the scaling doesn’t affect who wins. So assuming that the populace is large enough, the randomization more or less averages out and you elect the Borda winner.

Now, as mentioned, this is a highly unlikely model of voter behaviour – it’s probably just about valid if you have a very small number of candidates (3 or 4), but for larger number of candidates I would expect the number of votes actually cast to be bunched up around either \(0\) or \(N\) with fairly sharp drop-offs moving away from that. I don’t actually know if that’s how it works in practice, but whatever happens in practice it’s very unlikely to actually be uniformly at random.

So I present this mostly as a curiousity rather than as making any point about the actual merits of either system, but I thought it was interesting.