# A failure mode of single transferable vote

I’ve just realised a fairly bad failure mode of single transferable vote. It was obvious that it must happen once I saw the possibility, but I wasn’t previously aware of it and I had to throw my election generating machinery at it to find an example.

The property is this: If you increase the number of seats you are electing, candidates who were elected with the smaller number of seats may no longer be elected with the larger number of seats.

Here’s a small election demonstrating this:

• One vote of 2, 0, 1, 3
• Two votes of 0, 1, 2, 3
• Two votes of 1, 0, 2, 3
• Five votes of 2, 3, 1, 0

If you run this election (I tried it with the Gregory method and no restarts, but it doesn’t actually seem to matter which method) with two seats to be elected then you elect candidates 1 and 2. If you run it with three seats to be elected then you elect candidates 0, 2 and 3 – candidate 1 now fails to gain a seat at the larger table.

The reason this happens is as follows: In both cases candidate two clears the quota on first choice votes and is elected immediately. But with two seats rather than three the quota is higher, so they lose more of their vote. As a result, candidate 3 is penalised more harshly. In both cases again there is no second candidate who immediately gets a second seat, so someone has to drop out of the race. With two seats, the votes that are pushing for candidate 3 have lost more ground and candidate 3 no longer has enough strength to stay in the race and drops out, transferring the remaining votes from the large voting bloc to candidate 1. Meanwhile, with three seats candidate 3 beats out candidate 1, leaving it as the loser who drops out of the election at that point.

I haven’t really thought through the implications of this. It’s not news to me that STV has some weird edge cases, but this is a particularly annoying one of which I was not previously aware.

Updates: First off, I’ve actually been convinced that this is a feature not a bug. What essentially happens is that compromise candidates between coalitions who don’t have enough support for their favoured candidates come together to support a mutually acceptable less good candidate. When the number of seats available becomes larger they have more space to go their own way.

Secondly, here’s an example of this happening with only three candidates:

• Five votes: 0, 1, 2
• Four votes: 1, 0, 2
• Eight votes: 2, 1, 0

If you run this to elect one candidate (i.e. normal IRV) then you’ll elect candidate 0, because 1 drops out in the first round and transfers the needed number of votes to 0. If you run it to elect two candidates, both 0 and 2 immediately have enough votes to beat the quota and are elected immediately.

Notably, 1 is also the Condorcet winner in this election.

This entry was posted in voting on by .

## 2 thoughts on “A failure mode of single transferable vote”

1. Peter

Is this really a failure though? I’m not sure if your specific election is doing anything funny, but in general adding more seats can and should cause different people to be elected.

As a trivial example imagine a simplified UK where there is a party of the left (Labour), a party of the centre (the LibDems) and a party of the right (the Tories). Imagine a constituency where there were roughly the same number of left-leaning and right-leaning voters. If the constituency has one seat (AV) you wouldn’t be surprised if the LibDem got in. But if there were two seats you wouldn’t be surprised if there was one each of Labour and Tories.

1. david Post author

Yeah, someone else made this point on Twitter too, and I’ve decided I agree: It’s not a failure mode but an actively desirable feature that I just hadn’t thought about enough.