Mathematics of Voting: 3. Electing a candidate, Instant runoff voting

The method used in Australia to determine election to the House of Representatives is known to Australians as “preferential voting”; more formally it is called “instant runoff voting“, or IRV. Here the voter is required to indicate references for all the candidates: 1 for the most preferred candidate, 2 for the next, and so on.  IRV is designed to simulate the effects of a runoff (as described in my previous post) but with a single ballot.

To start off, the ballot papers are distributed to each of the candidates according to the first preferences. If no candidate obtains an absolute majority of ballot papers, then the candidate with the lowest number is excluded, and that candidate’s ballot papers are distributed to the other candidates according to the second preferences on those ballot papers. If still no candidate has obtained an absolute majority (of ballot papers), then again the candidate with the lowest number is excluded, and the ballot papers are distributed to the remaining candidates according to the highest preferences. This continues until one candidate has obtained an absolute majority.

To see how this works, we shall consider some examples.

Suppose that in a 60,000 voter electorate there are 3 candidates, and votes are counted as follows:

    \[ \begin{array}{ccc|c} A&B&C&\mbox{Votes}\\  \hline  1&2&3&10,000\\  1&3&2&10,000\\  2&1&3&11,000\\  3&1&2&11,000\\  2&3&1&11,000\\  3&2&1&7,000 \end{array} \]

10,000+10,000=20,000 first preference votes for A,

11,000+11,000=22,000 first preference votes for B,

11,000+7,000=18,000 first preference votes for C.

At this point C (having the lowest number of first preference votes) is excluded, and C‘s second preferences are considered. Of the 18,000 who voted 1 for C, 11,000 voted 2 for A and 7,000 voted 2 for B. The scores are now

20,000+11,000=31,000 for A

22,000+7,000=29,000 for B

at which point A has the absolute majority and so wins. (Note that on a simple plurality of first preferences, B would have won.)

Now consider a situation where ballot papers must be distributed more than once; a 100,000 voter electorate with four candidates standing. The distribution of ballot papers is as follows (the letters to the left of each different preferential orders are there for reference):

    \[ \begin{array}{ccccc|c} &A&B&C&D&\mbox{Votes}\\  \hline  \mbox{(a)}&1&2&3&4&28,000\\  \mbox{(b)}&2&1&3&4&32,000\\  \mbox{(c)}&2&3&1&4&15,000\\  \mbox{(d)}&4&2&1&3&10,000\\  \mbox{(e)}&3&4&2&1&5,000\\ \mbox{(f)}&4&2&3&1&5,000\\  \mbox{(g)}&2&3&4&1&5,000 \end{array} \]

These preferences can be listed according to which candidate has been given as first preference:

A:\,\mbox\{(a)\}\quad 28,000

B:\,\mbox\{(b)\}\quad 32,000

C:\,\mbox\{(c), (d)\}\quad 15,000+10,000=25,000

D:\,\mbox\{(e), (f), (g)\}\quad 5,000+5,000+5,000=15,000

At this point D is excluded. The ballot papers with preferences (e), (f) and (g) are distributed to the other candidates: papers (e) go to C, papers (f) go to B and papers (g) go to A. We now have:

A:\,\mbox\{(a), (g)\}\quad 28,000+5,000=33,000

B: \,\mbox\{(b), (f)\}\quad 32,000+5,000=37,000

C:\,\mbox\{(c), (d), (e)\}\quad 15,000+10,000+5,000=30,000

We can now eliminate C, and distribute papers (c), (d) and (e) to A and B. On papers (c), the highest remaining preference is 2 for A, so these papers go to A. On papers (d), the highest remaining preference is 2 for B, so these papers go to B. On papers (e), the highest remaining preference is 3 for A, so these papers go to A. We now have:

A:\,\mbox{(a), (f), (c), (e)}\quad 28,000+5,000+15,000+5,000=53,000

B:\,\mbox{(b), (g), (d)}\quad  32,000+5,000+10,000=47,000

and so A is the final winner.

The optional-preferential method

In some Australian state elections, voters are not required to give preferences to every candidate. Instead, a voter gives preferences to as many or as few candidates as he or she likes. Counting is done as above, except for one significant difference:  if a candidate is eliminated, and the ballot papers are to be distributed, there may be some papers that can’t be distributed, as the voters may not have indicated further preferences. In such a case those votes are called “wasted” votes and play no further part in the counting.

Problems with IRV

At first glance, IRV seems to ameliorate much of the problems with simple plurality. And in general it makes the voters happy, as they feel that they have fully expressed their concerns through the ballot (I can attest to this). Unfortunately, as a system, IRV has some major flaws.

We would expect any decent voting system to be monotonic. That is, if a change of ballots seems to favour a winning candidate, then that candidate should still win. But consider this IRV example (this example, and the following, are taken from here):

    \[ \begin{array}{ccc|c} A&B&C&\mbox{Votes}\\  \hline  1&2&3&27\\  2&3&1&42\\  3&1&2&24 \end{array} \]

First B is eliminated (fewest first place votes), and those ballots are all distributed to C, who wins with 66 ballots to A‘s 27.

Suppose now that some of the first 27 voters change their preferences to the second row. The ballots are now

    \[ \begin{array}{ccc|c} A&B&C&\mbox{Votes}\\  \hline  1&2&3&23\\  2&3&1&46\\  3&1&2&24 \end{array} \]

Now C, the previous winner, has more first place votes than before, so should win again, yes? Well, first A is eliminated, and those ballots go to B – which means B ends up winning with 47 votes to C‘s 46! We have here the extraordinary situation where votes which favour a candidate actually cause that candidate to lose! This is an example of non-monotonicity and is seem by some to be a major problem with IRV.

Another problem with IRV is that it is vulnerable to the no-show paradox. Suppose, as above, the ballots are

    \[ \begin{array}{ccc|c} A&B&C&\mbox{Votes}\\  \hline  1&2&3&23\\  2&3&1&46\\  3&1&2&24 \end{array} \]

Suppose that Mr and Mrs AlwaysLate missed the voting, and couldn’t cast their votes. As they would be equally happy with A or B as winners, they are happy with the outcome anyway. If they had voted, they would have cast their votes with the first row of preferences, so that the ballots would have been:

    \[ \begin{array}{ccc|c} A&B&C&\mbox{Votes}\\ \hline  1&2&3&25\\  2&3&1&46\\  3&1&2&24 \end{array} \]

Here B is eliminated, those ballots go to C – who wins! Here including votes which favour a candidate has the result of causing a different candidate to win.

Further comments

Voting systems have the curious propensity of arousing very strong passions amongst their adherents and detractors. You can read the “pro” side of IRV at FairVote.org, and the “con” side at RangeVoting.org, where IRV is denounced as being “idiotic”. You can also read one person’s negative take on IRV, and a response to it. In fact, it is very difficult to find a balanced view of IRV and its strengths and weaknesses; the wikipedia article makes a good attempt.

One thought on “Mathematics of Voting: 3. Electing a candidate, Instant runoff voting

  1. Oof, don’t get me started on the wikipedia stuff. FairVote supporters have had the page for “favorite betrayal criterion” (which IRV fails) deleted…twice; the page on the montonicity criterion is deliberately, misleading wishy-washy (good luck not getting reverted though).

    I guess I’ve outed myself as being on the “con” side, eh?

    Looking forward to the next part (don’t forget Bayesian regret!)

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