Having three options on the ballot paper is even more complicated than it soundsby Peter Kellner / July 16, 2018 / Leave a comment
Former Education Secretary Justine Greening. Photo: David Mirzoeff/PA Wire/PA Images Justine Greening has come out in favour of a People’s Vote—but with a twist. Instead of a binary choice—say, compromise deal versus no Brexit—she wants voters to judge three options: no Brexit, compromise or hard Brexit. It would certainly be novel. Other countries have occasionally held multi-choice referendums; but these have generally been first-past-the post choices, such as when Australia picked Advance Australia Fair as its national song in 1977. What Greening wants is for voters to give their first and second choices. If none of the options wins 50 per cent support, the least popular option would be eliminated, and the second choices of its supporters counted. There is nothing intrinsically wrong in constitutional innovation. However, Greening’s proposal raises an important question of democratic choice when more than two options are on the table. In short, how do we measure which is most popular when none of the options wins 50 per cent of first preference votes? Here is one scenario. It is emphatically not a prediction but it is, I believe, plausible. Suppose this is how the UK votes in a Greening referendum No Brexit: 38 per cent (Second preferences: Compromise 30 per cent, Hard Brexit 8 per cent) Compromise deal: 30 per cent (Second preferences: Hard Brexit 20 per cent, No Brexit 10 per cent) Hard Brexit: 32 per cent (Second preferences: Compromise 25 per cent, No Brexit 7 per cent) Here are three ways the nation’s choice could be calculated. First-past-the-post: No Brexit has the most votes, so the UK stays in the European Union Alternative vote: The compromise is the least popular option. Its supporters’ second preferences are counted. No Brexit gains an extra 10 per cent; added to its first-choice 38 per cent, it ends up with 48 per cent. Hard Brexit gains 20 per cent; added to its first-choice 32 per cent, it climbs to 52 per cent. The UK leaves the EU without a deal. Condorcet winner. This is named after the Marquis de Condorcet, who proposed that the winner of a multi-choice contest should be one that is the final head-to-head winner of every pair of choices. Let us apply this to our example: No Brexit v hard Brexit. As we have seen, if we eliminate the compromise option, Hard Brexit defeats No Brexit 52-48 per cent. No Brexit v compromise. For this, we count the second choices of Hard Brexit’s supporters. This gives us: compromise 55 per cent (30 per cent + 25 per cent), no Brexit 45 per cent (38 per cent + 7 per cent) Hard Brexit v compromise. Now we reallocate the second choices of no Brexit supporters. The result: compromise 60 per cent (30 per cent + 30 per cent), hard Brexit 40 per cent (32 per cent + 8 per cent) Under Condorcet rules, the UK votes for Theresa May’s compromise deal, because more people prefer that outcome to both of the other outcomes. There are, of course, many other things to be considered. The politics of the Brexit negotiations are plainly complex. There may in the end be no deal, agreed with the European Union, for either parliament or the electorate to vote on. However, if we are to consider seriously the option of a multi-choice referendum, we need to establish the democratic principles of how we conduct it. If three different ways of counting the same votes—all of which can be defended in theory—produce three different winners, then it may be hard to present the outcome as the settled, legitimate will of the people. A People’s Vote that offers a binary choice at least avoids that danger.