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I've been having trouble finding an authoritative definition of the efficiency gap when more than 2 parties are involved. Ballotopedia points to an article by Nicholas Stephanopoulos and Eric McGhee, citing that "any vote for a losing candidate is wasted by definition, but so too is any vote beyond the 50 percent threshold needed (in a two-candidate race) to win a seat." Intuitively, I imagine many more votes are wasted when cast for the winning party if the result is a landslide (so the second-place candidate receives a tiny fraction of the votes), or when there are more than 2 candidates competing.

What happens to the measure in elections with more than 2 candidates? The "50 percent threshold" feels a little silly if the winner receives well below half of the votes. Is it as simple as saying that no votes for the winning party are wasted if they receive less than half of the votes?

  • The Ballotopedia article you link to has a link to the paper that defines the concept, which would seem authoritative. At a quick glance it doesn't seem to address elections with more than 2 parties though. – Denis de Bernardy Oct 16 '17 at 4:11
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    I believe the correct generalisation is "vote which wouldn't change the result if not cast", i.e. the votes for the winner beyond the second candidate's vote plus one are wasted, but like you I can't find a reference. – origimbo Oct 16 '17 at 10:34
  • @origimbo that was my initial take on things, too, but in the example they mention in the article I linked, they use 50%+ 1 vote as the threshold above which votes are wasted. – Charlie Oct 16 '17 at 14:14
  • That's the right thing in a two horse race, since you want to gerrymander to win as many seats efficiently 51-49, and pack the rest of your opposition into seats where you aren't competitive, whereas in a "true" n-horse race, you can split the opposition vote. In principle that could reduce the threshold for an efficient win to (1/n), but I don't know of any author who assumes that amount of control over opposition voters, since Duverger's Law kicks in. – origimbo Oct 16 '17 at 15:50
  • I guess you could start from the bottom and count the smallest candidate's votes wasted and subtract them from the total until you got to the last two. It sounds silly calling the results 50%, so probably they just didn't even give third parties a thought (they both seem to be Americans), but the math would work out. – user9389 Oct 16 '17 at 16:47
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Taking the definition as "(Wasted votes for party A - Wasted votes for B)/total votes" generalises to multi-party elections:

party  A    B    C    D
votes  41  28   18   13    District A
wasted 13  28   18   13   

votes  82   2   10    6    District B
wasted 72   2   10    6

votes  22  27   26   25    District C
wasted 22   1   26   25

total
waste 108  31   54   44

There are 300 total votes, so the gaps are

A,B (108-31)/ 300 = 26%
A,C (108-54)/ 300 = 18%
C,D (54-44)/ 300 = 3%

and so on. This indicates that party B has a districting advantage over party A. If the boundries between districts B and C were drawn differently party A could have won all three districts. It also shows a feature: Party D was more efficient than party A, by getting few votes overall.

This can be further generalised to ranked voting systems, as then some of the first choice votes for parties C and D would be redistributed to winning candidates, and so wouldn't be counted as "wasted".

  • I like what you're getting at, but is there any precedent for defining wasted votes for the winning party as the difference between the winning party and the second-place party? Is there an advantage to measuring it this way vs a 50% threshold? – Charlie Oct 25 '17 at 15:28
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    It follows from the Stephanopolis McGhee paper, which defines a wasted vote as any that doesn't contribute to the winning candidate: All votes for losing candidates are therefore wasted, as are surplus votes for winning candidates who would have won anyway. It is just an application of their definitions. – James K Oct 25 '17 at 16:04
  • That's compelling. Accepted! The main bother to me, though, is that this interpretation is not consistent with the example on Wikipedia and Ballotopedia. Not sure if I should go change things there, too? – Charlie Oct 25 '17 at 21:01

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