From "The MMI's Curse" #22, in Apa:2001 176, March 1998. --George Flynn

I thought I might note (just for the heck of it) that there is more than one way to count preferential ballots, and that they don't necessarily yield the same result. I once did a lengthy writeup of this subject in Apa:NESFA (demonstrating with various counts of that year's Skylark results); this is the quick-and-dirty version.

I will illustrate with a completely artificial example, designed to magnify the differences between counting methods. But while this particular result is very improbable, be assured that similar differences can be obtained in real elections, if they're close enough. OK: Suppose you have five candidates, A through E, and 40 voters. Assume that E gets no first-place votes, but is everybody's second choice, and that the other four candidates divide the votes evenly, thus:

A | B | C | D | E | ||

First-place votes | 10 | 10 | 10 | 10 | 0 | |

Second-place votes | 0 | 0 | 0 | 0 | 40 | |

Third-place votes | 10 | 10 | 10 | 10 | 0 | (etc.) |

Now we'll see how the different counting methods treat these numbers.

I don't need to explain this, right? Here E has no first-place votes, and is thus immediately eliminated. (Then we have a four-way tie, but that's *really* unlikely in an actual case.) This method places a high premium on getting a significant number of first-place votes, and thus is not good for finding compromise choices. But it's in *Robert's Rules, *so it's the default.

Often used in sports polls, etc.: You count something like 3 points for first, 2 for second, 1 for third (or 5,3,1, or 5,4,3,2,1, or 8,7,6,5,4 as in the *Locus* Poll, etc.). Using any of these valuations, E wins here, since it has the highest *average* point count. For example, with the 3-2-1 method, we have (3X10) + (1X10) = 40 for A, B, C, D, but (2X40) = 80 for E. The point method is relatively easy to count, and it reduces the overweighting of first-place votes; however, the various weights it does use are arbitrary, and different choices can give different results. Worse yet, a voter can exert a disproportionate influence by bullet-voting (unlike the Hugo method, where bullet-voting throws away votes and *weakens* your influence).

(By the way, a standard *non-*preferential election is equivalent to the point method with zero weight for everything but first place.)

(This name may have been coined by Keith Fenske, a Canadian fan who once sent me a treatise on all this stuff.) This type of counting method uses *all* the information on the ballot, by considering the preferences in all the possible one-on-one pairings. There may be more than one way to do this, but here's the one I'm familiar with. (Excuse the algebra.) Let N(i,j) be the number of ballots on which candidate i does better than candidate j. Then for each candidate we calculate a score

S(i) = ∑j N(i,j) / ∑j [N(i,j) + N(j,i)] ,

which is simply the fraction of all one-on-one contests won by candidate i. So for the results above, E does better than each of the other candidates on 30 out of 40 ballots, i.e., N(E,j) = 30 and N(j,E) = 10 for all j; the other candidates have symmetrical votes, so all the other N(i,j) are presumably 20. Thus we have

S(E) = (4×30) / [(4×30) + (4×10)] = 120/160 = 0.75

S(i) = [(3×20) + 10] / [(3×20) + 10 + (3×20) + 30] = 70/160 = 0.4375

for i = A,B,C,D

and E wins again. I like this method because it gives the greatest likelihood of a victory by a compromise candidate like E, i.e., one relatively acceptable to the greatest number of voters. Unfortunately, it's also the most complicated to count. (I wouldn't care to try doing it during a meeting.) And as in the Hugo method, the runner-up for first place isn't necessarily the winner for second place: you have to do it all over again if you want a full ranking.