Post #508: Revisiting #3: Don’t split the vote

Posted on January 9, 2020

In the 2019 election I made a plea not to split the anti-MAC vote (Post #259).  That is, don’t field more candidates than there are open seats.  This is the 2020 version of that post.

Background and main conclusion

At the last Town Council meeting in 2019, Mayor DiRocco announced she would not run for re-election.  At the first meeting in 2020, Councilmembers Colbert and Springsteen announced that they will run for Mayor.  So this post looks back at the last election, and forward to the May 5 2020 election.

To cut to the chase:  Based on the arithmetic as I see it, if either side of the key development issue fields two equally-strong candidates for Mayor, that virtually guarantees a loss for that side in the Mayoral election.  If the “pro-MAC” forces do that, they lose 60%-20%-20%  If the “anti-MAC” forces do that, they lose 40%-30%-30%.   Fielding the equally-strong second  candidate for Mayor is very close to political suicide.

2019 election:  Best guess, 60%/40% split of voters

I summarized the 2019 election results in Post #266.

Briefly, in two graphics:

It was more-or-less tied with 2006 for the largest voter turnout this century.  If you account for the growth in registered voters, turnout was about 100 voters higher in this election that in the next-highest election (2006).

(Note:   The graph above was based on the immediately-published returns at the time.  Now that I have the final returns, I see that there were about 100 fewer voters than I had estimated.  That puts the 2018 election almost exactly tied with the 2006 election, after accounting for the growth in registered voters over the period.)

So the 2019 voter turnout was high, but no higher than in the 2006 election.  N.B., and relevant here, 2006 was the last time that the Mayoral race was contested.

Second, of total votes cast, 73% were for “anti-MAC” candidates.  But that should not be interpreted as a count of voters, because there were only two anti-MAC candidates. Likely some, but not all, “pro-MAC” voters cast their third vote to their least-objectionable anti-MAC candidate.

Any attempt to get from “votes cast” to “voters” is going to be crude and ad-hoc.  That’s never stopped me before, so let me take a few shots at it.

Three different SWAGs* suggest that actual voter sentiment in town runs roughly 60%/40% anti-MAC/pro-MAC.  I can get that number by taking the votes for just the two leading candidates.  Or, I can get that number by taking total votes, then netting out (my best guess**) for stray third-slot votes for anti-MAC candidates by pro-MAC voters.  Or I can appeal to my random-sample survey of Vienna residents, which showed roughly the same split of those actively disliking MAC  as a fraction of those who either liked or disliked MAC.

* Scientific wild-ass guesses. ** Pro-MAC forces in the last election urged their voters to vote for just two candidates.  I find that there were about 100 fewer votes cast, in total, than I would have expected, based on total count of voters.  If I take total candidate votes, and net out the votes for the highest pro-MAC candidate less 100, and take that as a fraction of total, I again come up with a roughly 60/40 split.  Did I not already say crude and ad-hoc?

The upshot of all that is that if I had to guess about the 2020 election, I’d guess that a) we’d see roughly the same turnout as 2019, and b) there’s maybe a 60%/40% split among voters, anti-MAC/pro-MAC.

An obvious conclusion:  You can’t split the vote and win.

I guess this needs to be said, but if either side of this issue fields two equally-strong candidates for Mayor, that virtually guarantees a loss for that side in the Mayoral election.  If the pro-MAC forces do that, they lose 60-20-20.  If the anti-MAC forces do that, they lose 40-30-30.   Fielding the equally-strong second  candidate is more-or-less political suicide.

What about the Town Council seats?  There, the arithmetic is harder — and the number of non-incumbent slots is less certain — but if I had to guess, I’d guess that more-or-less the same logic applies.  Arguably, both sides want as many strong candidates as there are seats, full stop.  No more, no fewer.

There may be more to the arithmetic there, in some special circumstances, but deking that out is beyond my capabilities.  In particular, that simple statement would have to be modified to the extent that there are a lot of “split-ticket” voters.  But in that case, I don’t think any simple arithmetic applies.  So, to the extent that I can guess at a rule, without going into it a lot more deeply than I am probably capable of, one-seat-one-candidate looks like it maximizes the chance of success.