Burlington Progressives Keep Mayoralty in Exciting 5-Candidate Race

On March 3, Burlington, Vermont held a mayoral election. The candidate were Progressive Bob Kiss (the incumbent), Republican Kurt Wright, Democratic Andy Montroll, independent Dan Smith, and Green Party nominee James Simpson. Kiss was re-elected. The election had been considered impossible to predict. The independent candidate, Dan Smith, had raised more money than any other candidate. The Democratic candidate, Andy Montroll, was an incumbent member of the City Council and had been endorsed by Howard Dean and the city’s daily newspaper. The Republican nominee, Kurt Wright, was the incumbent president of the city council.

Burlington uses Instant Runoff Voting. Wright led in the first count but no one had a majority of first place votes. In the second count the Republican also kept the lead. But after the votes cast for the Democratic nominee had been reassigned in a further count, Kiss won by 4,313 to 4,061 to the Republican nominee. Observers felt the election had been an excellent example to show that IRV does what it is supposed to do. In particular, the candidates were very courteous and respectful to each other during the campaign, since each candidate hoped to win second-choice or third-choice votes from various rivals. Also the city was spared the expense of a run-off election several weeks after the original March 3 event. For the election returns, see www.burlingtonvotes.org/20090303.


Burlington Progressives Keep Mayoralty in Exciting 5-Candidate Race — No Comments

  1. I find it odd that the Greens would run a candidate against a Progressive Party incumbent. Not that they don’t have the right to do so, of course. It just seems that there could hardly be someone more in agreement with their principles.

  2. But with instant runoff voting, they have the ability to stake out their own ground — and build their “brand” — without being criticized as spoilers.

  3. I see that there was only one candidate from each party. Were these party candidates nominated prior to the March 3 election? If so, how?

  4. One more extremist elected — claiming a mighty majority *mandate*.

    IRV is super-evil dangerous for single offices since it ignores most of the data in a place votes table — regardless of the armies of New Age math M-O-R-O-N-S hyping IRV.

    Approval Voting NOW for NONPARTISAN executive / judicial offices — vote for 1 or more, the highest win (more likely to be *moderates*) — pending major public education about head to head math.

    NO party hack primaries, caucuses and conventions are needed.

  5. Wow, Demo Rep may have some hostility, but Bob Kiss was a strong candidate. He’s third party, but not all third party candidates are “extremists.”

    As to party nominations, the parties did that privately through their own internal processes. You see that in various US elections — like the election for the US House vacancy created by Gillibrand going to the U.S. Senate in New York.

  6. I like the presentation — but I wish it had explained about “defeat was mathematically inevitable” elimination. I suppose it refers to something like this:

    A 1000
    B 900
    C 800
    D 400
    E 200
    F 100

    You could take three rounds and eliminate F, E, and D in that order. But you could also note that — even if D was the second choice of all those who voted for E or F first, D still couldn’t catch up to the next highest vote total (C’s) and would have to be eliminated anyway.

    Can someone please confirm or correct this?

  7. #6 Yes, that is correct.

    If you were doing a hand count, you would have sorted the ballots by their 1st preference, and counted the number of ballots in each bundle.

    If you then excluded the candidates one by one, you would take the 100 ballots in the F bundle, and distribute them based on their 2nd preference, some of which might be for D or E. You would then distribute the ballots in the E bundle (or possibly two sub-bundles of E and FE) based on their next preference.

    But if you exclude D,E, and F together, you can simply redistribute the bundles based on their next preference among A,B, or C. Likely as not, you would handle the F, E, and D bundles serially, but you still would skip the intermediate distributions.

    In the case of Burlington, they used optically scanned ballots, and the rankings of the individual ballots is online. It would probably be simpler to program by going through the intermediate distributions, but there may be legal requirements that the computer count be a replication of the hand count.

    It is actually feasible to use a programmable text editor to count an IRV election. Let’s say that you take each ballot and produce their rankings like this:

    CDABE_ (C 1st, D 2nd, A 3rd, B 4th, E 5th)
    C_ (C 1st, no others)
    DAC_ (D 1st, A 2nd, C 3rd, no others).
    _ (no preferences, undervote)

    You simply alphabetize all the “words” and count the number of words beginning with each letter. To redistribute ballots, you simply omit the first letter of the “word”, so DAC_ would become AC_ if D were being excluded.

  8. An oddity of Burlington is that they elect city councilors from single-member wards, but do not use IRV, which forces parties, candidates, and voters to game the elections.

    Under a conventional majority election, with a runoff if needed, there is generally little or no dropoff in voting for a top of the ballot race (see Louisiana for example). It is when there is no top of the ballot race in the runoff when everyone stays home. So actually, IRV is more useful for down ballot races such as for city councilor.

  9. Re: #8 — Thank you.

    The ability to program to eliminate multiple losing candidates in one step would seem to me to make it that much easier to *elect* multiple candidates with one preferential vote, too.

    Say you had four seats and more than four candidates. You could do normal IRV counting to find the “first winner” — and then count again, but eliminate that winner first and shift all her/his first-choice ballots to their second choices. Doing that would give you a “second winner” in the same general way you got the first one . . . and then you could count the ballots a third time, starting by eliminating both winners so far. Rinse and repeat until you have all the winners you’re looking for.

    Obviously, computers could still do it faster — but it should also almost as obviously still be reproducible by hand.

    (This might even offer one advantage as a citizens’ introduction to IRV/preferential voting. People would already be motivated to vote for more than one person in a multi-seat race. And it could well be a more manageable step from that to expressing a full slate of preferences than making the voters make the full cognitive leap all over one seat. But I’d welcome hearing what folks think of that possibility. . . .)

  10. An extreme IRV example —

    Times are tough – think 2009.
    The Middle is severely divided.

    49 HWS
    48 SWH
    1 WHS
    1 WSH

    1 2 3

    H 49 1 49
    S 48 1 50
    W 2 97 0

    IRV —
    W loses, H beats S 50-49 and claims a mighty IRV majority mandate — with a resulting Civil WAR.

    Head to Head —
    W beats H 50-49
    W beats S 51-48

    Approval Voting (assuming first 2 choices are YES votes) —
    H 50
    S 49
    W 99

    H Hitler, S Stalin, W Washington, George – American Hero 1775-1783 U.S.A. General in American Revolution, 1789-1797 First U.S.A. President.

    Regardless of ALL MORONS — Nonpartisan Approval Voting NOW for all elected executive offices and all judges.

    P.R. for legislative body elections.

    Way too difficult for the New Age Math MORONS who LOVE IRV type timebomb stuff to get leftwing / rightwing extremists into POWER — to enslave and kill to enforce their extremist agendas.

  11. #10 If one candidate receives a majority of first preferences in an IRV election, he is elected. This is actually just a case of it being “mathematically inevitable” that the majority-candidate would win. There is no way that the transferred votes of the other candidates could defeat him. Or alternatively, the total number of votes for the 2nd through nth candidates is less than that of the 1st place candidate.

    You could count ballots in a multi-seat election with ranked ballots doing it exactly like IRV, but stopping when N+1 candidates remain, with the Top N candidates being elected. In an election like the Burlington mayoral election, this would have been fairly reasonable.

    If the election had been for Tri-Mayor, you would have stopped after counting 1st preferences with the top 3 having 2951, 2585, and 2063 votes. There would have been no mathematical reason to redistribute the 4th and 5th place votes, nor the write-ins. But had you done so, the result would have been 3294, 2981, and 2554 which is an even more balanced result.

    In an election for co-mayors you would have stopped after the 2nd count, with candidates with winners with 3294 and 2981 votes, and last-standing loser with 2554 votes. Again, if you had actually redistributed those 2554 votes, the race between the top two would have been 4313 to 4061, which is slightly closer.

    But you could make an argument that some of the votes for the 1st place candidate were “wasted”, and could have helped the 3rd place candidate defeat the 2nd place candidate.

    The conventional way to do preferential voting for multi-seat races is Single Transferable Vote (STV), where each voter has a single vote, that can be transferred to different candidates based on their ranked preferences.

    In an STV election, there is a “quota” that a candidate must reach to be elected. Votes that the candidate receives in excess of the quota, the “surplus”, can be transferred to the other remaining candidates.

    If there are N candidates to be elected, the quota is ceil((V+1)/(N+1)). So if 3 candidates are to be elected, then the quota is 1/4 of the total vote, and each of the 3 winning candidates will end up with 1/4 of the vote, and the votes for the remaining candidates will be less than 1/4.

    There were 8976 valid votes in Burlington, so the quota in an STV election for tri-mayor would have been ceil(8977/4) or 2245. Both Wright, 2951; and Kiss, 2585 exceeded the quota and would have been elected. But their respective surpluses of 707 and 340, added to 4th place Smith’s 1st preferences and the 71 scattered other votes for a total of 2495, surpassing the 3rd place Montroll with 2063. So it would not be mathematically inevitable that Smith would be defeated.

  12. #12 — I do know *something* about how IRV works, and even STV. It’s just always seemed to me that it would be a lot easier to *explain* “my” method of multiple-seat election with one vote than go through more complicated math . . . and use words like “quota” that set some people off.

    (And you’d avoid the “wasted vote” argument, too — because you’d basically be saying, “Okay, Candidate A won as the first choice . . . so if she/he hadn’t been in this race, who would everybody have preferred as the first choice instead?” and so on.)

  13. #13 I think my way of counting a multi-seat IRV-like election was easier to understand.

    Let people imagine that they were going to stand next to their most favored candidate. Count noses.

    Eliminate the last placed candidate, and let his supporters go stand next to their next preference among the remaining candidates. If they don’t care for any of the remaining candidates, they go sit on the sidelines or go home.

    Repeat until only the number of desired candidates remain. They’re all elected. All the voters who have continued to participate are standing next to someone they support (perhaps not their first choice, but someone they favor more than the other winners).

    The only thing that would be different in a real election is that secret ballots would be used, and voters would express their preference ranking in advance. Explain that as matter of expediency so they don’t have to actually attend the vote counting.

    Now your way:

    It starts out the same way. Let people imagine that they were going to stand next to their most favored candidate. Count noses.

    But instead of identifying the least supported candidate, you identify the most supported candidate. He is elected. They all cheer.

    They are told to now go stand next to the candidate that they prefer from among the remaining candidates. Again count noses. Identify the winner.

    Repeat until enough are elected.

    But those who stood next to their candidate, only to see a mob of voters go from candidate to candidate electing each in turn, would figure out what was wrong. Some happy voters elected their 1st, 2nd, 3rd, and 4th preference. Certainly not a very pluralistic result.

  14. #13 Now STV, using rules based on Cambridge, MA counting rules.

    Candidates are placed in sedan chairs. If 4 candidates are to elected, it will take slightly over 20% of the voters to lift a sedan chair and carry the candidate through the victory arch. There will be one last faction with slightly under 20%, who won’t be able to lift their candidate.

    1) Everyone goes and stands next to their most favored candidate. Any candidates with enough supporters to lift their chair is elected. But, if there are more than are needed to lift the chair on their shoulders, the surplus may go help out someone else.

    We could let the candidate direct the excess to help out a friend; or we could ask for volunteers; or let those who are quickest, to rush off. But we choose a fairer way. We randomly select some supporters who then proceed to their next preference. For example, if we a candidate has 1100 supporters and only needs 1000 to lift him, then we choose 100 of the 1100 supporters to go to their next preference among the remaining unelected candidates. I selecting these 100 surplus supporters, we make sure that they have a 2nd preference. If not, we choose someone else who has expressed a preference for another candidate.

    We do the transfer in an orderly fashion. They go one by one. If a particular candidate accumulates enough supporters for victory, then he is elected. Any voters who would have switched to him are directed to their next preference (if they have no other preferences, we direct one of those already standing about – who does have another preference to transfer).

    Once all the surplus supporters from those elected on their first count have transferred, we check to see whether enough candidates have been elected.

    If not, we eliminate the least-supported candidate. His supporters go one-by-one to their next preference. As soon as a candidate has enough supporters he is elected, and any other transferees are directed to their following preferences.

    We repeat this step of eliminating the last candidate, just as we do under single-member IRV, or my quota-less multi-member IRV, until enough canidates have been elected.

Leave a Reply

Your email address will not be published. Required fields are marked *