On May 19, Vermont HB 744 took effect. Governor Phil Scott declined to sign it, but he didn’t veto it either, so it has become law. The bill lets Burlington use ranked choice voting for its own elections. Thanks to Fairvote for this news.
Ballot Access News
Edited By Richard Winger
RCV Condorcet fail.
THIS IS A PURGE LIST!
Vermont does not have municipal home rule. Burlington’s city charter is in state statute. To get a charter change the legislature must pass a law. Typically, Burlington will hold a referendum to demonstrate popular support for a change.
In this case the city council proposed a referendum that would have imposed RCV for mayoral, city council, and school board elections. The referendum would have been held in conjunction with the 2020 general election. The mayor vetoed the ordinance.
The city council then proposed using RCV for city council only. This was approved by city voters in the spring of 2021. The legislature is just now enacting a law conforming to the voters whims.
The charter does not “let” Burlington to use RCV for city elections. It requires that Burlington use RCV, but only for city council elections. Mayoral and school board will continue to be contested under the 40% rule, which requires a runoff only if the top vote-getter falls below 40% of the vote.
Meanwhile the fact that some Burlington wards have more than 3 times as many voters as others is ignored. Burlington is a small city. It would be better off using PAV.
Condorcet = RCV done right.
@az,
How do you do Condorcet for city council elections?
JR—
CONDORCET – ALL ELECTIONS
1. VOTERS HAVE
A. NUMBER VOTES 1, 2. 3, ETC. [RELATIVE VOTES] AND
B. YES/NO VOTES [ABSOLUTE VOTES – NO IS DEFAULT VOTE] – TO BREAK TIES BELOW] FOR EACH CHOICE.
2. ALL COMBINATIONS OF N TEST WINNERS VS ONE TEST LOSER VS OTHER TEST LOSERS.
N = NUMBER TO BE ELECTED
IF A TEST WINNER WINS/LOSES IN ALL COMBINATIONS, THEN IT IS A CONDORCET WINNER/LOSER.
IF SOME WINS/LOSES/TIES, THEN MOST YES VOTES USED TO FIND WINNERS.
3. LEGIS WINNERS WOULD HAVE A VOTING POWER EQUAL TO FINAL VOTES EACH RECEIVED.
[SUGGEST 5 PER DISTRICT – 5 LARGEST FACTIONS]
4. IF MULTIPLE EXECS/JUDICS, THEN M TOP VOTES COUNT IN COMBINATIONS MATH.
5. NEED COMPUTER IN ANY LARGE ELECTION.
SYSTEM AKIN TO MATH CALCULUS VS ESTIMATED AREAS UNDER MATH CURVES.
—–
BURLINGTON HAD RCV EXTREMIST MAYOR WINNER IN 2009 — CONDORCET LOSER.
https://en.wikipedia.org/wiki/2009_Burlington_mayoral_election
THE RCV CORPSE RISES FROM THE DEAD IN VT.
—–
PENDING CONDORCET –
LEGIS — SIMPLE PR —
TOTAL VOTES / TOTAL MEMBERS
= RATIO TO ELECT EACH MEMBER
SURPLUS VOTES DOWN.
LOWEST LOSER VOTES UP.
EXECS / JUDICS — NONPARTISAN APPV
— VOTE FOR 1 OR MORE — MAX YES VOTES WIN.
Uniform state law for local regimes >>> NOOOO *charters*
– extra layer for legal chaos
Local ordinances — esp burg names and number of city council folks.
@az,
So the ballot would have:
Allen, Ethan (Green Mountain) Rank ( ) Yes ( ) No ( )
Dean, Howard (Primal Scream) Rank ( ) Yes ( ) No ( )
Sanders, Bernard “Bernie” (Socialist) Rank ( ) Yes ( ) No ( )
???
Do I have to make my Yes/No consistent with rankings? Could I vote No on my #1 ranking? Yes on my #23 ranking? Do I have to rank all candidates? Use consecutive numbers? May I rank candidates equal?
I don’t understand item 2. Give me an example in a 15-candidate election between A, B, C, …, K, L, M, N, O, to elect 15.
How many members should Burlington have?
JR —-
Ballot form correct — with Rank ( ) Yes ( ) No ( ) LEFT of the names with a semi-wide vertical separation line between Rank ( ) and Yes ( ) No ( ) — with voter instructions on ballots.
NO consistency required in number ranks vs Yes/No.
NO equal number ranks — as in RCV [too many ties]
NO required number ranks for all candidates — unranked = default tied for last in number ranks.
NO required YES or NO — NO = default
Example — 15 candidates — elect 5 — all combinations of 5 Test Winners vs 1 Test Loser vs [ 15-5-1 = 9 ] Other Test Losers.
LOTS of Combinations !!!!!! – see stats combo factorial math.
Ranked votes from 9 Other Test Losers go to highest of the Test Winners or the Test Loser —
added to direct votes from voters for TW and TL.
YES votes to break ties.
But again — Legis — SUGGEST 5 winners PER DISTRICT – possible 5 LARGEST FACTIONS elect a person.
Legis Members Per Local Area – Circa Total Voters / 10,000 — rounded up — some max — 50 ??? — even in very large burgs — NY City, Chicago, etc. ???
A state Legislature — 20 Districts x 5 per district = 100 — possible 100 factions.
Short example —-
Final — Voting Powers — elect 5 per district
Dist 1 — J 25, R 20, D 15, M 26, Q 12 — Total 98
Dist 2 — A 19, Z 29, T 14, X 24, K 17 — Total 103
201 >>> 102 needed for YES majority
Simple spread sheet IF function for voting — If YES, then add Member’s VP to YES total.
—-
Factions will put out their lists — esp for marginal last joint seat(s).
OUR Super Party Faction LIst
1 YES J
2 YES R
etc.
Other folks
7 YES A
8 YES Z
Rest unranked and NO.
—
Possible Condorcet use for multi-offices only after Condorcet in one person exec or judic office.
Meantime — Simple PR and AppV
@az,
Is multi-candidate Condorcet described elsewhere?
Combinations of 5 from 15 = 15! / ( 10! × 5! ) = 3003. But you apparently divide the comination of 10 into 9 and one, so 30030 combinations. I do this brute force for all 30030 combinations.
If I have a combination (ABCDE)(FGHIJKLMN)(O) how do I determine if ABCDE are elected?
@az,
In 2021, Burlington had 34,338 registered voters. 9994 ballots were cast. How big should the city council be?
Ballot access must be a bit limited IF voters are doing the number votes.
NEXT TO EACH NAME — Perhaps numbered circles/ovals with numbers repeated under —
1 to 10
11 to 20
—–
JR — Is multi-candidate Condorcet described elsewhere?
Unknown — but some specialized Condorcet websites exist — Google “election methods”.
—–
JR — If I have a combination (ABCDE)(FGHIJKLMN)(O) how do I determine if ABCDE are elected?
If no ties then
ABCDE would defeat each F to O heads to head.
IE – more direct plus moved VOTES FOR ABCDE FROM EACH SET OF OTHER TEST LOSERS THAN FOR EACH TEST LOSER.
—-
Lower math examples
elect 5, 6 candidates
ABCDEZ
5 Test Winners – 1 Test Loser – NO Other Test Loser
Obviously the top 5 win – 2nd choice votes for losing candidate go to the winners
—–
elect 5, 7 candidates
ABCDEZX
5 Test Winners – 1 Test Loser – 1 Other Test Loser
A 24 – B 23 – C 22 – D 20 – E 12 — X 11 — Z 7 ALL 7 RANK X AS 2ND CHOICE
E 12 LOSES — X 11+ 7 = 18 TENTATIVE X WIN
DROOP QUOTA MATH INVOLVED.
IF CHOICE HAS FIRST PLACE DROOP QUOTA VOTES, THEN CHOICE WINS
24 + 23 + 22 + 20 + 12 + 11 + 7 = 109
DROOP QUOTA –
[109 / (5+1)] + 1 = 18.16 + 1 = 19.16
A,B,C,D ARE DROOP / CONDORCET WINNERS
ONE WINNER AMONG E, X, Z
Each loser vote goes to the highest ranked winner.
——————–
STANDARD ANTI-MATH / ANTI-SCIENCE MATH MORONS WHO BARELY UNDERSTAND PLURALITY MATH WOULD RAVE ABOUT CONDORCET HIGH TECH MATH AND USE OF COMPUTERS.
Correction part —
IF CHOICE HAS FIRST PLACE DROOP QUOTA VOTES, THEN CHOICE WINS
24 + 23 + 22 + 20 + 12 + 11 + 7 = 119
DROOP QUOTA –
[119 / (5+1)] + 1 = 19.83 + 1 = 20.83 >> 20 — only 5 choices can have initial 20 votes — 5 x 20 = 100
@az,
In the seven candidate example, ABCDE each defeated X and Z. You provided no information about relative ranking of X and Z.
But it then appears that you eliminated Z just as you would in conventional STV. But if E defeated X, he would still do so based on the transferred ballots.
E did not defeat X or Z on the ballots marked 1.Z 2.X, so there must be other ballots where E defeated both X and Z.
Ranking is hard.
Burlington elects 4 councillors and 8 councillors in alternating elections. In both cases they are elected from single member districts, either 4 “areas” or 8 “wards”. Each area consists of two wards, but that nesting has not always been used, it usec to be 7 wards, with an 11-member council.
So let’s assume that Burlington kept the 12 member-council electing 6 at a time. That might generate 15 candidates, with the D’s and P’s running 5 or 6, the R’s 3, and an independent or two. 15 is too many to rank.
So why not just use PAV?
A 24 – B 23 – C 22 – D 20 – each is Droop/Condorcet Winner
E 12 — X 11 — Z 7 ALL 7 RANK X AS 2ND CHOICE
E 12 LOSES — X 11+ 7 = 18 TENTATIVE X WIN
BUT could be all E 2nd choices vote for Z
Z 7 + 12 from E = 19 beats X
——
B council — should be odd number – all elected at same time.
Simple PR –
Total Votes / Total Members = EQUAL votes to elect
Pre-election CANDIDATE rank order lists of other candidates.
Surplus votes down
Lowest loser votes up
Example 100 Votes, Elect 5
Ratio = 100/5 = 20
—
Surplus Moved
C1 25-20 = 5 Surplus
C2 19+1 = 20
C3 14+4 = 18
—
Final
C1 20 = 20 Elected
C2 20 = 20 Elected
C3 18+2 = 20 Elected
C4 17+3 = 20 Elected
C5 15+5 = 20 Elected
Sum 90+10 = 100
Losers 10 are moved to elected persons.
——
another simple variant — multi-plurality – odd or even members
Each voter votes for 1
Top N are elected.
All loser votes go to 1 of the N winners – via candidate rank order lists.
Each winner has a Voting Power equal to votes received — direct from voters and via losers.
Defect — minority may be split – elects nobody.
BUT should be 2 or more districts – to NOT have ONE choice get a Voting Power majority.
Advanced math is due to obvious defects in plurality math.
A 34
B 33
C 32
99
34 A vs possible 65 B+C
—–
RCV has problems with divided middles.
34 AMZ
33 ZMA
16 MAZ
16 MZA
99
—-
Thus Condorcet noted that a 3rd choice may beat 2 existing choices head to head.
D>E
possible that F>D and F>E.
—
Possible circular tie
34 ABC
33 BCA
32 CAB
99
A>B>C>A
———-
Basic PR formula —
Party Seats = Party Votes x Total Seats / Total Votes
fraction problems
Seat = 1 vote or party votes problem.
Really simple PR Method – as mentioned some time ago.
ALL legis folks on ballots are *elected* — and have a Voting Power equal to the votes each receives in the body.
Extra pay for Legis body *leaders*. Reduced pay for low vote getters.
Top vote getters meet together when necessary — otherwise remote Zoom meetings.
Save Democracy from powermad monarchs-oligarchs.