The Command Ship
EEEGO Proposal #2: Tie-Breaking and Election Procedure
Posted by Vice Admiral Daniel Lerner (Procedures Guru / Personnel Director / Chief EGO) in EEEGO Proposal #2: Tie-Breaking and Election Procedure
This is second of the two proposals I discussed at the beginning of this post: https://www.star-fleet.com/core/command/command/posts/96625/. This proposal addresses the tie-breaking procedure for elimination rounds, and setting out in more detail the voting procedure.
Current tallying procedure
Other than saying this is an Instant Runoff Voting when there are more than two candidates, the bylaw is completely silent on what that means. However, the following procedure is currently used:
1) Voters rank candidates by preference.
2) If no one has more than 50% for first-choice votes, we use an elimination procedures. (In some elections, we keep doing eliminations even if a candidate has more than 50%, but that has not been consistent.) Elimination is done by removing the candidate has the least first-choice votes, moving votes up in rank as necessary, and re-tallying.
3) If there’s a tie of who got “last place”, we break the tie by random choice. This is not stated anywhere in the bylaw, which led to some confusion last year. It is also what EGO is proposing to change.
4) If there are two candidates left, and they are tied, EC breaks the tie.
Problems
Problem #1: We don’t actually define the voting procedure in the bylaw.
Problem #2: Is there a fairer way to break ties of “last place” other than random choice?
Proposal and Discussion
We are proposing to remove the sentence: If there are more than two candidates standing in the election, the Instant-Runoff Voting system is to be used..
The new language is as follows (the way it is worded, we no longer need to distinguish between two candidates or more than two candidates):
The following voting procedure shall be used:
- On each voter’s ballot, the voter will be asked to rank the candidates from first choice to last choice.
- The first choice of each ballot shall be tallied. If any candidate receives more than 50% of first-choice votes, that candidate shall be the winner.
- If there are more than two candidates left, and no candidate receives more than 50% of the first choice votes, the candidate with the least number of first-choice votes shall be eliminated. Any votes for the eliminated candidate at any rank on the ballots shall be disregarded and any subsequently ranked votes shall be deemed to move up one rank. Step (2) and this Step shall then be repeated.
- If there is a tie over which candidate should be eliminated in Step (3), the candidate with the lowest second choice votes shall be eliminated. If there is still a tie, this procedure shall be repeated for each subsequently lower rank of votes. If there is a tie in all ranks of votes, the Election Coordinator shall cast a tie-breaking vote on who should be eliminated.
- If there are only two candidates left and they are in a tie, the Election Coordinator shall cast a tie-breaking vote to determine the winner.
The Election Coordinator has the discretion to announce whether a tie-breaking vote was used during the election.
Step 4 is obviously the big change. Just to explain what is happening there. Instead of breaking ties for last place by random choice, we will move to the second-choice votes and see who got last place there. If there is still a tie, we go to third-choice votes, and so on. If the candidates are completely tied for last place, the EC decides who should be eliminated.
I can say as a former EGO EC, the breaking of last place in these early rounds can have an impact on the final outcome of the election (but I don’t want to get into specific examples here for obvious reasons). So doing breaking these ties needs to be done fairly, and access to the subsequent choices lets us determine the voters’ intentions.
Downside? TECH/Exodus mainly. Right now, for each round Exodus calculates the number of first choice votes, eliminates the last place person (using random draw to break a tie) and then re-tallies the first choice votes. Now, in each round, Exodus has to calculate the first-choice votes, the second-choice votes, the third-choice votes, etc. and then repeat that for each round. That’s a lot more work, and coding it will not be fun I’m guessing. There are workaround solutions: (1) revert back to manual tallying using a spreadsheet until this is figured out (if it is figured out), (2) have Exodus tally the first-choice automatically, but abort with a message if there is a tie for last place, so that ECs only have to do manual counting if there is a tie, (3) whatever else we can come up with. But we don’t think technical limitations should dictate the policy on the fairest voting procedure if we can do a workaround.
Here’s an example of how the tie breaking would work:
- A, B, C, D are the remaining candidates. No one has more than 50% of the first-choice vote, so we need to eliminate a candidate.
- B, C and D tie for last place for the first-choice votes (but together out-number candidate A - it is possible B, C or D could beat A after some eliminations).
- Instead of randomly eliminating B, C or D, their second-choice votes are tallied. B has more than C and D, but C and D are still tied.
- The third-choice votes are tallied for C and D, bu they are still tied (as is the fourth choice votes). Therefore, the EC decides who to eliminate to break the tie for that round, and the remaining three candidates move on.
For an example of how ties can show up and mess things up, please review this post from an EGO election last year: https://www.star-fleet.com/core/command/command/posts/48352/
Hope I’ve explained this proposal properly.
Daniel Lerner
I am going to be that very specific Jerk…
Is this example true?
Given
Geoff’s Ballot
A
B
C
D
Dan’s Ballot
B
A
C
D
Sarah’s Ballot
C
B
A
D
Rob’s Ballot
A
B
D
C
Katy’s Ballot
D
C
B
A
Then (3 Votes Needed to Win)
Round 1
A -2
B - 1
C - 1
D - 1
Tie Breaker:
B 2 number 2 choice
C 1 number 2 choice
D 0 number 2 choice
D is dropped
Round 2
A - 2
B - 1
C - 2
B is dropped
Round Three
A - 3
C - 2
A is the winner
Is that the process we are talking about?
~Geoff