- Ranked-choice voting requires a candidate to reach 50% support
- The first step is to count all first-choice votes
- If no candidate reaches 50% of 1st choice votes, the lowest candidate is eliminated, and those votes shift to 2nd or later choices
- This continues round-by-round until a candidate reaches 50% and wins
- The vertical “Rounds Slider” slider can be moved to show the round-by-round changes

Rounds Slider

R1 | R2 | R3 | R4 | ||||
---|---|---|---|---|---|---|---|

Shirley Chisholm | 3 | 0 | 3 | 2 | 5 | 3 | 8 |

Eleanor Roosevelt | 4 | 1 | 5 | 0 | 5 | 0 | 5 |

Fannie Lou Hamer | 3 | 0 | 3 | 0 | 3 | -3 | 0 |

Madeleine Albright | 2 | 0 | 2 | -2 | 0 | 0 | 0 |

Patsy Takemoto Mink | 1 | -1 | 0 | 0 | 0 | 0 | 0 |

Total | 13 | 13 | 13 | 13 | |||

Inactive Ballots | 0.00000 | 0.00000 | 0.00000 | 0.00000 |

- Use of mathematical tie-breaker formula - weights voter preferences from before rounds are calculated
- Use of random tie-breaker – because mathematical tie-breaker formula resulted in a tie

1st ch | 2nd ch | 3rd ch | 4th ch | 5th ch | |
---|---|---|---|---|---|

Fannie Lou Hamer | 3 | 3 | 1 | 3 | 3 |

Shirley Chisholm | 3 | 7 | 2 | 0 | 1 |

Madeleine Albright | 2 | 1 | 4 | 4 | 2 |

Eleanor Roosevelt | 4 | 1 | 4 | 1 | 3 |

Patsy Takemoto Mink | 1 | 1 | 2 | 5 | 4 |

Total Choices | 13 | 13 | 13 | 13 | 13 |

RCV123 on-line system handles ties among candidates facing elimination differently than any official RCV systems. (Other than tie-breaking, we use the WIGM RCV system that is the standard counting method.)

We vary from official RCV for tie-breaking because in elections with thousands or hundreds of thousands of voters, ties are very unlikely. But our mission is to make RCV helpful to anyone who wants to make a group decision – including smaller groups with perhaps only 25 voters in a classroom or small civic organization. In a small group election with five candidates and 20 voters, for example, there are very likely to be several ties as the rounds progress.

Official RCV uses random chance to settle any ties. We believe it would be unsatisfying for small voting groups to find that much of the outcome was determined by random chance.

So we developed a unique tie-breaking system that calculates a single number for each candidate based on their vote totals and the choice column they are in. The candidate with the highest tie-breaking number wins that tie. If that tie-breaker number winds up in a tie, then RCV123 resorts to random chance.

Each first-choice vote is worth 100, and each subsequent choice is worth 2/3 (.67) of the previous choice on a ballot. Then all the votes and weighting for each candidate in each column are totaled to determine an overall tie-breaker number. So in our method, for example, three 2nd place votes are worth very slightly more than two 1st place votes. But it would take 37 10th place votes to have the same weight as one 1st place vote.

Our tie-breaking method looks at all choice data from every ballot. This is different from the rounds of counting - which only looks at the data from each round as it is calculated. For example, in actual rounds of counting, a candidate with zero first-choice votes will be eliminated right away, and any 2nd or 5th or 10th place votes they may have does not matter at all.

If two candidates facing elimination have a tie, and have identical tie-breaker numbers, then RCV123 will use random chance to decide. We create a grid of randomly decided, head-to-head tie-breaking match-ups for each combination of candidates. That grid can be found on the results page of any election.

The use of the mathematical tie breakers will be noted in election results with a blue rectangle over vote totals in that round for the candidates involved. The use of the last-resort, random tie breaker will be noted by the color green.

We believe our tie-breaking system is a good compromise between not weighting the choice column of votes at all, and excessively weighting one choice column vs. another immediately adjacent.

This table shows the primary tie-breaker calculation. It starts with the raw ballot data from before any rounds were tabulated.

The total of all voter 1st choices for a candidate is multiplied by 100. Each successive set of total choices for a candidate ( 2nd, 3rd, 4th etc.) is assigned 2/3 (.67) of the weight given to the previous column of choice totals. Then all the columns are added together to arrive at a tie-breaker number for each candidate.

1st ch | x 1.00 | 2nd ch | x 0.67 | 3rd ch | x 0.45 | 4th ch | x 0.30 | 5th ch | x 0.20 | Candidate Tie-Breaker Number | |
---|---|---|---|---|---|---|---|---|---|---|---|

Fannie Lou Hamer | 3 | 3.00 | 3 | 2.01 | 1 | 0.45 | 3 | 0.90 | 3 | 0.60 | 6.97 |

Shirley Chisholm | 3 | 3.00 | 7 | 4.69 | 2 | 0.90 | 0 | 0.00 | 1 | 0.20 | 8.79 |

Madeleine Albright | 2 | 2.00 | 1 | 0.67 | 4 | 1.80 | 4 | 1.20 | 2 | 0.40 | 6.07 |

Eleanor Roosevelt | 4 | 4.00 | 1 | 0.67 | 4 | 1.80 | 1 | 0.30 | 3 | 0.60 | 7.37 |

Patsy Takemoto Mink | 1 | 1.00 | 1 | 0.67 | 2 | 0.90 | 5 | 1.50 | 4 | 0.81 | 4.88 |

Total Choices | 13 | 13 | 13 | 13 | 13 |

In the event our primary tie-breaking system can’t settle a tie among candidates with exactly the same number of votes and set of choice preferences, we have the computer generate a random list of all candidates. That order determines who will win any ties of the primary system.

Order | |
---|---|

Fannie Lou Hamer | 5 |

Shirley Chisholm | 1 |

Madeleine Albright | 4 |

Eleanor Roosevelt | 3 |

Patsy Takemoto Mink | 2 |