The simplest tiebreaker for ranked methods
Sass last edited by
If there are exactly two candidates tied after your convoluted calculations, just elect the one who beats the other head to head.
This seems stupendously obvious to an almost condescending degree, but I feel like this concept is often ignored when getting into the weeds with crazy Condorcet methods.
"If the they tie for the least greatest margin of loss, then elect the tied candidate with the least greatest number of winning votes from their pairwise oppositions." - a Minmax variation
It feels excessive at a certain point. Sometimes, the math for some methods creates more ties rather than fewer (see Schulze). I can't imagine being a voter knowing that a tie between my favorite and another candidate my favorite beat was broken in favor of the other candidate because of some crazy math algorithm I don't understand. I know most of these methods are more theoretical than practical, but to me, any tie-breaking method that doesn't naturally comply with this concept every time is fundamentally broken.
@sass I totally agree we should aim for simple ways to resolve Condorcet methods. I find Schulze (for instance) to be ridiculously complicated, and unnecessary so.
We need to be able to explain to people -- regular, everyday people -- how the method is tabulated, and why it is effective at arriving at the "will of the people." Schulze (and the like) don't come close.
However, a true Condorcet cycle requires there be at least 3, doesn't it? So just saying "pick the one that beats the other finalist head to head" has limited usefulness, I'd think.
There is also the case of two candidates having a "pairwise tie". That is, with 6 candidates, both candidates A and B can have 4 pairwise wins, each losing to one candidate, while all other candidates have fewer pairwise wins. If A beats B pairwise, yes, you can declare A the winner. (but keep in mind, the A over B win was effectively counted twice, first to make A a finalist, and then to make A the winner)
To me, the simplest way to resolve a pairwise tie is to choose the Borda winner (in the case of ranked ballets) or the Score winner (in the case of cardinal ballots) -- from that set of pairwise winners (i.e. the "Copeland set"). I believe you proposed one which picked the Borda winner out of the Copeland set, right? If so, I'm all in on that.
Sass last edited by
@rob All of the data was counted to arrive at a set of 2 tied candidates. Data will necessarily need to be double-counted to methodically break it.
Ranked Robin uses the tournament-style of Borda applied only to the tied candidates, which is equivalent to electing the tied candidate who beat the other in the 2-candidate case.
@sass said in The simplest tiebreaker for ranked methods:
All of the data was counted to arrive at a set of 2 tied candidates. Data will necessarily need to be double-counted to methodically break it.
I'm not sure what you are saying here. How do you arrive at a set of 2 tied candidates if there is a Condorcet cycle that has 3 candidates in it? Or is that not the sort of tie you are referring to.
A Former User last edited by
@rob There can be 2 Copeland winners
@brozai right there can be, but there isn't always. When there isn't a single Condorcet winner or Copeland winner, as often as not there are three that are tied.