Ranked voting: Difference between revisions
→Strength of preference
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=== Strength of preference ===
The first thing that should be mentioned is that ranked voting doesn't allow a voter to indicate weak preferences i.e. if a voter either slightly or strongly prefers one candidate over another. See [[rated ballot]] for information on this.
One criticism that can be made of ranked voting is that it creates a logical contradiction: if a voter ranks X>Y>Z, then the strength of their preference for X>Z must be stronger than their preference for X>Y or Y>Z, yet all 3 preferences are generally treated as equally strong in most ranked methods. This can most clearly be seen in many [[Condorcet methods]]: in the [[Head-to-head matchup|head-to-head matchups]], the voter is considered to give 1 vote in all 3 matchups, rather than giving less of a vote in the X>Y and Y>Z matchups and more of a vote in the X>Z matchup. ▼
▲One criticism that can be made of ranked voting is that it creates a logical contradiction: if a voter ranks X>Y>Z, then the strength of their preference for X>Z must be stronger than their preference for X>Y or Y>Z, yet all 3 preferences are generally treated as equally strong in most ranked methods. This can most clearly be seen in many [[Condorcet methods]]: in the [[Head-to-head matchup|head-to-head matchups]], the voter is considered to give 1 vote in all 3 matchups, rather than giving less of a vote in the X>Y and Y>Z matchups and more of a vote in the X>Z matchup. The [[Borda count]] resolves this issue if the point totals are kept the same i.e. a voter who gives 8 points to A, 7 to B, and 6 to C gives 1 point to A>B and 1 point to B>C, which adds up to equal the 2 points for A>C. This same criticism can be made for [[Rated pairwise preference ballot|rated pairwise preference ballots]] as well, since they allow (but do not force) voters to exaggerate all of their pairwise preferences.
[[Approval voting]] (and some [[Rated method|rated methods]] in general) can be thought of as a ranked method with constraints placed that fully resolve this contradiction: if an Approval ballot is thought of as a voter ranking one set of candidates equally 1st and above all others, then when a voter ranks an approved candidate above a disapproved candidate, they can't further indicate a preference between the disapproved candidates, thus ensuring that the strength of preference in each matchup is consistent with the strength in other matchups i.e. if they approve only X, then the strength of X>Y will be the same as X>Z, since the full preference is treated as X>Y=Z. In other words, from a [[pairwise counting]] perspective, if the voter gives 1 vote to X>Y, then they must give 0 votes to Y>Z, and when the number of votes given in both matchups is added up, this equals the 1 vote the voter gave to X>Z. If the voter's ranked preference is visualized as a [[beat-or-tie path]] from their 1st choice to their last choice, then the strength of their preference between any pair of candidates in the path will equal the strength of preference of each matchup between each pair of candidates starting from the first candidate of the pair vs the candidate sequentially after them, the candidate sequentially after them vs the candidate sequentially after the candidate sequentially after them, etc. all the way until the candidate sequentially before the second candidate in the pair vs the second candidate in the pair, added up. Another example would be a voter who approves A, B, and C, and disapproves of D, E, and F; this voter's Approval preference can be represented as A=B=C>D=E=F. If, for example, the B vs E matchup is analyzed, this voter is considered as giving 1 vote to B>E; this is equal to adding up the strength of their preference in the B vs C matchup (0 votes, because they ranked the two equally) plus their strength of preference in the C vs D matchup (1 vote, because they ranked C>D) plus the preference in the D vs E matchup (0 votes, because D=E). So, by starting at B and going sequentially one pair at a time down the beat-or-tie path that is the voter's ranked preference until you reach E, you can see that the strength of B>E is equal to the strength of the intervening matchups added up.
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