Sequentially Shrinking Quota: Difference between revisions
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Marylander (talk | contribs) (Explanation of how the method works) |
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===== Reweighting Step =====
Each round, a ballot loses weight equal to (Score contributed to the candidate elected this round / Max score) * (Quota / Total Score for that candidate this round).
== Notes ==
SSQ decays into [[D'Hondt]] in the [[Party list case|party list case]]. This is because it essentially searches for the highest divisor that each party can pay for its seats. 2-seat party list example:<blockquote>30 A
14 B</blockquote>The Hare quota is 22 ((30+14=44)/2). A has the most votes and gets the first seat, and then is left with 30-22=8 votes. So it becomes<blockquote>8 A
14 B</blockquote>Since no party can get 22 votes now, we look for the largest quota that anyone can pay now if earlier winners (the one A candidate) had paid that same amount.
We end up finding that if the quota is shrunken to 15:<blockquote>15 A (8 original votes + (22 (old quota) - 15 (shrunken quota) = 7 restored votes))
14 B</blockquote>A has the most votes, so they take the second seat, resulting in a [[Winner set|winner set]] of (2 A) i.e. (A, A). This is a notable contrast to most Hare quota-based methods that reduce to [[Largest remainder method|largest remainder methods]], such as regular [[Sequentially Spent Score]], which would've stopped at the second round, seen that B had 14 votes to A's 8, and thus elected (A, B).
== References ==
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