Talk:MARS voting: Difference between revisions
→Tie handling
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:The third case would be a true tie. But if you give the chance of a tie in a voting method a value T from 0 (never) to 1 (always) then the chance of such a tie occurring it the chance of a tie in score ''times'' the chance of tie in a 1 on 1 match. T(MARS) = T(score) x T(1on1). Both of those are already small. Therefor ties like this would be extremely rare.--[[User:Casimir|Casimir]] ([[User talk:Casimir|talk]]) 08:46, 24 February 2021 (UTC)
::Is the algorithm description missing a step, then? In the first example, why does A win by preference? Eliminating C still means A and B have the same unmodified score: 289. Is it because the cardinality of votes favors A (33 vs 32)? In your example -- B and C have the same cardinality with 32 votes each, so how do we know to drop C? We can't just drop the person with the lowest score, because they could still win with a modified score.
::To back up a step, consider the following two cases:
::{| class="wikitable" border="1" style="empty-cells: show"
|-
!# Voters
!A
!B
!C
|-
|4
|5
|3
|0
|-
|4
|0
|3
|5
|}
::'''Outcome''': Modified score goes to B. Case 2:
::{| class="wikitable" border="1" style="empty-cells: show"
|-
!# Voters
!A
!B
!C
|-
|4
|5
|2
|0
|-
|4
|0
|2
|5
|}
::'''Outcome''': B doesn't have enough popularity to win, but A & C tie in total scores, so this is a hard tie.
::I believe that you're saying there's another step or case in the tiebreaker determination: if two (leading) candidates have the same score, the one with the higher number of votes wins. If so, then the tiebreaker rules are:
::# The winner is the one with the highest modified score (HMS, score+modifier); or
::# If there's an HMS tie, then the winner is the leading candidate with the highest unmodified score (HUS, per MARS); or
::# If there's an HUS tie, then the winner is the leading candidate with the highest cardinality (number of votes); or
::# If there's a cardinality tie, then the election is a hard tie with no possible winner.
::Does that sound right?
::It feels like there's a step missing, though. Something like, if there's a cycle or tie, look at the score and drop the lowest candidate and iterate, like IRV. In the MARS algorithm, there is no candidate elimination step, so where would that go? --[[User:Ŝan|Ŝan]] ([[User talk:Ŝan|talk]]) 22:15, 24 February 2021 (UTC)
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