Bottom-Two-Runoff IRV: Difference between revisions
Moved clone independence failure to new section, populated the section, and added DMTCBR failure
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In a real election, of course, voters would show greater variation in the rankings they cast, which could influence the result.
==Passed and failed criteria==
==Notes==▼
BTR-IRV only requires eliminations to be done until one candidate remains who [[Pairwise counting#Terminology|pairwise beats]] all other uneliminated candidates, at which point that candidate can be declared the winner; this is because that candidate is guaranteed not to be eliminated in any remaining BTR-IRV pairwise matchups. This trick can be used to save time in counting if a pairwise comparison table has already been made, and also means BTR-IRV can be phrased analagously to [[Benham's method]], though in terms of BTR-IRV itself instead of IRV.▼
Like IRV, BTR-IRV fails [[monotonicity]] and [[summability]]. Unlike IRV, BTR-IRV passes the [[Smith criterion]].
BTR-IRV can be thought of as directly related to IRV in the sense that both focus on eliminating one of the two candidates with the fewest 1st choices in each round; the only difference is that BTR-IRV can eliminate the candidate with the 2nd-fewest 1st choices if they lose the pairwise matchup against the candidate with the fewest 1st choices, whereas IRV always eliminates the candidate with the fewest 1st choices.▼
If the voters don't produce any Condorcet cycles, then like every other Condorcet method, BTR-IRV is monotone and summable. However, this is not necessarily known in advance.
BTR-IRV is not immune to clones. Example:▼
=== Clone independence ===
{| class="wikitable"
|Chris Benham's BTR-IRV cloning-failure example (before cloning D). Winner is '''A''' after B,C,D eliminated in that order.
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|}
Note that the example requires two cases of the [[Condorcet paradox]] in the base case: b>a, a>c, c>b and also c>b, b>d, d>c, so it is unlikely to occur in practice.
=== Dominant mutual third candidate burial resistance ===
Unlike many other [[:Category:Condorcet-IRV hybrid methods|Condorcet-IRV hybrid methods]], BTR-IRV fails [[dominant mutual third candidate burial resistance]].
{{ballots|
4: A>B>C
2: B>A>C
3: B>C>A
2: C>A>B
}}
B has five first preferences, A has four, and C has two. A is the Condorcet winner with 36% of the first preferences, and thus the DMT candidate.
Let one B>A>C voter bury A under C:
{{ballots|
4: A>B>C
1: B>A>C
4: B>C>A
2: C>A>B
}}
This creates an ABCA cycle. BTR-IRV starts by determining which of the two Plurality losers (A and C) should be eliminated. Since C beats A pairwise, A is eliminated. In the second round, B beats C pairwise and wins.
Thus the burial benefited the B>A>C voter as the winner changed from A to B.
▲==Notes==
▲BTR-IRV only requires eliminations to be done until one candidate remains who [[Pairwise counting#Terminology|pairwise beats]] all other uneliminated candidates, at which point that candidate can be declared the winner; this is because that candidate is guaranteed not to be eliminated in any remaining BTR-IRV pairwise matchups. This trick can be used to save time in counting if a pairwise comparison table has already been made, and also means BTR-IRV can be phrased analagously to [[Benham's method]], though in terms of BTR-IRV itself instead of IRV.
▲BTR-IRV can be thought of as directly related to IRV in the sense that both focus on eliminating one of the two candidates with the fewest 1st choices in each round; the only difference is that BTR-IRV can eliminate the candidate with the 2nd-fewest 1st choices if they lose the pairwise matchup against the candidate with the fewest 1st choices, whereas IRV always eliminates the candidate with the fewest 1st choices.
There are likely to be many candidates tied for having the fewest 1st choices; one possible non-random tiebreaker is to look for those among the tied candidates that have the fewest 2nd choices, then 3rd choices, etc.
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