election_title: "Batch elimination empties the field — with a Condorcet winner sitting there"

scenario_description: |-
  The same three voters and the same three candidates as the cycle case
  (batch_all_out_cycle_c3_b3), with ONE ballot changed: the second voter now says
  Bruno>Amy>Clara instead of Bruno>Clara>Amy. That single swap breaks the cycle
  and hands the election an undisputed Condorcet winner — Amy beats Bruno 2-1 and
  beats Clara 2-1, so she is preferred head-to-head to everyone.
  It changes nothing about the instant-runoff count. First choices are still Amy
  1, Bruno 1, Clara 1; nobody has a majority; all three are still tied for fewest.
  Batch elimination removes the whole field again and reports a THREE-WAY TIE — in
  an election that has a clear winner by every pairwise measure.
  That is the price of the convention, and it is worth paying attention to,
  because the usual defence of the total batch does not apply here. In the cycle
  case a three-way tie is the only answer an anonymous, neutral rule CAN give: the
  profile is symmetric, so there is nothing to separate the candidates with. This
  profile is not symmetric. Ranked Robin elects Amy. Coombs elects Amy — its
  elimination rule reads the BOTTOM of the ballot, where Clara has 2 last-place
  votes to Bruno's 1 and Amy's 0, so Coombs cuts exactly one candidate and the
  count proceeds normally. A fair rule can absolutely name a winner here. Batch
  IRV declines to.
  Verified with pref_voting (Holliday & Pacuit): instant_runoff = {Amy, Bruno,
  Clara}, coombs = {Amy}, copeland = {Amy}.
  Lesson: 07_Concepts/topics/ties/batch_elimination.md

video_script: |-
  Same three voters as last time. I am going to change exactly one ballot.
  Voter two used to say Bruno, then Clara, then Amy. Now she says Bruno, then Amy,
  then Clara. One swap.
  That swap breaks the cycle. Amy now beats Bruno head-to-head, two to one. Amy
  beats Clara head-to-head, two to one. Amy beats everybody. There is nothing
  ambiguous left about this election.
  Now run instant-runoff on it. First choices: Amy one, Bruno one, Clara one.
  Nobody has a majority. Everybody is tied for last. Remove all of them, and the
  count reports a three-way tie.
  Last time a three-way tie was the only honest answer, because the ballots were
  perfectly symmetric and no fair rule could tell the candidates apart. That
  defence is gone. These ballots are not symmetric. Ranked Robin picks Amy.
  Coombs picks Amy — it looks at who voters rank LAST, and Clara is last on two
  ballots, so Coombs has somebody to cut and the count keeps going.
  So this is the real cost of eliminating everyone tied for the bottom. Not that
  it produces ties. That it produces them in elections that did not need one.

voting_method: RCV_IRV
num_winners: 1

ballots: |-
  Amy>Bruno>Clara
  Bruno>Amy>Clara
  Clara>Amy>Bruno

expected_winners:
- Amy