Solution
A Helpful Vote That Hurts
Show the problem again
Twenty-four voters rank candidates A, B and C. The ballots are: - 8 voters: A > B > C - 3 voters: B > A > C - 7 voters: B > C > A - 6 voters: C > B > A The election uses instant-runoff voting (IRV): if no candidate has a majority (more than half) of first-choice votes, the candidate with the fewest first-choice votes is eliminated and each of their ballots moves to its highest-ranked remaining candidate; this repeats until someone has a majority. Currently B wins. What is the smallest number of voters who can change their ballots (each to any ranking they like) so that C wins? A tie for elimination counts as a failure (no winner is declared).
Worked solution
Currently the first-choice tallies are A 8, B 10, C 6. C is eliminated, the 6 C-first ballots all go to B, and B wins with 16. For C to win, either A or B must be eliminated first, and C must then collect a majority. If A is eliminated first, A's voters (A > B > C) transfer to B, and B (who already has 10) only gains; C would lose. So B must be eliminated first with C then beating A. B needs strictly fewer first-choice votes than both A and C. Since B starts at 10, and A and C together have 14, B must fall to at most 7 (then A and C hold 17 between them, enough to both exceed 7: for example 9 and 8). That takes at least 3 voters switching away from B, and 3 changes suffice. Concretely, let the three B > A > C voters change: one to A > B > C and two to C > A > B. The tallies become A 9, B 7, C 8. B is eliminated, B > C > A's 7 ballots move to C, and C wins 15 to 9. A brute-force check of all changes of 0, 1 and 2 ballots confirms none works, so the answer is **3**. The lesson is the non-monotonic quirks of IRV: C wins only after voters abandon the candidate they were helping.
Source: Instant-runoff voting and its failures of monotonicity are discussed widely in the social-choice literature (see e.g. Brams and Fishburn); the electorate is invented for AxiomIQ. Statement written for AxiomIQ.