Imc · Probability & Brainteasers
Analyze Optimal Funding Allocations Across Election Districts
TrueInterview
October 7, 2026 · 2 min read
Prompt
Two players divide nonnegative integer units of funding among labeled electoral districts. A district is won by whichever player assigns more funding to it; if the amounts are equal, the district is tied. The player who carries more districts wins the overall election.
Constraints & Assumptions
- Each player must spend their entire budget across districts.
- When random allocations are considered, every weak composition of the given budget over the labeled districts is equally probable.
- For "possible to win," you may select the alignment most favorable to your allocation; for "cannot win," no alignment may give more district wins than losses.
- An opponent allocation that is as even as possible has district counts that differ by no more than one unit.
Clarifying Questions to Ask
- Are the districts symmetric, or do some have different weight? Treat them as symmetric for this problem.
- Does tying the overall election count as success? No.
- Does "minimum" refer to a possible favorable alignment, or to a guarantee against every alignment?
Part 1 — Three Districts and Equal Budgets
For three districts and four units per player, count the opponent's possible allocations. If the opponent picks one uniformly, compare your allocations (2,1,1) and (4,0,0), and explain why every rotation of (2,1,1) performs the same.
What This Part Should Cover
- Stars-and-bars enumeration for labeled weak compositions.
- Why rotated allocations are symmetric against a uniform symmetric opponent.
- Why concentrating all four units in a single district can win at most one district.
Part 2 — Minimum Budget for a Possible Win
There are eight districts. The opponent has nine units and spreads them as evenly as possible. Find the smallest budget that permits a possible win under a favorable alignment, and give an allocation and matchup that demonstrate feasibility.
What This Part Should Cover
- The opponent's multiset
(2,1,1,1,1,1,1,1), up to rotation. - A nine-unit allocation that yields strictly more wins than losses.
- A lower-bound argument showing eight units cannot achieve that.
Part 3 — Opponent Budget That Makes a Win Impossible
You have nine units across eight districts. The opponent again spreads units as evenly as possible. Find the smallest opponent budget that makes winning impossible for you even under the most favorable alignment.
What This Part Should Cover
- Compare opponent budgets of eleven and twelve.
- A favorable nine-unit allocation that still defeats the eleven-unit distribution.
- Why the twelve-unit distribution necessarily gives at least as many losses as wins.
Hint — Count wins and losses, not just majorities: Tied districts help neither side. A winning allocation may intentionally concede some districts and tie others rather than winning five outright.
What a Strong Answer Covers
- Exact combinatorial counts and explicit matched allocations.
- A feasibility construction together with a separate lower-bound or impossibility proof.
- A clear distinction among labeled rotations, random symmetry, and favorable alignment.
- Verification that every allocation vector uses the full stated budget.
Follow-up Questions
- How would you compute the best response for arbitrary numbers of districts and budgets?
- How would the answer change if an overall tie counted as success?
- What if districts had unequal electoral values instead of one vote each?
Overview: Analyze election-district funding games using weak-composition counting, favorable alignments, explicit allocations, and lower-bound proofs for possible and impossible wins under fixed budgets.