A "point runoff" voting system is one in which the votes are rank-orderings of the candidates, and we repeatedly eliminate the candidate with the lowest "score" where score is any weighted positional system – i.e. in the 3-candidate case, you get 1 point for being first ranked, 0 points for being last-ranked, and t points for being middle-ranked for some 0≤t≤1.
Important special cases: When t=0 the point-score runoff system becomes "Instant Runoff" Voting (IRV), when t=1 it becomes Coombs's system, and when t=½ it is the Nanson-Baldwin system.
In this web page we show that every point runoff system, in the 3-candidate case, automatically
This theorem is one of several good results in
John H. Smith: Aggregation of Preferences with Variable Electorate, Econometrica 41,6 (Nov. 1973) 1027-1041.Here's J.H.Smith's proof. He considers this 37-voter 3-candidate election:
| #voters | their vote |
|---|---|
| 10 | A>B>C |
| 9 | B>C>A |
| 8 | C>A>B |
| 4 | A>C>B |
| 3 | B>A>C |
| 3 | C>B>A |
He argues that in any runoff system,
C is eliminated then A wins.
(Scores: A=14+11t, B=12+13t, C=11+13t.)
Now suppose the pink votes are changed
CBA→CAB
and
BAC→ABC;
both these changes are
in A's favor.
Then in the new election
B is eliminated and C wins.
(New scores:
A=17+11t, B=9+13t, C=11+13t.)
Now consider
these 6 changes one by one.
The first one that stops A from
winning is a change of a single vote, in A's favor, which stops A from winning.
That proves non-monotonicity.
Q.E.D.
More generally, we can "iterate" essentially any voting system E to repeatedly eliminate the loser. Call the resulting system "E∞." Durand argued that every such rank-order voting system E∞ automatically must disobey "strict monotonicity" assuming the "input" system E obeyed it and was "symmetric" (neutral and anonymous):
Sylvain H. Durand: A note on monotonocity in iterated choice functions, Social Choice & Welfare 18 (2001) 129-134.
Durand simplified the argument very far. Monotonicity means making an adjacent interchange in one vote favoring X, cannot lessen X's winning chances. "Strict" monotonocity means making an adjacent interchange in one vote favoring X must increase X's winning chances if the situation before the change was a symmetry-forced perfect tie with X a co-winner.
Durand's Proof: In scenario 1 below
| #voters | their vote |
|---|---|
| 1 | A>B>C |
| 1 | A>C>B |
one can quickly see (from "unanimity," a property derivable from our assumptions) that both E and E∞ must elect A winner in this scenario. Now consider changing the first vote to "B>A>C" to get scenario 2:
| #voters | their vote |
|---|---|
| 1 | B>A>C |
| 1 | A>C>B |
This change is favorable to B. C cannot be scenario 2's winner since A is unanimously ranked ahead of C. So the new winner must be A, B, or an AB tie. If it is B or an AB tie, then changing the second vote to "A>B>C" (another B-favoring change) would by strict monotonicity result in B winning outright which contradicts the requirement from symmetry that this altered scenario 2 must yield an AB tie. So therefore A must still be the outright winner of the E-election (or using any strict-monotonic election system) in scenario 2. Therefore in E∞, A is not the one eliminated. Either B or C must be eliminated. Either way, we get (after the elimination) a scenario of the form
| #voters | their vote |
|---|---|
| 1 | X>A |
| 1 | A>X |
which by symmetry must yield an AB (or AC) tie.
But if E∞ were strict monotonic then we just argued that
its winner would have to be A outright in scenario 2. This is a contradiction which
proves
E∞ cannot be strict monotonic.
Q.E.D.
This result was proven by Warren D. Smith (no relation to John H. Smith) in 2007. Here is the proof.
THEOREM (Warren D. Smith 2007) Every point-runoff single-winner voting system exhibits "no show paradoxes" in which a set of co-feeling voters, by casting an honest A-last vote, cause A to win, whereas if they had abstained from voting, a better winner (in their view) would have won. Further, such a paradox exists with 3 candidates and ≤11 voters.
PROOF. Consider this 3-candidate 9-voter election:
| #voters | their vote |
|---|---|
| 1 | A>B>C |
| 3 | A>C>B |
| 3 | B>A>C |
| 2 | C>B>A |
In every point-score runoff system including IRV and Coombs, C is eliminated, then B wins 5:4 over A. That is because the scores are: A=4+3t > B=3+3t > C=2+3t.
Now add 2 extra C>B>A votes. In the new election B is eliminated and then A wins by 7:4 over C.
Thus, adding these 2 honest votes with A ranked last caused A to win.
Those voters would have been better off "not showing up."
Q.E.D.
THEOREM (Warren D. Smith 2007) Every point-runoff single winner voting system fails 3-candidate favorite betrayal, i.e. exhibits 3-candidate elections in which it is strategically better for a certain voter to "betray her favorite" than to cast any vote ranking her favorite honestly top.
PROOF. Let us first prove the theorem under the assumption that 0<t<1. Consider this 3-candidate 3-voter election:
| #voters | their vote |
|---|---|
| 1 | B>C>A |
| 1 | C>A>B |
| 1 | A>B>C |
The scores are A=B=C=1+t. This is a perfect 3-way tie. If an A>B>C voter decides to betray her favorite A by instead voting B>A>C then the scores are B=2 > C=1+t > A=2t, hence A is eliminated and B wins 2:1 over C, a result this voter (whom we may assume is really "A>B>>C") prefers versus the perfect 3-way tie.
Changing to the only other vote that does not betray A, namely A>C>B, causes scores A=1+t, B=1, C=1+2t so that B is eliminated whereupon C wins. This is a worse result (in fact the worst possible result) in the view of that voter. Hence betraying A was strategically forced.
The theorem is now proved when 0<t<1. What about t=0 and t=1?
When t=0 we have Instant Runoff Voting (the candidate with the fewest first-place votes is eliminated). Numerous IRV 3-candidate favorite-betrayal examples are already known, e.g.
| #voters | their vote |
|---|---|
| 8 | B>C>A |
| 6 | C>A>B |
| 5 | A>B>C |
In the above IRV election, B wins but if the six C>A>B voters insincerely switch to A>C>B ("betraying their favorite" C) then the "lesser evil" A becomes the winner, which in their view is a better election result. Only favorite-betrayal votes work.
The final case is when t=1. Then we have Coombs's voting system (the candidate with the most last-place votes is eliminated). Coombs exhibits favorite betrayal in this 3-candidate example:
| #voters | their vote |
|---|---|
| 9 | B>C>A |
| 6 | C>A>B |
| 8 | A>B>C |
A is eliminated (most last-place votes). Then B wins over C by 17:5. Suppose you-both ("you both" meaning 2 new co-feeling voters – you and your spouse) hate B.
If you-both vote honestly C>A>B then that does not alter the election – B still wins.
If you-both vote dishonestly A>B>C, maximally betraying your favorite C, then C is eliminated whereupon A beats B by 15:9. (The more-dishonest B>A>C has the same effect, i.e. also works.)
If you-both decide not to betray your favorite C, then you must vote either C>A>B or C>B>A (or one of each). Either way, A is eliminated and then your most-hated B wins over C by a huge margin.
So betraying C is strategically forced.
Q.E.D.
That is, there is an election situation in which X wins in district I, and X wins in district II, but Y wins in the combined country. (IRV example of that.)
This also was proved (actually he proves a much stronger theorem – a complete characterization of rank-order voting systems that obey district partitioning consistency) by J.H. Smith in the article we cited at top. Essentially the same theorem was also proved independently by
H.P.Young & A. Levenglick: A consistent extension of Condorcet's election principle, SIAM J. Applied Math. 35,2 (1978) 285-300.
If you answer this one, let me know and we'll post it here – warren.wds AT gmail.com. It's trickier because to fully settle this question you also will need to consider elections with more than 3 candidates. The question can be settled halfway with only 3-candidate elections (see below) and we conjecture we know the full answer (also below).
Partial Answer (under construction): If t>½ then vulnerable to clones In this situation, B beats A by 2 votes:
| #voters | Their Vote |
|---|---|
| 2N | A>B |
| 2N+2 | B>A |
Clone A into two clones A1 and A2. Every A-voter prefers both A-clones over B, and every B voter prefers B over both A-clones, and half of (each kind of) voter prefers A1 over A2, the other half prefer A2 over A1. The scores that result are A1 = N+2Nt+t, A2 = N+2Nt+t, and B = 2N+2. In other words (taking N sufficiently large) cloning A converts B's win into a loss (since B is eliminated in the first round) for each t>½.
Conjecture: If there are at least 3 distinct weights, then vulnerable to clones. Even with only two weights, I conjecture vulnerability to clones except for Instant Runoff Voting which is immune to clones (has only two distinct weights; t=0).