A "weighted positional" voting system inputs rank-orderings as votes. It elects the candidate with the greatest "score" where 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. (More generally for N-candidate elections with perhaps N>3, you get score W1 for being first-ranked in a vote, W2 for being second-ranked in a vote, and so on, with W1≥W2≥…≥WN≥0.)
Important special cases (in the 3-candidate case): The plurality voting system arises when t=0. Borda arises when t=½. AntiPlurality voting arises when t=1.
In this web page we show that every weighted positional voting system, in the 3-candidate case, automatically
On the bright side, though, all weighted positional voting systems trivially satisfy monotonicity (increasing your vote for X cannot decrease X's winning chances), "participation" (voting honestly is never strategically worse for you than not voting), and consistency when partitioning country into districts (if X wins in all districts, then X must win in combined country).
| #voters | Their Vote |
|---|---|
| 4 | M>W>B |
| 2 | W>B>M |
| 1 | B>M>W |
| 2 | M>B>W |
| 4 | B>W>M |
In this "Milk Wine Beer" election example by Don Saari (analysis added by Jack Rudd),
B scores 5+4t,
M scores 6+t,
and
W scores 2+8t.
Hence for 0≤t<1/3, M wins; for 1/3<t<3/4, B wins; and for 3/4<t≤1, W wins.
THEOREM: Every weighted positional voting system except anti-plurality voting exhibits favorite betrayal in some 3-candidate election situation.
Proof:
| #voters | Their Vote |
|---|---|
| N+1 | B>C>A |
| N | C>A>B |
| N | A>B>C |
In this (3N+1)-voter example
B wins
in every weighted positional voting system with 0≤t<1, because the
scores are
B=N+1+Nt > A=N+(N+1)t = C=N+(N+1)t.
But if the N C>A>B voters insincerely switch to A>C>B ("betraying their favorite" C)
then the "lesser evil" A becomes the winner under all these voting systems
(and is the Condorcet-winner), which in their view is a better election result.
(New scores:
A=2N > B=N+1+Nt > C=(2N+1)t where the ">"s are valid for all sufficiently large N.)
Any change which did not betray C (i.e. to C>B>A) would, however, not work, i.e,
B still would win.
Q.E.D.
Actually, we can say more – we can totally classify all weighted positional voting systems for any number of candidates that obey Favorite Betrayal:
THEOREM: For each number C≥3, every weighted positional voting system exhibits favorite betrayal in some C-candidate election situation except if the top two weights are equal.
Proof sketch: Similar to the above, except we now make a C-cycle rather than a 3-cycle, with N voters in each of the C cyclic classes except for N+1 voters in one class breaking the perfect tie so that A (say) wins the election. Now consider the class of N voters which consider candidate A to be third-ranked. These voters honestly rank (say) B second. If they promote B insincerely to first place, betraying their true favorite, then they (for all sufficiently large N, if the top two weights differ) will cause B to win, a result they prefer versus A winning. They can't do that unless they promote B to top (forcing favorite-betrayal, since they already honestly had B ranked second). Any insincere voting move that doesn't promote B will cause A or somebody even worse in their view, to win, so favorite betrayal was strategically forced. Of course this was a group decision to strategize not a single isolated voter decision, but by considering altering the votes one voter at a time in this group, the first one that changes the election result is a situation where it's a single voter's betrayal decision that does the job. Q.E.D.
Proof: In this situation, B beats A by 1 vote:
| #voters | Their Vote |
|---|---|
| N-1 | A>B |
| N | 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 all voters prefer A1 over A2. The scores that result are A1 = (N-1)(t+1), A2 = Nt, and B = N. In other words (taking N sufficiently large) cloning A converts A's loss into a win for A1 for each t>0.
In this situation, A beats B 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 A's win into a loss for both A-clones for each t<½.
The combination of both these examples has covered the entire range of allowable t,
namely all t with 0≤t≤1, showing that for each such t, cloning can either convert
a winner into two losers, or convert a loser into a winner,
or (which happens if 0<t<½) both.
Q.E.D.