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
W_{1} for being first-ranked in a vote,
W_{2} for being second-ranked in a vote,
and so on, with
W_{1}≥W_{2}≥…≥W_{N}≥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

- Exhibits favorite betrayal (except for antiplurality voting)
- Violates "clone immunity"

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 A_{1}
and A_{2}. Every A-voter prefers both A-clones over B,
and every B voter prefers B over both A-clones, and all voters
prefer A_{1} over A_{2}.
The scores that result are
A_{1} = (N-1)(t+1),
A_{2} = Nt,
and
B = N.
In other words (taking N sufficiently large) cloning
A converts A's loss into a win for A_{1}
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 A_{1}
and A_{2}. 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 A_{1} over A_{2},
the other half prefer A_{2} over A_{1}.
The scores that result are
A_{1} = N+2Nt-t,
A_{2} = 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.**