[EM] Voter Space Borda (VSB)
Forest Simmons
fsimmons at pcc.edu
Thu Feb 5 12:08:02 PST 2004
Here's another example of Voter Space Borda:
Start with the ballot set
x A>B>C
y B>C>A
z C>A>B
Assume that x>y, x>z, and that (x+y)>z, so that the three candidates A, B,
and C form a "Condorcet Cycle."
Condorcet would break the cycle at its weakest link (C beats A by only y+z
votes) making A the winner.
By way of contrast, plain Borda picks B when y > (x+z)/2 .
However, as we shall see, Voter Space Borda agrees with Condorcet:
Projecting this election into voter space we get x ballots of type
a1=a2=...=ax>b1=...=by>c1=...=cz, which we abbreviate as A^x>B^y>C^z.
Making use of similar abbreviations, we have the following voter space
ballot set:
x A^x>B^y>C^z
y B^y>C^z>A^x
z C^z>A^x>B^y
Applying the rules of Borda to this election we see that each of the A
voters gets a (relative) Borda score of x*(y+z)+z*y-y*(y+z)-z*z which
simplifies to a = x*(y+z)-y*y-z*z. Similarly each B voter gets a Borda
score of b = y*(x+z)-x*x-z*z, and each C voter gets a Borda score of c =
z*(x+y)-x*x-y*y.
The differences (a-b) and (a-c) simplify to
(a-b) = (x-y)*(x+y+z) and (x-z)*(x+y+z), respectively,
which are both positive since x>y and x>z.
So the A voters are tied with the highest Borda scores, and their ballots
elect A in agreement with Condorcet.
In all of the examples that I have given so far Voter Space Borda has
agreed with Condorcet.
However, it is easy to see that Voter Space Borda does not comply with the
Condorcet Criterion, since the Voter Space Borda winner has to be the
favorite on at least one ballot, whereas in some elections the CW may not
be ranked first on any ballot.
Some members of this EM listserv have advocated an ad hoc quota of first
place votes to help eliminate "weak CW's." Perhaps Voter Space Borda
would please these members as a more natural "seamless" way of handling
the "weak CW" problem.
Forest
On Mon, 2 Feb 2004, Forest Simmons wrote:
> If you just want to know the method, skip the next three paragraphs, and
> begin at "METHOD."
>
> Working with Joe Weinstein's weighted median idea has brought me around
> again to Richard's issue space idea.
>
> Richard says that if candidates were uniformly distributed among the
> voters, then Copeland and Borda would give excellent answers. He then
> suggests filling in the gaps between real candidates with virtual
> candidates, calculating the Copeland or Borda scores, and then taking the
> real candidate with the highest score.
>
> Playing around with Joe's idea has made me realize that the best way to
> get the virtual candidates uniformly distributed among the voters is to
> make all of the voters into virtual candidates.
>
> METHOD:
>
> Step 1. Use each ballot to rank the voters; the more their ballot agrees
> with yours, the higher you rank them. There are various ways to do this,
> but I suggest incorporating the following condition:
>
> Condition A. If your ballot shows a preference of candidate X over
> cadidate Y, then you should rank all voters that profess X as a unique
> favorite above all voters whose ballots show Y as a unique favorite.
>
> Step 2. The ballot of the voter with the highest average rank is used to
> pick the winner of the election.
>
> [End of Method Description.]
>
> Remarks:
>
> If issue space is one dimensional, and the voter rankings reflect this,
> then it is very likely that the "voter with the highest average rank" in
> step 2 above will be located at the "voter median" position in issue
> space, and therefore the method winner will be the "voter median
> candidate," i.e. the Condorcet Winner.
>
> Some readers may detect similarities with Voter Space Borda and Kemeny's
> method. Instead of the Kemeny distance between my suggested social order
> and yours, we use your rank as determined by my ballot. Instead of
> entertaining all possible social orders, we stick with the orders on
> actual ballots, which gives us a computationally feasible method, unlike
> Kemeny.
>
> If Condition A is adhered to, then our ranking distance removes the clone
> problem associated with the Kemeny distance.
>
>
> Let's see how Voter Space Borda fares on an example in which plain Borda
> would violate the Majority Criterion:
>
> 6 A>B>C
> 4 B>C>A
>
> Each of the six voters in the first faction would order all of the A
> voters before all of the B voters, and each of the four voters in the
> second faction would rank all of the B voters before all of the A voters,
> so we would have:
>
> 6 a1=a2=a3=a4=a5=a6>b1=b2=b3=b4
> 4 b1=b2=b3=b4>a1=a2=a3=a4=a5=a6
>
> Each A voter has 6*4=24 wins and 4*4=16 losses, which gives a Borda score
> of 24-16=8.
>
> Each B voter has 4*6=24 wins and 6*6=36 losses, which yields a Borda score
> of 24-36=-12.
>
> The A voters are all tied for first place.
>
> But this is no problem since they all agree that A should be the winner!
>
> Have fun!
>
> Forest
>
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