Reply #2: [EM] FWD: Borda Count by Paul Dumais

[Blake Cretney]

Paul Dumais wrote:
> > Here is the example that I think shows that Borda is unusable.
> > Imagine that there are two parties, the D's and the R's.  Now, assume
> > there are more R than D voters as in the following example.
> > 
> > 56 R D
> > 44 D R
> > 
> > Obviously, the R candidate wins, as one might expect.  But what
> > happens if the D party runs two candidates, D1 and D2.  The following
> > is a possible outcome.
> > 
> > 56 R D1 D2
> > 44 D1 D2 R
> > 
> > R 112
> > D1 144
> > D2 44
> > 
> > One of the D party candidates wins.  Note that there are no more D
> > voter's in the second example.  Running more candidates in Borda is a
> > very good strategy.  If a legislative body is choosing between
> > multiple proposals using Borda, it makes sense to submit as many
> > nearly identical copies of your proposal as possible.  This is both
> > absurd, and a real practical problem.
> 
> Paul's Borda Count reveals: 56 R, 44 D1D2 or 56RD1 44 D1D2
> R: 112     or   R:112  
> D1: 116         D1:144
> D2: 72          D2:44
> after first count
> eliminate d2 gives
> R:56 D1:44
> 

Should I assume by your use of Paul instead of Borda that you agree
that Borda does not handle this example acceptably?  If so, to what do
you attribute Borda's failure? 

> Using borda count to eliminate minor candidates reveals a fair result.
> The burying tactic is just another form of voting voting for minor
> candidates over a percieved favorite. It will always be a problem in
> voting methods. I feel however, that strategic voting will either be
> rare or self-moderating. When rare, it's effect is limited. When not
> rare, voting results will rebeal a closer race which will make it more
> difficult to chose who is favorite, second etc. If we use one or more
> steps with borda count (Paul's Borda count) to eliminate the minor
> candidates, the vote will always come down to "how many voters chose X
> over Y" in a single seat race. As you can see Borda Count converges with
> condorcet however, borda count has a higher degree of success due to no
> wasted votes.

You'll have to explain what you mean by that wasted votes comment. 
There are lots of methods that meet the Condorcet Criterion.  Do all
waste votes, or just some?  The method you describe as possibly
perfect, Nanson's method, actually meets the Condorcet Criterion.

Here are two reasons why I think Paul and Nanson are inferior to
another Condorcet Criterion method called Path Voting.

Reverse-Consistency -- This is what Markus calls Reverse-symmetry

If we think that our method is the best possible, then, if a group of
voters all vote sincerely the method should arrive at the best guess
for the best answer to whatever question is posed.

For example, if we pose the question, "Who is the best candidate?"
And the voters mark their ballots as follows:
45 A B C
35 B C A
20 C A B
Then by Nanson/Paul we get A 115 B 115 C 86 -> A 65 B 35 -> A wins
So, the answer to the question is A.

But, what if instead we asked, "Who is the worst candidate?"  By the
above assumption that everyone is voting sincerely, and since
everyone's order from worst to best is the reverse of best to worst,
we know that the ballots would look like this:

45 C B A
35 A C B
20 B A C

Then by Nanson/Paul we get A 90 B 85 C 125 -> A 55 B 45 -> A wins

But these results are inconsistent.  A can't be best guess for best
and best guess for worst.  Therefore, the first assumption, that the
method is giving us the best guess must be false. It must be giving
the wrong answer for one of these questions.

Compare this to Path Voting.  If we look at the pair-wise majorities
we have the margins
A>B 30
B>C 60
C>A 10

the best path from A to B is A>B 30.  The best path back is 
B>C 60
C>A 10
So, the minimum is 10.  30>10, so A is ranked higher than B.

the best path from A to C is
A>B 30
B>C 60
So, the minimum is 30
the best path from C to A is C>A 10. 30>10, so A is ranked higher
than C.
A is the winner.

But in the reverse,
B>A 30
C>B 60
A>C 10

C>B>A 30 > A>C 10
C>B 60   > B>A>C 10
So, C wins.  Path Voting is consistent, and ranks then A B C
---

Monotonicity

35 A B C
40 B C A
25 C A B

A 95 B 115 C 90 -> A 60 B 40 -> A wins

But consider if B fell in some voters opinion. The result could be

35 A B C
6  C B A -- used to be B C A
34 B C A
25 C A B

Now we get
A 95 B 109 C 96 -> B 69 C 31 -> B wins

So, a reduction in support for B among some people causes B to win.

Now, let's look at the examples in path voting
Before B reduction:
A>B 20
B>C 50
C>A 30

B>C 50    > C>A>B 20
B>C>A 30  > A>B 20
So, B wins.  When B is reduced:

A>B 20
B>C 38
C>A 30

B>C>A 30 > A>B 20
B>C 38   > C>A>B 20
B wins again.  But there's nothing inconsistent about this.

---
Blake Cretney
My Election Method Resource, which defines many of the popular methods and
criteria, is at
http://www.fortunecity.com/meltingpot/harrow/124/
This site strains to be objective

My page for advocating Path Voting is at
http://www.fortunecity.com/meltingpot/harrow/124/path
This site makes no such attempt