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

Wed Apr 21 16:42:10 PDT 1999

Blake Cretney wrote:
> 
> 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?

Yes, your assumtion is good. I would attribute Borda's failure to the
fact that it considers all candidates simultaneously in a fashion
similar to the question: Which candidate has the highest BP(X) given
that all candidates are considered? Borda count can fail here by using a
probability which considers all candidates. We should consider only the
top candidates when making the final decision. This might eliminate the
problem of running multiple candidates with the same platform. If this
doesn't work, then there might have to be a party limit on candidates.

	If we use the Nanson method, we reduce the number of candidates
considered so that we get something more fair. Notice that depending on
how many candidates we consider, we get answers to questions such as:
Which candidate is preferred when compared against all other candidates
(BP(X))? Or which candidate is preferred when compared against only the
top 3 candidates (BP(X)sub(3))?

> 
> > 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.

I was thinking of the only condorcet method I know of which is the one
that only chooses a winner if the candidate wins all of its pairings.
Votes are wasted in this method because each pairing involves the winner
wasting the margin (m) to win the pairing and the loser wasting all its
votes in the lost pairing. Borda does not waste votes by separating the
votes into pairings. That is why it can give a winner every time.
> 
> 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

I don't think Nanson's method works this way, if it does, then Paul's
does not. If the method says "drop the lowest candidate" then do so
based on the counting method. Don't change the method because the
question changed.
The candidated with the lowest score is eleiminated, which gives C 45 A
35 ->C 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.

I think this is a "fair" result. Nanson method has chosen the top two
candidates to compare. C has moved up while B has dropped. This caused A
to be dropped from consideration. When we use the Borda probabilities we
can understand this result. It was a close race. To determine the winner
we calculated which candidate is least likely to win when compared to
all other candidates. This one was eliminated. Then we calculated which
candidate was most like likly to win given the remaining candidates. So
a reduction in support for B and an increase in support for C caused B
to win. So in the final analysis, it was the gain in support for C that
made B win in a situation where this small increase in support puts C
into runner-up position.

> 
> 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

Thanks for the input. I'll try to make some time for critiquing your
path method. I have to admit I know little about it yet.

-- 
Paul Dumais