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

Thu Apr 22 08:29:43 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))?
> 
> I would attribute the problem slightly differently.  It seems to me
> that when Borda gets votes like
> 
> > > > > 56 R D1 D2
> > > > > 44 D1 D2 R
> > > > >
> > > > > R 112
> > > > > D1 144
> > > > > D2 44
> 
> the problem is that Borda acts as if every choice the voter makes is
> independent.  So, the decision to rank D1 over R and the decision to
> rank D2 over R are considered as if they were entirely separate
> unrelated decisions.  Often, however, they will not be.  In the
> previous example it was stated that D1 and D2 were of the same party,
> so a voters choice of D1 over R and D2 over R are not independent bits
> of information about the merits of R.

I agree (I think). The way I would say it is that we have some
information about D1 and D2 that connot be reflected in the ranking of
candidates by each vote. This information is that D1 and D2 are similar.
However, we must allow for the possibility that D1 and D2 are not
considered similar. That is why Borda or Nanson will always be a
compromise. In a way, we are forcing voters to make compromises instead
of voting strategicly. If we gave voters the choice of point allocation
for their candidates, they would give increase the effectiveness of
their vote by being incincere and allocating little or no points to a
popular second choice.

> 
> --snip--
> 
> > > 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

A 110 B 115 C 75 -> A 65 B 35
A 0.55 B 0.575 C 0.365 -> A 0.65 B 0.35

> > > 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.
> 
> Actually, I think we're both wrong.  The candidate to be dropped is
> of course B, not C.  But the result is
> 45 C A
> 55 A C

Interesting!

> 
> Remember to count the BAC people as voting A over C
> 
> So, A wins.  What do you think of this result?  Is it possible for
> the same candidate to be the most probable for both best and worst
> candidate?

I think you're onto something. I will try to find out what is happening.
Any ideas? It appears the assymetry appear when we drop candidates. If
borda was used alone, then the result would be sysmetrical. So it
appears that we should not drop candidates. Perhaps there's a feature of
path voting that can help here?

Does path voting always yield a winner? 
> 
> --snip--
> > >
> > > 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.
> >
> I think we can "understand" the result in the sense that we can
> describe what happened.  The fact remains, however, that B won solely
> because some people who thought that B was better than C changed their
> minds.  If you do not assume Nanson's method is the only way to
> interpret the data, is it possible to argue that some people deciding
> that C is better than B actually provided more evidence that B should
> win?

That can definitely be argued.

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

-- 
Paul Dumais