Fw: A further elucidation of my arguments

[Paul Dumais]

Hi David,

David Catchpole wrote:
> 
> > > A>B>C
> > > A>B>C
> > > A>B>C
> > > B>C>A
> > > B>C>A
> 
> 1/3- A vs. B (A wins)
> 1/3- A vs. C (A wins)
> 1/3- B vs. C (B wins)

You have to define what 1/3 is. If it is a probability, what probability
is it? The probability I've defined is: Borda Probability of X (BP(X)) -
Probability that candidate X will be preferred given a randomly selected
opponent. This is equivalent to candidate X being preferred over a
random opponent given a randomly selected vote.

So let's do the calculation. BP(X) = P(x)/P

where P(x) is the number of pairings that X wins and P is the total
number of opponents (pairings involving X)

P(x) = BC(x) which is the standard borda count for candidate X (total of
all votes)
P = V(C-1)
where V is the number of votes and C is the number of candidates (C-1 is
pairings involving X or # of opponents)

This gives BP(X) = BC(x)/V(C-1)

So in laymans terms the Borda Probability for candidate X is the
standard borda count divided by the number of votes which is then
divided by the number of opponents of X.

For the above example this gives BP(A) = 6/(5*2) = 0.6 (notice my
calculation error in my last post - sorry)
BP(B) = 7/(5*2) = 0.7 ; BP(C) = 2/(5*2) = 0.2
If we eliminate C then:
BP(A) = 3/(5*1) = 0.6 ; BP(B) = 2/(5*1) = 0.4

My apologies again for my mistakes in one of my earlier posts.
> 
> If we were to randomly select a pair, A would have the 66.7% chance of
> winning. Unless you're using Condorcet's model of "probability of
> rightness" (which obviously implies a Condorcet method), I don't see your
> point.

I'm not familiar with the "probability of rightness". I'm sorry if I
wasn't clear enough. Reading my original reply to your post, I realize
that I wanted to defend the merits of borda count by clearing up the
apparent paradox that you mentioned. The way I hope to clear up this
paradox is by converting the borda count result into a probability.
Since this conversion does not add new information, the probability and
the borda count are essentially equivalent.
	The paradox you mentioned was that Borda Count can give two different
preferences when a (minor) candidate is removed. Something is a porodox
only if accepted theory cannot explain it. By converting the borda count
to a probability (the probability that a random voter will prefer a
particular candidate over a randomly chosen one), we can use accepted
probability theory to remove the paradox. BP(X) will be different given
different numbers of candidates so if we change C we change BP(X) and
possibly the prefference given by the borda count. This realization led
me to what Nanson (?) has already discovered, that the borda count
paradox can be removed by modifying the standard borda count by removing
lower candidates.
	Since the Nanson (?) method chooses a condorcet winner where one
exists, this method should be preferred because it guarantees a winner
and has the added benefit of providing a useful scale against which all
other votes can be compared, namely - the probability BP(X) which is the
probabilty that a random voter will preffer candidate X over a random
opponent. If we use Nanson's (?) method, we get many such slightly
different probabilities PB(X)sub(y) which can be defined as the
probabilty that a random voter will preffer candidate X over a random
opponent given that only the y "highest" opponents are considered. When
y equals the number of available seats or choices + 1 (2 in the above
case) then we get a result that agrees with Condorcet.
> 
> >       So what does this mean? It means precisely what probabilty theory says
> > and nothing more: if we choose an opponent at random for a given
> > candidate, bananas are most likely to win. If the number of candidates
> > are reduced to two, apples are most likely to win. It is my oppinion
> > that borda count does not produce a parodox as long as voters understand
> > what the vote means. It seems reasonable that the family may want to
> > choose apples because they are preferred over bannanas. Therefore, I
> > propose a borda count with the following modification.
> >       Paul's borda count:
> 
> What you seem to mention below is a Nanson (?) count, developed by an
> Australian mathematician early this century. It is guaranteed to elect a
> Condorcet winner where one exists. There is a book from the 1930s whose
> authors' names evade me but whose title I think is "Proportional
> Representation" which goes into more detail.
> 
> >
> >       Do a standard borda count, then eliminate the lowest candidates. The
> > number of candidates kept is equal to the number of choices (seats, one
> > in this example) + 1. Do a second borda count. Eliminate the last
> > choice.
> >
> > This method should greatly reduce the possibility that the choice made
> > differs when comparing it to the first eliminated candidate only. The
> > "perfect" method would probably be to eliminate the lowest candidate one
> > at a time. I'm not sure what would be gained by this however.

Thank-you for the engaging debate.

-- 
Paul Dumais