Fw: A further elucidation of my arguments

[David Catchpole]

I'm afraid I can't except your argument along the lines of "Borda
Probability" because, as Saari is apt to be paraphrased, it deliberately
ignores a fact- in this case the fact of collective choice You're
choosing a model of randomly selecting a ballot rather than assessing the 
wishes of an entire society, which is dangerously far from approaching
the paradox I gave you.

Perhaps the thing that disturbsme most about Borda is that there is no
theory to support it which actually reflect any of the reasons a
particular voting system is better than another.

On Wed, 21 Apr 1999, Paul Dumais wrote:

> 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
>