[EM] Bucklin

[Forest Simmons]

On Wed, 7 Apr 2004, Forest Simmons wrote:

> On Tue, 6 Apr 2004, Rob LeGrand wrote:
>
> > Here's a question I thought about quite a bit a while ago but never posted
> > until now that there's talk of Bucklin on the list:  Which candidate should
> > win the following Bucklin election?
> >
> > 25:Brown>Jones>Davis>Smith
> > 26:Davis>Smith>Brown>Jones
> > 49:Jones>Smith>Brown>Davis
> >
> > Smith?  Jones?
>
> I'll add my two cents worth to responses given by Kevin and Alex:
>
> We think of Borda as giving the win to the candidate with the highest
> average rank, and we think of Bucklin as giving the win to the candidate
> with the highest median rank, which normally it does, but in this case
> both Smith and Jones have a median rank of two.
>
> Or do they?
>
> The answer to this question depends on how you define "median" for lumped
> data.
>
> Suppose that I have 49 numbers in the bin marked "one", 25 numbers in the
> bin marked "two", no numbers in the bin marked "three", and 26 numbers in
> the bin marked "four."
>
> If all the numbers inside a bin are exactly equal to the mark or label on
> the bin, then the median number is two.
>
> But suppose that the 25 numbers in the bin marked "two" are uniformly
> distributed between 1.5 and 2.5.  Then we can see that the median number
> is halfway between 1.5+(1/25) and 1.5+(2/25) or about 1.56 , a
> significantly better rank than two.
>
> Which median makes better sense in the context that Rob LeGrand has given
> us?
>
> The second type would definitely make more sense if the ranks were derived
> from four-slot CR ballots, since a wide range of CR values have to be
> compressed into only four "bins" if you will.
>
> I think Jones qualifies as the highest median rank candidate, whether or
> not he is the Bucklin winner.
>
> Here's another approach: for both Smith and Jones calculate the difference
> in the number of ballots above and below rank two (the rank of their
> common median ballot according to simple median).
>
> Jones: 49-26=23
> Smith: 0-25=-25
>
> Jones beats Smith.
>
> I think this is the simplest way to resolve median rank "ties."

The next simplest is to use quotients instead of differences:

Jones: 49/26 > 1
Smith: 0/25=0 < 1

so Jones beats Smith, again.


Forest

>
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