[EM] Bucklin

[Forest Simmons]

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."

Forest

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