[EM] Doesn't everybody see that IRV=BORDA?

Paul Kislanko kislanko at airmail.net
Mon Jan 26 12:45:18 PST 2004

My original post was an error, because the method I was comparing to Borda
was not IRV.

That wasn't obvious from this reply, but it was immediately made clear by
the next one. I apologize again for my mistake.

-----Original Message-----
From: James Gilmour <jgilmour at globalnet.co.uk>
To: 'EM List' <election-methods at electorama.com>
Date: Monday, January 26, 2004 2:11 PM
Subject: RE: [EM] Doesn't everybody see that IRV=BORDA?

>> >Paul wrote (here converted to plain text):
>> >> It is fairly easy to prove that IRV always selects the
>> Borda winner.
>I asked:
>> >Surely this statement is wrong?
>> >Consider 3 candidates and 7 voters:
>> > A>B>C: 4
>> > B>C>A: 2
>> > C>B>A: 1
>Paul replied:
>> In this case, there's a majority winner and no need to invoke
>> IRV or Borda or Condorcet. A wins 4>3 period.
>But that doesn't answer the question.  In any case, you cannot change the
counting rules part way
>through the count just because you've seen the first preferences.  A Borda
election is a Borda
>election: points are allocated to every preference and the result depends
on the summation of all
>those points.  In the example above Borda gives a different "winner" from
IRV, Condorcet and
>Plurality.  So, to go back to the question that was prompted by your
statement, how can you prove
>"that IRV always selects the Borda winner"?
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