Re: Bringing proportional representation to ranked pairs
Nov 16, 2000 22:14 PST
"Bram Cohen" <email@example.com>, on the subject of 'Re: Bringing
proportional representation to ranked pairs', is quoted as:
| ||Here's a quick rundown of how Weighted Ranked Pairs meets (or|
doesn't) various standards and criteria -
| ||Consistency Criterion - I'm pretty sure it passes this one just fine.
That seems unlikely. The Consistency criterion says that if you divide
the electorate into two arbitrary groups, and find the winner for each
group, and the winner is the same, then this should also be the winner
for the combined groups. That's obviously true of a method like
plurality, or in fact, any positional method. It is impossible to meet
both the Condorcet and Consistency criteria. It's pretty obvious which
I choose. I think I describe why in the Average Ratings section of my
Ranked Pairs site. Anyway, I'll sketch the outline of a proof that no
method can pass Condorcet and Consistency.
40 A B C
35 B C A
25 C A B
Now, the method has to select one of these 3 candidates. Let's say it
picks A. Notice that C beats A pairwise. Add votes like this,
51 A C B -- group 1 winner , candidate beating group 1 winner, other
49 C A B -- candidate beating group 1 winner, group 1 winner, other
When you add them together, you'll find that C is the CW. However, for
consistency A must win. To complete the proof, you have to form a new
group 2 for each other possible winner of group 1, which are B and C.
By proving that the winner can't be any of them, you know that Condorcet
and Consistency are in contradiction.
IRV also fails consistency in this example.