I gave the last faction as z C>B>A, when I meant z C>A>B. The entire corrected tableaux is
x A>C>B y B>C>A z C>A>B .
From: Forest Simmons <[EMAIL PROTECTED]> Subject: [EM] Re: lotteries
<snip>
However, even though it is non-deterministic, and highly manipulation resistant, it is not totally manipulation free:
(following Bart's critique on non-determinism...)
Suppose that there are three factions with sincere preferences
x A>C>B y B>C>A z C>B>A ,
Now for James' message
Date: Wed, 05 Jan 2005 21:17:42 -0500
From: "James Green-Armytage" <[EMAIL PROTECTED]>
Subject: Re: [EM] There is always a Condorcet Winner! (among all
lotteries of candidates :
To: "Jobst Heitzig" <[EMAIL PROTECTED]>,
election-methods-electorama.com@electorama.com<James wrote>
Dear Jobst,
Yes, it does sound like an intriguing idea, although non-deterministic
methods have never been an area of expertise for me.
In its present state, the proposal is still a little bit too abstract
for
me to grasp; as usual, it's easier for me to understand a tally method
once I am given an example of it in practice. So, could you please find
the Condorcet lottery for the following group of preferences, and show me
how you did it?35: A>B>C>D 25: B>C>A>D 30: C>A>B>D 10: D>C>A>B
thanks! James Green-Armytage
<Forest replies>
There are only three uncovered candidates here A, B, and C, and they form a cycle, so they also form the set W, and the winning "lottery" is the distribution p(A)=p(B)=p(C)=1/3 .
It's easy to get this once you realize that Lottery is just Spruced Up Random Candidate.
Forest ---- Election-methods mailing list - see http://electorama.com/em for list info
