[EM] Re: lotteries

[Forest Simmons]

First I have a typo correction in my example (inspired by Bart) showing 
that ordinal ballot methods satisfying neutrality and the Condorcet 
Criterion are manipulable, even when non-deterministic.

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 <simmonfo at up.edu>
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" <jarmyta at antioch-college.edu>
Subject: Re: [EM] There is always a Condorcet Winner! (among all
 	lotteries of 	candidates :
To: "Jobst Heitzig" <heitzig-j at web.de>,
 	election-methods-electorama.com at 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