Forest Simmons wrote:
And the obvious PAV solution would be 60 A's and 40 C's. Membership
in the cycle didn't guarantee B any positive probability.
It is interesting to me that in my other example
34 ABC
33 BCA
33 CAB
some respondents thought it was obvious that A, B, and C should have
nearly
From: Forest Simmons [EMAIL PROTECTED]
To: election-methods-electorama.com@electorama.com
Subject: non-determinism and PR.
From: Bart Ingles [EMAIL PROTECTED]
Subject: Re: [EM] non-deterministic methods
...
Wouldn't a random cycle-breaker provide strong incentive for a sure
loser in a cycle-free
