From: Brian Olson <[EMAIL PROTECTED]> Subject: Re: [EM] How to break this tie?
On Feb 10, 2005, at 8:06 AM, Forest Simmons wrote:
A1>A2>A3>B>C B>C>A2>A3>A1 C>A3>A1>A2>B
I don't suppose it would help to know that just about every system I've implemented answers "C", eh?
The A's form a clone set, which collapsed gives
ABC BCA CAB
So any method that gives C more probability than B is clone dependent.
To be clone independent the method would have to give both B and C probability 1/3 each.
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