At 2:02 PM -0500 1/6/03, Stephane Rouillon wrote:
A very interesting case. thank you.Please answer on the EM list, I cannot send anything there from the job...I tried with random numbers and obtained... a tie.
I find this case quite odd and would appreciate comments from people who have worked with Condorcet longer.
Input matrix
0 23 25 65
24 0 11 87
44 55 0 22
33 11 44 0
Initial defeats matrix:
0 0 0 65
24 0 0 87
44 55 0 0
0 0 44 0
Since SSD is equivalent to Beatpath and is also easier to work out by hand, here are my notes:
Details can be found at: http://electionmethods.org/CondorcetEx.htm
The ordered list of defeats for your matrix would be:
( A = 0, B = 1, C = 2, D = 3)
<winner column> / <defeat column> <magnitude of defeat>
B/D 87
A/D 65
C/B 55
C/A 44
D/C 44
B/A 24
By looking in the defeat column, we can see each option listed, so we eliminate B/A, which was the weakest defeat.
Each option is still listed in the defeat column, but there is an odd case since the next two defeats have a magnitude of 44. What should happen here? The algorithm throws them both out.
After these eliminations, we are left with:
B/D 87
A/D 65
C/B 55
Since both A and C are not defeated by anyone, the algorithm reports a tie.
However, we can look back at the original ordered list of defeats and see that C defeated A before...why isn't it the right thing to do to report C as the winner?
As with any method of voting, ties can and should be possible, I would just like to understand better why a tie must be reported in this case.
----
For more information about this list (subscribe, unsubscribe, FAQ, etc), please see http://www.eskimo.com/~robla/em
