# [EM] Condorcet Voting

Eric Gorr ericgorr at cox.net
Mon Jan 6 11:36:53 PST 2003

```At 2:02 PM -0500 1/6/03, Stephane Rouillon wrote:
>Please answer on the EM list, I cannot send anything there from the
>job...
>
>I tried with random numbers and obtained... a tie.

A very interesting case. thank you.

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.

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