<![CDATA[Condorcet Loser / Pairwise comparison / Preference Matrix]]> A,B,C,D,E 8: 3,0,1,1,1 5: 0,4,0,0,0 2: 0,1,3,2,1 1: 0,2,5,4,3

Who is the Condorcet Looser (CL) and Condorect Winner (CW)?
How to build the Preference Matrix (find CW and CL) - Pairwise comparison.

Assume that election officials tossed a coin ahead of time - and the following Candidates were selected as tie breakers:

1. A
2. B
3. C
4. D
5. E

Tiebreaker candidates are selected from the above list, preferring candidates with lower numbers.

We have 4 seats to win (with 5 candidates) - who should win this election?

]]>http://www.votingtheory.org/forum/topic/404/condorcet-loser-pairwise-comparison-preference-matrixRSS for NodeSun, 19 Apr 2026 11:09:39 GMTThu, 10 Aug 2023 17:39:56 GMT60<![CDATA[Reply to Condorcet Loser / Pairwise comparison / Preference Matrix on Sat, 12 Aug 2023 00:24:07 GMT]]>@masiarek A is the only Nash equilibrium, while not a Condorcet winner it is in my opinion the only sensible winner of the election (being the only weak Condorcet winner). Every other candidate has another candidate that is preferred over them by a majority.

Once A is elected, if they are removed from the running for second place, then B becomes a Condorcet loser and C becomes a Condorcet winner. If you continue with this process, either maximizing the rank of a remaining Condorcet winner and minimizing the rank of a remaining Condorcet loser, you arrive at the rank A,C,D,E,B. This process fails or is nondeterministic when there is a strong Condorcet cycle or more than one weak Condorcet winner.

That ranking might not be your favorite, but it’s the most stable in terms of game theory. It’s strange that some voters didn’t use the full range of scores but in a rank order system that doesn’t matter.

Another thing that one should keep in mind is that voting is for large populations. If the population is small, making social agreements is definitely way better for everyone!

]]>http://www.votingtheory.org/forum/post/2772http://www.votingtheory.org/forum/post/2772Sat, 12 Aug 2023 00:24:07 GMT<![CDATA[Reply to Condorcet Loser / Pairwise comparison / Preference Matrix on Thu, 10 Aug 2023 19:04:38 GMT]]>@masiarek There isn't a Condorcet winner or loser. A pairwise beats all candidates other than B and ties 8:8 with B. B loses pairwise against all candidates except A (they tie).

With four to elect, I'd say A, B, C, D is the obvious choice, with E the weakest candidate. E is Pareto dominated by C and D anyway, so C and D must be elected by any sensible measure.

E does pairwise beat B (the only thing they have going for them), but B looks generally stronger and would have to be elected if this is a PR election.

]]>http://www.votingtheory.org/forum/post/2769http://www.votingtheory.org/forum/post/2769Thu, 10 Aug 2023 19:04:38 GMT