Talk:Condorcet loser criterion

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"IRV fails CLC"

Consider the following sequence of ballots:






D is the (unique!) Condorcet loser (3-2 in all races) but wins by IRV (a,b, and c are eliminated in the first round since they have the fewest first place votes).

How is this not a proof that IRV **fails** to satisfy CLC? Aweyhau (talk) 03:49, 2 April 2010 (UTC)

Instant-runoff voting always eliminates one candidate at a time. Markus Schulze 08:47, 2 April 2010 (UTC)

First past the post:

FPTP only 'fails' this criterion because is does not have formal ranking of candidates. It is therefore subjective to say that therefore FPTP fails this criterion, as it would be more accurate to say that FPTP does not pass due to the criterions inapplicability. --jrleighton 04:35, 6 January 2006 (UTC)

FPTP is equivalent to a rank ballot method in which equal ranking is disallowed and only the top preference of each ballot is regarded, so that the candidate is elected who is the first preference on the greatest number of ballots. This is how e.g. Woodall analyzes it. I don't see what can be gained by analyzing it differently. KVenzke 23:50, 7 January 2006 (UTC)


I reworked this article significantly, since it seemed quickly written and incomplete.

I left in two added sections on specific methods, although I'm too impatient to actually read them in enough detail to see if they make sense!

I wondered about cases where there was no condorcet loser, but a sort of bottom smithset condorcet loser, a set of candidates who are mutually defeatable, and each will lose to ANY candidate outside the set. I believe such an expanded definition would exclude IRV (and all runoffs) from passing (since the final two might have been in this bottom set), and I don't know whether Borda would pass.

To me mostly this criterion seems cute more than useful EXCEPT to judge BAD Condorcet method rules for breaking pairwise cycles. Runoffs pass merely for being only able to elect a Condorcet runnerup loser.

Tom Ruen (talk) 00:49, 22 December 2007 (UTC)