Is a Condorcet Winner really most-preferred by the majority?

This is another “How high is up?” question. Let’s start by seeing if we can agree that a first-past-the-post election that has a majority winner does, in fact, reflect the voters’ true majority preference?

I think that we could then agree that a mere plurality result, where the preference of some large minority prevails over the majority, is not the true preference of the majority. If we’re looking for a majority-decision such a plurality result does not serve at all.

With FPTP, plurality wins only happen when we have more than two choices. So if we were to take each pair of candidates and hold an election between them, between them we will get either a tie, or one candidate would be the clear majority choice.

If we do this for all the possible pairings, it might turn out that there is one candidate who wins a majority in every such pair-by-pair election. The majorities in each case might involve slightly different voters, but still, in each case, we have a majority decision that between the given two candidates is either a tie, or one candidate is the clear preference over the other.

In such a circumstance, where there is a candidate who has won all the one-to-one elections, it would be inconceivable that some other candidate who must, therefore, have lost at least one, and might well not have won any, could in any way be considered the majority choice, or that such candidate could, in fact, win the overall election.

Yet, with first-past-the-post elections involving more than two candidates, where there is not a clear majority win, it is entirely possible that the first-past-the-post plurality winner loses some or even all of the pairwise elections.

With bare first past the post, it is entirely possible that the Condorcet loser, the candidate who loses each and every pairwise election, the candidate who is defeated on a majority basis by everyone in such pairwise match-ups, can, in fact, win the election as a whole.

It’s hard to argue that the candidate who, head to head, one-to-one, beats every other candidate, isn’t the choice on the whole most preferred by the voters. Such a candidate, who would beat each and every other candidate in head-to-head elections, is, by definition, the Condorcet winner.

It’s also similarly hard to argue that a Condorcet loser — losing every pairwise match-up — is not the candidate least-preferred by the voters.

The thesis here, then, if we accept that a first-past-the-post majority win is the correct majority decision in a given pairing, we must then accept that a Condorcet winner, if such a candidate can be determined, “should” be considered the candidate most preferred by the voters, and that if a Condorcet winner can be determined, that candidate should, in fact, win the election as a whole.

We must, similarly, also accept that a Condorcet loser, if such a candidate can be determined, “should” be considered least preferred by the voters and should NOT win the election as a whole.

In cases where there is a Condorcet winner, one Condorcet method is as good as any other to determine this candidate. In cases where there is no Condorcet winner, however, when no candidate wins every pairwise election, there is more room for ambiguity. This is where the various Condorcet methods differ.

Some methods, as noted earlier, resolve these non-Condorcet-winner decisions by reverting to a straight first past the post, or an IRV decision, or some other approach. Why not? – as long as the given secondary approach is nonetheless fair and attempts to make a reasonable determination of the majority will.

The Ranked-Pairs method, in such cases, honours the results of the pairwise elections as much as possible, with a predilection to keeping those results indicating stronger voter preferences over cases that indicate weaker preferences – in cases where they cannot both be retained. In my view, this approach best retains fidelity with the voters’ true collective intent.

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