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Condorcet method is an election system where voters rate their candidates in order of preference. As an example, for candidates A B and C, a valid vote can be [A > B > C], [C > B > A], [B > A > C] etc. If no candidate obtains the majority, the candidate who received the lowest number of votes is eliminated, and the 2nd choices of the voters that voted for her/him are redistributed among the existing candidates - as if it were first choices.

I would like to verify what is the best method to elect 2 candidates using the Condorcet method.

My preferred option would be asking the voters to provide 2 first choices, 1 second choice and 1 third choice. Then apply the Condorcet method to elect the two candidates.

In this case, my understanding is that the majority required to be elected should be determined by the following

[(valid votes cast\(seats to fill+1)] + 1

where the quota is an integer (Droop Quota).

This method could be assimiled at a Single Transferable vote where if a candidate has more votes than the quota, surplus votes doesn't need to be transferred to other candidates. In that case we would only elect the candidates that reached the majority required (Droop Quota) and eliminate the candidate that received the lowest number of first choices, redistributing the 2nd choices of the voters that voted for her/him. This operation should be repeated until a second candidate reach the majority, or until there is only two candidates left.

Would this method always work?

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  • Do you want proportional representation, where each candidate is a great representative of a faction of the population, or do you want to elect the two candidates who are the best representatives of the population as a whole?
    – endolith
    Jul 20, 2022 at 19:32
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    I can't find an English dictionary definition of the verb 'assimile', is it a technical term? Jul 22, 2022 at 15:34
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    //Condorcet method is an election system where voters rate their candidates in order of preference. As an example, for candidates A B and C, a valid vote can be [A > B > C], [C > B > A], [B > A > C] etc. If no candidate obtains the majority, the candidate who received the lowest number of votes is eliminated, and the 2nd choices of the voters that voted for her/him are redistributed among the existing candidates - as if it were first choices.// ----- That's not Condorcet, that's Hare. It's IRV, not Condorcet-consistent. Nov 25, 2022 at 20:13

2 Answers 2

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Would this method always work?

To do what? Comply with the Condorcet criterion? That's easy.

No

This method is not compliant with the Condorcet criterion. In fact, no method that drops people based on not having enough top votes is going to comply with the Condorcet criterion.

If no candidate obtains the majority, the candidate who received the lowest number of votes is eliminated, and the 2nd choices of the voters that voted for her/him are redistributed among the existing candidates - as if it were first choices.

This is not a description of "the" Condorcet method. There are a collection of Condorcet methods, but none of them work like this. What you are describing here sounds more like an Instant Runoff Voting (IRV) variant. Note that both Condorcet methods and IRV are ranked methods. The ballot looks the same. The scoring is different.

Example ballot totals:

42 A > B > C > D
26 B > C > D > A
15 C > D > B > A
17 D > C > B > A

IRV

In IRV, C would be eliminated as having the fewest first place votes (15). Leaving

42 A > B > D
26 B > D > A
15 D > B > A
17 D > B > A

But the last two are the same, so

32 D > B > A (15 + 17)

Now B has the fewest first place votes and is eliminated.

42 A > D
26 D > A
32 D > A

So D beats A 58 to 42.

Condorcet methods

In the Condorcet methods, we first rewrite it into pairwise comparisons:

42 A > D
26 D > A
15 D > A
17 D > A

which simplifies to 58 to 42, D beats A.

42 + 26 B > D
15 + 17 D > B

So B beats D, 58 to 32.

42 + 26 + 15 C > D
17 D > C

So C beats D 83 to 17.

42 + 26 B > C
15 + 17 C > B

B beats C 68 to 32.

42 A > C
26 + 15 + 17 C > A

C beats A 58 to 42.

42 A > B
26 + 15 + 17 B > A

Overall, B beat everyone. B wins. All Condorcet methods work like that.

Multiple winners

Now this just picks a first place winner. If you want a second place winner in a Condorcet-compliant method, you could then redo the election without the first place winner.

42 A > C > D
26 C > D > A
15 C > D > A
17 D > C > A

C and D both beat A by 58 to 42. C beats D by 83 to 17. C beats everyone, so C is the second place winner in a Condorcet-compliant method.

In an IRV variant, B would win, since B was the next to last to be eliminated.

So B is in the top two in both. C is also in the top two for Condorcet and D for IRV.

Real methods

I just made this up for the purpose of this post. But there are real Condorcet-compliant methods for multiple winners. Look at Schulze STV or CPO-STV.

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My guess for how to apply Condorcet reasoning to a multi-winner (M is the number of seats, N is the number of candidates) election would be to pick the top M candidates from the Condorcet-ordered list of preferences.

The top candidate in that ordered list is simply the Condorcet winner, what I would call the "Consistent Majority Candidate".

Then if you remove that top candidate from that ordered list from the set of candidates and run a Condorcet-consistent election with the remaining N-1 candidates, the Condorcet winner of the remaining set would be the second to the top in the Condorcet-ordered list of preferences.

If you remove the top two candidates and do this again, you will get the third most preferred candidate in the ordered list.

Etc.

No guarantee that this would be Proportional Representation, for which is why I only advocate Condorcet consistency for single-winner elections.

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