The Condorcet Win Criterion (Voting Theory)

An explication of the Condorcet Win Criterion (CWC), which requires a voting system to pick the Condorcet Winner if there is one.

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Information for this video gathered from The Stanford Encyclopedia of Philosophy, The Internet Encyclopedia of Philosophy, The Cambridge Dictionary of Philosophy, The Oxford Dictionary of Philosophy and more!

Information for this video gathered from The Stanford Encyclopedia of Philosophy, The Internet Encyclopedia of Philosophy, The Cambridge Dictionary of Philosophy, The Oxford Dictionary of Philosophy and more!

[tonymaloley]

I'm a proponent of Condorcet's method, and it took even me a while to interpret what looks like unfairness. Evidence that the Condorcet win criterion is not very intuitive.

The example of M winning only with Condorcet's method actually shows how the other methods are flawed. It is the interference of the less popular candidates, L and O, that creates the illusion that N has the most public support, when the truth is that M is preferred over N.

Creating a hybrid, consisting of Condorcet's and other methods, can clear up confusion. For example:
1. Eliminate one candidate having the fewest 1st ranks.
2. Eliminate one candidate having the fewest 1st and 2nd ranks combined.
3. Condorcet comparison of remaining candidates.

Or as a practical election method (Remember that most elections have far more than 11 voters):
1. Eliminate up to half the field by 1st ranks, but keep at least 3.
2. Condorcet comparison
3. Paradox or ties resolved by eliminating one candidate having the fewest 1st+2nd ranks.

[canoq]

I'm having trouble finding this criterion trivially desirable. Imagine each member of the society gave the candidates a number from 0 to 10 showing how much they like them (higher better) and then rank the candidates according these numbers. Assume a society of 3: ABC. A and B give candidate D a 2 and candidate E a 5 while C gives D a 5 and E a 2. Assume E wins. if we then change the valuation to A and B: 0 and C: 10 and 0 (for D, E respectively), if the system passes the Condorcet Criterion, E wins. but A and B clearly don't like their options too much, while C does. This criterion, in other words, cares about who you prefer, without worrying how much more you prefer him/her over your other option. I don't know if this is desirable. All least not trivially. Of course, maximizing happiness has its own problems, as it gives incentive to lie, and would not even be eligible as a system, as it does not give a list of candidates as input, but rather their valuations for each voter. But the question still stands, I think.

[russellwhisenant]

Cooms' rule Would choose M. in the example given. So this is not a counter example for Cooms' rule.

[Zyle]

At 3:23, the Borda Count for O is 14 not 13

[danielcappell]

the CWC is intuitively desirable but causes unworkable problems. Maybe this can be explained by Temkin's ideas on nontransitivty. Temkin says that a surviving option of many 1v1 contests shouldn't necessarily be the same winner as a single contest of many options. For instance, when you go to the dealership to buy a car you might be tempted to buy a fancier model than you originally intended. Then you may be tempted to get the slightly fancier model. Etc. Soon enough you'll be looking at a very expensive car that you would never have considered when you came in the door. Relatedly, you might not be entirely irrational to prefer candidate x to y, but then rank y over x when they are part of a larger set of options (e.g., you might make the ranking of "z>y>x>w" while you also prefer x over y when they are against each other one-on-one). Perhaps the CWC's concern with the winner being able to win each and every 1 on 1 contest is extraneous.

Temkin talks about some of his ideas on nontransitivity here:

[nienke]

If a>b, b>c, and c>a there's no Condorcet winner, and if a>b, b>c, and a>c, then a is the winner, but what is it considered when a=b, b>c, a>c (e.g. D: a>b>c, E: a>c>b, F: b>a>c, G: b>c>a), is that it then still considered that there is no Condorcet winner, or would we call it a tie between a and b in that case (or alternatively both a and b are winners)?

[nienke]

1:14 is there any evidence that there often is no Condorcet winner? Is this based on analysis of hypotheticals, or of real life scenarios?
Considering that in most elections candidates can generally be (roughly) ranked on an axis voters on the extremes will tend to choose their own extreme, then the centre, then the other extreme, and the centre will split some leaning one way and some another but centre first, which in practice often means that the centre/median (within the ideological space) tends to be the condorcet winner.
Only if the centre might be ruling long enough whilst being dissatisfying to voters, might people start to prefer anything non-establishment over the established centre, and thus choose extreme1>extreme2>centre, and then only if enough people have a preference like that, but also still enough stick with traditional preferences, would there reasonably arise a smith set rather than a single Condorcet winner.

Intuitively, I'd therefore expect that in real life election scenarios, having no condorcet winner is quite unlikely.

[QuintenCoret]

Either I'm misunderstanding IRV or you've made a mistake. The first round is done correctly: L has one vote, M three, N four, and O three. This means L is indeed eliminated. His vote still counts, though, and therefore goes to M. Second round results are therefore: M four, N four, O three. This eliminates O, whose votes all go to M, leaving him with 7 votes and N with 3 votes. There are indeed scenarios where IRV does not pick the concordet winner, but this is not one of them.

[tinakrats]

Why would Ranked Vote eliminate M and O at the same time (2:36)? Wouldn't tie-breaking rules dictate considering O to be weaker in 1st 2 rounds (or most commonly chose as the last choice)?

[shmendusel]

actually, dictatorship would fulfill the Condorcet win criterion since in a one on one race D picks N over M.