Can Condorcet methods pass Favorite Betrayal while remaining Condorcet-compliant?

One idea is to treat equal-rankings as favorably as possible so that in a Condorcet cycle, your vote would be treated as favoring whichever of the candidates you equally ranked in whatever way is necessary to ensure a Condorcet winner from the equal-ranked set of candidates. As an example:

#voters their vote
2 A>B>C
3 C>A>B
4 C=B>A
2 A>B>C

Defeats are A>B by 7:4, B>C by 4:3, and C>A by 7:4. There is an A>B>C>A cycle in which B>C is the weakest defeat (measured by either winning votes or by margins), so that C is elected .

Notice that the two A>B>C voters shown in blue on the bottom line can turn the “lesser evil” B into the Condorcet Winner by “betraying” their favorite “third party” candidate A and voting B>A>C or B>A=C or B>C>A.

Here, the A>B>C voters could vote A=B>C, and in the Condorcet cycle, the vote-counters would probably start by testing flipping their votes to B>A>C (because that would be more likely to change the pairwise matchup between A and B than voting A>B>C, which only increases A’s victory margin over B), and you’d find that B is now the Condorcet winner instead of C.

Improved Condorcet Approval, while passing Favorite Betrayal, isn’t fully Condorcet-compliant.

Can Condorcet methods pass Favorite Betrayal while remaining Condorcet-compliant?

Unfortunitly not.

Proof without equal rankings permitted:


Proofs with equal rankings permitted:


Such a method would technically not be a Condorcet method. This is because whenever a voter ranks 2 candidates equally the Condorcet criterion assumes that that voter would abstain in a head to head runoff between those two candidates.

If you use the definition of the Condorcet criterion that you are implying: for each opponent, a candidate has to have more then 50% voters ranking them higher then that opponent, not just more then the number of voters ranking their opponent higher then you (In the past I have called this criterion the weak Condorcet criterion but perhaps there is a better name for it), then yes, this is possible, and there are many favorite betrayal compliant methods that pass this modification of the Condorcet criterion here: (though I don’t think any of these methods also pass the anti-favorite betrayal criterion). I wonder if it’s possible to create such a method that passes this weaker modification of the Condorcet criterion, along with FBC, AFBC, reversal symmetry, and IIB (IIB = adding a ballot in which all the candidates were rated/ranked equally shouldn’t be able to change the results).

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Is the trick to beating this argument that essentially, the voter won’t know who else is Favorite Betraying, or in favor of whom? Essentially, enough other voters may Favorite Betray in favor of your favorite that if you Favorite Betray in favor of their candidate, and enough people do this sort of FB-swapping, then you will end up back at the symmetrical tie probability of any of the candidates winning?

I’m not really that keen on Venzke’s proof where equal rankings are allowed. If A and B are my co-favourites, is it really a favourite betrayal to have to rank one over the other? I mean, you could just define it that way, but it would be a pretty weak failure, and not something I would hold against Condorcet methods.

Also if you define it that way, then any method where equal ranks are not permitted would automatically fail with no need to appeal to any specific scenarios. Because if A and B are my co-favourites and I have to rank one above the other, it’s failed right away.

There was a discussion of this on the old Google Group.!msg/electionscience/RK7lLxiiGAU/r_-hto_gAwAJ

But basically unless it’s advantageous to rank your favourite or co-favourite below a candidate you like less, then while it might be a criterion failure by some definition, it’s a benign failure that I don’t think counts against the method.


It’s not a huge failure if the voter knows which of their co-favorites to rank above or below each other, but it seems unfair if they’ve got no idea, and end up losing because they guessed wrong or didn’t try to vote strategically or something like that. And considering most of these Favorite Betrayal-provoking situations are bizarre Condorcet cycles and whatnot, it might not be that easy to vote to ensure one of the co-favorites wins.

Isn’t his proof (the example I quote below) about voters who prefer A>B being forced to vote B>A? They aren’t co-favorites there.

It seems possible to do some form of algorithmic Asset Voting where the negotiators are allowed to change their preferences solely within the Smith Set (when it benefits them) in order to limit Condorcet’s Favorite Betrayal-incentivizing scenarios (almost?) only to tied situations, which seem too rare to be worth strategizing for.


Defeats are A>B by 7:4, B>C by 4:3, and C>A by 7:4. There is an A>B>C>A cycle in which B>C is the weakest defeat (measured by either winning votes or by margins), so that C is elected .

Notice that the two A>B>C voters shown in blue on the bottom line can turn the “lesser evil” B into the Condorcet Winner by “betraying” their favorite “third party” candidate A and voting B>A>C

If they were algorithmically negotiating, since A and B are in the Smith Set, the 2 A>B>C voters would be allowed to change their vote to B>A>C (or even, all 4 A>B>C voters could change it to A=B>C) and elect B. Note that this requires there to be a cycle resolution method used to decide who is leading in the Smith Set, so that the negotiators can strategize to pick someone they prefer.

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OK, but that’s not the bit I was reading on here

Now suppose the A=B>C voters change their vote to A>B>C, thus betraying their true co-favorite B. This turns A into the Condorcet winner, who would have to win with 100% probability since it is a Condorcet method. Votes A=B>C and A=B=C which do not betray B, cause either the original 3-way-tie scenario, or cause C to be the Condorcet winner which is even worse from this voter’s point of view. So: betrayal is strategically forced. Q.E.D.

It was probably from one of the other links I didn’t properly check out.

This one:

Smith//Score, while still vulnerable to Favorite Betrayal, seems to mitigate it a bit in this example. (The same reasoning that appears below can be applied to Condorcet//Score for this example, since every candidate here is in the Smith Set, and thus the Score winner is the same as the Smith//Score winner)

Supposing every voter scores their co-favorites a 5, their 2nd choices a 4, and their 3rd choices a 0, the scores are:
10 + 12 + 10 = 32 A
8 + 20 + 8 = 36 B
15 + 20 = 35 C

So that B is (very narrowly) the highest-scored candidate in the Smith Set and wins. A voters can even strategically maximize their support for B, in which case, B has 40 points (though B would anyways win in a Condorcet method if all A voters did this, since B would then be the CW). Not enough C voters have incentive and ability to strategically change the result, since the ones that ranked C>B already gave B a 0, and the ones who ranked C=B consider B a co-favorite.

FB appears to still be the optimal strategy for Smith//Score here though, in the sense that it better guarantees that B wins than any other strategy.

To expand, if a voter Favorite Betrays to elect their lesser evil, but some other voter does the same in favor of their lesser evil, who is the 1st voter’s favorite, then FB no longer makes sense, since they could’ve gotten their favorite by changing nothing. So this FB proof seems limited solely to cases where you can be confident more voters of a certain type will FB than other types such that it benefits you, rather than harms you.

I’d also mention that using Algorithmic Asset to flip preferences even outside the Smith Set seems to reduce Favorite Betrayal incentive here even further, because even if a voter does FB and makes their lesser evil a CW, the method will strategically flip other voters’ preferences to return to a tie.

This seems about like half of the argument that STAR voting encourages honesty: if other voters try to be “strategic”, then there is a good chance that you can just vote honestly and ride on their strategy. (The other half for STAR is that voting strategically is just as likely to backfire as work. Do you think this is true of that Condorcet method?)

If you mean Algorithmic Asset, I’d conjecture maybe. It basically does strategy on behalf of a voter, so if a voter honestly prefers A>B but strategically votes B>A, and the method is in a position to give that voter either one, then their strategic vote backfires. However, AAV can be done with some variations, such as trying to strategically optimize people’s votes even outside of the Smith Set i.e. starting Condorcet cycles in favor of your preferred candidates. This is way more computationally complex, definitely not precinct-summable, and possibly breaks the method, but maybe that makes strategy backfire even more.

Edit: Turns out the idea doesn’t seem to work as is.

I should also note @NoIRV that in Condorcet methods Favorite Betrayal might prevent you from getting your favorite when you could’ve had them, but it generally minimizes the greater evil’s chances. So it seems like a voter more worried about beating the other side would Favorite Betray or at least rank the lesser evil co-1st.

Something sort of amusing about Favorite Betrayal in a Condorcet cycle is that if you have a cycle

1 A>B>C
1 B>C>A
1 C>A>B

and every voter does FB, then you just get

1 B>C>A
1 C>A>B
1 A>B>C

(Actually it should be:
1 B>A>C
1 C>B>A
1 A>C>B

which is a mirror of the original election still a Condorcet cycle. The bigger concern with FB might be if voters started bullet voting the lesser evil or something to maximize the chances of reducing the election to a 2-viable candidate contest, so that no cycle could occur.
I also think that if a voter is unsure of who will win the cycle, or who the lesser evil is that they should be maximally strategically supporting, then FB can actually possibly backfire badly. For example, if your favorite actually is the CW, maybe by thinking you’re in a cycle you do FB and actually create a cycle that elects the greater evil.

Since Improved Condorcet Approval is being discussed here, it should be noted one Condorcet failure is when there is a cycle in a clone set. So one member of a clone set could be undefeated according to the ICA criterion if other members of the set were ignored, but if they are included, that candidate is no longer undefeated. Then possibly a non-Smith-candidate could be ICA-undefeated and win.