[EM] Strategy-resistant monotone methods

Kevin Venzke stepjak at yahoo.fr
Sun Feb 7 12:02:46 PST 2016

Hi Kristofer,

Interesting project. A few thoughts/questions on details:
1. I assume your ballots are cast with full, strict rankings.
2. Does the definition of susceptibility include changes caused by compromise strategy?
3. Did you try any issue space models simpler than 4D?
More below...

>De : Kristofer Munsterhjelm <km_elmet at t-online.de>
>À : EM <election-methods at lists.electorama.com> 
>Envoyé le : Dimanche 7 février 2016 8h03
>Objet : [EM] Strategy-resistant monotone methods
>(Furthermore, I've found out that if you want both monotonicity and
>reversal symmetry, it seems you have to pay for it by the method
>becoming a lot more vulnerable to strategy.
>The best nonlinear monotone and reversal symmetric method I found had
>about 65% susceptibility and a lot of ties on IC; now compare that to
>the stats below.
>I don't know why monotonicity and reversal symmetry gives such
>susceptibility to strategy, however. There's of course also the
>possibility that there's a magic monotone reversal-symmetric method out
>there but that my tests haven't found it.)

>If there's a cycle, then for each candidate, we can consider without
>loss of generality that candidate to be A, the one he beats pairwise to
>be B, and the one who beats him pairwise to be C. Then we can let that
>candidate's score be f(ABC, ACB, BAC, BCA, CAB, CBA) for some function
>f, where the variables give the number of voters who voted that way when
>the candidates have been thus relabeled. The candidate with the greatest
>score then wins.

>The best simple linear method I could find was this:
>f = fpA - fpC
>i.e. a candidate's score is the number of first preferences he has,

>minus the number of first preferences for whoever is beating him pairwise.

I think it's clear why this works: the candidate C who beats A doesn't get
"credit" for all C>A votes but only those dedicated to C as first preference.
So, the effect of strategic B>C>A votes (where sincere is B>A>C) is limited
to causing a cycle.

You could say the opposite approach to effecting this is in Condorcet//Approval,
where a strategic lower preference for C gets evaluated (when breaking the
cycle) with equal weight to the first preference for B.

I half think the answer to your reversal symmetry question is in there:
Perhaps reversal symmetry requires too much consideration of the middle rank.

>The linear method is not quite at Condorcet,IRV's level, but the
>nonlinear methods can do much better. The best one I found there was:
>f = A>B * min(C>A, A>B)/fpC
>so in an ABCA cycle:
>A's score: A>B * min(C>A, A>B)/fpC
>B's score: B>C * min(A>B, B>C)/fpA

>C's score: C>A * min(B>C, C>A)/fpB

So you get the strength of your win, times the strength of the weakest contest
you were part of, divided by the FPs of the candidate who beat you. That seems
to work in a similar way, emphasizing first preferences.

>From a mono-raise standpoint it does seem a little weird that if C>A drops,
it appears that that could hurt A. It's true that fpC would very likely drop
at the same rate, but I think this should still produce a slightly decreasing
ratio (i.e. the value of C>A / fpC), which A does not want.


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