[EM] My Favorite Deterministic Condorcet Efficient Method: TACC

[fsimmons at pcc.edu]

Chris wrote ...

> 31: A>B
> 32: B>C
> 37: C>A
>
> Approvals: B63, A68, C69. A>B>C>A.
>
> TACC elects A, but C is positionally the dominant candidate and
> pairwise beats A.
>
> For a Condorcet method with pretension to mathematical elegance,
> I don't
> see how that
> can be justified.
>
> Chris Benham
>
> PS: Could someone please refresh our memories: What is the
> "Banks Set"?

Forest Replies:

As you know C is the DMC winner, and would be a slightly better winner, given
that the ballots are sincere.  But DMC is not as burial resistant and truncation
resistant as TACC.  

It is interesting that DMC and TACC have opposite rules for which of the top two
approval members of the top cycle (of three) wins.  DMC awards the win to the
one (of these two) that beats the other.  TACC awards the win to the one that is
beaten by the other.

DMC is slightly better for elections where no voter would think of burial or
truncation.  But in the real world, I think the small sacrifice in goodness of
the result for sincere voting is worth the greatly increased
truncation and burial resistance.

I've come around to the belief that most Condorcet cycles in ordinary elections
are artificial, so chances are that this cycle was created from the burial of B
by the C faction.  Giving C the win only rewards this manipulation.  I think
that within the constraints of deterministic methods, TACC's resistance to
burial and truncation is more valuable than DMC's slightly better results for
sincere ballots.

With a stochastic analog of TACC that we could call RBOCC for Random Ballot
Order Chain Climbing, we get a better result for this scenario than either TACC
or DMC:

Prob(A wins)=37%
Prob(B wins)=31%
Prob(C wins)=32%

Since A has a higher winning probability than C, the C faction will regret
burying B provided that (to them) B was higher than about 46% (fraction
32/(32+37)) of the way between A and C in utility.  In other words if they
valued B more than a lottery in which there was only a 46 percent probability of
getting C, but a 54 percent probability of getting A, then they cannot
rationally bury B.

Banks:

An alternative B is in the Banks set iff there exists some subset S of
alternatives such that (1) S is linearly ordered by the pairwise "beats"
relation, and (2) no alternative outside of S pairwise beats all of the members
of S, and (3) B pairwise beats every member of S except itself.

In other words, B is the max member of a maximally extended chain.

It follows that every Banks alternative is uncovered, but there are cases where
some uncovered alternatives are not Banks alternatives.  So the category Banks
is more exclusive than the category Uncovered.  In particualr, if an alternative
A is merely uncovered, but not a member of Banks, then it would be irrational
for a player in the following Rivest Game
to bet on A:

Two players choose one alternative each.  Then a random ballot is drawn.  The
player whose choice is preferred over the other player's choice on the random
ballot must give a dollar to the other player.

Forest