[EM] ICC & AFB incompatible (with = allowed in votes, under weakened AFB definition version)

Abd ul-Rahman Lomax abd at lomaxdesign.com
Tue Jan 30 20:59:28 PST 2007

This one keeps nagging at me. I mentioned on the Range list one 
aspect of this problem, which is precedent action based on consequence.

For the umpteenth time, Warren's proof:

>Consider these 3 votes:
>By symmetry axiom 4 this is a perfect 3-way tie.
>However we shall argue under axioms 1-3 that A must win, which is a
>contradiction that establishes the proof.
>If A does not win, then B or C does.
>If B wins, then the C>A>B voter would betray C to vote A>C>B getting
>and then {B,C} is a clone set and hence by axioms 2 and 3 then A must win
>and hence the betrayal worked and hence we get a contradiction with axiom 1.

Now, suppose that the method, in the case of a tie, chooses randomly 
between the three candidates. Indeed, this is required. The tie will 
occur if the voters vote sincerely. They may indeed know how the 
other voters are going to vote (if they vote sincerely). But what 
they don't know is who will win the election.

So they cannot Favorite Betray based on the premise that B wins. They 
can, however, force a resolution of the election to trade a 
speculative maximization in on a certain one.

What this boils down to is that two voters may agree to elect one candidate.

The problem I have is with the statement "If B wins, then the C>A>B 
voter would betray C." But the C voter doesn't know that B will win. 
It's random.

If the C voter could know in advance that B would win, yes, the C 
voter can alter the vote as described.

Warren, I think you have stumbled across the time travel paradox....

Let's look at the expected utilities:

to the C voter, U(A) = U(B) + x and U(C) = U(A) + y, and we must 
assume that x and y are positive utility increments, nonzero.

To simplify things, let U(A) = 0; then U(B) = -x and U(C) = y.

The expected utility of the election without a betrayal is

y/3 - x/3.

This is satisfying because if the candidate rankings represent equal 
preference strengths, the expected utility is zero. I.e., the utility 
of the middle candidate.

Now, C may betray C to favor A and thereby force the election of A. 
The utility of this is zero. Whether or not this is a gain depends on 
how strong the preference is between C and A. If it is a weak 
preference, compared to the preference strength of A>B, then forcing 
the election of A is a gain. If it is a strong preference, then it 
would be a loss.

I think the contradictory assumptions are in

>method is deterministic aside from tiebreaks which (if any) are 
>random equally likely


If B wins, the C voter may alter his vote.

If C alters his vote to favor his second preference, that preference 
wins, and it is not B.

Imagine that I am the engineer on a train. I come to a track 
junction, and I can take track A, track B, or track C. These lead to 
three different destinations, and if the tracks were straight, with 
no complications, C would lead me to Paradise, A would lead me back 
where I started from, and B would lead me off a cliff.

Problem is, there is a roundhouse with a table that rotates and 
switches tracks. I have two reasonable choices. I can stay where I 
am, which is the same as following track A, or I can enter the roundhouse.

Now, what Warren has done is to say, "If I enter the roundhouse and 
the train goes over a cliff, I can instead 'betray my favorite' and 
leave the train where it already is."

"AFB is the criterion that it is never strategically forced
for any voter to rank his true favorite, strictly below topmost."

Above I examined the situation without the contradiction. I.e., the 
voter faces the election, anticipates the tie, wishes to avoid the 
election of B (but without knowing that B will win), and so makes the 
compromise and joins with the A voter to elect A. Whether or not this 
is a gain depends on preference strength, which is not expressed in 
ranked methods.

I'd claim that it is unreasonable to assume differing preference 
strengths from the ranked ballots shown. By allowing the assumption 
of different preference strengths, one creates the situation where it 
appears that Favorite Betrayal is an improvement, as game theory 
indeed would advise. If preference increments are equal, then there 
is no gain from the Favorite Betrayal.

The problem with ranked methods is precisely that they rest (and 
work) with the assumption that preference strengths are equal.

My intuition is that, if preference strengths of the ranks are equal 
and all ranks are full, Range will elect the Condorcet winner. Is this true?

Now, if we allow that preference strengths may vary, we have an 
interesting situation that leads directly to violation of ICC. For if 
we look at what happens as y in the example above becomes 
infinitesimal, A and C become clones. If we had the election only 
between the AC singlet and B, the singlet would win, because the A 
and C vote would not be split. Forcing a rank distinction between A 
and C based on an arbitrarily small preference difference creates an 
ICC violation.

Favorite Betrayal is a problematic criterion. It is not at all clear 
that it is even desirable. We don't like it, because "Betrayal" is 
Bad. But is compromise "betrayal"? The problem is partially resolved 
by allowing equal ranking, because equal ranking can represent 
compromise. "I may have a *personal* favorite C, but for the welfare 
of society and general satisfaction, I will rank A the same as C."

What we *really* don't like is the motivation in a system to rank 
insincerely. It grates against us. There has to be something wrong 
with a system designed to determine, through popular will, a winner, 
if it encourages the voters to essentially lie. How could good 
decisions come out of such a system?

Indeed, Range avoids the whole issue. Strategic considerations in 
Range may lead one to rank another candidate at top rating, along 
with one's favorite (I call this "magnification," and it is not 
insincere), but there is never a strategic reason to rank one's 
favorite *below* another candidate.)

"Magnification" isn't a good word; normalization in Range is really 
analogous to optical magnification, what happens is that the ratings 
hit the stops, so to speak. Between the stops, ratings are (sensibly) 
linear, but more than one candidate can be sitting at each extreme.

More information about the Election-Methods mailing list