[EM] More still on utilitarian methods

[Kristofer Munsterhjelm]

Suppose we have a two-candidate election. Assuming no second order 
effects (e.g. voters deliberately introducing randomness to keep 
candidates on their toes), every personal utility-maximizing voter would 
submit a preference of type:

(1, 0): the preferred lottery is one where A wins with certainty
(0, 1): prefers B to win with certainty
(1/2, 1/2): indifferent to the outcome (though such a voter probably 
wouldn't show up)

Then considering my categorization of cardinal methods: if we could 
somehow extract the utilities directly, and were okay with dictatorship 
of the strongest feelings, then each voter would rate the candidate 
according to utility on some common scale, and then the candidate who 
maximizes utility wins. But if that's not possible or desirable, and 
we're limited to affine scaling, then:

A method that's of type two (enforcing OMOV all the time at the cost of 
utility) would reduce to majority rule here (after the indifferent 
voters are removed).

A method that's of type three is like Range: in a two-candidate 
(Mushroom, Pepperoni) election, the meat-eating majority chooses to give 
up some power by rating Mushroom not at 0 but at, say, 0.8.

This, I *think* requires more than just lottery information; it requires 
a sort of common standard so that the majority knows that 1 is very 
tasty and 0.8 is sufficiently tasty[1]. In any case, strategic voters 
would not give away some voting power this way unless they're strategic 
in a much broader sense.

We could generalize Range to any p-norm, call it Lp-cumulative voting. 
In the "Range-ish" version, there is a threshold for max p-norm, and any 
vote that exceeds this norm is invalid. This is a type three method. 
Then there's the automatically renormalized version, where the voter's 
ballot (whatever its type) is normalized so that its p-norm is exactly 
the threshold. That's a type two.

For Range, the type-three method is Range itself. Since the norm for 
Range is the max norm, the automatically normalizing version adjusts the 
ratings so they exactly fit the scale (e.g. a 0 - 10 Range variant would 
have at least one candidate rated 10). Just how it does that isn't given 
- it could clamp values or scale linearly. If it's DSV, it would scale 
in such a way as to maximize the voter's personal utility.

As for Condorcet methods, as they reduce to majority in the 
two-candidate case, most cardinal Condorcet variants would be type two. 
It would be possible to make a type three Condorcet method by saying, 
suppose that a voter rates A at 1 and B at 0.8, then that voter only 
adds 0.2 of a vote to the pairwise contest of A vs B. But that again 
requires more than just lottery information.

Perhaps, then, three candidates is the minimum where type two cardinal 
methods start to differ from ordinal ones; at least if based on a 
Condorcet logic. But how? I would imagine there's a tension between 
strategy resistance (possibly modeled by Nash equilibria) and honest 
results. Suppose honesty. Then what lottery information is available? 
Each voter who is not indefferent to the outcome has some candidate 
where he's indifferent between electing that candidate with certainty, 
and a nontrivial lottery of the other two. Then what?

I'm kinda just tinkering, but one possible way is to use Lp-cumulative 
voting restricted to this tuple. Suppose we want to reconstruct a Range 
ballot for the three candidates, and suppose a particular voter's 
preference is A>B>C so that B is the candidate whose certain election is 
equivalent to a lottery of A and C. By normalization, we can set the 
ratig of A to 1 and that of C to 0. Then B's rating is p if he's 
indifferent between the lotteries (p, 0, 1-p), and (0, 1, 0). Then for 
any given three-tuple {A,B,C}, the winner is the Range winner of the 
thus normalized ballots.

If there exists a candidate W so that he wins any three-tuple he's in, 
then he's the "3-Condorcet" winner and wins outright. (Similar notions 
would be possible for the Smith set.) Honesty might not be the best 
policy here, but it would mitigate >3-candidate Burr dilemmas. It 
wouldn't fix two-candidate ones, though... so more thinking may be 
needed.[2]

Still, I think I've at least determined a bit more about the shape of 
this particular elephant :-)

[1] The drawback of such a common standard, even if the voters are 
completely honest, is that later experiences may recalibrate it. E.g. 
suppose someone has only eaten mass-produced frozen pizzas and rate them 
1 because they're the best he knows; then later he visits Italy and gets 
a proper pizza. Now he'll have to readjust what 1 means. The voters may 
even be confused about which standard others are using, which is the 
problem of incommensurability.

[2] I'm thinking that the way to solve three-candidate Burr dilemmas 
would have to rely on a plain Condorcet back-up - i.e. that the STAR 
proponents have got at least that right. But then the {A,B,C} outcome 
can't simply be the Range winner of the three with reconstructed 
ballots. But we wouldn't want to simply reduce entirely into normal 
ordinal Condorcet either. What's the nature of this tradeoff?