[EM] Election-Methods Digest, Vol 103, Issue 1

[Kristofer Munsterhjelm]

On 01/04/2013 04:14 PM, Fred Gohlke wrote:
> Good Morning, Andy
>
> Your response appears to be missing from the list. I'll quote the
> paragraph I'm commenting on:
>
> re: "The voters' grades do matter. If one voter changed his
> grade from D to B, then one more C vote falls down into
> the bottom half of the votes, so his tie-breaking value
> is 67199/155781 instead of 67198/155781, or 43.1368%
> instead of 43.1362%"
>
> The process you describe seems to be a rather complicated way of finding
> the top or bottom half of the votes. The fact that 'B' is higher than
> 'D' and pushes a 'C' vote into the bottom half of the votes is nothing
> more than a Yes/No decision. It helps you decide whether a candidate got
> more than one-half the votes, but is devoid of additional value. A
> simple Yes/No ballot yields precisely that result with no mathematical
> constructs.

As someone thinking MJ could be a good idea, I suppose I should explain 
the reasoning.

MJ uses a grade ballot instead of a yes/no ballot because it encourages 
comparison to a common standard. That is, if the voter is faced with the 
question of assigning each candidate a grade, and the grades are well 
defined (each person knows what a B means), then the voters will grade 
according to that common standard. This means that MJ passes IIA: since 
the grades are common, a candidate dropping out doesn't change the 
grades of the other candidates.

In contrast, Balinski & Laraki has experimental evidence that suggests 
that when the ballot is a yes/no (Approval style), voters are much more 
likely to act in a comparative manner - which is to say that they 
compare the candidates to each other, rather than comparing every 
candidate to a common standard. When they do that, B&L says, Arrow's 
impossibility theorem creeps back into the picture and you lose IIA 
compliance.

> If a voter grades a candidate as 'B' rather than 'A', the voter has
> detected some flaw in the candidate and is expressing it in the grade.
> To treat that voter's vote as simply above or below the median is to
> debase it. Why should the voter take the trouble to assign a grade if
> it's only use is to place the vote in the higher or lower half of the
> votes cast?

There are three reasons for this.

The first, let's call it "flexibility of meaning". Taking the median 
means that the system doesn't impose any particular meaning of the 
grades. It doesn't matter to the method *how much better* an A is than a 
B, because it'll still work. Thus, the method doesn't require the voter 
to rate the candidate. He can simply choose from the grades.

The second is strategy resistance. By taking the median, anybody voting 
higher than the median will accomplish exactly the same thing by voting 
highest possible, and anybody voting lower than the median will 
accomplish exactly the same thing by voting lowest possible. So in that 
sense, there's no reason to vote bottom or top; and if honesty is the 
default position, it will take very many voters indeed to make strategy 
actually have an effect, compared to rated-ballot systems. A 
strategizing voter might still get some probability of a better outcome 
by voting top on some candidates and bottom on others, or by voting a 
particular way if he knows what the median is, but these qualifying 
"ifs" would keep most ordinary voters from doing that. (Some other 
people on this list disagree. Different priors, apparently.)

Third, the median satisfies majority by grade. If a majority says X has 
grade B, and B is the highest grade used, then X wins. This is a 
protection against "crankiness", as someone (I don't remember his name) 
proposing median voting for budget calculation problems said. If you use 
the mean, then someone who is very loud will get his say 
disproportionate to his number. Range voting advocates say this is an 
advantage, but in a political system, if a majority gets overruled, it 
could easily try to regain its "right" by less peaceful means. Thus, if 
point #2 is about resistance to deliberate strategy, this is about 
resistance to outliers otherwise.

> I'm sorry we disagree on this point, but if the grading system is to
> have significance in the electoral process, the higher ranks must be
> more valuable than the lower ranks.

The higher grades (or ranks) have a greater implicit value.

Say you have a Condorcet system, and a voter submits the vote:

A > B > C.

Now, first rank A has greater value not by itself, but because it beats 
two candidates whereas second rank B beats only one. Contrast that to 
Borda, where first rank has an absolutely greater value than second 
rank: first rank gives the candidate three points, second only gives it two.

Then MJ is like Condorcet. A greater grade has greater value, but only 
in its implied power, which is that it pushes the median up in more 
situations. If your vote is A, it will push the median up in every 
situation except for when it's already A; but if your vote is C, it will 
push the median up if the median without your vote was C, D, E, or F, 
but not if it was A, B, or C.

So yes, whether your vote actually pushes the median up or down is a 
binary matter. However, it is only truly a binary decision if you knew 
what the median would have been without your vote.