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.
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