[EM] Seven +/- Two
Richard Moore
rmoore4 at home.com
Tue Sep 18 21:39:56 PDT 2001
Forest Simmons wrote:
> What's the best way to use the two bit per candidate ballot?
This is sort of the e-mail equivalent of thinking out loud,
but as an initial though: Maybe it could be used in a
collapsed-vote version of Martin Harper's Universal Approval?
If a voter votes, e.g.,
A 10
B 11
C 01
D 00
E 01
F 10
then the Universal Approval counts for that voter could be
inferred from the information on the ballot. For instance,
for the set ABD, the voter approves A and B (because the
first position is marked for those candidates but not for
D). For the set ABF, the voter approves B (because the first
position is the same for all three, but only B has the
second position marked). For the set CD, there is no
preference expressed, so the voter registers no approval for
either in this race.
Unfortunately, Universal Approval following Martin's
definition might be very hard to tabulate and count for
large numbers of candidates, even with the preference
collapsing forced by this type of ballot. But there might be
an easier variation.
Let's find out how many subsets there are in which candidate
A will get Approval votes from the above voter. That is the
total number of Approval votes this candidate will get from
this voter across all subsets (though you can't find the
total number of Approval subset wins this way).
A will get Approval votes in any subset that contains A plus
at least one candidate that has a "0X" vote. These
candidates are C, D, and E. One could picture (!) a
6-variable Karnaugh map of the Boolean function A & (C + D +
E) which should have 7/16 coverage, if I'm not mistaken.
Thus candidate A gets 7/16 points from this ballot (and so
does candidate F, whom the voter has given the same marks).
Repeating for all 6 candidates:
A 10 A & (C + D + E) -> 7/16
B 11 B & (A + C + D + E + F) -> 31/64
C 01 C & !(A + B + F) & D -> 1/32
D 00 0
E 01 similar to C -> 1/32
F 10 similar to A -> 7/16
Scaling by a factor of 64, we see that the candidates get
A=28, B=31, C=2, D=0, E=2, and F=28 points from this voter.
Ratings are pushed towards the extremes, so in effect this
method of scoring is strategizing (?) for the voter. The
more candidates there are, the more pronounced this effect
will be (and you may wish to counteract it by adding more bits).
Add up the scores for each candidate from all the voters.
Since you can't tell from the summation how many subset
elections a candidate would have won in UA, this is not the
equivalent of Martin's method. How close does it approximate
it, I wonder?
Also, take note of the fact that although the ratings on
this ballot are symmetrically distributed (1 "11", 2 "10"s,
2 "01"s, and 1 "00"), the resulting scores are not (31, 28,
28, 2, 2, 0). The reason is that the subsets (A), (F), and
(AF) are not counted as votes for A and/or F. The subsets
(C), (E), and (CE) are not counted as votes for C and/or E.
All four candidates have their scores reduced as a result.
Perhaps these cases should count as half a point each. Then
you would also do this for the other single-member sets (B)
and (D). The new scores would be distributed symmetrically,
as we would expect: 31.5, 29, 29, 3, 3, 0.5.
I'm not sure if I like this method, but there may be one
benefit: It does seem to make it easier on the voter who is
unsure whether to fully approve candidate X or not. By
"partially approving" (with a 10 vote), he expresses almost
as much support as "fully approving", but with a hedge that
is calculated in relation to how he marks the other
candidates on his ballot. The hedge may reduce the potential
for regret if a person's statistical information is wrong or
insufficient. This might reduce the agonizing over
non-zero-info Approval strategies that bothers some people.
Richard
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