On Jun 6, 2009, at 11:10 AM, Paul Kislanko wrote:
The number of possible votes is not the same as the amount of
information in
a single ballot. With 3 candidates, there are indeed 8 possible
ballots, but
any one ballot can be encoded in 3 bits, since any particular choice
requires only that many to represent it.
Ranked ballots require 2 bits per alternative (01 = 1st, 10 = 2nd,
11 = 3rd)
so the minimum ballot representation is six bits, twice as much
information
as is contained in an approval ballot.
If we disallow truncation and equal-ranking, we have 3! or 6 ballots,
or ~2.6 bits.
If we allow truncation but not equal-ranking (except as implied by
truncation), we have:
A > B [> C]
A > C [> B]
A
...and the symmetrical cases ranking B or C first, for 9 possible
ballots. Add one for the empty (no preference) ballot and we have 10
ballots, requiring ~3.3 bits to represent.
If we allow equality but not truncation (which would be redundant) we
have:
A = B = C
A > B > C
A > C > B
B > A > C
B > C > A
C > A > B
C > B > A
A > B = C
B > A = C
C > A = B
A = B > C
A = C > B
B = C > A
Do I have them all? 13 possible ballots, ~3.7 bits.
There are 7 possible approval ballots (if we consider approve-all and
approve-none equivalent): ~2.8 bits.
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