[EM] Some myths about voting methods

[Jonathan Lundell]

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.