[EM] Some myths about voting methods

[Paul Kislanko]

Besides the obvious problem with the notion of a fraction of a bit, you're
still confusing the number of possible ballots with the amount of
information conveyed by a single ballot.

If there are 3 candidates, in approval a ballot only needs 3 bits. Ranked
ballots need to carry the order the voter selected, and that requires 2 bits
per alternative. I.e for ballots with ABC, you need 11, 01, 10 to indicate
B>C>A. You cannot do that with fewer than 6 bits, even though it only takes
3 bits to count the 3! = 6 possible ballots. 

-----Original Message-----
From: election-methods-bounces at lists.electorama.com
[mailto:election-methods-bounces at lists.electorama.com] On Behalf Of Jonathan
Lundell
Sent: Saturday, June 06, 2009 1:44 PM
To: Paul Kislanko
Cc: election-methods at electorama.com
Subject: Re: [EM] Some myths about voting methods

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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