[EM] IRV vs Plurality (back to the pile count controversy)

[robert bristow-johnson]

On Jan 22, 2010, at 3:57 AM, James Gilmour wrote:

> robert bristow-johnson  > Sent: Friday, January 22, 2010 12:25 AM
>> On Jan 21, 2010, at 7:05 PM, James Gilmour wrote:
>>
>>>>> N	Unique Preference Profiles
>>>>> 2	4
>>>>> 3	15
>>>> ...
>>>>
>>>> then your calculation is mistaken.  the fact that you
>> ostensibly need
>>>> 4 piles when there are only two candidates should serve as a clue.
>>>>
>>>
>>> If there are two candidates, A and B, then the possible unique
>>> preference profiles are:	
>>> A	
>>> B	
>>> A>B	
>>> B>A
>>>
>>
>>
>> what, on a ballot, is the consequential difference in meaning between
>> "A" and "A>B"?  what effect does a ballot marked "A>B" have over one
>> marked just "A" (or vise versa) in *any* election method that uses
>> ranked ballots?
>
> In terms of "preference profiles" the question is completely  
> irrelevant.  "A" and "A>B" are two different preference profiles.  So
> the possible numbers of preference profiles for given numbers of  
> candidates are, I think, correctly stated in the table in my
> earlier post.
>
> How the STV counting rules handle the two preference profiles "A"  
> and "A>B" is a different matter.  Some STV counting rules handle
> these two profiles identically.  But for some other STV counting  
> rules the profiles "A" and "A>B" are handled differently.  This
> second set of rules are those that prescribe the transfer of votes  
> "to the bitter end", i.e. even after the winners have all been
> determined.

so what consequential difference is that?

>   Under this rule a ballot marked "A" would be treated differently  
> from a ballot marked "A>B": at the last possible
> transfer, the "A" ballot would become 'non-transferable  
> (exhausted)', but the "A>B" ballot would be transferred to A.

now remember in the case we're discussing here, there is only two  
candidates.  again, what consequence to the outcome of the election  
(that is, who of A or B wins) occurs whether a ballot is marked  
"A" (and B is last by default) or is marked "A>B"?

there is none.

i'm not going to discuss this any more with Abd, because he's not a  
straight shooter, but James, if you want to get into this, it's  
pretty much "cut and dried" from the POV of Information Theory (a.la.  
Claude Shannon).  it's a discipline within my purview (being an  
electrical engineer that does signal processing), but the fundamental  
fact comes early.  for information to be transmitted from one  
location (the voter and his ballot) to another location (the ballot  
counters), it is necessary that such information does not exist at  
the destination prior.  Information Theory is about the measure of  
how much information is contained in a message and how many bits one  
bests commit to a message.

the first three statements following provide a non-zero amount of  
information (providing the destination was originally ignorant of the  
same).  the fourth statement (given the first three) has a measure of  
information of precisely zero (which means that it is inconsequential  
whether the message is transmitted or not).

1.  there are three eligible candidates, A, B, and C.
2.  a particular voter has A as his/her first preference.
3.  the same voter has B as the second preference.
4.  the same voter has C as the last preference.

when i taught Information Theory (just once, almost two decades ago),  
i was sorta enamored of some of George Carlin's humor and found an  
interesting example i used to illustrate a similar point.  Consider a  
weather forecast (say on TV).  If it were to say "tonight's forecast:  
snow", that would have some real information and the number of bits  
needed to encode it is greater than zero.  But now consider Carlin's  
Hippie-Dippie Weatherman, Al Sleet: "Tonight's forecast: Dark.  
Continued dark throughout most of the evening, with some widely- 
scattered light towards morning." If you're not considering Joshua in  
the Old Testament, how much information is in that forecast?

in general, the number of bits inherent to a particular message,  
given the probability of occurrence of the message, is:

     I(m) = -log2( p(m) )    (base 2 log and p(m) is the probability)

if p(m)=1, there is no information content in the message.   
committing bits or words or time to "saying it" is redundant.


again, the number of piles *necessary* (for recording and  
transmitting the information) when there are precisely N distinct  
candidates is

            N-1
     P(N) = SUM{ N!/n! }
            n=1

not

            N-1
     P(N) = SUM{ N!/n! }
            n=0

the latter redundantly (and unnecessarily) divides N! piles into  
twice that number with no difference of information between the two  
piles of each pair.


this is not social science.  it's not politics.  it's not opinion.   
it's just math.

i think i am now going to bow out of this.

it's similar to the alien abduction controversy.  no matter how many  
people claim to be abducted by extra-terrestials and can provide  
vivid and detailed information of they're alleged abduction (and even  
scars, where they stuck the needles in), even if it is hundreds of  
witnesses and i am only one, i know that what they're saying is  
untrue.  you can bring up the fact that all other planets in our  
solar system are able to support such life naturally, they'll say ET  
came from another star system.  you can bring up the known value for  
c and what we know about special relativity and they'll deny its  
validity (or invoke wormholes or something).  eventually you just  
have to walk away.

--

r b-j                  rbj at audioimagination.com

"Imagination is more important than knowledge."