[EM] IRV ballot pile count (proof of closed form)

[robert bristow-johnson]

On Feb 4, 2010, at 8:29 PM, Kathy Dopp wrote:

> On Thu, Feb 4, 2010 at 8:18 PM, robert bristow-johnson
> <rbj at audioimagination.com> wrote:
>>
>> On Feb 4, 2010, at 7:51 PM, Kathy Dopp wrote:
>>
>>> The general formula for the number of possible rankings (for strict
>>> ordering, without allowing equal rankings) for N candidates when
>>> partial rankings are allowed and voters may rank up to R candidates
>>> (N=R if voters are allowed to rank all candidates) on a ballot is
>>> given on p. 6 of this doc:
>>>
>>>
>>> http://electionmathematics.org/ucvAnalysis/US/RCV-IRV/ 
>>> InstantRunoffVotingFlaws.pdf
>>
>> the only issue, Kathy, is whether the lower limit is i=0 or i=1.   
>> you have
>> to defend your use of i=0 for the case illustrated below (using  
>> the rules in
>> Burlington VT and Cambridge MA).
>
> False Robert. Starting the index at 0 or 1 is completely irrelevant.

so you're saying that

     N-1              N-1
     SUM{ N!/i! }  =  SUM{ N!/i! }  ?
     i=0              i=1

that the N!/0! term is equal to zero?


> Any formula is easily adjusted to either initial index.

i ain't talking about any substitution of dummy variable, i, and  
changing the limits.

> You are obviously not a mathematician.

i guess not.  just a Neanderthal electrical engineer who does signal  
processing algs for a living.


>> in Burlington it was 5 in both 2006 and 2009.  N was also 5 (not  
>> counting
>> any write-in).
>
> My formula gives the general case for R equals anything, as I said.
>
>>
>> but Kathy, suppose N=R=3 and it's the "regular-old" IRV rules that  
>> do not
>> require any minimum number of candidates ranked and do not allow  
>> ties.  to
>> be clear, i need to also point out that only *relative* ranking is  
>> salient
>> (at least in Burlington).  if a voter only ranks two candidates and
>> mistakenly marks the ballot 1 and 3, the IRV tabulation software  
>> will close
>> up the gaps and treat that precisely as if it was marked 1 and 2.
>
> So?  What's your point?

the point is that a ballot marked with 1 and 3 goes on the same pile  
as a "properly marked" ballot marked with the same two candidates as  
1 and 2.

>>
>> now, given those parameters, are you telling us that the 9 tallies  
>> shown on
>> Warren's page:
>>  http://rangevoting.org/Burlington.html (i have very similar  
>> numbers, no
>> more different than 4), are not sufficient to apply the IRV rules and
>> resolve the election?
>
> Obviously you did not read my email. I'll read and respond to yours
> after you've tried to read and understand my points. Otherwise I am
> not wasting my time responding to you.

no need to, but...

>>
>>    1332  M>K>W
>>     767  M>W>K
>>     455  M
>>    2043  K>M>W
>>     371  K>W>M
>>     568  K
>>    1513  W>M>K
>>     495  W>K>M
>>    1289  W
>>
>> is there any reason those 9 tallies could not have been summed from
>> subtotals coming from all 7 wards of Burlington?
>>
>> please tell us why those 9 piles are not enough, given the  
>> parameters stated
>> above?


... you're running away from the salient question.  are those 9 piles  
good enough to resolve the IRV election with 3 candidates or not?   
was salient information lost when the M>K pile was combined with the  
M>K>W pile, enough that could cause the IRV election (with the rules  
above) to be decided differently?

--

r b-j                  rbj at audioimagination.com

"Imagination is more important than knowledge."