[EM] IRV vs Plurality (Dave Ketchum) (Kathy Dopp)

[robert bristow-johnson]

On Jan 16, 2010, at 5:02 PM, Kristofer Munsterhjelm wrote:

> robert bristow-johnson wrote:
>> On Jan 16, 2010, at 3:41 PM, Kathy Dopp wrote:
>
>>> Exactly as I tried to point out to you, you were either disallowing
>>> voters to rank only two candidates or to rank all three.
>> no, it has nothing at all to do with allowing or disallowing the  
>> voters to
>>>   I see I was
>>> correct and you are disallowing voters to rank only two  
>>> candidates and
>>> have, as Abd ul also pointed out to you, left 3 choose 2 or 6  
>>> possible
>>> choices out of your list.
>> because all unmarked candidates are tied for last place, when  
>> there is only one unmarked candidate, there is *no* consequential  
>> difference between leaving that candidate unmarked or marking that  
>> candidate last.
>
> Is that true in IRV? Consider a vote of the sort:
>
> A > B
>
> where A and B are eliminated. Then this would be an empty vote, I  
> think, and so be removed from the count, whereas if it had been
>
> A > B > C
>
> it would count as one point for C.

now lemme see, if there are three candidates, how are two of them  
eliminated before the IRV final round?

and what counts in the IRV final round?  let's say that it's A  
eliminated before the IRV final round.  it doesn't matter if B is 1st  
or 2nd, if B ranked above (or is the only candidate left that's  
marked), it counts as a vote for B in the final round.  doesn't  
matter if C is marked below B or not marked at all.

think about it little bit, Kristofer, it *is* a useful fiction to  
leave the 2 bottom candidates (of 5) out of consideration (so one can  
get a grip of what happened in Burlington VT in 2009), but once  
you've done that (and you're considering only what happens between  
the remaining 3), the 9 numbers that are the only tallies you need to  
consider *any* counting scenario, IRV, Condorcet, Plurality of 1st  
choice, tallies for 1st or 2nd choice (some people in Burlington have  
suggested that as the number to use to determine the weakest  
candidate to eliminate in an IRV round), whatever, are:

>>   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

with exactly those three candidates in consideration, what  
consequential difference would it make in IRV (or any other rule of  
tabulation) if the [1332  M>K>W] pile was split into two piles;  
[M>K>W] and [M>K] that totaled 1332?  those 9 numbers could certainly  
be determined in individual precincts and meaningfully summed at City  
Hall or the campaign headquarters of either candidate.

--

r b-j                  rbj at audioimagination.com

"Imagination is more important than knowledge."