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

[Kristofer Munsterhjelm]

robert bristow-johnson wrote:
> 
> 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?

Ah, I see. The only way for that to happen is if both B and C tie for 
last, in which case A wins by default.

Thus it only matters if there are more than three candidates, i.e. when 
both "A>B>C" and "A>B" votes are truncated.

> 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.

I'm not sure if it would be equivalent in Borda, however. Some ways of 
extending Borda to incomplete (truncated) ballots suggest that you give 
the last candidate 1 point, the next to last 2, etc, so that for

A > B

B gets one point and A two, whereas for

A > B > C

A gets three.

It would also make a difference if the election method in question uses 
"no opinion" the way Warren's Range extension does: that "no opinion" is 
to provide no information at all (alters neither the numerator nor 
denominator of the average).