[EM] “Monotonic” Binomial STV

[Richard Lung]

On 28/02/2022 00:45, Richard Lung wrote:
>
>  Thanks for your thoughts, Kevin,
>
> In this simple instance, the election and exclusion quotas cancel. But 
> I would be lost without it, in multi-member PR cases of involved 
> transferable voting. There are a few examples in my e-books, (The 
> Super-Vote supercharged..., Elect and Exclude..., FAB STV...) free 
> from Smashwords, in epub format, and pdf versions free from archive. 
> org where putting "Richard Lung" in quotes in the text box should come 
> up with about 19 titles.
>
> The square root may not be strictly necessary, which may be why I keep 
> forgetting it. But it keeps the average keep values on a par with the 
> election and exclusion keep values. The square root is for the correct 
> form of the geometric mean, -- an important average.
>
> Yes, you are right, there is some other rule not stated -- All the 
> abstentions are counted. in more complex elections, they have to be, 
> so as not to distort the relative importnce of the election and 
> exclusion counts. It follows that if the abstentions add up to a 
> quota, a seat is not taken. This provides an incentive to nominate 
> good candidates, who work for the voters rather than their nominees.
>
> So, a candidate is not necessarily electable. More-over a large enough 
> quota like Hare, with a small number of seats would also be 
> prohibitive of election, given the voters free choice.
>
> Regards,
>
> Richard Lung.
>
>
>
> On 27/02/2022 19:30, Kevin Venzke wrote:
>> Hi Kristofer/Richard,
>>
>> I wonder not just about the square root, but also if the quota has some additional
>> role in the method, perhaps when there are 4+ candidates.
>>
>> Because this expression:
>> ( quota / keep ) * ( exclude / quota )
>> Appears to simplify to:
>> ( exclude / keep )
>>
>> This creates the appearance that the quota has no effect on the outcome.
>>
>> Richard stated that final values below unity are electable. It looks like there
>> will always be an electable candidate, unless it's a complete tie, or perhaps if
>> there is some other rule not yet stated here.
>>
>> It seems to me that the 3-candidate 1-winner case of this method is monotone.
>> It would help to see a four-candidate election resolved, too.
>>
>> Kevin
>>   
>> Le dimanche 27 février 2022, 07:41:20 UTC−6, Kristofer Munsterhjelm<km_elmet at t-online.de>  a écrit :
>>> On 27.02.2022 14:04, Richard Lung wrote:
>>>> Thank you, Kristofer,
>>>>
>>>>
>>>> for first example.
>>>>
>>>> The quota is 100/(1+1) = 50.
>>>>
>>>> Election keep value is quota/(candidates preference votes)
>>>>
>>>> Exclusion keep value equals quota/(candidates reverse preference vote):
>>>>
>>>> Geometric mean keep value ( election keep value multiplied by inverse
>>>> exclusion keep value):
>>> Geometric mean:
>>>   
>>> A: square root of (50/51 x 2/50) ~ 0.198
>>> B: square root of (50/49 x 1/50) ~ 0.143
>>> C: ~= infinity (or very high)
>>>   
>>> So B wins, having the lowest keep value. Is this correct?
>>>   
>>> (You seem to have omitted the square root in your calculations, but it
>>> shouldn't make a difference. Without the square root, A and B's values
>>> are 0.0392 and 0.0204 respectively.)
>> Hi Kristofer/Richard,
>>
>> I wonder not just about the square root, but also if the quota has some additional
>> role in the method, perhaps when there are 4+ candidates.
>>
>> Because this expression:
>> ( quota / keep ) * ( exclude / quota )
>> Appears to simplify to:
>> ( exclude / keep )
>>
>> This creates the appearance that the quota has no effect on the outcome.
>>
>> Richard says final values below unity are electable. It seems like there will
>> always be an electable candidate, unless it's a complete tie, or perhaps if there
>> is some other rule not yet stated here.
>>
>> Kevin
>>
>>   
>> Le dimanche 27 février 2022, 07:41:20 UTC−6, Kristofer Munsterhjelm<km_elmet at t-online.de>  a écrit :
>>> On 27.02.2022 14:04, Richard Lung wrote:
>>>> Thank you, Kristofer,
>>>>
>>>>
>>>> for first example.
>>>>
>>>> The quota is 100/(1+1) = 50.
>>>>
>>>> Election keep value is quota/(candidates preference votes)
>>>>
>>>> Exclusion keep value equals quota/(candidates reverse preference vote):
>>>>
>>>> Geometric mean keep value ( election keep value multiplied by inverse
>>>> exclusion keep value):
>>> Geometric mean:
>>>   
>>> A: square root of (50/51 x 2/50) ~ 0.198
>>> B: square root of (50/49 x 1/50) ~ 0.143
>>> C: ~= infinity (or very high)
>>>   
>>> So B wins, having the lowest keep value. Is this correct?
>>>   
>>> (You seem to have omitted the square root in your calculations, but it
>>> shouldn't make a difference. Without the square root, A and B's values
>>> are 0.0392 and 0.0204 respectively.)