[Election-Methods] D(n)MAC

[raphfrk at netscape.net]

 > P: A>C>B
> Q: B>C>A,

I wonder if it would be worth repeating the draw if no compromise appears.

The chances of each side winning outright would be

P-win = P^n/(P^n+Q^n)
Q-win = Q^n/(P^n+Q^n)

If both are 50/50, then that is the same as a random lottery.

This system favours the larger side.? If n=4 and P was 0.6 and Q was 0.4, then
A would win in 83% of cases.? This would mean that P voters would need to rate
C > 8.3 out of 10 before they will compromise.

The advantage is that it allows recovery if one of the ballots drawn is from a
small hold out group, so compromise fails.

Another option is to have a limited/finite number of redraws.? Maybe this 
has basically the same effect as modifing n as it sets the 'chances of
no compromise needed probability'.? Also, in practice, it might be needed
as you could be drawing forever, but that would require no candidate to 
be approved by n voters.? 

If there was 1 re-draw, then that doubles the odds of an A or B outright win.? 
This means the compromise must be more popular to ensure a win.

There are ofc some issues if there is more than 3 candidates and deciding who to
approve.? 

A feature of this system is that it reverts to random ballot for the 2 candidate case.? 
This may result in candidates who would be discouraged from running due to spoiler 
effects in plurality being encouraged to run due to 'civic duty'.


 


Raphfrk
--------------------
Interesting site
"what if anyone could modify the laws"

www.wikocracy.com

 



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