[EM] Democratic Lottery Enhancement
raphfrk at netscape.net
raphfrk at netscape.net
Thu Jun 8 15:04:24 PDT 2006
Jobst Heitzig <heitzig-j at web.de>
>> This is not independent of cloning ... in a big way :).
>
>What makes you think so? Assuming we add a clone y of x so that all
>voters have x and y neighboured in their ranking and so that the new
>initial proportions of papers naming either x or y is the same as the
>original proportion of papers naming x, it seems that each voter will
>return exactly the same number of papers naming either x or y as they
>returned papers naming x originally. This seems to indicate that the
>procedure is clone-proof all right, isn't it?
Yeah, I misread the procedure. I missed the doubling the ballots step.
If there are 10 candidates with 9 from party B and 1 from party A
all voters return 10 papers not 5 as I thought.
'A' supporters return 4 A's and 6 from B
'B' supporters return 10 B's
The new odds are 4/20 = 20%, this is better for A than the original
lottery (10%) and presumably in the limit will end up 50/50 as required.
>> If so, there is an odd number of winners in that case, which might
>> prove problematic.
>This is not obvious to me. Can you prove it?
I guess I just assumed as all results I have ever seen are 3 or 5 in
the tie.
It might be true though :).
Some simple examples elections:
2 candidates
A>B
B>A
all candidate tie
3 candidates
A>B>C
B>C>A
C>A>B
result: A>B>C>A (circle)
4 candidates
A>B>C>D
B>C>D>A
C>D>A>B
D>A>B>C
Result: total tie, removing 1 vote gives 3 way circular tie.
I can't see an obvious way to get the 4 to tie.
___________________________________________________
Try the New Netscape Mail Today!
Virtually Spam-Free | More Storage | Import Your Contact List
http://mail.netscape.com
More information about the Election-Methods
mailing list