[EM] Democratic Lottery Enhancement

[raphfrk at netscape.net]

Hmm, this didn't seem to make it to the archive, so re-sending.

Also, my comment on the 3 way tie being a problem was based
on my (wrong) assumption that there was no doubling step.  The
odd number of ballots could not have been divided in 2 evenly.

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.
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