# [EM] Democratic Lottery Enhancement

raphfrk at netscape.net raphfrk at netscape.net
Fri Jun 9 01:35:00 PDT 2006

```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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