[EM] 48 C, 24 B, 28 A > B example

[Forest Simmons]

In this example, Borda, IRV and Margins go with C, while wv and Bucklin 
say B.

Besides this disagreement, there is the problem that if the win is given 
to B for this ballot set, then if the 24 faction voters' true preferences 
were B>A>C, they would be sorely tempted to truncate so that B would win 
instead of A.

I think that the best way out of this dilemma is appropriate randomization 
in "situations like this".

To randomize this example I would use "symmetric depletion" as opposed to 
symmetric completion, to split the last faction in two:

   48 C,  24 B,  14 A,  14 not(C).

Then I would cancel the 14 not(C) against 14 of the 48 C to get

   34 C,  24 B,  14 A.

Then I would subtract 14 from each of these factions to get

   20 C, 10 B.

Finally, I would roll a die and pick B if a one or a two came up, or C in 
the case of 3, 4, 5, or 6.

An equivalent way of finding these 2:1 odds is to subtract the least 
Borda Count from the other two, and take the ratio of the two differences.

In this example, the A>B, B>C, and C>A margins are 4, 4, and 20, 
respectively, so the relative Borda Counts for A, B, and C are
(4-20)=-16, (4-4)=0, and (20-4)=16, which are proportional to
-1, 0, and 1, respectively.  When we subtract the negative one (by adding 
its opposite) from the other two scores we get 0, 1, and 2, respectively.
So the required odds are 2:1.

Suppose that the members of the second faction secretly rate A, B, and C 
at x, 1, and 0, respectively.  Then with this randomization each member of 
this faction has expectation of 1/3.  If they don't truncate, then there 
is no randomization, and A (with rating x) wins.

So it is to their advantage to truncate only if x is less than 1/3.  In 
that case, candidate A  SHOULD BE the loser!

The full payoff matrix for A and B with sincere preferences

    48 C, 24 B > A, 28 A > B   is


          \ B
         A \____truncate_____|__show preference__
           |                 |
   truncate|      (0,0)      |    ( 1, x )
           |                 |
         ------------------------------------
   show    |                 |
   pref.   |    (y/3, 1/3)   |    ( 1, x )


   where y is the third faction's rating of B.


Note that for A, the strategy "show pref." dominates, no matter the value 
of x or y.  Since B knows this, he crosses row one out.  Then he chooses 
"truncate" iff 1/3 is greater than x.

I think that this is a good way out of the dilemma.  For those that don't 
like randomization, let them contemplate that different methods give 
different winners, so both B and C have a legitimate claim.  Giving the 
win to C cheats B in the case when B sincerely doesn't like A at all. 
Giving the win to B cheats A in the case when A is the sincere CW.

I would rather have a clean, up front randomization in a case like this, 
then leave the voters with this dilemma.

Exactly which ballot sets should be considered appropriate for 
randomization, I don't know... perhaps whenever there is a beat cycle, 
perhaps only when margins and wv disagree, or perhaps whenever Black and 
wv disagree.


Forest