[EM] R. B. MacSmith

[Forest Simmons]

On Wed, 2 Mar 2005, Jobst Heitzig wrote:

> Dear Forest!
>
> When I understand you right, you propose to just strike out all strongly 
> covered candidates and then use Random Ballot on the rest, right?
>
> But then there must be some error in your proof of monotonicity, I fear -- 
> look at the following example:
>
>     Situation 1:  Situation 2:
>     |             |
> Ballots:
>  2  A>B>>C>D      ditto
>  2  B>C>>D>A      ditto
>  2  C>D>>A>B      ditto
>  2  D>A>>B>C      ditto
>  1  B>D>>A>C      D>B>>A>C
>     ---           ---
> Defeats:
>     A>B>C>D>A>C   ditto
>     and B>D       and D>B
>
> Approval:
>     A B C D       ditto
>     4 5 4 5
>
> Covering relation:
>     B>C           A>B
>
> Strong covering relation:
>     B>C           none
>
> Random Ballot results (probabilities times 9):
>     A B C D       A B C D
>     2 3 2 2       2 2 2 3
>  (monotonic)
>
> R.B.MacSmith results (probabilities times 9):
>     A B C D       A B C D
>     2 5 0 2       2 2 2 3
>  (monotonic)
>
> Your proposal results (probabilities times 9):
>     A B C D       A B C D
>     2 3 0 4       2 2 2 3
>  (not monotonic: raising D decreases its probability!)
>


At least raising D relative to B preserved D's positive probability and 
increased the ratio of their probabilities from 4/3 to 3/2, as in the 
Condorcet Lottery kind of monotonicity.  Maybe something can be salvaged 
here.


>
> That's a pity since I like your proposal. But I have seen so many rules seem 
> to be monotonic at first glance and then turn out not to be monotonic that 
> I'm always quite suspicious. I hope my own proof that R.B.MacSmith is 
> monotonic really holds...
>

I wonder about the "Needle Point" method. Is it monotonic?

Remember that a needle is a maximal chain such that no member is beaten by 
any preceding member either approvalwise or pairwise, but every member 
beats every predecessor either approvalwise or pairwise or both.

A needle point is the sharp end of a maximal needle.

At least the needle point method survives the above example:

Both before and after D is moved up, the only needle points are A, B, and 
C.

Note that AC, BC, and DA are needles both before and after the change, 
while BD is a needle before and DB is a needle after.  Candidate C is not 
the point of any maximal needle in either case.  So "random ballot needle 
point" yields the respective winning probabilities 2, 3, 0, and 4 per 
nine in both cases.

If we iterate needle, eventually we end up with B alone in the first case 
and D alone in the second case.

In general it may not be a good idea to iterate needle, but in this 
example it seems to amplify the effect of changing B>D to D>B.

Unfortunately needle doesn't punish the B faction defection in the example

49 C
24 B>>A
27 A>B>>C

If the B supporters truncate A, then the needle points go from A and B 
(before the truncation) to C and B (after), which is a reward to the 
truncators.

But if the A supporters take the precaution of raising the approval 
cutoff, then A and C become the only needle point candidates whether or 
not B truncates.

If we iterate in the case of B truncating, then the winner becomes C.

If we iterate in the case of C not truncating (but A raising the bar) then 
the needle point candidates remain A and C.

"Random ballot among not strongly covered" seems to do better at 
discouraging insincere ballots.  Do we have to sacrifice monotonicity to 
some degree for that advantage?

Forest