[EM] forms of "equality" and "efficiency"

[Jobst Heitzig]

Dear EM folks!

I thought some while about "equality" and "efficiency" as a basic 
principles of voting. It seems that both terms can mean quite different 
things that we should perhaps try to distinguish. First, there is 
"formal equality" (=anonymity), "equality of outcomes", "equality of 
expected outcomes", and "equality of power". On the efficiency side, 
there is "total utility efficiency", "social welfare efficiency", 
"Condorcet efficiency" and so on.

To start a discussion, I have divised a simple example which I use to 
distinguish these kinds of equality and efficiency, and which therefore 
also leads to quite different results under various voting methods. At 
the end of this post, I also present a new method designed to provide 
both "equality of power" and some sort of efficiency.


[please read with a monospaced font!]

Example preferences:
--------------------               mean utility
1 voter  A> > > > > >C>G|E>T>M>B   26/7=3.7
1 voter  B> > > >M>T>G| >E> >C>A   33/7=4.7
1 voter  C>T> >G>M> | > >E> >B>A   40/7=5.7
         | |   | | | | | | | | |
utility 11 10  8 7 6 5 4 3 2 1 0

(> designates a utility gap of 1,
 | shows the approval cutoff under above-mean strategy)


Analysis:
---------
               Approval  Borda  median   total    Gini social
option  beats  score     score  utility  utility  welfare
C       all    2         12 !   5        17       31/9
T       ABEMG  2         11     6        18 !     38/9
G       ABEM   3 !       11     5        17       43/9 !
M       ABE    2          9     7 !      15       33/9
E       AB     0          7     3         9       27/9
B       A      1          7     1        12       14/9
A       -      1          6     0        11       11/9

C is both the Borda and the beats-all (Condorcet) winner.
The beats are transitively C>T>G>M>E>B>A.

M maximizes median utility.

T maximizes total utility.

G maximizes Gini social welfare (=expected minimum utility of two 
randomly drawn ballots) and is the Approval winner with above-mean 
approval strategies. In a more realistic scenario, the third voter would 
rather approve of C and T only since that would change the Approval 
winning set to {C,T,G}.

If we want equal utilities for all voters, we must elect E.

If we want equal expected utilities for all voters, one solution is to 
elect A,M,E with probabilities 6/s,11/s,(s-17)/s, for some arbitrary 
s>=17. (There also exist other solutions with three possible winners.)

If we give each voter equal voting power, i.e. let her distribute 1/3 of 
the winning probability, the winner is A,B,C each with probability 1/3. 
For the B and C voters, it would increase their expected utility when 
they gave their probability share of 2/3 all to M rather two B and C. 
However, under Random Ballot, this is a prisoner's dilemma since, given 
the other's strategy, it is always best to vote sincerely A or B instead 
of M. Therefore, such a cooperation won't probably occurr with Random 
Ballot. In this respect, Random Ballot is quite inefficient in terms of 
total utility and/or social welfare.


A method providing both "equality of power" and efficiency:
-----------------------------------------------------------

The latter observation is the reason why I studied more cooperative 
variants of Random Ballot for a while. The variant I consider best at 
this moment is  the simple "Draw Two / Most Approved Compromise" (D2MAC).

In D2MAC, each voter marks one option as "favourite", and marks as "also 
approved" any number of additional options she considers promising 
compromise options. Then two ballots are drawn at random (with 
replacement). The winner is the most approved option of those options 
which are marked on both ballots (if any such exist), or else the option 
marked as "favourite" on the first ballot. In this way, each voter can 
be sure that her share of the probability goes to one of the options she 
marked, and only goes to a non-favourite if many other voters approved 
of her compromise options as well. Hence voters can cooperate safely to 
transfer probability to good compromise options.

In our above example, the B and C voters can safely mark A or B as 
"favourite" and M as "also approved", thereby changing the winning 
probabilities to 1/3 A, 4/9 M, 1/9 B, 1/9 C. (M wins iff both of the two 
ballots drawn at random are different from the first ballot.) If only 
one of the two voters approves of M, the probabilities remain 1/3 A,B,C, 
hence no prisoner's dilemma occurs in this case with D2MAC.


Jobst