[EM] non-determinism and PR.

[Forest Simmons]

From: Forest Simmons <simmonfo at up.edu>
To: election-methods-electorama.com at electorama.com
Subject: non-determinism and PR.

> From: Bart Ingles <bartman at netgate.net>
> Subject: Re: [EM] non-deterministic methods
...

> 
> Wouldn't a random cycle-breaker provide strong incentive for a sure
> loser in a cycle-free election to try to create a cycle?

I think I can show that this wouldn't work if the "sure loser" were the 
Condorcet Loser in the three candidate case that concerns us in spruced up 
methods.

Suppose that L is the Condorcet Loser, and that W is the Condorcet Winner.

Only a faction that ranks L above W would want to try to give L a chance 
at winning at the expense of W.

There are five cases: (1) L>X>W, (2) X>L>W, (3) L>W>X, (4) L=X>W, and
   (5) L>X=W.

In order to create a cycle, L must be raised relative to at least one 
other candidate, otherwise L will remain a Condorcet Loser, and there will 
be no cycle.

Furthermore, in order to create a cycle, W must be lowered relative to at 
least one candidate, otherwise W will remain a Condorcet Winner, and there 
will be no cycle.

So far we have shown that in order to create a cycle, L must be raised AND 
W must be lowered.

But in the odd cases L cannot be raised, and in the even cases W cannot be 
lowered. Therefore it is impossible for any faction that ranks L above W 
to induce a cycle by itself.

In summary, the only factions that could possibly create a cycle would be 
the ones that rank the Condorcet Winner W above or equal to the Condorcet 
Loser L, and these factions would stand to lose as much as they might 
gain.

Also, in my last message I made the point that non-deterministic methods 
don't necessarily give every member of a cycle a positive probability of 
winning.  A case in point is Rob LeGrand's ballot-by-ballot Declared 
Strategy Voting method.  The ballots are shuffled before applying the 
method, yet in many cases of Condorcet cycles, the method gives the same 
winner with certainty, regardless of the order of the ballots (whenever 
there is a unique Approval Equilibrium candidate, for example).

In a previous message we treated the cyclic example

3000 A
3000 A=B
4000 B>C

We found that if the last faction disapproved C, then PAV based 
randomization would give the win to A 43% of the time and B 57% of the 
time.

On the other hand, if C were approved by by the last faction, then PAV 
based randomization would give A and C all of the probability with 60% for 
A and 40% for C.

So should C be approved or not?  Well, if A has zero utility for the last 
faction, and C has more than 70% of the utility of B, then it would be to 
their advantage to approve C, otherwise not.

This kind of calculation could be done automatically if the voters 
submitted Cardinal ratings ballots.

Note also that Rob LeGrands DSV method would give B the win with certainty 
in this example, whether or not the third faction raised C to the level of 
B.

Somebody should look at the

49 C
24 B>A
27 A>B

election with the PAV probabilities in mind for the various approval 
options, to see if they exacerbate or ameliorate the Prisoner's Dilemma 
problem, e.g. the temptation for the second faction to "defect" by 
truncating.

Forest