From: Forest Simmons <[EMAIL PROTECTED]> To: election-methods-electorama.com@electorama.com Subject: non-determinism and PR.
From: Bart Ingles <[EMAIL PROTECTED]> Subject: Re: [EM] non-deterministic methods
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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 ---- Election-methods mailing list - see http://electorama.com/em for list info
