From: Bart Ingles <[EMAIL PROTECTED]> Subject: Re: [EM] non-deterministic methods
Forest Simmons wrote:
What if we tossed two coins, and gave the win to B if they both came up heads, to C if they both came up tails, and to A otherwise.
Wouldn't a random cycle-breaker provide strong incentive for a sure loser in a cycle-free election to try to create a cycle?
To address this issue the method would have to give little or no probability to the weakest candidate in the cycle. Random ballot would certainly work better than random candidate in this regard. I'm looking for something better than random ballot.
Here are two analogous questions concerning a cycle in which A beats C beats B beats A:
3000 A 3000 A=B 4000 B>C
(1) Suppose that A, B, and C represent parties, and that based on these ballots, we are supposed to allocate 100 seats in congress to the parties A, B, and C. How many seats should each party get?
(2) Suppose that A, B, and C represent individual candidates, and that based on these ballots we are supposed to put 100 marbles in a bag for a drawing to determine the winner. How many marbles should each candidate get?
In other words, we should be able to use ideas from Proportional Representation to get ideas for non-deterministic single winner methods.
In the case of Approval ballots we can use PAV to determine the winner. Assuming the C voters also approve B, we can always get a larger PAV count by using a B in place of a C, so the approval ballots boil down to
3000 A 3000 AB 4000 B
For a hundred winner election the max PAV score is for a set with 43 A's and 57 B's. On this basis, the respective probabilities for A, B, and C should be 43%, 57%, and zero, even though C was a member of the cycle.
If B were not approved, then the PAV count would always be improved by using an A in place of a B, so the ballots would boil down to
6000 A 4000 C
And the obvious PAV solution would be 60 A's and 40 C's. Membership in the cycle didn't guarantee B any positive probability.
It is interesting to me that in my other example
34 ABC 33 BCA 33 CAB
some respondents thought it was obvious that A, B, and C should have nearly equal chances, while others thought it was obvious that the deterministic winner A should be chosen 100 percent of the time.
Some thought my hypothetical coin toss was way out of line in one direction, while others thought it was way out of line in the other.
Those who completely pooh poohed chance probably don't realize how much random element is introduced just by disinformation, slipshod election practices, and insincerity due to misguided attempts at strategy. If clean randomization can help cut down on these biased random influences, then it will serve a useful purpose.
Those who hazarded an opinion about probabilities always gave B and C equal probabilities even though B and C differ in Borda count as much as A and B do.
Consider the perfectly symmetrical ballot set
33 ABC 33 BCA 33 CAB.
The probabilities should be equal in this case.
Now add one ABC. It seems to me that this additional ballot should increase A's probability and decrease C's probability. By how much? I do not know. If I could answer that question, then I could also tell you whether or not it should also decrease B's probability, and by how much.
Of course the best randomization of all is the one advocated by Joe Weinstein, choosing a random sample of (a few thousand) voters to attend a convention, study the candidates carefully and cast their ballots. That would remove most (if not all) of the recount fiascos, and other dirty sources of randomness, while giving enough voting power to the voters to actually take the trouble to study the candidates seriously. The hundreds of millions of dollars wasted on campaigning would be replaced by debates in front of the actual voters, since blanket TV ads would be a thousand times more pointless than they already are.
Forest ---- Election-methods mailing list - see http://electorama.com/em for list info
