[EM] non-determinism and PR.

[Forest Simmons]

> From: Bart Ingles <bartman at netgate.net>
> 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