[EM] Re: Approval Strategy in the Three Competitive Party case.

[Forest Simmons]

> From: Russ Paielli <6049awj02 at sneakemail.com>
> Subject: Re: [EM] Rewording Strategy A (BF(1st))

> Forest Simmons simmonfo-at-up.edu |EMlist| wrote:
>
>> Departing from Strategy A, we offer the following refinement in the same
>> spirit:
>>
>> For each candidate C, if you think the winner is more likely to come
>> from the set of candidates that are worse than C than from the set of
>> candidates that are better than C, then approve C, else don't.
>

<snip>

>
> You've stated what is perhaps the best strategy for Approval, but how
> effective is that strategy? Well, in some cases it may be very
> effective, but in other cases it may not be.
>
> For example, what if *three* parties are equally popular? Let's take the
> classic Democrat (D), Republican (R), and Green (G) case. Suppose they
> are approximately equally likely to win. And suppose your own order of
> perference is G>D>R. Who do you approve?
>

It is very unlikely that G and R would be exactly equal in perceived 
likelihood of winning at the same time that D's utility is exactly half 
way between G and R.  But even if that were the case for several G>D>R 
voters, it would not be so for the vast majority of them.

So the negligible set of extreme symmetry victims should consult with 
their more numerous friends, and reinforce their friends' votes by 
following suit on their decision about D.

As to the efficacy of the ballot of a voter isolated from friends in this 
case of extreme symmetry:

Suppose that G, D, and R are equally likely to win by the best estimates.

Then your ballot can make a difference only if two or more of these 
candidates would be within one vote of each other without your vote.

It is extremely unlikely (even relative to this unlikely case that we are 
now considering) that all three candidates would be within one vote, so 
let's consider the cases where two are within one vote of each other:

Let M be the max approval of any candidate when your ballot is removed 
from the mix.

The only remotely likely cases where your approval ballot (G or GD) could 
possibly make any difference are the equally likely cases

(1a) g=d=M, (1b) g=d-1=M

(2a) g=r=M,  (2b) g=r-1=M,

(3a) d=r=M,  and  (3b) d=r-1=M,

where g, d, and r are the respective approvals of G, D, and R, not taking 
into account your ballot.

If you approve only G, then you will improve the result as far as one vote 
possibly could in the first four cases, by either breaking the tie (in 
(1a) or (2a) ) or making a favorable tie (in part b of the cases).

If you approve both G and D, then you will improve the result as far as 
possible for one voter in the last four cases.

Suppose that you flip a coin to decide which whether or not to approve D. 
Then there are six equally likely possibilities:

(GD,1), (GD,2), (GD,3), (G,1), (G,2), (G,3).

Of these, only in the first (GD,1) and last (G,3) would your ballot not 
improve the outcome of the election.

Two out of three isn't bad.  What other method can beat it?

Remember that methods that don't satisfy the FBC have the potential of 
actually tipping a tie in a direction opposite to the wishes of the voter.

Non-monotone methods might even work against you if you vote sincerely.

So the analysis of other methods is not as simple, but my intuition is 
that Approval is as efficacious as any of them, even in this high symmetry 
case.

You asked about the case of four equally matched candidates.  That's 
easier.  You should approve the two you like the best, but not the other 
two.

Then there are C(4,2)=6 cases of pairwise ties (or near ties) to consider:

(C1,C2), (C1,C3), (C1, C4), (C2,C3), (C2,C4) and (C3,C4)

Suppose that your two preferred of the four are C1 and C2.

Then by approving them (but not the others) your ballot will be effective 
in all cases except the first (because both are approved) and the last 
(because neither is approved).  The the two thirds efficacy still applies 
in the four candidate near tie case.

Can you produce as convincing results for some other method?

One other comment.  If (say) three candidates are truly equally likely to 
win (under a fair method like approval) then what really decides the 
winner, if not chance?

In that case (of statistically indistinguishable likelihoods of three 
candidates winning) the accidents that side track voters on the way to the 
polls, a facial expression of a newscaster, and a multitude of other 
random variables act as effectively as the roll of a die to determine the 
outcome of the election.

And what difference does it make to the electorate as a whole? If the 
candidate support is so close that the winner is determined largely by 
incidental noise external to the method (no matter how good an election 
method might be), then what does it matter who wins?

It might matter to individual voters, but two thirds of the voters must be 
disappointed no matter who wins in a case like this.

The cases that matter the most are the ones where Approval shines the 
most, but it does better than most methods even in these extremely 
symmetrical cases that are going to be decided by incidental noise anyway.

The main drawback of Approval is the (one or two election) time lag of 
locking on to the voter median candidate in the case of disinformation.

If the choices are changing rapidly compared to the time between 
elections, and if disinformation is a big problem, then Approval will 
perform suboptimally, but still better than IRV, Borda, Plurality, and all 
but the best Condorcet methods (in my opinion).

Forest