[EM] A "top 3" to replace the "top 2"

[Dick Burkhart]

It is well known, of course, that in certain situations Borda will not give good results if one side votes strategically while the other side does not. This would not happen in the normal course of partisan politics. 

Take your example:

55%: A>B>C
35%: B>C>A
10%: C>B>A

If B and C represent moderate and right wing Republicans, while A represents the Democrats, then the Republican parties are voting strategically by listing the Democratic candidate last. However the Democrats are ignoring the strength of their principle foe, the moderate Republicans, by listing their candidate second instead of last. Actually, if the Dems do a little math, they would know that they could win easily by instructing their followers to randomly rank the B and C candidates second (2 points) or third (1 point), so that on the average they get 1.5 points.

Then the points for the Democrats are proportional to 0.55 * 3.0 + 0.35 * 1 + 0.10 * 1 = 1.650 + 0.450 = 2.100
And for the moderate Republicans we get                        0.55 * 1.5 + 0.35 * 3 + 0.10 * 2 = 0.825 + 1.250 = 2.075 
And for the right wing Republicans we get                        0.55 * 1.5 + 0.35 * 2 + 0.10 * 3 = 0.825 + 1.000 = 1.825

So A wins.  More generally, we can devise a formula for splitting the Dems vote between B an C to guarantee that the Dems will always win in this kind of situation:

frac/pts  A   B    C
x               3  p   3-p
y               1   3   2
z                1  2    3

Here x > 0.5 is A's fraction of the vote, y < 1 - x is B's fraction, z = 1 - x - y is C's fraction, 
and p is the fractional # points (the vote split) assigned to candidate B by the Democrats.

Then 1.0 < p < 4 - (y+1)/x guarantees that A wins. This is derived by computing
the Democrats sum of points  as   Sum_A= 3x + y + z           = 1 + 2x,
the moderate Republicans as         Sum_B= px + 3y + 2z       = 2 + (p-2)x + y, 
and the right wing Republicans as Sum_C= (3-p)x + 2y + 3z = 3 - px - y

and then requiring that Sum_A > Sum_B and Sum_A > Sum_C.

The second inequality reduces to px > 2 - 2x - y > 1 - x, or p > 1/x - 1 which is always less than 1 since x > 1/2, so any p > 1 will do.
The first inequality reduces to px < -1 + 4x - y, or p < 4 - (y+1)/x. Note that if x = 2/3 and y = 1/3, then any value of p < 2 would do.

Of course if the Democrats lack a majority in this situation, they would do well to form an alliance with a progressive third party, so that they could still win if the Democrats plus the progressives were a majority. Thus third and fourth parties would be strengthened, along with voter turnout and less mudslinging.

Dick Burkhart
4802 S Othello St,  Seattle, WA  98118
206-721-5672 (home)  206-851-0027 (cell)
dickburkhart at comcast.net

-----Original Message-----
From: Kristofer Munsterhjelm [mailto:km_elmet at t-online.de] 
Sent: September 18, 2014 12:16 AM
To: Dick Burkhart; 'EM'
Subject: Re: [EM] A "top 3" to replace the "top 2"

On 09/16/2014 05:40 AM, Dick Burkhart wrote:
> Oregon is considering a "top 2" primary, a very regressive move in my 
> opinion.
>
> So I thought what would be simplest "top 3" method that would give 
> voters more choices, especially in regard to independent or minority 
> candidates or parties. Less partisanship would be an added bonus.
>
> Below is a draft answer. Effective and simple, except for the modest 
> complexity needed to counter strategic voting.

At first, this sounds like Borda. As you mention, Borda is known for its susceptibility to strategy, particularly teaming. So to handle this, you switch to a variant of Baldwin's method until the number of candidates is reduced to three.

Why not use straight Baldwin (Borda-elimination) all the way down to the winner? Most likely, the variant is not cloneproof nor monotone (just like Baldwin), but by using Baldwin all the way until one candidate remains, you can have Condorcet. Baldwin is simpler than the point redistribution method you're using, as well: just repeatedly eliminate the Borda loser. And if you use Nanson instead (repeatedly eliminate candidates with average or below-average Borda score), you get reversal symmetry as well. Furthermore, Nanson has already been used (in Michigan), a point you could use when advocating for it.

Also note that Borda, even restricted Borda, can fail majority. For instance (taken from Wikipedia):

55%: A>B>C
35%: B>C>A
10%: C>B>A

and B wins. B still wins even if the shares are 63%, 27%, and 10% respectively, or for that matter if the shares are 66%, 33%, and 1%. 
That means that a near two thirds majority vote for A can be overridden by the B-voters.

Perhaps this could happen if A was a Democrat, B was a Republican, and C was a further-right figure. The Democrats prefer the lesser right-wing candidate B, the Republicans here prefer the right-wing to the "leftist" 
Democrat, and the right-wing candidate's supporters prefer the Republican to the Democrat.

In any case, if something like this happens, you'll have to be prepared to defend why the person who had a considerable majority of the first place votes failed to win. In contrast, if you use Baldwin or Nanson all the way to a single candidate, A wins because both of those methods pass Condorcet, which implies Majority; in the particular case above, C is eliminated first, then A wins with a majority preference against B.

If you'd like a top-three second round, just use Baldwin or Nanson in the first round until three candidates are left, gather ballots for the second round, and apply Baldwin or Nanson to the second round ballots to find the winner.

Now you might say that if there are more parties, then the elimination stage will remove the similar right-wing candidates; but that still leaves the problem in place if there happens to be only three (say D + R 
+ further-right independent).