[EM] Instant Strategic Approval

[Raph Frank]

On Sun, Aug 15, 2010 at 11:55 AM, Kristofer Munsterhjelm
<km-elmet at broadpark.no> wrote:
> After talking about that Approval should be DSV, I thought of this method.
> It is:
>
> - Voters submit ranked ballots.
> - First count as in Plurality. Candidates that are tied at top rank may
> either get one point each, or 1/k if there are k tied candidates - I don't
> know which would be better.
> - Call the Plurality winner "X" and the runner-up "Y".
> - Using the Plurality data and the original ballots, count each ballot as if
> he voter employed approval strategy A. That is, approve all candidates
> ranked ahead of X, and X if he is ranked ahead of Y.
> - The candidate with the highest approval score wins.
>
> Would that method be any good? I don't know. It probably wouldn't be
> monotone, but it is quite simple. Since it's "instant", it would have the
> same problems with regards to ordinary strategic approval as IRV has with
> regards to exhaustive runoff.

You might need multiple rounds to get it to stabilise.  Ofc, if there
are no condorcet winners, then that rule could cause it to oscillate.

It doesn't help with the Burr dilemma.

26: D1>D2>R
25: D2>D1>R
49: R>D1=D2

R(X) + D1(Y) win round 1, so round 2 is

26: D1 + D2
25: D1 + D2
49: R

This gives a D1+D2 tie.

The reason is that it has estimated the expected winner wrongly.

>From round 2, D1 and D2 are the expected winners, so the vote would go

26: D1
25: D2
49: R + ?(D1 or D2)

This flips it back so the Republican votes pick the winner.

What about using a method that sums up the votes over may rounds,
using the running totals to estimate the expected winners.

So round 1: (plurality)

D1: 26
D2: 25
R: 49

X = R , Y = D1

Round 2

D1: 51
D2: 51
R: 49

Total:
D1: 77
D2: 76
R: 98

X = R , Y = D1 (still)

(The next few rounds could be done by computer and don't need a
physical recount, as X and Y don't change).

Round 12 (10 more rounds)
D1: 77 + 10*51 = 587
D2: 76 + 10*51 = 586
R: 98 + 10*49 = 588

Still X = R, Y = D1

Round 13
D1: 587 + 51 = 638
D2: 586 + 51 = 637
R: 588 + 49 = 637

X = D1, Y = R (ahem, pretend it is really round 12.5 :) )

Votes become
26: D1
25: D2 + D1
49: R

D1: 689
D2: 662
R: 686

Now, D1 + R are the top-2 and that will not change, so D2 can never catch up.

However, I think the above is just a complex way of saying pick the
condorcet winner and the method would only stabilise if there is one.