Hi again.
There is still another slight improvement which might be useful in
practice: Instead of using the function 1/(5-4x), use the function
(1 + 3x + 3x^7 + x^8) / 8.
This is only slightly smaller than 1/(5-4x) and has the same value of 1
and slope of 4 for x=1. Therefore, it still encourages unanimous
cooperation in our benchmark situation
50: A(1) > C(gamma) > B(0)
50: B(1) > C(gamma) > A(0)
whenever gamma > (1+1/(1+(slope at x=1)))/2 = 0.6, just as the other
methods did.
The advantage of using (1 + 3x + 3x^7 + x^8) / 8 is that then there is a
procedure in which you don't need any calculator or random number
generator, only three coins:
**
** Method 3-coin-FAWRB
** ---------------------
** Ballots: Approval with one option marked "favourite".
** Tally:
** 1. Determine the approval winner, X.
** 2. Draw a ballot at random;
** if it does not approve of X, its favourite, Y, wins.
** 3. Otherwise, toss three coins;
** if they show no heads, X wins.
** 4. If it's one head, draw one further ballot;
** if it's two heads, draw another seven ballots;
** if all three show heads, draw another eight ballots.
** 5. If all drawn ballots approve of X, she wins;
** otherwise, Y wins.
**
Isn't this guaranteed fun?
Jobst
I wrote:
Dear Forest,
well - thanks.
Anyway, there is still room for improvement.
Our last version was this: Let x be the highest approval rate (=approval
score divided by total number of voters). Draw a ballot at random. With
probability 1/(5-4x), the option with the highest approval score amoung
those approved on the drawn ballot wins. Otherwise the favourite of that
ballot wins.
We saw that this method performs well in a large number of situations.
But it seems to me that, with more than three options, it can be hard to
find the optimal strategies because approving a non-approval-winner can
be bad.
For example, consider this case:
33: A > B >> C=D=E
34: C > B=D >> A=E
33: E > D >> A=B=C
Here the C faction can either cooperate with the A faction to give B a
high probability of winning, or with the E faction to give D a high
probability of winning. But when the A faction approves of B but the C
and E factions approve D, it would have been better for the A faction to
have bullet-voted.
The following even simpler method, however, makes it safe to approve of
an option which does not turn out the approval winner:
**
** Method FAWRB (Favourite-or-Approval-Winner Random Ballot):
** -------------------------------------------------------------
** Everyone marks a favourite and may mark any number of "also approved"
** options. The approval winner X and her approval rate x are
** determined. A ballot is drawn at random. If the ballot approves of X,
** X wins with probability 1/(5-4x). Otherwise, or if the ballot does
** not approve of X, its favourite option wins.
**
FAWRB is again monotonic and solves the original challenge problem in
the same way as the other methods we discussed recently. But in the
above situation it makes it safe for the A and C factions to approve of
B and D since only one of the two factions will actually partially
transfer their winning probability from their favourite to the
compromise option.
I guess it should be possible to analyse FAWRBs strategic implications
in detail since the method is so extremely simple!
I'm pretty sure already that with FAWRB you will never have an incentive
to misrepresent your favourite, and seldom or never to approve of one
option while not approving of all more preferred options as well. With
the other methods these variations of "order reversal" would occur more
often I think.
Yours, Jobst
PS: I have not yet thought much about your most recent proposals. Only
it seems that they won't elect any compromise option that's not the
favourite of anyone, right?
[EMAIL PROTECTED] schrieb:
Jobst wrote
...
What do you think about this?
I think you have the "golden touch!"
Forest
----
Election-Methods mailing list - see http://electorama.com/em for list info
----
Election-Methods mailing list - see http://electorama.com/em for list info
