[EM] Square Vote
Peter Barath
peb at freemail.hu
Fri Dec 12 05:08:06 PST 2008
On 2008 December 04 Thursday 23.46, Peter Barath wrote:
> - The voting authority makes an estimation for the d density
> and makes it public.
> - Every voter declares which option she wants and the value for
> that.
> - The option with the bigger sum wins.
> - If a voter is on the winning side, and she declared value a,
> she must pay to the treasury d*a^2
By thinking a little more I think now:
This method can be transformulated: everybody declares, how much
she is willing to pay if her side wins, and when counting the
votes the voting authority divides every vote with d (the
predeclared estimated probability density) and take the square
root of the result.
Of course, the constant divider does not influence the result,
so it can be left out of the play.
So the method is:
- Every voter declares how much she wants to pay.
- The voting authority sums the square roots.
- The bigger sum wins.
(There would be voters asking: If an issue is twice as important
for me, why to pay four times more? I would answer: twice for
the twice more importance, and twice for the twice more probability
to influence the result.)
This way, the burden of estimating the probability density (the
probability of the margin being between $0 and $+1) moves
from the authority to the voter. (But it was there already: only
with the option that she simply believe the "official" one.)
The payments will be simply the declared ones. But who is to pay?
Now, I don't thik this is a very simple question, but here is what
I think: There are three possibilities:
- Everybody pay.
- The winners pay.
- The winners pay, the losers get their declared money.
Let's see in the first case, wher everybody pays, what the
rational payment (which in this case can be paid when the
actual voting takes place) is.
Let d be that probability density which were real if the voters
declared the square of their V values, V^2 Actually, they
will declare a different (much smaller) sum. Let c be the constant
averaging this effect so that the voters usually pay c^2 * V^2,
so their counted sum is c*V
But this way the real probability density will change to d/c
Look at a single voter now, what her rational decision for the
payment would be? Let z be her personal constant so she
pays z^2 * V^2 counts z*V
What is her expected value if she pays z^2 * V^2 instead of nothing?
- z^2 * V^2 for she paid that amount, plus the bet: the probability
multiplied with the value, that is the probability density multiplied
with the counted sum multiplied with the value.
density: d/c
counted: z*V
Value: V
So her expected value is: (d/c)*z*V*V - z^2*V^2
The maximum is where z = d/(2*c)
The equilibrium (is this the proper concept here?) is where the
usual constant equals the usual contant, that is where z = c
so the stable c is when c = d/(2*c) that is c^2 = d/2
So the rational payment for a voter with V value is (d/2)*V^2
So the counted rational sum is the square root of this, so is
proportional to the value of the question, so this voting is
sincere.
If only the winners pay, the rational payment is (d/2)*V^2/P
where P the probability of her side to win. So this is not
sincere, it motivates the less likely winning side to "overstate"
their value.
If the losers get their declared money, the result is even worse.
The likely losers will grossly overstate their values in hope of
getting that money.
Generally, this is my reaction to Abd: if losers get real
compensations, I don't know how to defend against false claims.
So my Square Vote proposal gets this simple:
- Every voter pays whatever she wants to the treasury.
- The square roots are counted.
- Biggest sum wins.
Of course, this way we can vote for more than 2 options, but
I don't have enough math to decide whether sincerity remains.
Peter Barath
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