[EM] (no subject)

[MIKE OSSIPOFF]

Forest was the first to mention the Better-Than-Expectation strategy for 
Approval--the strategy whereby a voter votes for the candidates who are 
better than his/her expectation for the election, better than the value of 
the election. So the voter using that strategy votes for a candidate if that 
candidate is so good that s/he would rather have that candidate in office 
than hold the election.

One can come up with situations in which that isn´t optimal. But it 
maximizes one´s utility expectation if certain approximations or assumptions 
are made. One usual assumption is that there are so many voters that one´s 
own ballot won´t change the probabilities significantly. By one approach, 
it´s also necessary to assume that the voters are so numerous that ties & 
near-ties will have only 2 members, and that Weber´s Pij = Wi*Wj, the 
product of the win-probabilities of i & j.

But, instead of the last 2 assumptions named in the previous paragraph, it 
would also be enough to assume that when your vote for a candidate increases 
his win-probability, it decreases everyone else´s win-probability by a 
uniform factor.

That´s the approach that Russ used, except that he didn´t state that 
assumption.

Russ, don´t take any of this as criticism--I´m just telling you so that 
you´ll know.
In your derivation-description, you stated the goal "to keep the sum of all 
probabilities at unity without changing the probability ratios." But keeping 
the other candidates win-probabilities in the same ratios isn´t a goal of 
the derivation; it´s an  assumption by which Better-Than-Expectation 
maximizes the voter´s utility expectation. It´s important to state 
assumptions, and that particular assumption is really key to the derivation.

You said that delta Pj the increase in candidate j´s win-probability if the 
voter votes for j. But later you say that actually that increase is 
different from delta Pj. Preceding a quantiy by delta usually indicates a 
change in that quantity. You started out that way, but then, in your 
derivation,  delta Pj no longer represented that change.

One could let delta Pj really always stand for the increase in j´s 
win-probability. That makes for a more direct derivation:

Let E = your expectation in the election if you don´t vote for j.

Let Pj = candidate j´s probability of winning if you don´t vote for j.

Let delta Pj = the amount by which j´s win-probability increases if you vote 
for j.

Let Uj be j´s utility for that you.

So, if the you vote for j, j´s win-probability will be Pj + delta Pj.

So, if the you vote for j, then j´s contribution to the expectation is (Pj + 
delta Pj)Uj.

When the you vote for j, what´s the combined contribution of the other 
candidates to the expectation? Well, first, what´s their contribution if you 
don´t vote for j? It´s E minus j´s contribution if you don´t vote for j. 
That´s E - PjUj.

Now you want their combined contribution to the expectation when you make 
their wins less probable by voting for j. Aside from the fact that it turns 
out to accomplish our goal, it´s a reasonable simplifying assumption to 
assume that we reduce their win probabilities by a uniform factor.

Then, the combined contribution of the non-j candidates to your expectation 
is it´s initial value,
E - PjUj, mulitplied by the factor by which you reduce their win 
probabilities.

If you don´t vote for j, the probability that a non-j candidate will win is 
1 - Pj.

If you vote for j, the probability that a non-j candidate will win is 1 - Pj 
- delta Pj. That´s one minus j´s new probability of winning.

So, multiply the initial non-j candidates´ expectation contribution, E - 
PjUj, by the ratio of the probabilities in he previous 2 paragraphs:

(E - PjUj) * ( (1 - Pj - delta Pj)/(1 - Pj)) That´s the j candidates´new 
contribution to your expectation.

So, add j´s new contribution to the expectation and that of the non-j 
candidates:

(Pj + delta Pj)Uj + (E - PjUj)(1 - Pj - delta Pj)/(1 - Pj)

We want the new expectation to be more than the initial expectation:

E < (Pj + delta Pj)Uj + (E - PjUj)(1 - Pj - delta Pj)/(1 - Pj)

Solve that for Uj. You get:

Uj > E

Because of the directness of that derivation, it doesn´t need summations to 
be written, and that makes it acceptable to people who wouldn´t want to look 
at something with summation notation.

Also, because delta Pj continues to mean the increase in Pj, this derivation 
is straightforward and direct.

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