[EM] expectation with sincere voting, table.
nkklrp at hotmail.com
Sat Apr 1 22:03:31 PST 2000
>MIKE OSSIPOFF wrote:
> > B Approval Plurality IRV Borda
> > .4 1.6 1.6 4/3 3
> > .6 1.6 1.4 4/3 3
> > .25 1.75 1.75 4/3 3
> > .75 1.75 1.25 4/3 3
> > .1 1.9 1.9 4/3 3
> > .9 1.9 1.1 4/3 3
But I changed the IRV entry to 5/3.
>I don't know about the IRV figures, but the Borda figures don't seem
>right to me. For example, in the u(B) = 0.9 case, the Borda figure
>should probably be 2, if calculated in the same way as the others.
I was about to post here tonight to say that maybe there's a
confusing ethical issue about how to weight Borda's performance
in that test, for fair comparison to the other methods. I still
don't know how Borda's entry should be weighted for fair
comparison. Borda's different from all the other methods that
I mentioned for testing this way, because you can vote different
vote-differences between different pairs of candidates.
But let me tell how I got my Borda result, and why I began to
have doubts about the entry's weighting:
Each term of the expectation-improvement ideally has, mulitplied
together, the probability that that particular candidate-pair
are the frontrunners, & so close together that you & one other
same-voting voter change the winner, and the vote-difference
that you're voting between that pair, and the utility-difference
of those 2 candidates. For the probability, I, without considering
it, used the probability that I'd used for the other methods--
the probability that 2 same-voting voters can change the winner
by voting a vote-difference of 1 between that pair. Maybe that's
wrong, but that's how I did it.
Of course I didn't write the probabilities, or know what they
are, since it depends on how many voters there are. I just let
that be an unwritten factor preceding every term in the
Ok, so here's what I got:
1(Ua-Ub) + 1(Ub-Uc) + 2(Ua-Uc)
Some of the authors who described these calculations said that
it's reasonable to assume that doubling the vote difference
that you vote between the 2 candidates doubles the probability
that you'll change the outcome there.
Combining terms, you get:
Since A = 1, & C = 0, that's expression equals 3.
But, again, now I have some doubts about weighting of Borda's
table entry for fair comparison to the other methods.
I used the same unwritten factor in front of each term as with
the other methods: the probability that those 2 candidates are
the 2 frontrunners, and so close that 2 same-voting voters voting
a vote-difference of 1 vote between those 2 candidates will change
the winner between those two candidates.
Now I doubt that that's the right way to do for Borda, but I
don't know what is the right way. So disregard my Borda column
in that table.
I have to go somewhere, so I'll finish this reply tomorrow,
earlier than this. I have more to say tomorrow.
I make a few quick comments below.
>I am assuming the probability of a tie between AB, AC, and BC are equal,
>and that the probability of a three-way tie is negligible. For u(A) =
>1.0, u(B) = 0.9, and u(C) = 0.0, the utilities in the case of a tie
>(broken randomly) if the voter doesn't participate are shown in the
>Outcome utility for voter with u(A)=1, u(B)=.9, u(C)=.1
>where participation breaks a two-way tie:
>Non-participation Plurality Approval Borda
>AB .95 1.00 .95 1.00
>AC .50 1.00 1.00 1.00
>BC .45 .45 .90 .90
With Borda, the utility improvement of breaking an AB tie
is only .1, or so it seems to me at first glance.
>mean(x3) 1.9 2.45 2.85 2.90
>Utility gains for ideal strategies:
>Non-participation Plurality Approval Borda
>AB 0 .05 0 .05
>AC 0 .50 .50 .50
>BC 0 0 .45 .45
>mean(x3) 0 .55 .95 1.00
Well the actual utility gain is quite small due to the large
number of voters. Ideal Borda strategy with 0-info is to rank
in order of preference.
The argument below is something that has occurred to me too, but
I don't think it changes the strategies that one would calculate.
Or the relation between 2 methods' table entries in this test.
We could say it's you & your faction rather than just you &
one same-voting voter. That changes the probabilities, increasing
them, but it increases them for all the methods and all the
pair-comparisons. So I think it's ok to use the 2-voter approach,
and its results seem applicable to calculating strategies &
comparing methods even if it's really a whole faction that votes
>Calculating the strategy for a tie-breaking voter is one way of
>analyzing strategy characteristics of a method, but I don't think it's
>the only way. This approach only looks at a single voter in isolation,
>who only has the power to make or break a tie, and not to put a favorite
>into front-runner territory to begin with.
>Another approach would be to assume that the faction to which the voter
>belongs will behave as a unit (albeit an average-sized one). Since the
>faction has a common goal, there is no need for one member to know who
>the others are or to communicate with these members in order for them to
>act in unison (but must he must assume the others are using the same
>logic). "Faction" in this case would mean a group of voters who have
>incentives to use the same strategy in a given election.
>I believe that the tie-breaker approach would fail to show any incentive
>for truncation or order-reversal under Borda or pairwise methods, while
>the second approach would at times.
My initial impression is that it's the same either way. But
of course there's an issue about whether strategy would happen
with 0-info. Maybe it's unlikely enough that it's safe to ignore
it. I hope so.
A more complete reply tomorrow.
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