[EM] expectation with sincere voting, table.

MIKE OSSIPOFF nkklrp at hotmail.com
Sat Apr 1 22:03:31 PST 2000

Bart 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
methods' determinations.

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
>first column.
>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
with you.

>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.

Mike Ossipoff

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