[EM] How to find the voters' honest preferences (Kevin Venzke)
fsimmons at pcc.edu
Mon Sep 9 16:05:56 PDT 2013
> (Well, both methods have strategy there anyway, since the ballots either
> must conform to each other or are interpreted as conforming to each other.
> I don't think you stated it was required for the second ballot to not
> reverse preferences of the first ballot.)
The second ballot, being non-strategic will perforce sometimes reverse
preferences that are implicit in the three slot ballot. However, the
second ballot is only used where the first ballot shows no preference.
This usage gives no incentive for insincere order (even though it does not
always require sincere order). So even though the second ballot will be
sinere if the voter has any impulse for honesty, no matter how small, the
winner between finalists may not be the true preference of the two, since
some of the preferences in that pairwise contest may come from the
That's the price we pay in order to discourage burial.
> If I understand this correctly then this means that if you rate one
> candidate 100%, one candidate 99%, and then a dozen candidates at 0%, you
> will approve only the candidate at 100%. Can that be right?
When the last cup fills only part way, you have to round to either zero or
100%, which ever is closer.
In the three slot case it would work something like this:
Suppose the original cardinal ratings are on a scale of zero to two, but
with high resolution:
A(2), B(1.9), C(1.6), D(1.3), E(1.2), F(1), G(0.7), H(0.2 ), I(0.1), J(0),
Then in B through E where the ratings are strictly between 1 and 2, we add
the fractional parts to get .9 +.6 +.3 +.2 = 2. This tells us that
exactly two of these four ratings should be rounded up: B(2),
Then in G, H, and I which have ratings strictly between zero and one, the
fractional parts add up to one, so exactly one of these gets rounded up.
Putting all of this together we get
A(2), B(2), C(2), D(1), E(1), F(1), G(1), H(0), I(0), J(0), etc.
In this example, the fractional parts that were added together have whole
sums. When that does not happen, we round to the nearest whole (with round
to even tie breaker), etc as in the following example:
A(2), B(1.3), C(1.25), D(1.2), E(1.15), F(0), G(0), etc.
The fractional parts sum to .9, which rounds to one, so the modified
A(2), B(2), C(1), D(1), E(1), F(0), G(0), etc.
I'll try to give the rationale for this in my remaining time on this public
Think of it this way: suppose I use Bernoulli random variables x1, x2, x3,
x4 with expected vales equal to the respective fractional parts .3, .25,
.2, and .15 :
B(1+x1), C(1+x2), D(1+x3), and E(1+x4).
Use a random number generator to get values for x1, x2, x3, and x4. [A
Bernoulli random variable takes on a value of one with probability equal to
its expectation, and zero otherwise.]
If we use these values and all voters do the same kind of thing, the total
cardinal ratings will turn out the same statistically in the case of large
numbers of ballots. But wait, what if x3 and x4 turn out to be one's,
while x1 and x2 turn out to be zeros? Wouldn't it be better to trade those
one's into the the preferred positions?
Now we ask ourselves, what is the expected number of one's? The answer is
the sum of the individual expectations, namely .3 + .25 +.2 +.15 = .9 .
Instead of using the Bernoulli experiments, just use this expected number
of one's as the appropriate number of one's. We expect approximately one
bernoulli variable to be a one. So we round up only the first rating (of
those with ratings strictly between one and two).
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