Up to Elections: Results and Voting systems
# Bias in Allocation Rules for Party List PR:

## by Mike Ossipoff

### To write to Mike write nkklrp before the "@" sign, and then write hotmail.com after the "@" sign.

I've elsewhere described Webster's method & Jefferson's method
Jefferson's method is the same as Webster's except that, where
Webster rounds to the nearest whole number, Jefferson rounds
each party *down*. Rounding down by a particular amount will
hurt a small party more, percentage-wise. So Jefferson's method
is biased against small parties, biased in favor of large parties.
John Quincy Adams' method is the same as Webster's method except
that, where Webster rounds to the nearest whole number, Adams
round each party *up*. Rounding up by a particular amount will
help a small party more, percentage-wise. So Adams' method is
biased in favor of small parties.
Webster's method doesn't consistently round up or down, and for
that reason it doesn't favor or disfavor small parties.
Because it rounds to the nearest whole number, Webster rounds
a party up if it's above the halfway point between the 2 adjacent
whole numbers. Obviously it will, on average, round up half the
time, & down half the time, with no preference for 1 or the other.
That's why I say Webster doesn't consistently round up or down.

Largest Remainder, when parties below the inclusion threshold
aren't counted when calculating the quota, is unbiased (though
it has other problems which I've described elsewhere). [Maybe
here should be a link to the article where I define the methods
& mention Largest Remainder's problems].

## Bias of Hill's method:

First I'd like to show it by example, and then algebraically for
the general case.
Say we're talking about parties for which, when we divide their
votes by the common divisor, we get a result between 1 & 2. Whether
we round up or down depends on whether that party's result is above
or below the "round-off point". Where is the roundoff point for Hill's
method, between the whole numbers 1 & 2?:
As I said, Hill's principle is to put each party's seats as
close as possible to the result of dividing its votes by the
common divisor, which is also Webster's principle. The difference
is that Webster goes by closeness in terms of _difference_, while
Hill goes by closeness in terms of _ratio_.
So, while Webster rounds to the nearest (by difference) whole
number, Hill rounds to the nearest (by ratio) whole number.
So, for parties with division result between 1 & 2, where is
the round-off point?
Suppose a party's division result is equal to the square root
of 2. I'm going to use "1.4" to stand for the square root of 2,
even though of course 1.4 isn't *exactly* the square root of 2.
Since 1.4... is the square root of 2, if you multiply it by
itself you get 2. So 1.4... differs from 2 by a factor of 1.4...
Of course it also differs from 1 by a factor of 1.4...
So 1.4... differs from 1 & 2 by the same factor. Anything above
1.4... will differ from 2 by a smaller factor than the factor
by which it differs from 1, and that would round a party up.
Well, 1.4... is of course less than 1.5, the midpoint between
1 & 2. So, in the long run, a party is more likely to have a
division result above 1.4 than below it, when that party's
division result is between 1 & 2. Therefore, a party with
division result between 1 & 2 is more likely to be rounded up
by Hill's method than it is to be rounded down.
It's reasonable that this would also be true for other adjacent
pairs of whole numbers: In general, Hill's round-off point is
lower than the midpoint of those numbers, and so, in general,
Hill is more likely to round up than down.
Furthermore, this tendency to favor rounding up is more pronounced
for lower pairs of adjacent whole numbers than for higher ones,
because the numbers in those lower pairs, like 1 & 2 differ from
eachother by a larger factor. That's plausible, but we needn't
prove it. The important thing is that Hill tends to round up.
Just as always rounding up favors small parties in Adams' method,
rounding up more often than down also favors small parties in
Hill's method, though to a lesser degree, since Hill doesn't
always round up.

But I should show that that tendency to round up is true
in general, because the round-off point is always lower than
the midpoint between 2 adjacent whole numbers:
Let a & b be 2 successive whole numbers.
c is a number that differs from a & b by the same factor,
meaning that c/a = b/c
If c/a = b/c, then, multiplying both sides of the equation by
ac, then:
c squared = ab Taking the square root of both sides of the equation:
c = square root of ab
c is what is called the "geometric mean" of a & b.
So that geometric mean of a & bc is the roundoff point. It's
necessary to show that the geometric mean of 2 successive whole
numbers, a & b, is always lower than their midpoint, the point
halfway between them.
The midpoint between a & b is:
(a+b)/2 To show why that is, let d = the midpoint, equidistant
from a & b:
d-a = b-d Adding a & d to both sides of the equation:
2d = a+b
d = (a+b)/2
So it's necessary to show that:
(a+b)/2 > square root of ab [ ">" means "is greater than"]
If the above "inequality" is true, then, squaring both sides of the
equation:
(a+b) squared /4 > ab Multiplying both sides by 4:
(a+b) squared > 4ab
[I'm going to start representing "squared" by the FORTRAN notation
"**2]
Since a & b are successive whole numbers, b = a + 1
Substituting that for b in the above equation:
(2a+1)**2 > 4a**2 + 4a [because a+a+1 = 2a+1, and
a(a+1) = a**2 + a]
(2a+1)(2a+1) > 4a**2 + 4a Multiplying out the left side of the
equation:
4a**2 + 4a + 1 > 4a**2 + 4a Subtracting 4a**2 + 4a from both sides
of the equation:
1 > 0
A true statement. Had we started out with an inequality in the
opposite direction, had we begun with the "less than" sign, "<"
then we would have gotten the false statement "1 < 0".
Of course had we started with an "=" sign, we'd have gotten
the false statement "1 = 0".
So the original statement is correct:
(a+b)/2 > square root of ab

So the geometric mean of any 2 successive whole numbers is less
than their midpoint, and so Hill will round up more likely than
down, and so Hill is biased in favor of small parties, because
rounding a small party up by a particular amount helps it more,
percentage-wise, than rounding a larger party up by the same amount.
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