Up to Elections: Results and Voting systems

Bias in Allocation Rules for Party List PR:

by Mike Ossipoff

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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. --