[EM] Joe: Could you explain the question?

Joseph Malkevitch malkevitch at york.cuny.edu
Tue Jan 23 12:38:58 PST 2007

Dear Mike,

It turns out that Huntington showed that corresponding to the divisor  
methods one can think of these methods as "rank index" methods, or  
what I like to call table methods. One constructs a table from the  
original population where the table involved depends on the method.  
(This is the way the Census Bureau describes Huntington-Hill on its  
web page.) For Webster the table is constructed by dividing the  
populations by a + 1/2 where a = number of states, starting with a =  
0 for the first row, a= 1 for the second, etc. . So one obtains rows  
for the table by dividing by 1/2, 3/2, 5/2, etc. (In this way of  
thinking one can easily give each state one seat, or whatever, at the  
start.) However, since all that matters for assigning h seats is to  
assign the seats to the entries in the table ordered by size  (after  
we give each state one seat if required) we can divide instead by 1,  
3, 5, etc. rather than the numbers above. This is the way that  
Webster (St. Lague) is described in Europe. Viewed in this framework  
the presence or lack of the factor 1/e makes no difference. One would  
get the same apportionment either way. This means there are are  
"equivalence classes" of formulas which for the rank index point of  
view one can use and which will give the same apportionment.

That is what I was trying to get at.

Huntington called his method the Method of Equal Proportions which is  
a "clever" name but what it really does is optimize some measure of  
fairness for transfer between pairs of states as well as some  
"axiom,"  while Webster optimizes a different measure as well as some  
axioms. When it comes to bias, there are many different kinds of bias  
that one might talk about and differing ways to measure that bias.  
Calling something "bias free" does not shed extra light without more  



On Jan 23, 2007, at 8:37 AM, Michael Ossipoff wrote:

> Joe--
> You asked what is the role of 1/e in the formula for Bias-Free's  
> round-off
> point.  I tried to answer by saying that ((b^b)/(a^a))(1/e) is just  
> the
> result I got when looking for the round-off point that would make  
> cycles'
> s/q = 1. That's all I can say about where the formula came from.
> What was the meanng of your question? In what sense was "role"  
> intended?
> Unless I made an error (and I probably didn't make an error), the  
> formula
> for Weighted Bias-Free doesn't contain "e". Weighted Bias-Free is  
> gotten by
> a/q-1 and b/q-1 by a weight that approximates the frequency density of
> states at a particular q value. I approximate that density function  
> by:
> B/(q+A), which has the same general behavior as the distribution  
> function,
> though of course it will be a rough approximation.
> Mike Ossipoff
> ----
> election-methods mailing list - see http://electorama.com/em for  
> list info

Joseph Malkevitch
Department of Mathematics
York College (CUNY)
Jamaica, New York 11451

Phone: 718-262-2551 (Voicemail available)

My new email is:

malkevitch at york.cuny.edu

web page:


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