[EM] Clarifying Bias-Free definition
Joseph Malkevitch
malkevitch at york.cuny.edu
Sun Jan 21 10:54:30 PST 2007
Dear Mike,
1. What is the value of your expression when a = 0?
2. What role do you see the constant 1/e as playing?
Thanks.
Joe
On Jan 21, 2007, at 12:11 PM, Michael Ossipoff wrote:
>
> Bias-Free is a divisor method, in the strictest sense of the word.
> You know
> how the divisor methods work, so I won't go into that, except to
> say that
> they all choose a rounding point between consecutive integers a &
> b, by a
> formula that specifies a function of a and b.
>
> For example, with Webster the function is (a+b)/2. With Hill it's
> sqr(a*b).
> With Jefferson it's b. With Adams it's a.
>
> So, with Bias-Free, it's ((b^b)/(a^a))(1/e).
>
> 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:
http://www.york.cuny.edu/~malk
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