[EM] Clarifying Bias-Free definition

[Joseph Malkevitch]

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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