[EM] counting ballots

[Paul Kislanko]

Jobst already provided the correct answer, which is part of the mathematical
literature. No need to publish yours.

 

> -----Original Message-----
> From: election-methods-bounces at electorama.com 
> [mailto:election-methods-bounces at electorama.com] On Behalf Of 
> Warren Smith
> Sent: Wednesday, December 14, 2005 7:02 PM
> To: election-methods at electorama.com
> Subject: [EM] counting ballots
> 
> >Kislanko:
> The number of full ranked ballots is just the number of 
> permutations of N
> alternatives = N!
>  
> If equal ranknigs are allowed, it's N! + 2^N - 1
>  
> If truncation is allowed it is approximately N! * e
>  
> And if both truncation AND equal rankings are allowed it's 
> approximately N!
> * e + 2^N - 1
> 
> 
> --the statements containing "2^N"  are false.
> 
> The true answers are in one of the chapters of my (as yet unpublished)
> book.   Anyhow, letting E(N) denote the answer, there are 
> various direct and indirect
> expressions for E(N).
> E(N) = #ballots for N candidates with equal rankings allowed. 
>  Truncation forbidden.
> E(0)=E(1)=1, E(2)=3 E(3)=13 E(4)=75 E(5)=541
> Amazingly, E(N) is the nearest integer to
>    N! / (2*ln(2)^(N+1))    when  N=0,1,...,15,16
> but this is false when N=17.  It is true asymptitically for N large.
> 
> Generating function:
> sum  E(N) * x^N / N! = 1/(2-exp(x))
> N>=0
> 
> 
> 
> I also have the exact result if truncatio is allowed too, but 
> it is messier.
> wds
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