27 May
2015
27 May
'15
1:17 p.m.
So the expected value of F(N) ought to be Sum[ Binomial[N-1,Q-1]*(k-1)^(N-Q)*k^(1-N)*F(N), {N,Q,infinity} ] and whenever the sum is doable you will get a closed form.
--Sorry, that sounded good when I wrote it, but it yields wrong answers. F(N)=1 ==> sum/k=1. F(N)=N ==> sum/k=kQ. F(N)=(N-1)*N/2 ==> sum/k=kQ (kQ+k-2)/2. F(N)=(N choose 3) ==> sum/k=k*Q * (k^2*Q^2+(k-2)*k*3*Q+?) / 6 where ? is a certain quadratic function of k, and these answers are the WRONG answers to Propp's problem. Jeez. This problem evidently is out to get me. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)