27 May
2015
27 May
'15
6:41 p.m.
Well, ok, let me just try to solve the Q-version of Propp/Pitman problem by brute force. The chance the Nth dice roll yields for the first time Q equal rolls, is k^(-N) times the number of N-letter words from k-size-alphabet containing exactly one Q-subset including the last latter, of all-equal letters, namely Binomial(N-1,Q-1) * (k-1)^(N-Q) * k^(1-N). 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. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)