My first approach was to use a Markov chain: in an n point match, you start at position (n, n) in the MET. After each finished game, you either move up k_W positions (a won game) or you move k_L positions to the left (a lost game), with k_W and k_L being 1, 2 or 3 times the value of the cube. (Respecting the borders of the MET of course). A new game is started until position (0, m) or (m, 0) is reached.
If the probabilities for each k_W and k_L for each position in the MET were known, the problem can easily be solved using a Markov chain. Unfortunately, I don't know them, but their information should somehow be incorporated in the MET.
Pck's answer, P(I win n-m) = 2 * [P_MET(n-away / (n-m)-away) - P_MET(n-away / (n-m-1)-away)], fulfills two requirements: it uses the MET, and the n probabilities are positive and add up to 1. However, I am not convinced for three reasons:
- Using the Rockwell-Kazaross MET, I calculated the probabilities for a 25 point match and it showed that a 25-0 end score is 30 times as likely as a 25-23 end score (resp. 6.2% vs. 0.21%). This sounds a bit unrealistic for players of equal level;
- A MET consists of a pre-Crawford and a post-Crawford part and the latter is not used;
- P_MET(n-away / (n-m)-away) gives us the probability of final scores n-m, n-(m+1), ..., n-(n-1) from match score 0-m, and P_MET(n-away / (n-m-1)-away) gives us the probability of final scores n-(m+1), n-(m+2), ..., n-(n-1) from match score 0-(m+1). Since P_MET(n-away / (n-m)-away) and P_MET(n-away / (n-m-1)-away) are conditional on different match scores, I am not sure that probabilities in red will cancel out.
None of my arguments refute pck's answer though... I'll return to the drawing board as well!