A golf match consists of 18 independent holes worth 1 point. If a player wins a hole outright they get 1 point, if the hole is tied, each player gets .5 points. The match is over when one player has accumulated at least 9.5 points, which can occur before the 18th hole.
Given that Player A leads or trails by N points with X holes left to play, and player A has probability of winning a hole W, tying a hole T, and losing a hole L (these sum to 1), how can you calculate player A's probability of winning the match, or that the match ends in a tie?
Example: Player A leads by 1 point with 5 holes to play (14th hole is next), with probabilities of hole outcomes (W = .29, T = .42, L = .29)
So far, here is what I'm thinking on the example problem above...
p(win match) = p(win match on 14) + p(win match on 15) + p(win match on 16) + p(win match on 17) + p(win match on 18)
p(win match on 14) = 0 since the largest lead that could be accumulated is 2 with 4 left to play
p(win match on 15) = 0 since the largest lead that could be accumulated is 3 with 3 left to play
p(win match on 16) = p(having 3 point lead going into 16) * p(W + T) + p(having 2 point lead going into 16) * p(W)
p(having 3 point lead going into 16) = p(W)^2 (have to win twice to get to this point)
p(having 2 point lead going into 16) = 2 * p(W)*p(T) (have to win and tie to get to this point)
so p(win match on 16) = p(W)^2 * p(W + T) + 2 * p(W)*p(T) * p(T)
p(win match on 17) = p(having 2 point lead going into 17) * p(W + T) + p(having 1 point lead going into 17) * p(W)
p(having 2 point lead going into 17) = 3 * p(W)*p(T)^2 + 3 * p(L)*p(W)^2 (these are only two ways to gain 1 point over span of 3 holes)
p(having 1 point lead going into 16) = 3 * p(T)^3 + 3 * p(W)*p(L)*P(T) (these are only two ways to gain 0 points over span of 3 holes)
so p(win match on 17) = (3 * p(W)*p(T)^2 + 3 * p(L)*p(W)^2) * p(W + T) + (3 * p(T)^3 + 3 * p(W)*p(L)*P(T)) * P(W)
similar logic applies to 18. Player must be up 1 and win or tie, or be up 0 and win.
Is there a general formula I can use to calculate win or tie probabilities given any status of the match?
Rather than try to enumerate all possibilities, it is much simpler and less error-prone to use recursive equations. Let $p(N,X)$ be the probability that player $A$ wins, given current state $(N,X)$. We have $p(N,X)=1$ if $N \ge 9.5$ and $p(N,X)=0$ if $N \le -9.5$. By conditioning on the next hole, we obtain recursion $$p(N,X) = p(N+1,X-1) \mathbb{P}(W) + p(N,X-1) \mathbb{P}(T) + p(N-1,X-1) \mathbb{P}(L).$$