Probability question about a game I'm making I am really not sure what to do, and I hope someone can help to tell me, how I should calculate my problem. I'm currently trying to make my own board game. When both player 1 and player 2 are playing optimal strategies, player one will on average earn 1/3 of a point for each round played. There are 45 rounds in total, and player 1 needs to win 15 points within for 45 rounds to win the game. My question is, what are the chances that they make it? Or what are the chances they don't make it? I don't know if I have put out enough information so solve the problem. Just in case, here are a little more information:
For each round there are 4 outcomes:  There´s a 1/18 chance player 1 will lose 3 points, a 2/18 chance player 1 wins 2 points, a 5/18 chance player 1 gets 1 point, and 10/18 chance nothing happens. On average a 1/3 point to player 1 per round.
Hoping so much someone can guide me to solve the problem. I know a little bit of Java if its necessary to use that to solve the problem. Thanks.
 A: Preface: It appears the OP wanted the probability that player $1$ wins given that the game ends when the margin in either direction reaches $\pm 15$ points. The answer below considers only the probability that player $1$ wins with at least $15$ points at the end of all $45$ rounds.

The exact answer can be obtained using generating functions (and, preferably, a computer algebra package to do the dirty work of expanding the enormous power series). A pretty good approximation can be obtained by using the normal approximation. The score per round $X$ has a mean, as you say, of $1/3$, and it has a variance of
\begin{align}
\sigma^2 & = \overline{X^2}-\left(\overline{X}\right)^2 \\
         & = \frac79-\left(\frac13\right)^2 = \frac23
\end{align}
Over a total of $45$ rounds, the mean score will be $45(1/3) = 15$, with a variance of $45(2/3) = 30$, giving a standard deviation of $\sqrt{30} \approx 5.48$. Employing a continuous approximation, we want the probability that a normally distributed variable $Y$ with a mean of $15$ and a standard deviation of $5.48$ is equal to at least $14.5$. Normalizing, we have, for a standard normal variable $Z$,
$$
P\left(Z > \frac{-0.5}{5.48}\right) \approx 0.536
$$
Give me a couple of minutes and I'll compute the exact value for comparison. ETA: The exact probability is
\begin{align}
P(X \geq 15)
& = \frac{41532220960430477945855577769472253664557916175161550215}
         {76771979439341965575488257683639040242866370026397499392} \\
& \approx 0.541
\end{align}
Given that we're close to the center of the distribution, that level of accuracy makes sense for a sum of $45$ samples.
A: Brian Tung's answer is exactly right if it's required that Player 1 have a score of 15 or higher at the end of 45 rounds. It's possible, however, that the OP intends that Player 1 can win instantly by attaining a score of 15 or higher at any point. In that case, the probability of winning will be higher than the 54.1% given by Brian Tung, since it includes games where Player 1 reaches a score of 15 or more somewhere in the middle but then loses ground by the end.
There are theoretical ways to calculate such probabilities, but really I think a simulation works great (and can be implemented in Java): simply run ten million (say) simulated games and count the number of such games won by Player 1. When I did so for the first rule (score $\ge15$ after 45 rounds), I got about 54.1% as expected. When I did so for the second rule (score $\ge15$ at any point during the game), I got about 59.7%.
