Expected number of points in k turns? I'm not too familiar with expected values so I'm wondering if people could verify if I'm on the right track with my thought process here.
If the probability of winning one point in a turn is 1/9, what is the expected number of points we will have in k turns?
Is that just k/9, Or am I looking at it in the wrong way? How would I express that in terms of an expected value?
 A: Exactly. This is a textbook example of a binomial distribution with a success probability of $p=\frac{1}{9}$ and $n=k$ trials. The expected total number of successes (i.e., points), is $pn=\frac{k}{9}$. The Wikipedia page gives much more information (like the variance etc.), or you can browse through the binomial tag.
A: Answer avoiding binomial distribution and focused on expectation.
You are looking in the right direction.
I preassume that in one turn at most one point can be won.
Expectation is linear wich means the we have: $$\mathbb EcX=c\mathbb EX$$ and $$\mathbb E(X+Y) =\mathbb EX+\mathbb EY$$ for constand $c$ and random variables $X,Y$ defined on the same probability space.
For turn $i\in\{1,\dots,k\}$ you can define a random variable $X_i$ that takes value $1$ if a point is won in that turn and takes value $0$ otherwise. These random variables have equal distribution (hence equal expectation) and we find:
$$\mathbb EX_1=1\cdot\text{Pr}(X_1=1)+0\cdot\text{Pr}(X_1=0)=\text{Pr}(X_1=1)=\frac19$$
Now observe that $X:=X_1+\cdots+X_k$ is exactly the number of points that will be won in $k$ turns. Application of the mentioned linearity then gives:
$$\mathbb EX=\mathbb E(X_1+\cdots+X_k)=\mathbb EX_1+\cdots+\mathbb EX_k=\frac{k}9$$
