Calculating Simple Functions of Discrete Random Variables Say I have a random variable $X$ that has some distribution (e.g. the value obtained when rolling a standard dice). I'm trying to find out whether the probability distributions of functions of that variable are calculated by simply taking the probability for $x_1$, $x_2$, $x_3$.... $x_n$ and assigning them to $f(x_1),~ f(x_2),\dots,~f(x_n)$ or whether there is a more complicated process, and that depends perhaps on the functions and types, number of etc r.v.'s themselves?
For example, say the player of a game scores points based on the value of the above dice they roll (whose r.v. is $X$), according to:
$$f(X) = X^2 + 2X + 2 $$
For the dice: $X = 1, 2, 3, 4, 5$ or $6$. And of course $\Bbb P(X = x_{1-6}) = 1/6$
Calculating $f(X)$ for the dice's values, the possible points you can score are: $5, 10, 17, 26, 37$ and $50$. 
Is then the probability distribution for the points you can score simply: $$\Bbb P(f(X) = 5) = 1/6; \Bbb P(f(X)=10) = 1/6,\dots, \Bbb P(f(X) = 50) = 1/6 $$? Would it be accurate too to say that the score $f(X)$ is itself a random variable?
I'm hoping to extend this concept to things like the addition of different r.v.'s - i.e. to understand what it means to add two r.v.'s $X + Y$ - but I've heard this is much more complicated, so I thought I'd start with this.
Many thanks for all your insight, really appreciate it.
 A: Yes, a function of a random variable is a random variable as well. There are a few cases where the function can "remove all randomness" (e.g. $X$ is $-1$ or $1$ with equal probability and $f(X)=X^2$), but even in these cases $f(X)$ can still be considered a random variable.
Your probabilities for $f(X)$ with the die roll example are correct. Generally, for a discrete random variable the calculation is identical: you just plug in the $X$ values to get the $f(X)$ values and carry over the same numerical probability for each outcome. Of course you have to be a little careful if multiple $X$ values give the same function output, and, in this case, you need to combine the probabilities appropriately.
There is a general result for continuous random variables. You need to know calculus to apply this method though. See the Wikipedia page here. You have to be careful with functions that don't satisfy certain properties (they need to be one-to-one), e.g. $f(X)=X^2$ can be tricky. 
For two random variables (or three more random variables, and so on), you need the concept of a joint probability distribution. This can become more complicated though. For multiple discrete random variables, you can generally calculate the probabilities similarly to the way you have already done.
A: Yes, that is indeed how it works in this case, and that $f(X)$ is indeed a random variable in its own right.
To see this more clearly, we can calculate
\begin{align} \\
\ \Bbb P(f(X)=5) & = \Bbb P(X^2+2X+2=5) \\
\ & = \Bbb P(X^2+2X-3=0) \\
\ & = \Bbb P((X+3)(X-1)=0) \\
\ & = \Bbb P(X=-3)+\Bbb P(X=1) \\
\ & = 0 + 1/6 \\
\ & = 1/6
\end{align}
Similarly for the others.
A: Comment: Consider $Y = g(X) = (X - 2.5)^2,$ just to see what happens when two $X$-values can yield a single $Y$-value. Here is a graph of the four $Y$-values and their
probabilities. (Horizontal green lines at $0, 1/6, 2/6.$)

Also, notice that $E(X) = 3.5,$ but $E(Y) = 47/12 \approx 3.9167,$ not $(3.5-2.5)^2 = 1.$
