Suppose $P (1) + P (2) = 3P (3)$ and $P (2) + P (3) = P (1)$ for a sample space $(S,P)$ . Find the probability function. So we have the following equations:
$P(1) + P(2) = 3P(3)$
$P(2) + P(3) = P(1)$
I know that by definition, $P(1) + P(2) + P(3) = 1$
I'm not entirely sure what I'm looking for as my answer though. Are there any other equations hiding that could be useful here? Hints are fine.
I've worked out that I can end up with $4P(3) = 1$ by using those three equations.

Here's one way to solve this problem.
Since we know that $P(1) + P(2) + P(3) = 1$ by definition, we have the following three equations:
$P(1) + P(2) = 3P(3)$
$P(2) + P(3) = P(1)$
$P(1) + P(2) + P(3) = 1$
Plugging the first equation into the third, we get $4P(3) = 1$. So $P(3) = \frac{1}{4}$.
From the last equation, we get $P(2) = 1 - P(3) - P(1) \Rightarrow P(2) = \frac{3}{4} - P(1)$.
Now substitute that into the second equation to find that $P(1) = \frac{1}{2}$.
Lastly, we can return to the first equation to find that $P(2) = \frac{1}{4}$.
So we have:
$P(1) = \frac{1}{2}$
$P(2) = \frac{1}{4}$
$P(3) = \frac{1}{4}$
More formally, for all $x \in S = \{1, 2, 3\}$, 
$$P(x) =
\begin{cases}
\frac{1}{2},  & \text{$x$ = 1} \\[2ex]
\frac{1}{4}, & \text{$x \in \{2, 3\}$}
\end{cases}$$
