Question about finding joint PMF 
A particle starts at $(0,0)$ and moves in one-unit independent steps with equal probabilities of $\frac{1}{4}$ in each of the four directions: north,south,east and west. Let S equal to the east-west position and T the north-south position after three steps. 

Question:Define the joint PMF of S and T..
I have hard time understanding the problem, say the particle moves in the west three times, then the probability is $\frac{1}{64}$. I just have hard time categorizing kinds of situations and need help on that. Appreciate it!
 A: Let $(S_0,T_0)=0$ and define $\{(S_n,T_n):n=0,1,2,\ldots\}$ by the transition probabilities
$$
\mathbb P((S_{n+1},T_{n+1}) = (i',j') \mid (S_n,T_n) = (i,j) = \begin{cases}
\frac14,& |i'-i| + |j'-j| = 1\\
0,& \text{otherwise}.
\end{cases}
$$
By symmetry, 
$$
\mathbb P((S_1,T_1) = (1,0)) = \mathbb P((S_1,T_1) = (0,1)) = \mathbb P((S_1,T_1) = (-1,0)) = \mathbb P((S_1,T_1) = (0,-1)) = \frac14. 
$$
For the distribution of $(S_2,T_2)$, there are three cases. First, the case where two steps are made in the same direction:
$$
\mathbb P((S_2,T_2) = (2,0)) = \mathbb P((S_2,T_2) = (0,2)) = \mathbb P((S_2,T_2) = (-2,0)) = \mathbb P((S_2,T_2) = (0,-2)).
$$
These probabilities are given by
\begin{align}
\mathbb P((S_2,T_2) = (2,0)) &= \mathbb P((S_2,T_2) = (2,0)\mid (S_1,T_1)=(1,0))\mathbb P((S_1,T_1)=(1,0))\\
&= \left(\frac14\right)^2\\
&= \frac1{16}.
\end{align}
Second, the case where one horizontal step is made and one vertical step is made:
$$
\mathbb P((S_2,T_2) = (1,1)) = P((S_2,T_2) = (-1,1)) = P((S_2,T_2) = (1,-1)) = P((S_2,T_2) = (-1,-1)).
$$
Since the steps could have been made in two different orders, these probabilities are $2\cdot\frac1{16}=\frac18$.
Third, the case where $(S_2,T_2)=(0,0)$. There are four ways this can happen, so the probability is $4\cdot\frac1{16}=\frac14$.
The distribution of $(S_3,T_3)$ may be found by a similar analysis - the probabilities will be multiples of $\frac1{4^3}=\frac1{64}$ depending on how many paths there are that end at a given point.
A: 
I have hard time understanding the problem, say the particle moves in the west three times, then the probability is $1/64$.

That is to say, $\mathsf P\{S{=}{-3}, T{=}0\} = 1/4^3$, and that is the case.   It seems that you do understand.
Now do $\mathsf P\{S{=}{-2}, T{=}1\}$, $\mathsf P\{S{=}{-2}, T{=}{-1}\}$, $\mathsf P\{S{=}{-1},T{=}2\}$, $\mathsf P\{S{=}{-1},T{=}0\}$, $\mathsf P\{S{=}{-1},T{=}{-2}\}$, ... and so on.
$\star$ However, remember that some destinations can be reached by more than one path. 
