The distribution of cards may have been realized as follows. First we give $13+13$ cards to N+S. If we do not have exactly eight $\spadesuit$ cards among them, we ignore this case, it is not contributing to the conditional probability. Else we go on. There are $26=21+5$ cards remaining.
Let us count then the number of ways to split the five $\spadesuit$ cards for the EW axis. We have the possibilities, and the corresponding number of ways to realize them, determined by the E hand:
- $5=5+0$, totally $\binom 55\cdot \binom {21}8$ distributions,
- $5=4+1$, totally $\binom 54\cdot \binom {21}9$ distributions,
- $5=3+2$, totally $\binom 53\cdot \binom {21}{10}$ distributions,
- $5=2+3$, totally $\binom 52\cdot \binom {21}{11}$ distributions,
- $5=1+4$, totally $\binom 51\cdot \binom {21}{12}$ distributions,
- $5=5+0$, totally $\binom 50\cdot \binom {21}{13}$ distributions.
Each "split" has to be weighted with the corresponding number of distributions. So the probabilities are:
sage: for k in [0..5]:
....: p = binomial(5,k)*binomial(21,13-k)/binomial(26,13)
....: print "5=%s+%s :: probability %s ~ %s" % (k, 5-k, p, p.n())
....:
5=0+5 :: probability 9/460 ~ 0.0195652173913043
5=1+4 :: probability 13/92 ~ 0.141304347826087
5=2+3 :: probability 39/115 ~ 0.339130434782609
5=3+2 :: probability 39/115 ~ 0.339130434782609
5=4+1 :: probability 13/92 ~ 0.141304347826087
5=5+0 :: probability 9/460 ~ 0.0195652173913043
(The $5=5+0$ cases will probably feel not so rare, because they remain a long time in the memory.)
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.
Best Answer
The event $E_{-2,-1} = \{S=-2,T=-1\}$ means that in the three steps you chose, one had to be down and two had to be to the left, in any order. So, the number of ways to choose the particular step when you went down was $\binom{3}{2}$.
At the same time, at each particular step, the probability of going down is $p_D = 1/4$, and the probability of going left is $p_L = 1/4$ each. Therefore, $$ \mathbb{P}\left[E_{-2,-1}\right] = \binom{3}{2} \cdot p_D \cdot p_L^2 = \binom{3}{2} \left(\frac{1}{4}\right)^3 $$
To find the marginal PMFs of $X$ and $Y$, you can find their joint PMF using a similar construction to the one above, and then sum across one of the variables, which will give you the PMF of the other one.