Probability of a Particle. Let there be a $17 \times 17$ grid. Also, there exist a particle at the center. Every
minute, the particle moves horizontally or vertically. If the particle reaches the edge of the grid, it is absorbed.


*

*Calculate the probability that after 8 minutes the particle is absorbed.  

*Let $A_i$  be the event “the particle is at the center of the grid after $i$ minutes.”
Calculate $\Pr[A_4]$ (knowing that $A_0$ is certain, by assumption).


My thoughts:


*

*I think it is just $P=4 \cdot(1/4)^8$  

*I am very confused on this one.

 A: For 1, you are correct.  You can move any direction the first move, but then have to keep going in that direction every time.
For 2, use symmetry.  After $2$ minutes, the particle can only be in $9$ locations, but there are only three classes:  the center, two cells away from the center in a straight line, or diagonally next to the center.  Calculate the chance that it is in each of them.  Then calculate the chance of starting in each one and returning to the center.  For example, the chance that after $2$ minutes it is at the center is $4 \cdot (1/4)^2=1/4$ because you can move in any direction the first move, but then must reverse the move to get back.  This gives a chance of $(1/4)^2$ of returning to the center at both $2$ minutes and $4$ minutes.  Add this to the other routes and you are there.
A: The position of the particle $n \mapsto (X_n,Y_n)$ is a (finite) Markov chain, with transition probabilities:
$$
   \Pr\left(X_{i+1}=x, Y_{i+1}=y \mid X_{i}=x_0, Y_i=y_0\right)= \frac{1}{4} \chi\left( |x-x_0| + |y-y_0| = 1 \right)
$$
where $\chi(A)$ is the indicator function. 
For the 1, there is a unique path leading to one of the four absorbing boundaries in 8 moves, thus your result is correct. 
For the 2, denote by $P$ a conditional transition probabilities matrix with elements $$P_{(x_0,y_0),(x_1,y_1)} = \mathbb{P}\left(X_{i+1}=x_1, Y_{i+1}=y_1 \mid X_{i}=x_0, Y_{i}=y_0 \right)$$
Then, the probability you seek is $\left(P^4\right)_{(0,0),(0,0)}$ which equals $\frac{9}{64}$.

Alternative
A single step distribution is equally distributed between 4 available moves. Consider its probability generating function:
$$
    \sum_{x,y} u^x v^y \mathbb{P}\left(\Delta X = x, \Delta Y=y\right) = \frac{1}{4} \left( u + \frac{1}{u} + v + \frac{1}{v} \right)
$$
After $n$ moves, the position of the particle is going to be a sum of these shits. The probability generating function of the sum of iid random variables equals:
$$ 
   \frac{1}{4^n} \left( u + \frac{1}{u} + v + \frac{1}{v} \right)^n
$$
For question one, $n=8$ and you are interested is the total of coefficients by $u^8$, $v^8$, $u^{-8}$ and $v^{-8}$
In[42]:= With[{pgf = ((u + 1/u + v + 1/v)/4)^8}, 
 Coefficient[pgf, u^8] + Coefficient[pgf, v^8] + 
  Coefficient[pgf, u^(-8)] + Coefficient[pgf, v^(-8)]]

Out[42]= 1/16384

For question two, $n=4$ and you are interested in the coefficient by $u^0 v^0$:
In[46]:= With[{pgf = ((u + 1/u + v + 1/v)/4)^4}, 
 Coefficient[Coefficient[pgf, u, 0], v, 0]]

Out[46]= 9/64

A: For part $2$, $A_4$ represents particle is at center after four moves.  Possible paths are:
$1.)$ move back twice and then forward twice $\times 4$ (four directions possible)
$2.)$ move forward, turn left, turn back, turn right $\times 16$ (try to visualize)
$3.)$ move forward, move back,move forward, move back $\times (4\times 4)$ because after coming back for first time , you can go again in $4$ direction for $3^{rd}$ and $4^{th}$ move.
Since each path has equal probability $=(\frac{1}{4})^4$, therefore total probability $=36(\frac{1}{4})^4=\frac{9}{64}$
A: 1.You are right 
2.One straightforward way-
Number of ways to get to centre times $(\frac {1}{4})^4$
Number of ways=($nA_2^2+4+8+8$). The four is for two steps ahead and then back in four directions. The 8 for the square motions(clock and anti-clock). $nA_2$(=4) is number of ways to come back to centre in two steps. Another 8 for L shaped motions.
So, 9/64.
