What could be the general formula be in this case? A runner can take a step forward, backward,left and right with equal probability. Find the probability that after $n$ steps they will be just 1 step away from their initial point.
This is what I obtained as the general formula after a series of manipulations.
If the total number of moves is (2n+1)
, the total number of different ways is given by
$$4\sum_{r=0}^n\binom {2n+1}{r+1,r,n-r,n-r}=4\sum_{r=0}^n \frac {(2n+1)!}{(r+1)!r!(n-r)!(n-r)!}=4\binom {2n+1}n^2$$
and the probability is given by
$$\frac {4\binom {2n+1}n^2}{4^{2n+1}}=\frac {\binom {2n+1}n^2}{4^{2n}}$$
Now if the question was say...
"....after taking $n$ steps they will just be $m$ steps away from their initial point."
Could someone help me how would the general formula turn out?
 A: The probability that, after $n$ steps, you end up at $(x,y)$, is
$$
\binom{n}{\frac{n+x+y}{2}} \binom{n}{\frac{n+x-y}2}2^{-n}\qquad \text{if $x+y+n$ is even.}
$$
When $x+y+n$ is odd, the probability is zero. Here is a proof: Number of N step Random walks on 2D infinite grid or lattice.
We just need to sum this equation over all $(x,y)$ which are $m$ steps from the origin. The set of these points resembles a square with vertices $(0,\pm m)$ and $(\pm m,0)$. Due to rotational symmetry, we can compute this by summing over one side of this square, and multiplying by four. The points in the quadrant where $x\ge 0$ and $y>0$ are given by $(k,m-k)$ for $k\in \{0,\dots,m-1\}$.
Therefore, as long as $n+m$ is even, the probability that you end up $m$ squares from the origin is
$$
4\sum_{k=0}^{m-1}\binom{n}{\frac{n+k+(m-k)}{2}}\binom{n}{\frac{n+k-(m-k)}{2}}2^{-n}=2^{-(n-2)}\binom{n}{\frac{n+m}{2}}\sum_{k=0}^{m-1}\binom{n}{\frac{n-m}{2}+k}
$$
There is no way to simplify the summation in the final expression.
