TITLE: The 9 step problem
QUESTION [6 upvotes]: A man can take a step forward, backward,left and right with equal probability. Find the probability that after 9 steps he will be just 1 step away from his initial point.
I have done similar questions with movement restricted to forward and backward only ,but this one just blows my mind.

REPLY [3 votes]: Here's a new combinatorial approach. 
Assume that there are $2n+1$ steps.    
Denote directions $H$ (Horizontal) and $V$ (Vertical) and polarities $+$ and $-$,
(ie. $H+$ is right, $H-$ left, $V+$ forward, $V-$ backward).
Of the $2n+1$ steps, choose any $\color{red}{n}$ steps. Number of ways: $\binom {2n+1}n$. 

For the chosen steps, assign polarity "$+$".  
For the remaining $n+1$ steps, assign polarity "$-$". 

$$\begin{array}{&c|&c &c &c | &c}
\hline
&  &+ &- &\text{Total} \\
\hline
H&  &? &?\\
V&  &? &?\\
\hline
& &\color{red}{n} &n+1  &2n+1\\
\hline\end{array}$$
Again, of the $2n+1$ steps, choose any $\color{orange}{n}$ steps. Number of ways: $\binom {2n+1}n$. 

For these $n$ steps,  mark $H$ for those with polarity "$+$" and mark $V$ for those with polarity "$-$".  
For the remaining $n+1$ steps, do the opposite, i.e. 
mark $V$ for those with polarity "$+$" and 
mark $H$ for those with polarity "$-$". 

$$\begin{array}{&c|&c &c &c | &c}
\hline
&  &+ &- &\text{Total} \\
\hline
H&  &\color{orange}{r} &?\\
V&  &? &\color{orange}{n-r}\\
\hline
& &\color{red}{n} &n+1  &2n+1\\
\hline\end{array}
\hspace{3cm}
\begin{array}{&c|&c &c &c | &c}
\hline
&  &+ &- &\text{Total} \\
\hline
H&  &\color{orange}{r} &r+1 & 2r+1\\
V&  &n-r &\color{orange}{n-r} & 2n-2r\\
\hline
& &\color{red}{n} &n+1  &2n+1\\
\hline\end{array}$$
This ensures that $n$ matched polarity step-pairs, with one unmatched step, in this case $H-$. 
Multiply by $4$ for all four directions $(H+, H-, V+, V-)$
This gives total number of combinations as $4\large\binom {2n+1}n^2$.
Hence probability of ending up one step away from original position is $$\frac {4\large{\binom {2n+1}n} ^2}{4^{2n+1}}=\frac {\large\binom {2n+1}n ^2}{4^{2n}}$$

NOTE also that the total number of ways to end up one step away from the original position after $2n+1(=2N-1)$ steps is the same as the total number of ways to end up back at the original position after $2n+2(=2N)$ steps, where  $N=n+1$. 
Assume that there are $2N$ steps.    
Denote directions $H$ (Horizontal) and $V$ (Vertical) and polarities $+$ and $-$,
(ie. $H+$ is right, $H-$ left, $V+$ forward, $V-$ backward).
Of the $2N$ steps, choose any $\color{red}{N}$ steps. Number of ways: $\binom {2N}N$. 

For the chosen steps, assign polarity "$+$".  
For the remaining $N$ steps, assign polarity "$-$". 

$$\begin{array}{&c|&c &c &c | &c}
\hline
&  &+ &- &\text{Total} \\
\hline
H&  &? &?\\
V&  &? &?\\
\hline
& &\color{red}{N} &N  &2N\\
\hline\end{array}$$
Again, of the $2N$ steps, choose any $\color{orange}{N}$ steps. Number of ways: $\binom {2N}N$. 

For these $N$ steps,  mark $H$ for those with polarity "$+$" and mark $V$ for those with polarity "$-$". 
For the remaining $N$ steps, do the opposite, i.e. 
mark $V$ for those with polarity "$+$" and 
mark $H$ for those with polarity "$-$". 

$$\begin{array}{&c|&c &c &c | &c}
\hline
&  &+ &- &\text{Total} \\
\hline
H&  &\color{orange}{r} &?\\
V&  &? &\color{orange}{N-r}\\
\hline
& &\color{red}{N} &N  &2N\\
\hline\end{array}
\hspace{3cm}
\begin{array}{&c|&c &c &c | &c}
\hline
&  &+ &- &\text{Total} \\
\hline
H&  &\color{orange}{r} &r & 2r\\
V&  &n-r &\color{orange}{N-r} & 2N-2r\\
\hline
& &\color{red}{N} &N  &2N\\
\hline\end{array}$$
Total number of combinations is 
$$\binom {2N}N\cdot \binom {2N}N=\binom {2N}N^2$$
Note that this equal to 
$$\binom {2n+2}{n+1}^2=4\binom {2n+1}n^2$$