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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?

Mike Earnest
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Elizabeth Huffman
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1 Answers1

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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.

Mike Earnest
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