Probability two symmetric random walks are at the same point after $n$ steps Consider two independent symmetric random walks starting at the origin and go left one step with probability $1/2$ and right one step with probability $1/2$. 

What is the probability that after $n$ steps they are at the same
  place?

 A: Call our random walkers A and B. They are at the same place after $n$ steps precisely if they have taken the same number of steps to the right. 
By symmetry, the probability that B has taken $k$ steps to the right is the same as the probability she has taken $k$ steps to the left.  Imagine walking as A did, and appending B's path, reversed. 
The net effect is the same as a single walker C, taking $2n$ steps. The probability that A and B are at the same place after $n$ steps is the same as the probability that after $2n$ steps, C is at the origin. This probability is
$$\binom{2n}{n}\frac{1}{2^{2n}}.$$  
A: Let $A_k=a_1+\cdots+a_k$ and $B_k=b_1+\cdots+b_k$ be two independent random walks where $a_i, b_i$ iid and $P(a_i=-1)=P(a_i=1)=1/2$. Notice that $a_i/2+1/2 \sim \text{Ber}(1/2)$ and therefore $A_k/2+k/2$ and $B_k/2+k/2 \sim \text{Bin}(k,1/2)$. It follows that $P(B_n/2+k/2=i)=P(A_n/2+k/2=i)=\binom{n}{i}(1/2)^n$. Therefore, because they are independent $P(A_n/2+k/2=i, B_n/2+k/2=i)=\binom{n}{i}^2(1/2)^{2n}$. It follows that
$$P(A_n=B_n)=\sum_{k=0}^n(1/2)^{2n}\binom{n}{k}^2$$
Note that this is the same as André's answer, just calculated differently.
