Contributing Editor Anirban DasGupta writes the solution to puzzle 24:

Congratulations to the four student members who sent in correct answers—some more complete than others. They are Prakash Chakraborty, Purdue University; Sihan Huang, Columbia University; Kumar Somnath, The Ohio State University; and Andrew Thomas, Purdue University.

Now for the solution. Denote the number of steps required to reach the point $(n,n)$ by $S_n$. The quickest that the particle can reach the point $(n,n)$ is in $2n$ steps, which happens if exactly $n$ heads and $n$ tails are produced in $2n$ tosses of our fair coin. This has probability $\frac{(\genfrac{}{}{0ex}{}{2n}{n})}{2^{2n}}$.

Next, for any given integer $k\ge 1,P(S_n=2n+k)=(\genfrac{}{}{0ex}{}{2n+k-1}{n-1})2^{-2n-k+1}$.

Thus, ${\mu }_n=E(S_n)=2n+\sum _{k=1}^{\mathrm{\infty }}k(\genfrac{}{}{0ex}{}{2n+k-1}{n-1})2^{-2n-k+1}=2n(1+\frac{(\genfrac{}{}{0ex}{}{2n}{n})}{2^{2n}})$, with a little bit of calculation. In particular, ${\mu }_3=\frac{63}{8}=7.875$, and on using Stirling’s series for $n!$, we get
${\mu }_n=2n+\frac{2\sqrt{n}}{\sqrt{\pi }}-\frac{1}{4\sqrt{n\pi }}+O(n^{-3/2})$.