**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 $\frac{{2n \choose n}}{2^{2n}}$.

Next, for any given integer $k \geq 1,

P(S_n = 2n+k) = {2n+k-1 \choose n-1}\,2^{-2n-k+1}$.

Thus, $\mu _n = E(S_n) = 2n+\sum_{k=1}^\infty k\,{2n+k-1 \choose n-1}\,2^{-2n-k+1} = 2n\,\bigg (1+\frac{{2n \choose n}}{2^{2n}}\bigg )$, 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})$.

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