How long the walk lasts

Idea. The expected duration satisfies a recurrence whose second difference is constant, so it is a quadratic in the starting point. The two boundary conditions pin it down.

Step 1: recurrence. Conditioning on the first step, $$m_i=1+\frac{1}{2}{m}_{i-1}+\frac{1}{2}{m}_{i+1},m_0=m_{10}=0,$$ which rearranges to $m_{i+1}-2m_i+m_{i-1}=-2$.

Step 2: solve. A constant second difference of $-2$ means $m_i$ is a quadratic with leading coefficient $-1$. The quadratic that vanishes at $0$ and $10$ is $$m_i=i(10-i).$$ Check: $(i+1)(9-i)-2i(10-i)+(i-1)(11-i)=-2$.

Step 3. From $i=3$, the expected number of steps is $3\cdot 7=21$.

Remark. The formula $i(N-i)$ is worth remembering: starting in the middle of $[0,N]$ gives $N^2/4$, which shows the walk spreads like the square root of time.