Consider integers ${1,2,...,10}$. A particle is initially at $1$. It moves to an adjacent integer in the next step. What is the expected number of steps it will take to reach $10$ for the first time?
When the particle is at $1$, it can only move to $2$ since there's no $0$. But from $2$, you can either move to $1$ or $3$ with $50%$ of probability and so on. Clearly making a tree diagram will take a long time, but I'm not sure how to approach this more mathematically.
Let $X_n$ be the location of the particle at time $n$, then $\{X_n:n=0,1,\ldots\}$ is a Markov chain on $\{1,2,\ldots,10\}$ with transition probabilities $$ P_{ij} =\begin{cases} 1,& (i,j)=(0,1)\text{ or } (i,j) = (10,10)\\ p,& j=i+1, 2\leqslant i\leqslant 9\\ 1-p,& j=i-1, 2\leqslant i\leqslant 8\\ 0,& \text{ otherwise}. \end{cases} $$ Here we allow $10$ to be an absorbing state for convenience. Let $$ E_i = \inf\{n>0: X_n=10\mid X_0=i\} $$ for $1\leqslant i\leqslant 10$ (trivially, $E_{10}=0$ with probability $1$). This gives the linear system of equations \begin{align} E_1 &= 1+E_2\\ E_i &= 1 + (1-p)E_{i-1} + pE_{i+1},\ 2\leqslant i\leqslant 9, \end{align} and solving these yields $$ E_1 = \frac2{p^8}(1 - 6 p + 17 p^2 - 28 p^3 + 30 p^4 - 20 p^5 + 10 p^6 + p^8). $$ When $p=\frac12$, we have $E_1 = 82$.