Let $X_0, X_1 , \dots$ be independent identically distributed Bernoulli random variables such that
$$ \mathbb{P} [ X_k = 0 ] = \mathbb{P} [ X_k = 1] = 1/2, \ k \geq 0 $$
Let us define
$$ \tau_{110} = \inf [ k \geq 0 : (X_k , X_{k+1}, X_{k+2} ) = (1,1,0) ] $$
and $\tau_{010}$ similarly. I would like to calculate $\mathbb{P} [ \tau_{010} < \tau_{110}]$ and $\mathbb{E}[ \tau_{110}] $, $\mathbb{E} [ \tau_{111} ]$. We can define
$$ Y_k = ( X_k , X_{k+1}, X_{k+2} ) , \ k \geq 0. $$
Then $Y$ is a Markov chain with certain transition probabilities that could be defined as an oriented graph on eight vertices. Finally, we need to calculate the absorption probabilities for the chain $Y$ stopped at $\tau_{010} \wedge \tau_{110}$. However, I am kind of lost in the graph.
For $\Pr[\tau_{010} < \tau_{110}]$, first observe that if there's any point at which $X_k = X_{k+1}=1$, then $\tau_{110}$ is guaranteed to happen first. (As long as we keep getting $1$, we make no progress toward either $\tau_{010}$ or $\tau_{110}$; once we get $0$, we've hit $\tau_{110}$.
In particular, with probability $\frac14$, we get $X_0 = X_1 = 1$ and then $\tau_{010} > \tau_{110}$.
Otherwise, either $X_0=0$ or $X_1=0$. From the moment where we see that first $0$, we can wait until we see a $1$. Now we have $X_k = 0$, $X_{k+1}=1$, and there are two possibilities:
Therefore $\Pr[\tau_{010} < \tau_{110}] = \frac34 \cdot \frac12 = \frac38$: first, we must get either $X_0 =0$ or $X_1=0$, and second, we must get a $0$ after the first occurrence of $(0,1)$.
Both $\mathbb E[\tau_{010}]$ and $\mathbb E[\tau_{011}]$ are standard hitting-time problems, but they don't have to be done in the $8$-state chain of all possible pairs $(X_k, X_{k+1}, X_{k+2})$. Instead, we can work in the Markov chain with states $\{0,1,2,3\}$. We are in state $i$ when $i$ is the largest integer such that the last $i$ random variables seen match the first $i$ values of the pattern we want to see; in particular, reaching state $3$ means we see the pattern exactly.
For more details, see The average time before the appearance of a sequence of length 3 of Bernoulli r.v.