What is the probability this Markov chain does not reach state $r$? Consider a random walk on the non-negative integers.
You start at $0$, and in each step you either move $1$ higher, or $2$ lower (but can't go below $0$). The direction is picked w.p. $1/2$ independently at any step.
From $0$ you can either stay at $0$, or go to $1$.
From $1$ you flip a coin and go either to $0$ or to $2$.
From $i>1$, you go to $i-2$ or to $i+1$.


*

*What is the probability that during $n$ steps, the process never reaches state $r$?
If this does not have a simple form (as a function of $n,r$), can we derive an upper bound for it?
This seems to be solvable by a recurrence formula, but I couldn't reach an explicit expression.
 A: Since it is a Markov process, these probabilities satisfy recursive equations. Namely, since in your case reaching $r$ is the same as reaching any number from $r$ till $\infty$ (upwards move are of size $1$), you solve a reachability problem for the interval $[r, +\infty)$. In general you have $V_0(x) = 1_A(x)$ and 
$$
  V_{n+1}(x) = 1_A(x) + 1_{A^c}(x) \cdot PV_n(x)
$$
where $1_A$ is an the indicator function of $A$ and $P$ is the stochastic matrix. So over $A$ the solution is obviously $1$ for all $n$, and we can simplify this for $x\notin A$: for those
$$
  V_{n+1}(x) = \sum_{y\notin A}p(x,y)V_n(y) + \sum_{y\in A^c}p(x,y)
$$
here $p(x,y)$ is the probability of going from $x$ to $y$. In your case only two values are probable when departing from $x$, let's say $x^+$ and $x^-$, each with probability $\frac12$. So you get
$$
  V_{n+1}(x) = \frac12\left(V_n(x^-) + V_n(x^+)\right)
$$
with the following boundary conditions: $V_0(x) = 1_A(x)$ and $V_n(x) = 1$ for all $x\in A$. That should be enough for you to carry on computations.
P.S. Just realized you look for the probability of not reaching $r$. In that case, you can compute what I have proposed and subtract it from $1$. Or, if you want to have a neater solution, just substitute $U_n := 1 - V_n$ in the equations above: you'll get an even simpler recurrent scheme.
A: As I mentioned in my comment, if $n>r$, then we can have atmost $r-1$ steps that move up. So the probability that after $n$ steps, we do not reach $r$ is,
$$P_{n,r} = \sum_{i=0}^{r-1} {n \choose i} {1 \over 2^{n}}$$
To include the constraint that we remain above 0, notice that if we move up $p$ times up, we have to move $q$ times down such that $p+q=n$ and $p-2q>=0$. Solving for minimum value of $p$, we get $p>={2n \over 3}$. This also puts a contraint that $r>=2n/3$. Adjusting the value of $i$ in our original eqution, we get
$$P_{n,r} = \sum_{i={2n \over 3}}^{r-1} {n \choose i} {1 \over 2^{n}}$$
