Two "racing" sequences, what is the probability and expected stopping time that one will "catch up" to the other? Let $a_n$ and $b_n$ be two sequences with $a_0 = 0$ and $b_0 = c$ with $c > 0$. For simplicity, we can assume that $c$ is some small, positive integer, such as 3. Then, at every step, we flip a fair coin, and let $X_n$ denote the result of the coin flip.
If $X_n = H$, then
$$ 
a_{n+1} = a_n + 2\\
b_{n+1} = b_n + 1
$$
Otherwise if $X_n = T$, then
$$ 
a_{n+1} = 0\\
b_{n+1} = b_n
$$
So the $a_n$ sequence gets "zeroed out" whenever we flip a $T$.
My interest is in the probability that $a_n$ catches up to $b_n$, i.e. $a_n \geq b_n$, at step $n$. In other words, what is $P(a_n \geq b_n)$? In addition, if we define the stopping time $\tau$ to be the first $n$ such that $a_n \geq b_n$, I am also interested in $P(\tau \leq k)$, i.e. what is the probability that $a_n$ will have caught up to $b_n$ by step $k$?
After running some simulations, I'm pretty sure this $\tau$ has infinite expectation, i.e. $E(\tau) = \infty$. I think showing that $P(\tau = \infty) > 0$ may be easier than finding the probabilities above (I am still interested in those probabilities though). However, I haven't been able to find a way to prove even this yet.
 A: I am going to try to establish your conjecture of infinite expectation of the length of time until $a_n$ catches $b_n$.
By the way my language below may reflect my visualization of this problem of A (a rabbit) trying to catch B (a turtle), but at every naptime (both A and B take a nap when a tails occurs), a magical being transports A back to $0$, while leaving B unaffected.
During a run of heads, either: (a) A catches B sometime during the run; or (b) A tail ends the run of heads before A catches B. In case (b), A goes back to position $0$, and B ends farther from $0$ than it was before the run started (farther by the length of the run of heads).
Also note that once a tail occurs, further consecutive tails have no effect on A and B.
Thus the key to A catching B lies in the lengths of consecutive head runs only. And after each such run of heads, if A has not caught B, then B is farther out than at the beginning of the run.
Also if at a certain instant, $b_k=t$ and $a_k=0$, the probability of A catching B on the next run of heads is given by $\frac{1}{2^{t-1}}$. (The extra $-1$ in the exponent is because A gets the first step for free, since a run of heads always starts with a head.)
Suppose B starts at position $2$ (if $B$ starts at $1$, $B$ will definitely get caught with the first occurrence of a single head). Then $P({\rm A\; catches\; B\; on\; the\; first\; run\; of\; heads})=\frac12$. So the probability that there isn't catch on the first run of heads is $1-\frac12=\frac12$.
If B survives the first run, B will definitely be at position $3$, and the conditional probability that B survives the second run of heads is $\frac34$. If B does survive the second run, B is at least at position $4$, and so the conditional probability that B survives the third run of heads is at least $\frac78$; and so on.
The probability that B survives forever is therefore at least
$$\frac12\cdot\frac34\cdot\frac78\cdot\cdots\cdot \frac{2^k-1}{2^k}\cdot\cdots$$.
This is a convergent infinite product (value is approximately $.288765$), giving a lower bound for the probability of never being caught. If B starts somewhere farther out than $2$, the chance of never being caught increases. Also longer runs of heads (without B being caught) also increase the chance of B never being caught).
For more on the infinite product above, see Evaluate $\prod_{n=1}^\infty \frac{2^n-1}{2^n}$
