Consider a one-dimensional random walk whose steps are $+2$ and $-1$ with probabilities $p$ and $1-p$ respectively, starting from $0$ and in the interval {$-n$, $n$}. The walk ends at $-n$ or $n$ or $n+1$. Let $m$ be the number of integers "jumped" during the walk.
Is there a limit for the ratio $\frac{m}{2n+1}$ for $n \rightarrow \infty$?
Three examples to clarify:
1) n=15 p= 1/2 Steps = {-1, 2, -1, 2, -1, 2, -1, 2, -1, -1, 2, 2, -1, -1, 2, 2, 2, 2, -1, \ -1, 2, -1, -1, 2, -1, 2, -1, -1, 2, 2}
Positions = {0, -1, 1, 0, 2, 1, 3, 2, 4, 3, 2, 4, 6, 5, 4, 6, 8, 10, 12, 11, 10, \ 12, 11, 10, 12, 11, 13, 12, 11, 13, 15}
Missed (jumped) positions = {-15, -14, -13, -12, -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, 7, 9, 14}
m = 17; r = m/(2 n +1) = 0.548387
2) n= 15 p=1/2 Steps = {2, 2, 2, 2, -1, -1, 2, 2, 2, -1, -1, -1, -1, 2, 2, -1, -1, -1, 2, \ -1, 2, -1, -1, 2, 2, 2}
Positions = {0, 2, 4, 6, 8, 7, 6, 8, 10, 12, 11, 10, 9, 8, 10, 12, 11, 10, 9, 11, \ 10, 12, 11, 10, 12, 14, 16}
Missed positions = {-15, -14, -13, -12, -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 1, 3, 5, 13, 15}
m =20; r = m/(2 n +1) = 0.645161
3) n=20 p=1/2 Steps = {-1, 2, -1, 2, 2, 2, -1, -1, -1, -1, -1, -1, 2, -1, 2, 2, 2, -1, 2, \ -1, 2, -1, -1, -1, -1, -1, 2, -1, -1, -1, 2, -1, -1, -1, 2, -1, 2, \ -1, 2, 2, -1, -1, -1, -1, -1, -1, -1, 2, 2, -1, -1, 2, 2, -1, 2, -1, \ 2, 2, 2, -1, 2, 2, 2, 2}
Positions = {0, -1, 1, 0, 2, 4, 6, 5, 4, 3, 2, 1, 0, 2, 1, 3, 5, 7, 6, 8, 7, 9, \ 8, 7, 6, 5, 4, 6, 5, 4, 3, 5, 4, 3, 2, 4, 3, 5, 4, 6, 8, 7, 6, 5, 4, \ 3, 2, 1, 3, 5, 4, 3, 5, 7, 6, 8, 7, 9, 11, 13, 12, 14, 16, 18, 20}
Missed positions = {-20, -19, -18, -17, -16, -15, -14, -13, -12, -11, -10, -9, -8, -7, \ -6, -5, -4, -3, -2, 10, 15, 17, 19}
m =20; r = m/(2 n +1) = 0.560976
A simulation with the range {-2000,2000}, iterated 1000 times provides r as 0.572958.
The question is: Is there a limit for n -> Infinity based on: (n, p, base steps {-1,2}) ?