Landing on Consecutive Spaces with Dice Rolls Imagine you are continuously rolling one six-sided fair die, and moving a token the amount on the die on a board with 1000 spaces. Assuming the token starts at zero, is there a general formula that describes the probability you will land on any three consecutive spaces? For instance, what is the probability you will land on either the 4th, 5th, or 6th spaces. What is the probability you will land on the 998th, 999th, 1000th spaces? 
I know it converges to a probability by running simulations, but I don't know how to mathematically describe it.
 A: Let's start by defining $r(n)$ to be the probability that the sum is ever exactly $n$.  
Let's express the probability that the sum will ever be $a$, $a+1$, or $a+2$ using $r$.  Let's call that probability $P$.
For a set of positive integers $S$, define $p_s$ to be the probability that the sum will ever be in $S$.
Using the inclusion exclusion principal, we have
$$
P = p_{\{a\}}+p_{\{a+1\}}+p_{\{a+2\}}-p_{\{a,a+1\}}-p_{\{a,a+2\}}-p_{\{a+1,a+2\}}+p_{\{a,a+1,a_2\}}.$$
We can calculate all of these probabilities using the function $r$.
We have 
$$
p_{\{a\}} = r(a), p_{\{a+1\}} = r(a+1), \text{ and } p_{\{a+2\}} = r(a+2).$$
Further,
$$p_{\{a,a+1\}} = r(a) \frac{1}{6}$$
and 
$$p_{\{a+1,a+2\}} = r(a+1) \frac{1}{6}.$$
Since we can get to $a+2$ from $a$ in two ways, we have
$$
p_{\{a,a+2\}} = r(a)\frac{1}{6}+ r(a)\frac{1}{36} = \frac{7}{36}r(a).
$$
Finally, 
$$p_{\{a,a+1,a_2\}} = r(a) \frac{1}{36}.$$
Putting this all together, we have
$$
P = \frac{2}{3}r(a)+\frac{5}{6}r(a+1)+r(a+2).
$$
Now, here's the complicated part.  The closed form for $r$ is 
$$
r(n) = \frac{2}{7} + \frac{1}{7} \sum_{i=2}^{7} A_i^n
$$
where the $A_i$ are complex numbers inside the unit circle.  From this, we know that as $n$ goes to infinity, $r(n)$ goes to $\frac{2}{7}$, and so, as $a$ goes to infinity, $P$ approaches $\frac{5}{7}$.
If we really want to calculate $P$ exactly for some specific $a$, we can use the relation $\frac{2}{3}r(a)+\frac{5}{6}r(a+1)+r(a+2)$ given earlier and then go about calculating these $r$ values using the recurrence
$$
r(n) = \frac{1}{6}\sum_{i=1}^6 r(n-i)
$$
for $n>6$ 
with 
$$r(1)= \frac{1}{6}
$$
$$
r(2)=\frac{1}{6}+\frac{1}{6} r(1) = \frac{7}{36}
$$
$$
r(3) = \frac{1}{6} + \frac{1}{6} r(2) + \frac{1}{6} r(1) = \frac{49}{216} 
$$
$$
r(4) = \frac{1}{6} + \frac{1}{6} r(3) + \frac{1}{6} r(2) + \frac{1}{6} r(1) = \frac{343}{1296}
$$
$$
r(5) =  \frac{1}{6} + \frac{1}{6} r(4) + \frac{1}{6} r(3) + \frac{1}{6} r(2) + \frac{1}{6} r(1) = \frac{2401}{7776}
$$
$$
r(6) = \frac{1}{6} + \frac{1}{6} r(5) + \frac{1}{6} r(4) + \frac{1}{6} r(3) + \frac{1}{6} r(2) + \frac{1}{6} r(1)
= \frac{16807}{46656}
$$
This way, we can calculate, for instance, that the probability of ever having a sum of $4$, $5$ or $6$ is exactly
$$
\frac{343}{432}
$$
and of ever having a sum of $1000$, $1001$ or $1002$ is
approximately $\frac{5}{7}+4.139\times10^{-138}$ (the exact fraction has
too many digits in its numerator and denominator to display here).
