Suppose you start at position 0. You then roll 2 6-sided dice. You move to the integer, call it z, that is the sum of the two dice. You then roll again. If the result of the roll is z', you move to z+z'. You then continue in this fashion. I am looking for formulas (recursive or non-recursive) for the probability of eventually landing on spot n (where n is a positive integer). For small n, n < 10 say, these probabilities are relatively easy to compute by just checking all cases. Note, as n grows, the functions will converge relatively quickly. So, for say n > 40, this may not be a very interesting question. 



Probabilities and rolling 2 dice
The probabilities do converge to 1/7. One way to see this is to start from Tony Huynh's comment: the probability that $n$ is hit is the coefficient of $t^n$ in $$f(t) = {1 \over (1-(t+t^2+t^3+t^4+t^5+t^6)^2/36)}$$. The denominator is a polynomial of degree 12; its roots are $t = 1$ and eleven points $r\_1, \ldots, r\_{11}$ which are outside the unit circle. Thus we can write
$$ f(t) = {A \over 1-t} + \sum\_{k=1}^{11} {C\_k \over 1 - t/r\_k} $$
where $A$ and $C\_1, C\_2, \ldots, C\_{11}$ are (complex) constants. This is just the usual partial fraction expansion of a rational function. Taking the $z^n$ coefficient of both sides of the above equation gives
$$p(n) = A + \sum\_{k=1}^{11} C\_k r\_k^{-n}.$$
If we want to show that $\lim\_{n \to \infty} p(n) = 1/7$, we just need to show that $A = 1/7$. This is easy: $A = \lim\_{t \to 1} f(t) (1-t)$. The denominator in $f(t)$ is divisible by $1-t$, so do the polynomial division and substitute $t = 1$.


I wouldn't call the closed form above a "nice" closed form for $p(n)$, though.


