Probability of stepping on n-th field in game where you move using dice 
We play a game where in each round we roll a six-sided dice and move the appropriate number of squares. How likely are we to step on the $n$-th field during the game, if we started with the number $0$?
The answer is sufficient in the form of a recurrent formula, which refers to a constant number of values for the lower one. Can you use only $2$ values for lower?

I am kinda lost here. I did bit know how even approach this.
Thanks to anyone who can help me guide to right direction
I made this:
| 1 | 1                                             |  
| 2 | 1+1,2                                         |  
| 3 | 1+1+1, 2+1, 3                                 |  
| 4 | 1+1+1+1, 1+1+2, 1+3, 2+2, 4                   |  
| 5 | 1+1+1+1+1, 1+1+1+2, 1+2+2, 1+1+3, 1+4, 2+3, 5 |

But I can not see any pattern or how I can calculate probability from it. Does order of numbers matter ?
 A: For $n=1\dots 6$ we have:
$$
p_n=\frac16+\frac16\sum_{k=1}^{n-1}p_{n-k}\implies p_n=\frac16\left(\frac76\right)^{n-1}\tag1
$$
Indeed (1) obviously holds for $p_1$ and by induction:
$$
p_n=\frac16+\frac1{6}\sum_{k=1}^{n-1}\frac1{6}\left(\frac76\right)^{k-1}
=\frac16+\frac1{36}\frac{\left(\frac76\right)^{n-1}-1}{\frac76-1}=
\frac16\left(\frac76\right)^{n-1}.
$$
For $n>6$ the probabilities can be computed by the following formula:
$$
p_n=\frac16\sum_{k=1}^{6}p_{n-k}=\frac{7p_{n-1}-p_{n-7}}6.\tag2
$$
The rightmost expression in $(2)$ is valid for $n>7$. It can be used for $n=7$ with the convention $p_0=1$.
A: To find the number of ways of reaching $n$ observe the last roll that can happen it can be either 1,2,3 ... 6 thus $$ H_n = H_{n-1} + H_{n-2} + \cdots + H_{n-6}\\
H_{n+1} = 2H_{n}-H_{n-6} \ \forall n \ge6  $$
A: At each step you can move by $x_k = 1,2, \ldots, 6$, so after $m$ steps you have moved away from $0$ of $x_1+x_2+ \cdots + x_m$ units.
You can reach to $n$ in min $\left\lceil {n/6} \right\rceil $ and max $n$ steps.
Once you reach to it in $m$ steps you will pass beyond it in the next steps.
At each step you have $6$ equally probable outcomes, after $n$ steps they are $6^n$.
The number of outcomes that bring you to $n$ im steps are the number of
compositions of $n$ into $m$ parts with each part in the range $[1,6]$,i.e.
$$
\eqalign{
  & {\rm No}{\rm .}\,{\rm of}\,{\rm solutions}\,{\rm to}\;\left\{ \matrix{
  {\rm 1} \le {\rm integer}\;x_{\,j}  \le 6 \hfill \cr 
  x_{\,1}  + x_{\,2}  + \; \cdots \; + x_{\,m}  = n \hfill \cr}  \right. =   \cr 
  &  = {\rm No}{\rm .}\,{\rm of}\,{\rm solutions}\,{\rm to}\;\left\{ \matrix{
  {\rm 0} \le {\rm integer}\;y_{\,j}  \le 5 \hfill \cr 
  y_{\,1}  + y_{\,2}  + \; \cdots \; + y_{\,m}  = n - m \hfill \cr}  \right. =   \cr 
  &  = N_b (n - m,5,m) \cr} 
$$
where $Nb$ is given by
$$
N_b (s,r,m)\quad \left| {\;0 \leqslant \text{integers  }s,m,r} \right.\quad  =
\sum\limits_{\left( {0\, \leqslant } \right)\,\,k\,\,\left( { \leqslant \,\frac{s}{r+1}\, \leqslant \,m} \right)} 
{\left( { - 1} \right)^k \binom{m}{k}
 \binom
 { s + m - 1 - k\left( {r + 1} \right) } 
 { s - k\left( {r + 1} \right)}\ }
$$
as explained in this related post.
Concerning the recurrence, it is clear that $N_b$ satisfies the following (among others)
$$
N_b (s,r,m) = \sum\limits_{i\, = \,s - r}^s {N_b (i,r,m - 1)} 
$$
I leave to you to continue with adapting the above to your case (summing over $m$, ..).
A: For large $n$:

*

*For $n$ that is large enough, each field have the same probability
to be landed.

*The expected value for a 6-sided dice is $3.5$
Thus, the probability to land on $n$ in the limit is
$$\lim_{n \to \infty} p_n=\frac{1}{3.5}= 0.285$$
I am not an expert, so take my solution with a grain of salt.
