Question: > 
> 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 ?

Answer: 
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 $p\_0=1$.
For large $n$:

1. For $n$ that is large enough, each field have the same probability
to be landed.
2. 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.