Find amount of steps. I need to calculate the amount of steps the person needs to take to have a probability of 50% to be aleast 10m away from his starting point (in both directions). He has a probability of 50% of moving in either direction.
Could someone tell if I'm thinking the right way or set me in the good direction. That would be very helpfull.
Thanks in advance!

Answer to first posed question which was later edited

The target will be reached in an odd number of steps.
Let $S_{m}$ denote the number of steps that separate the drunkard
from the target after taking $1+2m$ steps.
Then $S_{0}=2$ and $S_{1}\in\left\{ 0,2\right\} $ so a walk that
ends with $S_{m}=0$ will have shape: $$\left(S_{0},S_{1},\dots,S_{m-1},S_{m}\right)=\left(2,2,\dots,2,0\right)$$
After taking exactly one step we are in status $S_{0}=2$ and there
are $3$ possible outcomes for the next two steps:


*

*It is a loop of $2$ steps that hits the original starting point.
This has probability $\frac{1}{2}$ to occur.

*It is a loop of $2$ steps that does not hit the original starting
position. This has probability $\frac{1}{4}$ to occur.

*It is an arrival at the target. This has probability $\frac{1}{4}$
to occur.
So - taking the first two bullets together - when the first
step is taken and the next $2$ steps are taken then there is probability $\frac{3}{4}$ on making a loop and probability $\frac{1}{4}$ on reaching the target.
That means that the probability that the target is reached in exactly
$1+2m$ steps equals: $$\frac{1}{4}\left(\frac{3}{4}\right)^{m-1}$$ for $m=1,2,\dots$.
Then the probability of reaching the target in at most $1+2m$ steps
is: $$\sum_{k=1}^{m}\frac{1}{4}\left(\frac{3}{4}\right)^{k-1}=1-\left(\frac{3}{4}\right)^{m}$$
To be found is the smallest $m$ that satisfies $1-\left(\frac{3}{4}\right)^{m}\geq0.95$ which is $m_0:=11$ 
So the final answer is: $$n_0=1+2m_0=23$$

edit (answer to edited question)
For $n=0,1,2,\dots$ let $X_{n}$ denote the distance of the drunkard
from his starting point.
Then to be found are expressions for $P\left(X_{n}=i\right)$ for
$i=0,1,2,3$.
After finding them we can go for finding the smallest $n$ that satisfies:
$$1-P\left(X_{n}=0\right)-P\left(X_{n}=1\right)-P\left(X_{n}=2\right)-P\left(X_{n}=3\right)\geq0.95$$
or equivalently: $$P\left(X_{n}=0\right)+P\left(X_{n}=1\right)+P\left(X_{n}=2\right)+P\left(X_{n}=3\right)\leq0.05$$
Under the convention that $\binom{n}{k}=0$ if $k\notin\left\{ 0,1,2,\dots,n\right\} $ we find for nonnegative integer $n$:


*

*$P\left(X_{n}=0\right)=2^{-n}\binom{n}{\frac{1}{2}n}$

*$P\left(X_{n}=1\right)=2^{-n}\left[\binom{n}{\frac{1}{2}n-\frac{1}{2}}+\binom{n}{\frac{1}{2}n+\frac{1}{2}}\right]$

*$P\left(X_{n}=2\right)=2^{-n}\left[\binom{n}{\frac{1}{2}n-1}+\binom{n}{\frac{1}{2}n+1}\right]$

*$P\left(X_{n}=3\right)=2^{-n}\left[\binom{n}{\frac{1}{2}n-\frac{3}{2}}+\binom{n}{\frac{1}{2}n+\frac{3}{2}}\right]$
I leave the rest to you.
