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!
 A: 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.
