Lets write a "word" made by $A$ and $B$, each $A$ represents an head on the coin and each $B$ a tail. So each "word" gives us a sequence of moves. The idea is to count the number of "word" which bring us from $(0;0)$ to $(n;n)$ and divide it by the number of "word" that we can make in total. To reach $(n;n)$ we have to do $n$ steps north and $n+k$ east or $n+k$ steps north and $n$ east. So we have to count how many are the words made by $n+k$ $A$ and $n$ $B$ which are $\displaystyle \frac{(2n+k)!}{n!(n+k)!}$ and the words made by $n$ $A$ and $n+k$ $B$ which are also $\displaystyle \frac{(2n+k)!}{n!(n+k)!}$. So the number of "good" words is \[2\frac{(2n+k)!}{n!(n+k)!}\] Now lets count how mach words there are in total, this number is $\displaystyle 2^2n+k$. So the probability asked is \[\frac{\displaystyle 2\frac{(2n+k)!}{n!(n+k)!}}{\displaystyle 2^{2n+k}}\]. BONUS QUESTION! For which $k$ the probability to reach $(n;n)$ take the maximum value?

MathBaron wrote: . So we have to count how many are the words made by $n+k$ $A$ and $n$ $B$ which are $\displaystyle \frac{(2n+k)!}{n!(n+k)!}$ Almost, the word with $n+k$ A's and $n$ B's must end with a $B$, otherwise the sequence will stop before we get to $(n,n)$, the number is is $2\frac{(2n+k-1)!}{(n-1)!(n+k)!}=2\binom{2n+k-1}{n-1}$ This now satisfies $\sum_{k=0}^{\infty} \frac{1}{2^{2n+k-1}} \binom{2n-1+k}{n+k} \to 1$

Yes, ocha you're right, i didn't considered "exactly" in the question.