TITLE: Probability of landing on the nth stair.
QUESTION [11 upvotes]: Initial Question: We begin climbing a staircase beginning at stair zero.  We  choose to take either 1,2, or 3 steps at a time, where each number of steps have an equal chance of being chosen.  What is the probability of hitting the fourth stair?
How I proceeded: In lieu of recognizing what type of problem I was looking at, I formed a tree, and found that there are seven ways to hit the fourth stair.  Taking into consideration the weights of each of these outcomes I found the probability of hitting the fourth stair to be 37/81. 
Deeper Question: While I found that finding the number of ways to hit the $n$th step was not too difficult (I think it is the sum of the ways of hitting the three previous steps, $k(n-3)+k(n-2)+k(n-1)$), I was not able to use this successfully to find general rule for the probability of hitting the $n$th step.
I am looking for some advice classifying this type of probability problem, or perhaps a link to a similar problem.  Could this be viewed as some sort of random walk?  Is there any hope for a closed form (explicit) solution?

REPLY [7 votes]: Let $p_{n,k}$ be the probability of hitting the $n^\text{th}$ stair after exactly $k$ steps, and let $p_n$ be the probability of hitting the $n^\text{th}$ stair at some point. Then $p_{n,k}$ is the coefficient of $x^n$ in the polynomial $(\frac13x+\frac13x^2+\frac13x^3)^k$. Therefore, $p_n=\sum_{k=0}^\infty p_{n,k}$ is the coefficient of $x^n$ in 
$$
\begin{align}
\sum_{k=0}^\infty \left(\tfrac13x+\tfrac13x^2+\tfrac13x^3\right)^k
&=\frac1{1-(\tfrac13x+\tfrac13x^2+\tfrac13x^3)}
\\&=\frac{1/2}{1-x}+\frac{1/4}{1+x/(1+i\sqrt{2})}+\frac{1/4}{1+x/(1-i\sqrt{2})}
\end{align}
$$
The last equality can be derived using the usual partial fractions method.  Expanding out each of these fractions into their Taylor series, you get
$$
\boxed{p_n = 1/2+\frac{1/4}{(-1-i\sqrt{2})^n}+\frac{1/4}{(-1+i\sqrt{2})^n}}
$$
So, you get an exact answer involving complex numbers magically canceling, along with an asymptotic result that the probabilities converge exponentially quickly to $1/2$.