# $1$D bidirectional random walk question

In a $1$D random walk on x axis a particle can turn left with probability $\frac{3}{4}$ and right with probability $\frac{1}{4}$. What is the probability that $|x|\leq 1$ for $1\leq t\leq 4$ total steps.

I am not looking for answers, but ways to think about it. How to approach this kind of a question? What is a good source that discusses random walks from the basics.

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I assume you are starting at the origin. Draw a tree diagram. With such a small range of values of $t$, you can calculate all the relevant probabilities with not much effort. – André Nicolas Jun 6 '11 at 17:58

You start the particle at some state in the set $A=\{ -1, 0, 1\}$. Let $Q$ be the matrix of transition probabilities from this set into itself, i.e., $$Q=\pmatrix{0&1/4&0\cr 3/4&0&1/4\cr 0&3/4&0}.$$
Then $Q_{ij}^n$, the $(i,j)$th entry in the $n$th power of $Q$, is the probability that the walk is in state $j$ at time $n$, starting at state $i$, without leaving the set $A$.
Since you don't want the answer, I will simply say that the row sums of $Q^4$ give you what you want, each row sum corresponding to one of the three possible initial states.