Harry the Hamster is put in a maze, and he starts at point $S$. The paths are such that Harry can move forward only in the direction of the arrows. At any junction, he is equally likely to choose any of the forward paths. What is the probability that Harry ends up at $B$ ?
(A) $\frac{2}{3}$
(D) $\frac{1}{3}$
(B) $\frac{13}{18}$
(C) $\frac{11}{18}$
(E) $\frac{1}{4}$

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-112.jpg?height=553&width=436&top_left_y=1119&top_left_x=1278)
We label the five junctions as $V, W, X, Y$, and $Z$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-205.jpg?height=550&width=437&top_left_y=1411&top_left_x=882)

From the arrows which Harry can follow, we see that in order to get to $B$, he must get to $X$. So we calculate the probability that he gets to $X$.

To get to $X$, Harry can go $S$ to $V$ to $W$ to $X$, or $S$ to $V$ to $Y$ to $X$, or $S$ to $V$ to $X$ directly. At $V$, the probability that Harry goes down any of the three paths (that is, towards $W, X$ or $Y$ ) is $\frac{1}{3}$.

So the probability that Harry goes directly from $V$ to $X$ to $\frac{1}{3}$.

At $W$, the probability that Harry turns to $X$ is $\frac{1}{2}$, so the probability that he goes from $V$ to $W$ to $X$ is $\frac{1}{3} \times \frac{1}{2}=\frac{1}{6}$.

At $Y$, the probability that Harry turns to $X$ is $\frac{1}{3}$, so the probability that he goes from $V$ to $Y$ to $X$ is $\frac{1}{3} \times \frac{1}{3}=\frac{1}{9}$.

Therefore, the probability that Harry gets to $X$ (and thus to $B$ ) is $\frac{1}{3}+\frac{1}{6}+\frac{1}{9}=\frac{6+3+2}{18}=\frac{11}{18}$.