4. [4] An ant starts at the point $(1,0)$. Each minute, it walks from its current position to one of the four adjacent lattice points until it reaches a point $(x, y)$ with $|x|+|y| \geq 2$. What is the probability that the ant ends at the point $(1,1)$ ?
Answer: $\frac{7}{24}$ From the starting point of $(1,0)$, there is a $\frac{1}{4}$ chance we will go directly to $(1,1)$, a $\frac{1}{2}$ chance we will end at $(2,0)$ or $(1,-1)$, and a $\frac{1}{4}$ chance we will go to $(0,0)$. Thus, if $p$ is the probability that we will reach $(1,1)$ from $(0,0)$, then the desired probability is equal to $\frac{1}{4}+\frac{1}{4} p$, so we need only calculate $p$. Note that we can replace the condition $|x|+|y| \geq 2$ by $|x|+|y|=2$, since in each iteration the quantity $|x|+|y|$ can increase by at most 1 . Thus, we only have to consider the eight points $(2,0),(1,1),(0,2),(-1,1),(-2,0),(-1,-1),(0,-2),(1,-1)$. Let $p_{1}, p_{2}, \ldots, p_{8}$ be the probability of reaching each of these points from $(0,0)$, respectively. By symmetry, we see that $p_{1}=p_{3}=p_{5}=p_{7}$ and $p_{2}=p_{4}=p_{6}=p_{8}$. We also know that there are two paths from $(0,0)$ to $(1,1)$ and one path from $(0,0)$ to $(2,0)$, thus $p_{2}=2 p_{1}$. Because the sum of all probabilities is 1 , we have $p_{1}+p_{2}+\ldots+p_{8}=1$. Combining these equations, we see that $4 p_{1}+4 p_{2}=12 p_{1}=1$, so $p_{1}=\frac{1}{12}$ and $p_{2}=\frac{1}{6}$. Since $p=p_{2}=\frac{1}{6}$, then the final answer is $\frac{1}{4}+\frac{1}{4} \cdot \frac{1}{6}=\frac{7}{24}$