smc

Firstly, notice that if Fiona jumps over the predator on pad $3$, she must land on pad $4$. Similarly, she must land on $7$ if she makes it past $6$. Thus, we can split the problem into $3$ smaller sub-problems, separately finding the probability Fiona skips $3$, the probability she skips $6$ (starting at $4$) and the probability she doesn't skip $10$ (starting at $7$). Notice that by symmetry, the last of these three sub-problems is the complement of the first sub-problem, so the probability will be $1-\text{the probability obtained in the first sub-problem}$. In the analysis below, we call the larger jump a $2$-jump, and the smaller a $1$-jump. For the first sub-problem, consider Fiona's options. She can either go $1$-jump, $1$-jump, $2$-jump, with probability $\frac{1}{8}$, or she can go $2$-jump, $2$-jump, with probability $\frac{1}{4}$. These are the only two options, so they together make the answer $\frac{1}{8}+\frac{1}{4}=\frac{3}{8}$. We now also know the answer to the last sub-problem is $1-\frac{3}{8}=\frac{5}{8}$. For the second sub-problem, Fiona must go $1$-jump, $2$-jump, with probability $\frac{1}{4}$, since any other option would result in her death to a predator. Thus, since the three sub-problems are independent, the final answer is $\frac{3}{8}\cdot \frac{1}{4}\cdot \frac{5}{8}=\boxed{\frac{15}{256}}$.