2023 AMC 12A Problem 17

Let $a_n$ be the probability that Flora ever lands exactly on $n,$ with $a_0=1.$ Conditioning on the first jump, $$a_n=\sum_{k=1}^{n}\dfrac{1}{2^k}\,a_{n-k}.$$ We prove by induction that $a_n=\tfrac12$ for every $n\ge 1.$ The case $n=1$ is immediate. If the claim holds through $n-1,$ then $$\begin{aligned} a_n&=\frac{1}{2^n}a_0 +\sum_{k=1}^{n-1}\frac{1}{2^k}\cdot\frac12\\ &=\frac{1}{2^n} +\frac12\left(1-\frac{1}{2^{n-1}}\right)\\ &=\frac12. \end{aligned}$$ Hence the probability of landing on $10$ is $\dfrac12.$ Thus, the correct answer is E.