2. [5] Jerry and Neil have a 3 -sided die that rolls the numbers 1,2 , and 3 , each with probability $\frac{1}{3}$. Jerry rolls first, then Neil rolls the die repeatedly until his number is at least as large as Jerry's. Compute the probability that Neil's final number is 3 .
Answer: $\square$
Solution: If Jerry rolls $k$, then there is a $\frac{1}{4-k}$ probability that Neil's number is 3 , since Neil has an equal chance of rolling any of the $4-k$ integers not less than $k$. Thus, the answer is
$$
\frac{1}{3}\left(1+\frac{1}{2}+\frac{1}{3}\right)=\frac{11}{18}
$$