Alice rolls a standard 6-sided die. Bob rolls a second standard 6 -sided die. Alice wins if the values shown differ by 1 . What is the probability that Alice wins?
(A) $\frac{1}{3}$
(B) $\frac{2}{9}$
(C) $\frac{5}{18}$
(D) $\frac{1}{6}$
(E) $\frac{5}{36}$
Since each of Alice and Bob rolls one 6 -sided die, then there are $6 \times 6=36$ possible combinations of rolls.

Each of these 36 possibilities is equally likely.

Alice wins when the two values rolled differ by 1 . The possible combinations that differ by 1 are $(1,2),(2,3),(3,4),(4,5),(5,6),(2,1),(3,2),(4,3),(5,4)$, and $(6,5)$.

Therefore, there are 10 combinations when Alice wins.

Thus her probability of winning is $\frac{10}{36}=\frac{5}{18}$.

Answer: (C)