Answer
$P(F)=\frac{1}{9}\approx0.1111$
If somebody plays the game 900 times, we would expect he or she will lose on the first roll about 100 times.
Work Step by Step
See figure 2, page 260.
The sample space = {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}. So, $N(S)=36$
Let $F$ be the event "2, 3, or 12 on the first roll" = {(1,1),(1,2),(2,1),(6,6)}. So, $N(E)=4$
Using the Classical Method (page 259):
$P(F)=\frac{N(F)}{N(S)}=\frac{4}{36}=\frac{1}{9}\approx0.1111$
