In a board game played with a die, our pawn is four spaces away from the finish. If we roll at least a four, we reach the finish. If we roll a three, we will definitely be in with the next roll.

What is the probability that we will reach the finish in only two more rolls?
Solution. More than two throws are needed when the sum of the first two throws is less than 4. This occurs in three cases:

1) both of the first two throws are 1;
2) the first throw is 1 and the second is 2; or
3) the first throw is 2 and the second is 1.

In each of these cases, the probability of the throws is: $\frac{1}{6} \cdot \frac{1}{6}=\frac{1}{36}$.

Since these are mutually exclusive events, the probability that we reach the goal with more than two throws is: $3 \cdot \frac{1}{36}=\frac{1}{12}$

We did not assume that we can only reach the goal by rolling exactly 4, because we could not calculate this probability so simply.