In this game, you and your friend each throw a dart at a $3$ foot line drawn on a wall. His will be thrown at the left half and yours at the right. Suppose that the darts always land on their halves of the line in a random, uniformly distributed way. What is the probability that the darts are less than $1$ foot apart?
So I am at a loss as to how to solve this problem. I was hoping to find some measure of average distances away. I know that the probability would change based on where the left dart lands. For example, if the left dart was more than one foot left of center, the probability of the darts being within a foot would be $0$. On the other hand, if the left dart was right at center, the probability of the darts being within a foot would be $\frac{12}{18}$ ($12$ inches in a foot, so $18$ on each half.)
Let the three-foot line be the interval $[0,3]$, and let $X$ and $Y$ be the positions of the left and right darts respectively. \begin{align} P(Y-X < 1) &= \int_0^{1.5} P(Y - X < 1 \mid X = x) p_X(x) \mathop{dx} \\ &= \int_{0.5}^{1.5} P(Y < 1+x) p_X(x) \mathop{dx} \\ &= \int_{0.5}^{1.5} \frac{(1+x) - 1.5}{1.5} \cdot \frac{1}{1.5} \mathop{dx}. \end{align} Can you take it from here?