Consecutive Times for Runners Say we have runners $R1, R2$ choosing to start running at random  times $tr_1, tr_2$ in interval $[0,a]$, a a Real number, for a period of time $t_i < a-tr_i ; i=1,2$ respectively. I want to find the probability of R2 starting immediately after R1 is done.
My work. I represent intervals $(tr_1, tr_1 + t_1), (tr_2, tr_2 + t_2)$ in the  square $ [0,a] \times [0,a]$ . I assume coordinates to begin, end: $(t_i, t_i+tr_i); i=1,$are chosen at random (uniformly) in [0,a] .Then the condition of $R2$ starting as $R1$ ends is satisfied when $tr_1+t1= tr_2$ . So I then use an integral :
$\int_{tr1}^a dtr_2\int_0^a dtr_1\int _{tr1+t1}^{tr_2} dt_2 \int_0^a dt_1$
But this seems absurdly long; I think a double integral should be able to do it. Any hints, please?
 A: Ok, sorry, my phone died out and I don't have any way of importing pictures into the answer. 
  So we have $A,B$ , both on $[0,1]$ choosing intervals $(a_1,a_2),(b_1,b_2)$ respectfully. We want pairs of intervals with $a_2=b_1$. This is represented by all pairs $(a_1,a_2),(a_2,1)$ over all values $a_1,a_2 \in [0,1]$. In other words, we want to compute the integral $$ \int_0^1(a_2-a_1)da_1 \int_0^1 (1-a_2)da_2 = 1/12$$
How can I check if this is right?
EDIT: We have A arriving at $t=a_1$ and staying in until $t=a_2$, and B arriving at $t=b_1$ and staying in until $t=b_2$, where time is in the interval $[0,1]$ for
both $A,B$, i.e. $ 0 \leq a_1,a_2,b_1,b_2 \leq 1 $. So we want to know the total area of events where $a_2=b_1$, or events where A comes in at t=a1, stays in until t=a2 AND B comes in at $t=a_2$ and stays in until b2. Each of these is represented by a rectangle of sides $(a_2-a_1), (b_2-a_2)$ with $ 0 \leq a_1<a_2<b1 \leq 1 $. But for every choice $(a_1,a_2)$ for $A$ , any choice $(a_2,1)$ satisfies our requirements. This will be a rectangle of size $(a_2-a_1)(1-a_2)$ for $a_2,a_1$ both ranging over $[0,1]$
