probability - Determine the expected time until the first arrival.

Two people agree to meet at a specified place between 3:00

P.M. and 4:00 P.M. Suppose that you measure time to the nearest
minute relative to 3:00 P.M. so that, for instance, time 40
represents 3:40 P.M. Further suppose that each person arrives
according to the discrete uniform distribution on {0, 1, ..., 60}
and that the two arrival times are independent. Determine the
expected time until the first arrival.

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Here I've tried setting :

$X_i$=person i's arrival time with i=1,2

$$P_X(x_i)= \begin{cases} 1/61, & \text{if $x_i$ $\in[0,1,2,3...60]$} \\ 0, & \text{otherwise} \end{cases}$$

I create a function Z=min{x,y}

$$E(z)=\sum_{z=0}^{60}zP_Z(z)=\sum_{z=0}^{60}z[P(x_1=z,x_2>z)+P(x_1>z,x_2=z)+P(x_1=z,x_2=z)]$$

And that is where I'm kind of stuck because
I've tried rewrittin $P(x_1=z,x_2>z)$ as $P(x_2>z)P(x_1=z|x_2>z)$ and drawing a little pmf table but still no luck.

If anybody by any chance knows a much easier way to solve this problem I would be happy to hear from them.

Answer

A reusable trick . . .

Since the values of $z$ are nonnegative integers, it follows that
$$E(z) = p_1 + p_2 + p_3 + \cdots$$
where $p_k = P(z \ge k)$.

Hence we get
$$E(z) = \sum_{k=1}^{60} \left(\frac{61-k}{61}\right)^2 = \frac{1210}{61}\approx 19.84$$

To explain the trick . . .

For each positive integer $k$, let $q_k=P(z=k)$.

Then we have
$$\begin{align*} p_1 &= q_1 + q_2 + q_3 +\cdots\\[4pt] p_2 &= \phantom{q_1 +\; }q_2 + q_3 +\cdots\\[4pt] p_3 &= \phantom{q_1 + q_2 + \;}q_3 + \cdots\\[4pt] \vdots\\[4pt] \end{align*}$$
hence, summing the columns, we get
$$p_1 + p_2 + p_3 + \cdots = 1q_1 + 2q_2 + 3q_3 + \cdots \qquad\qquad\;\;\;\;\;$$