【Example 2】There are $m$ white balls and $n$ black balls $(m>n)$ in a jar. The balls are taken out one by one, without replacement. If at some point during the process, the number of white balls taken out equals the number of black balls taken out, it is said that a coincidence has occurred. Find the probability that at least one coincidence occurs.
Let the event of at least one coincidence be denoted as $E$, the event that the first ball drawn is white be denoted as $A$, and the event that the first ball drawn is black be denoted as $B$.
It is easy to see that $B \subset E$, and
$$
P(B)=\frac{n}{m+n}
$$

Therefore, we only need to find the probability $P(E A)$, which is the probability that the first ball drawn is white and at least one coincidence occurs.

To find $P(E A)$, we will use the approach considered in Example 1. Let $y_{0}=0$,
$$
y_{j}-y_{j-1}=\left\{\begin{aligned}
1, & \text { if the } j \text {th ball drawn is white } \\
-1, & \text { if the } j \text {th ball drawn is black }
\end{aligned}\right.
$$
$$
j=1,2, \cdots, m+n
$$

Thus, $y_{m+n}=m-n$. If in a Cartesian coordinate system we sequentially connect the points $\left(j, y_{j}\right), j=0,1,2, \cdots, m+n$, then each way of drawing the balls corresponds to a broken line starting from $(0,0)$ and ending at $(m+n, m-n)$. The slope of each segment of the broken line is either 1 or -1, and the endpoint $(m+n, m-n)$ is above the $x$-axis.
$\Omega$ consists of all different broken lines, and it is easy to know that $|\Omega|=C_{m+n}^{m}$. Event $E$ consists of all broken lines that have a common point with the $x$-axis (excluding the point $\left(0, y_{0}\right)$). The broken lines in event $B$ all pass through the point $(1,-1)$, while the broken lines in event $E A$ all pass through the point $(1,1)$ but have a common point with the $x$-axis. If we reflect the part of the broken line in event $E A$ to the left of its first common point with the $x$-axis (point $T$ in Figure 18) about the $x$-axis, we obtain a broken line passing through the point $(1,-1)$; conversely, any broken line passing through the point $(1,-1)$ has a common point with the $x$-axis, and if we reflect the part of the broken line to the left of its first common point with the $x$-axis about the $x$-axis, we obtain a broken line passing through the point $(1,1)$ and having a common point with the $x$-axis. This shows that the broken lines in event $E A$ and $B$ can form a one-to-one correspondence, so $|E A|=|B|$, and thus $P(E A)=P(B)=\frac{n}{m+n}$, leading to
$$
P(E)=P(E A)+P(B)=\frac{2 n}{m+n}
$$

Interestingly, in calculating the probability $P(E)$, we did not determine the number of broken lines in event $E$. Instead, we can use the probability $P(E)$ to find $|E|$. In fact,
$$
|E|=|\Omega| \cdot P(E)=C_{m+n}^{m} \cdot \frac{2 n}{m+n}=2 C_{m+n-1}^{m}
$$

Now let's look at some more examples. In Example 8 of Section 8, we discussed the problem of the number of shortest paths from point $A$ to point $B$ in Figure 19. Each segment of such a move is either to the right or upwards, which can be called a non-decreasing move. Now we will further examine problems related to non-decreasing moves.