【Example 6】There are $a$ white balls and $b$ black balls $(a \neq b)$ in a jar, from which all the balls are drawn one by one without replacement. Try to determine which of the following two events has a higher probability:
Event $A$: At some point, the number of white balls drawn equals the number of black balls drawn,
Event $B$: At some point, the number of white balls left in the jar equals the number of black balls left in the jar.
Try to find the probabilities of these events.
Let's assume $a > b$. The phrase "at some moment" should be understood as not including the final moment; otherwise, event $B$ would be a certain event, because the number of remaining white balls and black balls would both be 0 (of course, they would be equal).

We will sequentially examine the difference in the number of white and black balls taken out, and represent the results with points $\left(k, y_{k}\right)$ on a plane, where $y_{k}$ is the difference in the number of white and black balls taken out after the $k$-th draw. For $0 \leqslant k \leqslant a+b$, connect the points on the plane in the order of increasing $k$ to form a broken line, as described in Example 1.

Each draw sequence corresponds to a path, and different draw sequences correspond to different paths. Each broken line starts at point $(0,0)$ and ends at point $(a+b, a-b)$, with a total of $C_{a+b}^{a}$ distinct paths.

According to the description in Example 2, when event $A$ occurs, there is at least one coincidence, meaning the corresponding path intersects the $x$-axis; and every path that intersects the $x$-axis belongs to event $A$. Therefore, $|A|$ is the number of broken lines that intersect the $x$-axis, and event $A$ is the event $E$ in Example 2. There, it has been found that $P(A) = (2b) / (a+b)$, or written as
$$
P(A) = \frac{2 \min \{a, b\}}{a+b}
$$

And $|A| = 2 C_{a+b-1}^{a}$, or written as
$$
|A| = 2 C_{a+b-1}^{\max \{a, b\}}
$$

Now, let's find $P(B)$.
When the difference in the number of white and black balls taken out at some intermediate moment is exactly $a-b$, event $B$ occurs. This means the path intersects the line $y = a - b$ before the final moment.
Reflect the part of each path to the left of the first intersection with the line $y = a - b$ about the line $y = a - b$, and you get a path starting from point $(0, 2(a-b))$ and ultimately reaching point $(a+b, a-b)$. For any path starting from point $(0, 2(a-b))$, intersecting the line $y = a - b$ at some intermediate point, and ultimately reaching point $(a+b, a-b)$, if the part to the left of the first intersection is reflected about the line $y = a - b$, then you get a path starting from $(0,0)$, intersecting the line $y = a - b$ at some intermediate point, and ultimately reaching point $(a+b, a-b)$. Therefore, $|B|$ is the number of paths starting from point $(0, 2(a-b))$, intersecting the line $y = a - b$ at some intermediate point, and ultimately reaching point $(a+b, a-b)$.
If a path starting from point $(0, 2(a-b))$, intersecting the line $y = a - b$ at some intermediate point, and ultimately reaching point $(a+b, a-b)$ is shifted downward by $a - b$, then it becomes a path starting from point $(0, a - b)$, intersecting the $x$-axis at some intermediate point, and ultimately reaching point $(a+b, 0)$; the reverse is also true.
Therefore, $|B|$ is the number of paths starting from point $(0, a - b)$, intersecting the $x$-axis at some intermediate point, and ultimately reaching point $(a+b, 0)$. It is easy to see that the number of such paths is equal to the number of paths starting from point $(0,0)$, intersecting the $x$-axis at some intermediate point, and ultimately reaching point $(a+b, a - b)$.
Thus, $P(B) = P(A)$.