Example 3: In a table tennis match between A and B, the score is tied at $14: 14$. How many score sequences (denoted as $M$ in total) are there in the game process, where A is leading except for one exact tie during the game?
Let the score sequence $m$ in $M$ be $x_{k}: y_{k}(k=1,2,3, \cdots, 28)$, and let $b_{k}=x_{k}-y_{k}$. Then $b_{28}=0$. Let $m$ correspond to a broken line $l$ connecting the following sequence of points:
$$
(0,0),\left(1, b_{1}\right),\left(2, b_{2}\right), \cdots,\left(27, b_{27}\right),(28,0)
$$

Let $A$ be the set of all broken lines, except the starting point $(0,0)$ and the endpoint $(28,0)$, that intersect the $x$-axis exactly once and lie entirely in the upper half-plane. Then $l \in A$, and by the above method, $M$ and $A$ are in one-to-one correspondence, so $|M|=|A|$.

Let $B$ be the set of all broken lines, except the starting point $(0,0)$ and the endpoint $(28,0)$, that do not intersect the $x$-axis and lie entirely in the upper half-plane. Since the broken lines in $B$ must pass through the points $(1,1)$ and $(27,1)$, by the corollary of the reflection principle (2), we can obtain

Next, we prove $|B|=|A|$. Let $l \in A$, and suppose $l$ intersects the $x$-axis at $x=2k$, dividing $l$ into two segments $l_{1}$ and $l_{2}$ (Figure 1-8). Remove the first segment of $l_{2}$, translate $l_{1}$ to connect at point $(2k+1,1)$, and then add the removed segment to the line connecting $(0,0)$ to $(1,1)$. This results in a broken line $l^{\prime}$ that belongs to $B$. Let $l$ and $l^{\prime}$ correspond, and it is easy to see that this is a one-to-one correspondence between $A$ and $B$, so $|A|=|B|$. Therefore, $|M|=|A|=|B|=\frac{1}{14} C_{26}^{13}$.