Example 14 (2003 Tianjin Competition Question) There are 20 cards each written with a number $1,2, \cdots, 19,20$. They are placed in a box, and 4 people each draw one card. The two people who draw the two smaller numbers are in one group, and the two people who draw the two larger numbers are in another group. If two of them draw 5 and 14, respectively, then the probability that these two are in the same group is ( ).
A. $\frac{1}{2}$
B. $\frac{2}{51}$
C. $\frac{5}{51}$
D. $\frac{7}{51}$
Solution: Choose D. Reason: Since two people draw cards numbered 5 and 14, the other two need to draw from the remaining 18 cards, which results in $18 \times 17$ possible scenarios. If the two who drew 5 and 14 are in the same group, there are two cases.
(1) 5 and 14 are the smaller numbers, and the other two need to draw from the 6 cards numbered $15 \sim 20$, with $5 \times 6$ possibilities;
(2) 5 and 14 are the larger numbers, and the other two need to draw from the 4 cards numbered $1 \sim 4$, with $3 \times 4$ possibilities.
Thus, the probability that the two who drew cards numbered 5 and 14 are in the same group is $\frac{5 \times 6+3 \times 4}{18 \times 17}=\frac{7}{51}$.