4. There are 20 cards each inscribed with the numbers $1,2, \cdots, 19,20$. They are placed in a box, and 4 people each draw one card. The two people who draw the smaller numbers are in one group, and the two people who draw the larger numbers are in another group. If two of them draw 5 and 14, the probability that these two are in the same group is ( ).
(A) $\frac{1}{2}$
(B) $\frac{2}{51}$
(C) $\frac{5}{51}$
(D) $\frac{7}{51}$
4.D.

Since two people have drawn the cards 5 and 14, the other two need to draw from the remaining 18 cards, which results in $18 \times 17$ 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 $15 \sim 20$, which gives $5 \times 6$ scenarios;
(2) 5 and 14 are the larger numbers, and the other two need to draw from the 4 cards numbered $1 \sim 4$, which gives $3 \times 4$ scenarios;

Therefore, the probability that the two who drew the cards 5 and 14 are in the same group is $\frac{5 \times 6 + 3 \times 4}{18 \times 17} = \frac{7}{51}$.