(14) There are 20 cards, each with a number from $1,2,3, \cdots, 20$, placed in a box. Four people draw one card each without replacement. 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 $\qquad$. (Answer in the simplest fraction form)
14 Since two people have already drawn the cards 5 and 14, the other two people need to draw from the remaining 18 cards, which gives $\mathrm{A}_{18}^{2}$ possible situations. If the two people who drew 5 and 14 are in the same group, there are two scenarios: if 5 and 14 are the smaller two numbers, then the other two people need to draw one card each from the 6 cards numbered $15 \sim 20$, which gives $\mathrm{A}_{6}^{2}$ possible situations; if 5 and 14 are the larger two numbers, then the other two people need to draw one card each from the 4 cards numbered $1 \sim 4$, which gives $\mathrm{A}_{4}^{2}$ possible situations. Therefore, the probability that the two people who drew the cards 5 and 14 are in the same group is $P=\frac{\mathrm{A}_{6}^{2}+\mathrm{A}_{4}^{2}}{\mathrm{~A}_{18}^{2}}=\frac{7}{51}$.