14. In a promotional sales event with prizes, there are $n$ $(n \geqslant 4)$ people who win the third prize. The third prize consists of four different items, and each winner of the third prize can freely choose one of the four items. The probability that one of the items is not chosen is $\qquad$
$14 \cdot 4\left(\frac{3}{4}\right)^{n}-12\left(\frac{1}{2}\right)^{n}+\frac{3}{4^{n-1}}$.
Since each person has 4 different choices for the prize, the total number of different possible choices is $4^{n}$.
One prize is not chosen, meaning that three prizes are chosen by someone.
（1）Only one prize is chosen, clearly, there are $\mathrm{C}_{4}^{1}=4$ different results.
（2）Two prizes are chosen. The number of ways to choose two prizes from four prizes is $\mathrm{C}_{4}^{2}$. For the two chosen prizes, there are $2^{n}$ different choices for $n$ people, but we need to exclude the two cases where only one prize is chosen, so the number of different choices where two prizes are chosen is
$$
C_{4}^{2}\left(2^{n}-2\right)=6 \times 2^{n}-12 .
$$
（3）Three prizes are chosen. Similarly, when three prizes are chosen, the number of different choices is
$$
3^{n}-C_{3}^{2}\left(2^{n}-2\right)-3=3^{n}-3 \times 2^{n}+3
$$

different choice results, so the number of different results where three prizes are chosen, i.e., one prize is not chosen, is
$$
\mathrm{C}_{4}^{3}\left(3^{n}-3 \times 2^{n}+3\right)=4 \times 3^{n}-12 \times 2^{n}+12
$$

different results. The corresponding probability is
$$
\frac{4 \times 3^{n}-12 \times 2^{n}+12}{4^{n}}=4\left(\frac{3}{4}\right)^{n}-12\left(\frac{1}{2}\right)^{n}+\frac{3}{4^{n-1}}
$$