Three. (25 points) A city has built an urban subway to facilitate the travel of its citizens, with a total of 20 stations (including the starting and terminal stations). The subway company is preparing to submit a pricing scheme to the Price Bureau. There are two pricing schemes.

【Scheme One】Fare for each passenger = the actual number of stations the passenger travels (excluding the boarding station) × 0.5 yuan;

【Scheme Two】Three fixed fares: 2 yuan (up to 3 stations), 5 yuan (up to 10 stations), 7 yuan (full journey).

The company has conducted a statistical analysis, finding that the number of people boarding at each station is basically consistent, and assuming that each passenger has an equal probability of getting off at any subsequent station.

If you were the leader of the company, to maximize the company's profit, which scheme would you submit to the Price Bureau? Please explain your reasoning.
Three, Plan Two.
Assume the number of people boarding at each station is $k$.
According to Plan One:
The average number of stations traveled by passengers boarding at the $i(i=1,2, \cdots, 19)$ station is
$$
\frac{1}{2}[1+(20-i)]=\frac{1}{2}(21-i) .
$$

Therefore, the ticket revenue is
$$
\begin{array}{l}
k \times 0.5 \times \frac{1}{2}(20+19+\cdots+2) \\
=\frac{k}{4}(10 \times 21-1)=\frac{209 k}{4}(\text { yuan }) .
\end{array}
$$

According to Plan Two:
Passengers boarding at the 1st station buy 2 yuan, 5 yuan, and 7 yuan tickets in proportions of $\frac{3}{19}, \frac{7}{19}, \frac{9}{19}$; passengers boarding at the 2nd station buy 2 yuan, 5 yuan, and 7 yuan tickets in proportions of $\frac{3}{18}, \frac{7}{18}, \frac{8}{18} ; \cdots \cdots$ passengers boarding at the 9th station buy 2 yuan, 5 yuan, and 7 yuan tickets in proportions of $\frac{3}{11}, \frac{7}{11}, \frac{1}{11}$; passengers boarding at the 10th station buy 2 yuan and 5 yuan tickets in proportions of $\frac{3}{10}, \frac{7}{10}$; passengers boarding at the 11th station buy 2 yuan and 5 yuan tickets in proportions of $\frac{3}{9}, \frac{6}{9}, \cdots \cdots$ passengers boarding at the 16th station buy 2 yuan and 5 yuan tickets in proportions of $\frac{3}{4}, \frac{1}{4}$; passengers boarding at the 17th, 18th, and 19th stations all buy 2 yuan tickets.
Therefore, the ticket revenue is
$$
\begin{array}{l}
k\left[2\left(\frac{3}{19}+\frac{3}{18}+\cdots+\frac{3}{4}+3\right)+\right. \\
5\left(\frac{7}{19}+\frac{7}{18}+\cdots+\frac{7}{10}+\frac{6}{9}+\frac{5}{8}+\cdots+\frac{1}{4}\right)+ \\
\left.7\left(\frac{9}{19}+\frac{8}{18}+\cdots+\frac{1}{11}\right)\right] \\
=k\left[6\left(\frac{1}{19}+\frac{1}{18}+\cdots+\frac{1}{4}\right)+6+\right. \\
35\left(\frac{1}{19}+\frac{1}{18}+\cdots+\frac{1}{10}\right)+ \\
5\left(6-\frac{3}{9}-\frac{3}{8}-\cdots-\frac{3}{4}\right)+ \\
\left.7\left(9-\frac{10}{19}-\frac{10}{18}-\cdots-\frac{10}{11}\right)\right] \\
=k\left[99+(6+35-70)\left(\frac{1}{19}+\frac{1}{18}+\cdots+\frac{1}{11}\right)+\right. \\
\left.(6+35) \times \frac{1}{10}+(6-15)\left(\frac{1}{9}+\frac{1}{8}+\cdots+\frac{1}{4}\right)\right] \\
=k\left[99+\frac{41}{10}-29\left(\frac{1}{19}+\frac{1}{18}+\cdots+\frac{1}{11}\right)-\right. \\
\left.9\left(\frac{1}{9}+\frac{1}{8}+\cdots+\frac{1}{4}\right)\right] \\
>k\left(99+4-29 \times \frac{9}{11}-9 \times \frac{6}{4}\right) \\
>k(99+4-24-14)=65 k>\frac{209 k}{4} \text {. } \\
\end{array}
$$

Therefore, Plan Two should be chosen.