For the New Year's celebration in 2014, Hongqi Mall is hosting a shopping and voucher return event. For every 200 yuan spent, a customer can participate in one round of a lottery. The lottery rules are as follows: take out the 13 hearts cards $(A, 2, 3, \cdots, Q, K)$ from a deck of cards, shuffle them, and randomly draw one. The winning conditions: drawing a $K$ or $Q$, and not drawing three consecutive $Q$s in this round, results in winning a 100 yuan voucher, and the customer can continue to draw. If the card drawn is not a $K$ or $Q$, or if three consecutive $Q$s are drawn, it is considered a non-winning draw (the awards for the first two $Q$s are not revoked), and this round of the lottery ends. After each draw, the card is returned and the deck is reshuffled. Calculate the expected number of vouchers that can be won in one round of the lottery.
Solve $\frac{365}{2014}$.
If the first $n$ draws in a round are all $K$ or $Q$, and do not contain three consecutive $Q$s, then at least $n$ coupons are won, with a probability of $\frac{a_{n}}{13^{n}}$, where $a_{n}$ is the number of sequences of length $n$ consisting only of $\mathrm{K}$ and $\mathrm{Q}$ and not containing three consecutive $Q$s.
The expected number of coupons won is $\sum_{n=1}^{+\infty} \frac{a_{n}}{13^{n}}$.
Next, we calculate the number of such sequences.
When $n>3$, if the last term of the sequence is $\mathrm{K}$, there are $a_{n-1}$ possibilities; if the last two terms are $K Q$, there are $a_{n-2}$ possibilities; if the last three terms are $K Q Q$, there are $a_{n-3}$ possibilities. These three cases cover all such sequences without overlap. Therefore,
$$
a_{n}=a_{n-1}+a_{n-2}+a_{n-3} \text {. }
$$

And $a_{1}=2, a_{2}=4, a_{3}=7$, so we can write out this sequence.
Let $S=\sum_{n=1}^{+\infty} \frac{a_{n}}{13^{n}}$.
Then $13 S=\sum_{n=1}^{+\infty} \frac{a_{n}}{13^{n-1}}=2+\sum_{n=1}^{+\infty} \frac{a_{n+1}}{13^{n}}$,
$169 S=\sum_{n=1}^{+\infty} \frac{a_{n}}{13^{n-2}}=30+\sum_{n=1}^{+\infty} \frac{a_{n+2}}{13^{n}}$,
$2197 S=\sum_{n=1}^{+\infty} \frac{a_{n}}{13^{n-3}}=397+\sum_{n=1}^{+\infty} \frac{a_{n+3}}{13^{n}}$.
(4) - (3) - (2) - (1) gives
$2014 S=365 \Rightarrow S=\frac{365}{2014}$.