Question 200: Suppose a bag contains one red, one yellow, and one blue ball. Each time a ball is drawn from the bag (the probability of drawing a ball of any color is the same), its color is determined and then it is put back. The process continues until a red ball is drawn twice in a row. Let the number of draws at this point be $X$. Try to find the mathematical expectation $\mathrm{E}(\mathrm{X})$ of $X$.
Question 200, Solution: Classify based on the color of the ball drawn first. Let the expected number of ways to draw the ball with colors red, yellow, and blue for the first time be $E(a)$, $E(b)$, and $E(c)$, respectively. Then, by symmetry, we know $E(b)=E(c)$. Since the probability of drawing a red, yellow, or blue ball first is the same, we have
$$
E(X)=\frac{E(a)+2 E(b)}{3}
$$

First, consider the case where the first ball drawn is red. If the second ball drawn is also red, the operation ends; otherwise, the first ball is red and the second ball is yellow or blue. Ignoring the first ball, the remaining drawing method can be considered a new method (i.e., the first ball is yellow or blue), so
$$
E(a)=\frac{1}{3} \times 2+\frac{2}{3}(1+E(b))
$$

Next, consider the case where the first ball drawn is yellow or blue. Ignoring the first ball, the remaining drawing method can be considered a new method, so
$$
E(b)=E(X)+1
$$

Combining (1), (2), and (3), we find $E(X)=12$.
Note: Let the probability of not drawing two consecutive red balls in $n$ draws be $p_{n}$, then $p_{0}=1$, $p_{1}=1$, $p_{2}=1-\left(\frac{1}{3}\right)^{2}=\frac{8}{9}$. And $p_{n}=\frac{1}{3} \cdot \frac{2}{3} \cdot p_{n-2}+\frac{2}{3} \cdot p_{n-1}$ (if the $n$-th draw is a red ball, then the $(n-1)$-th draw is not a red ball, and the first $n-2$ draws did not have two consecutive red balls; if the $n$-th draw is not a red ball, then the first $n-1$ draws did not have two consecutive red balls).
$$
\begin{array}{l}
\text { Thus, } P(X=n)=\left\{\begin{array}{c}
0, \text { when } n=1 \\
\frac{1}{9}, \text { when } n=2 \\
\frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} \cdot p_{n-3}, \text { when } n \geq 3
\end{array}\right. \text { . Therefore, } \\
E(x)=2 \times \frac{1}{9}+\sum_{n=3}^{+\infty} \frac{2 p_{n-3}}{27} \cdot n=\frac{2}{9}+\frac{2}{27} \sum_{n=0}^{+\infty} p_{n} \cdot(n+3)=12
\end{array}
$$