Expected number of trials until all 3 (independent?) events happened Suppose there are three boxes, each has $9$ white balls and $1$ colored ball:
Box $1$: $9$ white and $1$ red.
Box $2$: $9$ white and $1$ green.
Box $3$: $9$ white and $1$ blue.
Now I am drawing balls from the boxes with replacement. In each trial, I draw $1$ ball from each of the three boxes. If any of them is red, I increase my red ball counter by $1$, then I put all balls back to their respective boxes. Similarly, I increase the corresponding counter if I get a green ball or a blue ball. The trials end only if my counters for all red, green and blue balls are at least $1$ (i.e., I get balls of each color at least once over all the trials).
The question is, how do I calculate the expected number of trials until I hit the end condition?
I know, if I only have $1$ box, then it is a simple geometric distribution, with $p=\frac{1}{10}$, and hence the expected number of trials will be $10$. But now, although the three events (get a red ball, get a green ball and get a blue ball) are independent, the combined event is not. And also since it is possible to have more than one event happen in each trial (e.g., getting red from box $1$ and green from box $2$ in the same trial), I cannot consider the events to happen one after another.
I ran a quick simulation and it seems like the expected number of trials is around $17.9$. I would like to know if there is a mathematical way to justify this number.
 A: For $i \in \{1, 2, 3\}$, let $X_i$ denote the number of trials needed to take the non-white ball from box $i$ (for the first time). Then we are interested in finding $E(X)$ where $X := \max\{X_1, X_2, X_3\}$.
For any non-negative integer $c$, we have
\begin{align*}
P(X \leq c) &= P(\max\{X_1, X_2, X_3\} \leq c) \\
&= P((X_1 \leq c) \: \cap \: (X_2 \leq c) \: \cap \: (X_3 \leq c)) \\
&= P(X_1 \leq c) \cdot P(X_2 \leq c) \cdot P(X_3 \leq c)
\end{align*}
due to the fact that $X_1$, $X_2$, and $X_3$ are independent. We can easily find that $$P(X_i \leq c) = 1 - P(X_i \geq c+1) = 1 - (9/10)^c, \quad \quad i \in \{1, 2, 3\}.$$
Substituting back, we get
$$P(X \leq c) = (1 - (9/10)^c)^3.$$
Now, we can compute the expectation of $X$:
\begin{align*}
E(X) &= \sum_{c=1}^{\infty} P(X \geq c) \\
&= \sum_{c=1}^{\infty} 1 - P(X \leq c-1) \\
&= \sum_{c=0}^{\infty} 1 - P(X \leq c) \\
&= \sum_{c=0}^{\infty} 1 - (1 - (9/10)^c)^3 \\
&= 3 \sum_{c=0}^{\infty} (9/10)^c - 3\sum_{c=0}^{\infty} (9/10)^{2c} + \sum_{c=0}^{\infty} (9/10)^{3c} \\
&= 30 - \frac{300}{19} + \frac{1000}{271} \\
&\approx 17.90056.
\end{align*}

Remark: The method to write $P(X \leq c) = P(X_1 \leq c) \cdot P(X_2 \leq c) \cdot P(X_3 \leq c)$ is useful in situations when you are dealing with a variable that is defined as the maximum of other variables that are independent. Here, we see that it can be used to find the probability density function and the expectation of $X$.
A: Couldn't you do it much simpler like this, using only high-school math?
Denote the expected additional trials given $j$ number of colored balls has
appeared for the first time by $x_{j}$ ($j=0,1,2$).  Then
\begin{eqnarray*}
x_{0} &=&1+3\ast \frac{1}{10}\frac{9}{10}\frac{9}{10}x_{1}+3\ast \frac{1}{10}%
\frac{1}{10}\frac{9}{10}x_{2}+\frac{9}{10}\frac{9}{10}\frac{9}{10}x_{0} \\
x_{1} &=&1+2\ast \frac{1}{10}\frac{9}{10}x_{2}+\frac{9}{10}\frac{9}{10}x_{1}
\\
x_{2} &=&1+\frac{9}{10}x_{2}
\end{eqnarray*}
, Solution is: $\left[ x_{0}=\frac{92\,170}{5149},x_{1}=\frac{280}{19}%
,x_{2}=10\right] $
