Finding the expected value of a geometric random variable 
A monkey is sitting at a simplified keyboard that only includes the keys “a”, “b”, and “c”. The monkey presses the keys at random. Let X be the number of keys pressed until the monkey has pressed all the different keys at least once. For example, if the monkey typed “accaacbcaaac. . . ” then X would equal $7$ whereas if the monkey typed “cbaccaabbcab. . . ” then X would equal $3$

What is the expected value of $X$?
I know that to get the expected value of a geometric random variable, its just $\frac{1}{p}$.
However, $X$ is just the number of keys pressed until all characters have been pressed. What is the probability of $X$ to begin with?
 A: Let $C$ be the set of keys that have not been pressed. At start $C = \{a,b,c\}$. 
Let $T_1$ be the time $|C| = 2$, let $T_2$ be the time $|C| = 1$ and let $T_3$ be the time that $|C| = 0$ (i.e. the last key is pressed).
We know that $T_1 = 1$ (nonrandom). In addition, $T_2-T_1$ is geometric with parameter 2/3 given $T_1$ (2/3 keys are still in $C$) and $T_3-T_2$ is geometric with parameter 1/3 given $T_2$ (1/3 keys are still in $C$). Thus
$$ \mathbb{E}(T_3) = \mathbb{E} (T_2 + \mathbb{E} (T_3 - T_2 | T_2) ) = \mathbb{E} (T_2) + 3 = T_1 + \mathbb{E} ( \mathbb{E}(T_2 - T_1 | T_1) ) + 3 \\
= 1 + 3/2  + 3/1 = 11/2.$$
In general, for a keyboard with $n$ keys, by the same method the expected time becomes
$$ \mathbb{E}(T_n) =  \sum_{k=1}^n \frac{n}{k}. $$
A: We are looking for the probability for the monkey to press two from the three keys at least once each among $x-1$ presses, then the third key on press number $x$.
Assuming the monkey displays no bias in selecting a key on each stroke, then we just need to compare the count of favoured outcomes to the count of outcomes for the sample space (of any $x$ keypresses).
There are $2^{x-1}-2$ ways to press two keys at least once among $x-1$ presses.   There are also $\binom 32$ ways to select two from three keys.   The size of the favoured event is $3(2^{x-1}-2)$ outcomes.
There are $3^x$ ways to press three keys however often among $x$ presses.
$$\mathsf P(X=x) ~=~ \frac{(2^{x-1}-2) }{ 3^{x-1} } \quad\Big[x\in\{3,4,\ldots\}\Big]$$
Then $$\mathsf E(X) = \sum_{x=\color{red}3}^\infty x\left({\left(\frac{2}{3}\right)}^{x-1}-\frac 2 {3^{x-1}}\right)$$

Alternatively due to linearity of expectation, we just need to find the sum of expected times: to press any key, to press a different key after that, to press the third key after that.   As these are a sequence of three Geometric Processes with rates $1, 2/3, 1/3$ then:

 $$\mathsf E(X) = 1+ \tfrac 32+\tfrac 31 = \tfrac {11}2$$

