Randomly choose letters, from the word CHOOSE until both O's have been obtained. Find E(x) Letters are chosen without replacement. 
I get it that if I was to choose, lets say the letter C, then my E(x)=(1/6)(1+2+3+4+5+6). Because I have an equal chance to choose the letters for C.
But when I have two O's in the word, why isn't E(x)=(1/6)(1+2+3+4)+(1/3)(1+2). 
The answer in the back of the book says 14/3. No matter how I format this E(x) I stated, I can't get it. I was hoping to just find the correct equation and be able to figure out why it works. 
 A: $P$(it takes 5 choices) = $P$(choosing a string of five letters containing two O's with an O at the end)
$$= \frac{\textrm{total number of such strings which are distinct}}{\textrm{total number of distinct strings of six letters}}$$ 
Total number of such strings which are distinct
= 4 (total number of ways of ordering the letters C,H,S,E)
= $4 \cdot 24 = 96, $
where the factor of $4$ corresponds to the the number of ways of positioning the first O among the first three letters chosen from C,H,S,E.
Total number of distinct strings of six letters = $\frac{6!}{2} = 360$.
So $P$(it takes five choices) = $\frac{96}{320} $.
Using the same method you can calculate the probabilites $P$(it takes $n$ choices) for $n \in \{2,3,4,5,6\}$.
You can then calculate the expectation using the formula:
$$E(X) = \sum_{n=2}^6 n P(\textrm{it takes $n$ choices}) \ . $$
General formula:
$P(X=n) = \frac{(n-2)! \cdot (6-n)! \cdot (n-1)}{320}$ for $n=2,3,4,5,6.$
The $(n-2)!$ corresponds to the non-O letters before the second O.
The $(6-n)!$ corresponds to the letters after the second O.
The $(n-1)$ corresponds to the placing of the first O among the letters preceding the second O.
