Probability of selecting four letters from ENCYCLOPAEDIA From a maths textbook: "Four letters are randomly selected from the word ENCYCLOPAEDIA. Find the probability that one letter E will occur in the selection of 4 letters."

Clarification - as I understand the question, it is asking about the probability of getting combinations with exactly one E. Samples would be with replacement - I'm picturing a bag with 13 lettered tiles in it and replacing the tiles after each sample.

In the textbook we have worked through questions involving permutations with repeated values, but not combinations. And there are a number of good examples on the web about combinations with repeated values, but this problem has limited repeated values (2 A's, 2 C's, 2 E's). I actually put the letters in alphabetical order and started making my own list of 4 letter combinations and discovered it was tricky with the repeats (and going to take a while).

It seems that $^{13}C_4$ is a good place to start for the total number of combinations (for the denominator of the probability result), but that doesn't adjust for the duplicate letters.

Since I've recently been doing Javascript in my IT class I went ahead and put together a webpage that created a list of all 715 combinations and then tried to work out how to eliminate the ones that were duplicates (more than half of the starting list).

Eventually I got the Javascript created list down to 321. While it created the list it also counted the ones with one E and that total was 100.

So problem solved: 100/321 is the probability of getting one E when randomly selecting four letters out of ENCYCLOPAEDIA. By the way, this does not match the answer in the textbook, but I suspect it is far too complicated for the Yr 11 class I teach.

I just wonder if there is a way to answer this without creating a complete list of all the possible combinations? The stuff I read about combinations with repeats is not something we do in the class (or something I have ever seen done before), so is this question possible to do "without a computer"? And what about words with different numbers of repeated objects?
Let us imagine that the letters are written on slips of paper, one letter to a slip. So we have $13$ slips of paper. It is convenient to paint letters that repeat using different colours, so we have $2$ distinct $E$'s, $E_r$ (blue) and $E_r$ (red).

It is not clear whether the question asks for the probability of exactly $1$ $E$, or the probability of at least $1$ $E$. I would lean to the exactly $1$ interpretation.

We interpret "at random" to mean that all selections of $4$ slips of paper from the $13$ are equally likely. There are $\binom{13}{4}$ ways to choose $4$ slips of paper from the $13$. 

Of these, $\binom{11}{3}$ have $E_b$ but not $E_r$, and $\binom{11}{3}$ have $E_r$ but not $E_b$. Thus the required probability is
$$\frac{2\binom{11}{3}}{\binom{13}{4}}.$$

Remark: At least $1$ is easiest to do by finding first the probability of the complement, no $E$. There are $\binom{11}{4}$ ways to choose $4$ slips, none of which is $E_b$ or $E_r$.
Probability would be (No. of Selections with E) / (All Possible no. of selections)

First, lets calculate 'All Possible no. of selections'

Letters of the word ENCYCLOPAEDIA are (E,E),(C,C),(A,A),N,Y,L,O,P,D,I
i.e. 3 duplicates and 7 singles.
For selections, the duplicates and singles should be considered as independent groups.

The four letters selected could have:
(a) 4 singles
(b) 2 duplicates
(c) 1 duplicate and 2 singles

(a) can happen in 10C4 ways = 210
4 out of E,C,A,N,Y,L,O,P,D,I   

(b) can happen in 3C2 ways = 3
2 out of EE,CC,AA  

(c) can happen in 3C1 x 9C2 ways = 108
1 out of EE,CC,AA and then
2 out of (N,Y,L,O,P,D,I and 1 each from 2 unselected doubles)

All Possible no. of selections = 210 + 3 + 108 = 321 ways

On similar principles, lets calculate 'No. of Selections with E'

Since one 'E' is always included and the other is always excluded,
we need to select 3 letters out of (C,C),(A,A),N,Y,L,O,P,D,I
i.e. 2 duplicates and 7 singles.  

The three letters selected could have:
(a) 3 singles
(b) 1 duplicate and 1 single

(a) can happen in 9C3 ways = 84
3 out of C,A,N,Y,L,O,P,D,I

(b) can happen in 2C1 x 8C1 ways = 16
1 out of CC,AA and then
2 out of (N,Y,L,O,P,D,I and 1 from the unselected double) 

No. of Selections with E = 84 + 16 = 100 ways

Hence, Probability = 100/321