1st Yr Probability: what is wrong with my approach? (using indicator RVs) 
Joe’s iPod has $500$ different songs, consisting of $50$ albums of $10$ songs each. He listens to $11$ random songs on his iPod, with all songs equally likely and chosen independently (so repetitions may occur).
A pair of songs is a “match” if they are from the same album. If, say, the 1st,
3rd, and 7th songs are all from the same album, this counts as $3$ matches. Among
the $11$ songs he listens to, how many matches are there on average? (Simplify.)

My attempt
Let $X = $ # of matches among the $11$ songs
Let $X = X_1 + ... X_{50}$  where $X_i =$ # of matches among the $11$ songs from album $i$
$$ E[X] = E[X_1 + ... + X_{50}] = 50E[X_1] $$
To find $E[X_1]$ I will condition on if there are at least $2$ matches of the first album in the $11$ songs.
$$ E[X_1] = E[X_1 \mid \text{at least 2}] P(\text{at least 2}) +  E[X_1 \mid \text{0 or 1}] P(\text{0 or 1}) $$
So the second term goes to zero since each $X_i$ is in $[2,11]$.
$$ P(\text{at least 2}) = 1 - P(\text{first album not in 11 songs}) - P(\text{first album shows up once in 11})$$
$$ =1 - \left( \frac{49}{50} \right)^{11} - 11 \left( \frac{1}{50} \right) \left( \frac{49}{50} \right)^{10} \approx 0.01951263$$
$ E[X_1 \mid \text{at least 2}] = 2 + Y $ where $Y \sim Binomial(11-2, 1/50)$ and so
$$ E[X_1 \mid \text{at least 2}] = 2 + 9/50 $$
Finally
$$ E[X] = 50 ( 2 + 9/50) 0.01951263 \approx 2.1268768591 $$
However the answer is $1.1$. The book solution below uses the $11$ choose $2$ possible pairs as indicator RVs... berry clever... it makes sense to me and is superior to what I'm doing. But what is wrong with my approach above? Thanks.
My question: What is wrong with my approach, where am I going off the rails?

Book solution:

 A: Your method works if you use the correct definition of "match" (though it is harder than the official method).
From the official solution, we see that a "match" means "a pair $(i,j)$ such that the $i^{th}$ and $j^{th}$ song are from the same album.  Thus if songs $(i,j,k)$ are all from the same album we have three matches since $(i,j), (i,k), (j,k)$ are all matches.  
Note:  the example given in the problem statement is misleading.  It would be better to look at the case where the four songs $(i,j,k,l)$ are all on the same album.  In that case there are $6$ matches:  $(i,j), (i,k), (i,l), (j,k), (j,l), (k,l)$.  In general, if there are $n$ songs from a given album there are $\frac {n(n-1)}2$ matches.
To use your method we compute $E[X_i]$ via the binomial distribution. We get $$E[X_i]=\sum_{n=0}^{n=11} \frac {n(n-1)}2\times \binom {11}n\times \left( \frac 1{50}\right)^n\times \left( \frac {49}{50} \right)^{11-n}=.022$$
The answer is then $$E=50\times .022=1.1$$ confirming the official solution.
