7 pairs of socks, 7 different colours. Expected value of the number of days, in which 2 socks of the same colour are picked. 
A student has 7 pairs of socks of 7 different colours. During 7 days he randomly picks 2 socks from the drawer (not necessarily of the same colour) and then doesn't put them back. Find the expected value of the number of days, in which 2 socks of the same colour are picked.

My work so far: There are 7 colours, so on the first day the student has a $\frac{1}{7}$ chance to pick the first sock, and $\frac{1}{13}$ chance to pick another sock of the same colour, giving $\frac{1}{91}$ chance to pick a pair of matching socks on the first day. Since drawing of all socks is equally likely, this can be multiplied by 7 to give $\frac{7}{91}$. Multiplying it by the seven days of the week, we have $\frac{49}{91}$, which comes out to $0.5385$ days (question asks for 4 decimal places). Are there any flaws in my logic, and is the answer correct (or close)?
Edit: corrected wrong decimal approximation
 A: Thanks everyone for comments! For clarity, I will write an answer that summarizes all the comments written.
We call a day "good" if two socks of the same color are chosen on that day. As far as i understand, it is required to calculate the expected number of "good" days. The indicator method works well here. Let $\xi$ be a random variable that represents the number of good days. Enter the following event
$$A_j = \left \{  j\text{-th day was good} \right\}, \quad j=1,...,7.$$
Then
$$\xi = I_{A_1} + I_{A_2} + ... + I_{A_7} \text{ where } I_{A_j} = \begin{cases} 1, \text{ if event }A_j \text{ happened},\\ 0, \text{ otherwise}  \end{cases}$$
Hence,
$$E \xi = E\left(\sum\limits_{j=1}^{7} I_{A_j}  \right)= \sum\limits_{j=1}^{7} E\left(I_{A_j} \right) =  \sum\limits_{j=1}^{7} P(A_j) = 7 \cdot \frac{C_7^1}{C_{14}^2} =  \frac{7}{13}, \quad (\text{here } C_n^k = \left( \begin{gathered} n \\ k\end{gathered} \right)). $$
A good question here is why $P(A_j) = \frac{1}{13} \quad \forall j \in \left\{1,...,7 \right\}$. It follows from symmetry as @JMorawitz wrote or we can verify this, using conditional probability (but this take long time). Anyway, the probability $P(A_2)$ can be calculated quickly
$$P(A_2) = P(A_2|A_1) \cdot P(A_1) + P(A_2|\overline{A_1})\cdot P(\overline{A_1}) =$$
$$ = \underbrace{\frac{C_6^1}{C_{12}^2} }_{12 \text{ socks left of which there are } 6 \text{ pairs}} \cdot \frac{1}{13} + \underbrace{\frac{C_5^1}{C_{12}^2}}_{12 \text{ socks left of which there are } 5 \text{ pairs}} \cdot \frac{12}{13} = $$
$$ = \frac{1}{11}\cdot \frac{1}{13} + \frac{10}{12\cdot 11} \cdot \frac{12}{13} = \frac{1}{11}\cdot \frac{1}{13} + \frac{10}{11\cdot 13}= \frac{1}{13}.$$
