Stuck on a probability question about pairs of socks The question is as follows:

You have 10 pairs of socks (i.e., 20 socks in total) with each pair in a different color.  You put all the socks into the washing machine but it “eats” four of the 20 socks at random.


What  is  the  expected  number  of  complete  pairs  left  in  the  washing machine?

I approached this question like this:
There are 3 possible outcomes:

*

*8 pairs of socks survive

*7 pairs of socks survive

*6 pairs of socks survive

I then tried to find the probability of each of these events occurring.
The ways that 8 pairs of socks survive is:

*

*Eating 2 different colour socks then eating their pairs


*Eating first sock, then eating it's pair, Eating the second sock, then eating its pair
P(8 pairs) = $1*\frac{18}{19}*\frac{2}{18}*\frac{1}{17}\\+1*\frac{1}{19}*1*\frac{1}{17}$
The ways that 7 pairs of socks survive is:

*

*Eating three socks of different colours, then eating any one of their pairs

*Eating two socks of different colours, then eating any one of their pairs, then drawing another sock of a different colour

*Eating any sock, eating another sock of the same colour, then eating two other socks of different colours

P(7 socks)=$1*\frac{18}{19}*\frac{16}{18}*\frac{3}{17}\\+1*\frac{18}{19}*\frac{2}{18}*\frac{16}{17}\\+1*\frac{1}{19}*1*\frac{16}{17}$
The ways that 6 pairs of socks survive is:

*

*Eating 4 socks of different colours

P(6 pairs)=$1*\frac{18}{19}*\frac{16}{18}*\frac{14}{17}$
I am confident that my solution is incorrect.
First of all, the probabilities sum to $\frac{2906}{2907}$ when it should sum to 1.
Secondly, I simulated the process in the question on my computer and the probabilities that I obtained were:
P(8 pairs)=$0.0269986$
P(7 pairs)=$0.4319539$
P(6 pairs)=$0.5410475$
Compared to my answers which were:
P(8 pairs)=$0.0093$
P(7 pairs)=$0.2972$
P(6 pairs)=$0.6935$
Would there be another simpler way of solving the question?
 A: There are $\binom{20}{16}$ ways to select the $16$ remaining socks, all of which we assume are equally likely.  Let $$X_i = \begin{cases}
1 \qquad \text{if pair i survives intact} \\
0 \qquad \text{otherwise}
\end{cases}$$
for $1 \le i \le 10$.   So
$$P(X_i = 1) = \frac{\binom{18}{14}}{\binom{20}{16}} =\frac{3060}{4845}$$
By linearity of expectation, the expected value of the number of surviving pairs is
$$E \left( \sum_{i=1}^{10} X_i \right) = \sum_{i=1}^{10} E(X_i) = 10 \times \frac{3060}{4845} = \boxed{6.31579}$$
A: I would go for more direct approach.
All possibilities of four socks being eaten is of type $abcd$ (six pairs survive), $aabc$ (seven pairs) and $aabb$ (eight pairs). Here each symbol stands for one of left or right sock of a pair.
Then expected number of survivals is
$$ \left[ 6 \cdot \binom{10}{4} \cdot 2^4  + 7 \cdot \binom{10}{3} \cdot \binom{3}{1} \cdot 2^2 + 8 \cdot \binom{10}{2} \right] \Big/ \binom{20}{4} $$
$$=\frac{120}{19} $$
which is roughly $6.316$ pairs.
A: The order of selection does not matter, just which socks are selected.
There are $\binom{20}{4}$ ways to select the four socks which are "eaten".
Eight pairs of socks survive if two full pairs are eaten.  There are $\binom{10}{2}$ ways to select two full pairs of socks.  Hence, the probability that eight pairs of socks survive is
$$\frac{\dbinom{10}{2}}{\dbinom{20}{4}}$$
Seven pairs of socks survive if one full pair of socks is eaten and one sock each from two of the other nine pairs are eaten.  There are $\binom{10}{1}$ ways to select which pair of socks is eaten, $\binom{9}{2}$ ways to select which two of the other pairs lose a sock, and two ways to select a sock from each of those pairs.  Hence, the probability that seven pairs of socks survive is
$$\frac{\dbinom{10}{1}\dbinom{9}{2}\dbinom{2}{1}^2}{\dbinom{20}{4}}$$
Six pairs of socks survive if one sock each from four of the ten pairs is eaten. There are $\binom{10}{4}$ ways to select the four pairs from which a sock is eaten and two possible choices for each of those pairs.  Hence, the probability six pairs of socks survive is
$$\frac{\dbinom{10}{4}\dbinom{2}{1}^4}{\dbinom{20}{4}}$$
Thus, the expected number of pairs of socks which survive is
$$\frac{6 \cdot \dbinom{10}{4}\dbinom{2}{1}^4 + 7 \cdot \dbinom{10}{1}\dbinom{9}{2}\dbinom{2}{1}^2 + 8 \cdot \dbinom{10}{2}}{\dbinom{20}{4}}$$
