Multivariate hypergeometric distribution

20 individuals consisting of 10 married couples are to be seated at 5 different tables, with 4 people at each table.

If 2 men and 2 women are randomly chosen to be seated at each table, what is the expected number (E(X)) of married couples that are seated at the same table.

I did it by finding the probability of wife j being seated with husband j, which is $\frac{C_1^1C_1^9C_1^9}{C_3^{19}}$ (Multivariate hypergeometric distribution) where the first 1C1 is to pick the husband, the two 9C1's are to pick 1 woman and 1 guy out of the rest of the people. I get 0.834 for E(X) at the end.

I know my answer is wrong but where did I made a mistake?

I am not that far yet that I can answer your real question ('where did I make a mistake?') but could not withhold myself from this:

Let's say that we start by choosing $2$ men to be seated at some table.

Then $2$ of the women are chosen to join them at that table. Call them $W_1$ and $W_2$.

For $i=1,2$ define $X_{i}=1$ if woman $W_i$ is married to one of the $2$ men and $X_{i}=0$ otherwise.

Then $X_{1}+X_{2}$ is the number of couples that come to sit at this table.

Here $P\left\{ X_{i}=1\right\} =\frac{2}{10}=\frac{1}{5}$ so that $\mathbb{E}X_{i}=\frac{1}{5}$ and leads to: $$\mathbb{E}\left(X_{1}+X_{2}\right)=\mathbb{E}X_{1}+\mathbb{E}X_{2}=\frac{2}{5}$$

This gives the expected number of couples at exactly one table. If you want the total number of couples that are placed at the same table then it is enough to multiply with $5$ i.e. the number of tables.