To get the probability, you'll need
- the total number of combinations of $16$ people taken $4$ at a time, and
- the total number of those combinations that have no married couples among them.
You have the first part: ${16 \choose 4}$.
Rather than count up the number of combinations for the second part directly, you can count the complement, and subtract from the total.
So if you do have at least one married couple in the chosen, you can have one couple, or two.
If you have two married couples in the chosen group, then there are ${8 \choose 2}$ possible combinations.
If you have (exactly) one married couple, first you pick the married couple in the group: ${8 \choose 1}$. Now, from the remaining $14$ people, you pick two that aren't married to each other. So pick the two couples that you're going to draw from -- ${7 \choose 2}$ -- and then pick the particular spouse from each one -- ${2 \choose 1} \cdot {2 \choose 1}$.
Now, you subtract the number of combinations that do have at least one married couple from the total number of combinations to get the number of combinations that have no married couples. This gives as your probability:
$$P = \frac{{16 \choose 4} - {8 \choose 2} - {8 \choose 1}{7 \choose 2}{2 \choose 1}{2 \choose 1}}{{16 \choose 4}} = \frac{8}{13} \approx 0.61538.$$
Pick the couple: $\binom 61 =6$ ways
Pick the other couples that you're going to choose one partner from: $\binom 54 = 5$
Pick which partner from those couples: $2^4=16$
Answer : $6\cdot 5\cdot 16 = 480$
Your mistake is in overestimating the number of ways to pick the four without including a couple. As you can see here there are only $5\cdot 16=80$ ways to do that.
You can also get the $80$ options of picking $4$ people without any couples from $5$ couples by inclusion-exclusion, starting with your initial "free" choice of $4$:
$$\binom {10}{4} - \binom {5}{1}\binom {8}{2} + \binom {5}{2}\binom {6}{0} = 210 - 5\cdot 28 + 10\cdot 1 = 80$$
where the first subtraction is all the ways of choosing one couple and any two other people, and the final addition remedies the "oversubtraction" of cases where two couples were chosen.
Best Answer
Since there seems to be some confusion among the other answers, I'll expand my comment into an answer.
You can specify each such group of four people by choosing which three of the six couples are to be represented, then which one of the three couples is to appear in its entirety, then which one of the two people from the second couple is included, and finally which one of the two people from the third couple is included.
This makes for $${6\choose 3}\cdot{3\choose 1}\cdot{2\choose 1}\cdot{2\choose 1}=20\cdot3\cdot2\cdot2$$ favorable outcomes. Dividing by the ${12\choose4}=33\cdot15$ total outcomes gives $16/33$.