jmc

There are $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$ ways to pick people to fill the first two seats, and 3 ways for those two people to be a couple, for a probability $3/15=1/5$ that the first two seats are a couple. If a couple is seated successfully, there are $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ ways to seat people in the next two seats, and 2 ways for those two people to be a couple (you could pick either of the two remaining couples), for a probability $2/6=1/3$ that those two seats are a couple. If the first two couples are seated successfully, the last two seats will be guaranteed to go to the last couple. So, the probability that everything goes right is $1/5\cdot 1/3=\boxed{\frac{1}{15}}$.