lottery to pick a group while respecting pairs I am running an event that will be oversubscribed, so I'd like to use a lottery to randomly pick the participants that will be accepted.  (For example, 29 people want to attend, but I can accommodate only 17.)
The twist is that there are some couples among the people who want to attend, so I must either accept both of them or neither of them.
I would like to run this lottery so that any single person has an equally likely chance of being selected, regardless of whether that person is part of a couple or not.
Is it even possible to make the lottery fair in this manner?  If so, how should I run the lottery?  If not, what would be the fairest way to run the lottery?
I have thought of a few things that don't work.
Suppose you take all the combinations (choose 17 from 29 in my example), throw out the ones that do not violate any of the couples constraints, and pick one of the remaining combinations at random with uniform probability.  In general, the number of combinations with a member of a couple is not the same as the number of combinations with an individual.
Another simple scheme would be to write each individual's name on a slip of paper, write both names of each couple on a single slip of paper, throw all the pieces of paper in an urn, and draw slips of paper until the total number of names reaches the capacity of the event.  Let's say that if the slip you draw for the last spot has two names of a couple, you throw it out.  This also does not provide the same odds of an individual and one member of a couple being selected.  
Perhaps there is a way of weighting the combinations in such a way to provide fair odds to an individual?  (Though when there are several couples this could get complicated...)
 A: You can start by choosing the number of couples to invite.  Say you have five couples and nineteen singles.  Each person should have a chance $\frac {17}{29}\approx 0.5862$ of attending, which is between $\frac 25$ and $\frac 35$.  If you invite two couples with probability $x$ and three couples with probability $1-x$, the chance of each couple attending is $0.4x+0.6(1-x)$.  Equating that to $\frac {17}{29}$ gives $x=\frac 2{29}$.  Choose two couples randomly from the five with probability $\frac 2{29}$ and three randomly from the five with probability $\frac {27}{29}$.  Then choose the requisite number of singles from that pool.
A: Let individuals in couples be given tickets with probability $\beta$ of being selected, and let individuals not in couples be given tickets with probability $\alpha$ of being selected. An individual in a couple is selected if his or his partner's or both of their tickets are selected, so with probability $2\beta-\beta^2$. For the fairness requirement we set this equal to $\alpha$. Since the the couples' ticket probabilities and individuals' ticket probabilities must sum to 1, $\alpha$ and $\beta$ are determined in terms of the number of couples and individuals.
