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Suppose that I have $n$ tickets for an event that I want to distribute fairly among $N > n$ people. In this simple case, a lottery suffices. But suppose certain groups of people want to attend the event together or not at all, that is, there are a-priori declared groups whom either all will win or none will win. (To avoid edge problems, assume that the size of all groups is $<< n$.) We want the distribution to be "fair" in the sense that every person has the same probability of getting a ticket regardless of how people are grouped (or not).

It seems to me that an equivalent but simpler formulation is that we want to randomly put $N$ people in a linear order subject to the constraints (1) each person has a $1/N$ chance of being in any particular position in the order and (2) certain relatively small ($<< N$) groups must appear in consecutive positions of the order. Given a solution to this problem, we would allocate the $n$ tickets to the people in the first $n$ positions in the order. (Requiring special case handling if a group spans positions $n$ and $n+1$.)

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    $\begingroup$ A solution to this seems as though it should give a solution to the knapsack problem? $\endgroup$ Commented Feb 22, 2015 at 3:17
  • $\begingroup$ I intend to avoid the knapsack problem by requiring the groups to be $<<N$ and allowing a small amount of inequality in chances near the beginning of the order, the end of the order, and the cut point $n$. $\endgroup$
    – Dale
    Commented Feb 22, 2015 at 16:59
  • $\begingroup$ The groups are disjoint and $k=n$, right? $\endgroup$
    – domotorp
    Commented Feb 22, 2015 at 19:52
  • $\begingroup$ Do you want exactly $n/N$ chance of winning for each person? Then you definitely need more constraints. Imagine that $N$ is even, everyone is in a group of size two and $n$ is odd. $\endgroup$
    – domotorp
    Commented Feb 22, 2015 at 19:58
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    $\begingroup$ I think the easiest approach is to take a uniformly random permutation of the groups, then give out tickets in this order until you run out of tickets. The extent to which this is a bad or non-random approach is essentially the extent to which you run into knapsack and boundary problems. So if you are assuming those effects are small, this should be fine. $\endgroup$
    – usul
    Commented Feb 23, 2015 at 1:12

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