I'm trying to organize a raffle that has two prizes — prize X and prize Y:
the $N_X$ X-fans want prize X and don't want prize Y,
the $N_Y$ Y-fans want prize Y and don't want prize X,
the other $N_{XY}$ are the XY-fans who would be happy with any of the two prizes.
If a XY-fan would have a higher chance of winning than a X-fan, that seems unfair.
I want to know how to randomly choose the winners in such a way that:
- is fair (everybody has the same chance of winning)
- nobody receives a prize that they do not want
- nobody receives two prizes
- there are as many winners as possible*
- their probabilities of winning are as high as possible*
(if there is a way, or maybe there might be more than one way)
*I think that in some cases (depending on $N_X$, $N_Y$ and $N_XY$) the solution (if there is one) might have a nonzero chance that one of the two prizes will not be given to anyone. For example if $N_X=9$, $N_Y=4$ and $N_{XY}=0$ then I think that each X-fan would have to have a $\frac{1}{9}$ chance of winning, and for it to be fair then each Y-fan would then have to also have a $\frac{1}{9}$ chance of winning, but then (since there are only 4 Y-fans) there would be a $\frac{5}{9}$ chance that nobody wins prize Y.
Perhaps it might be better to try to solve the more general problem with an arbitrary number of prizes instead of focusing on two prizes.
UPDATE: This is now a full solution
The interesting twist in this question is that everybody must have the same probability of winning (some prize). Let this probability be $p$, and we want to maximize $p$.
Without loss lets assume $N_X \ge N_Y$.
Case $1$: $N_X \ge N_Y + N_{XY}$.
$p = \frac{1}{N_X}$. Obviously no higher $p$ is possible.
First do a random draw to give away the X prize among the X-fans.
Next do a random draw to give away the Y prize among the Y-fans, the XY-fans, and $K$ number of dummy people where $K = N_X - N_Y - N_{XY}$. If a dummy person wins this draw, then the Y prize is not awarded.
Case $2a$: $N_X < N_Y + N_{XY}$ and $N_X + N_Y + N_{XY}$ is an even number, say $= 2D$.
$p = \frac{2}{N_X + N_Y + N_{XY}} = \frac{1}{D}$. Obviously, no higher $p$ is possible.
First randomly pick a $K$-subset of XY-fans to join the X-fans to create an X-pool. The rest of the XY-fans join the Y-fans to create a Y-pool. $K = D - N_X$ s.t. each pool has $D$ fans.
Do a random draw to award the X prize among the $D$ fans in the X-pool.
Do a random draw to award the Y prize among the $D$ fans in the Y-pool.
Case $2b$: $N_X < N_Y + N_{XY}$ and $N_X + N_Y + N_{XY}$ is an odd number, say $2D+1$.
UPDATE: General solution for case 2b (which also works for case 2a)
The general solution for 2b (and 2a) works by listing all possible outcomes. An outcome is an assignment of who gets which prize. There are $4$ types of outcomes.
(type-a) X-fan gets X-prize, Y-fan gets Y-prize. There are $N_X N_Y$ such outcomes, and let $a$ be the probability of picking a specific such outcome.
(type-b) X-fan gets X-prize, XY-fan gets Y-prize. There are $N_X N_{XY}$ such outcomes, and let $b$ be the probability of picking a specific such outcome.
(type-c) XY-fan gets X-prize, Y-fan gets Y-prize. There are $N_{XY} N_Y$ such outcomes, and let $c$ be the probability of picking a specific such outcome.
(type-d) XY-fan gets X-prize, another XY-fan gets Y-prize. There are $N_{XY} (N_{XY} -1)$ such outcomes, and let $d$ be the probability of picking a specific such outcome. (Note that swapping prizes between two XY-fans are two different outcomes.)
So we have four unknowns. They satisfy four equations, so they can be solved. Here $p = \frac{2}{N_X + N_Y + N_{XY}} = $ an individual's win prob.
Total probability: $N_X N_Y a + N_X N_{XY} b + N_{XY} N_Y c + N_{XY} (N_{XY} -1) d = 1$
X-fan individual win prob: A specific X-fan has $N_Y$ type-a outcomes and $N_{XY}$ type-b outcomes where he/she wins. So: $N_Y a + N_{XY} b = p$.
Y-fan individual win prob: Similarly, a specific Y-fan has favorable type-a and type-c outcomes: $N_X a + N_{XY} c = p$.
XY-fan individual win prob: A specific XY-fan has favorable type-b, type-c, type-d outcomes. Note that there are $2 (N_{XY} - 1)$ favorable type-d outcomes for him/her (who he/she shares with, and which prize goes to him/her). So: $N_X b + N_Y c + 2(N_{XY} - 1) d = p$.
Once we solved for $a,b,c,d$, then simply draw one outcome among all four types, with these probabilities. E.g. pre-assign each outcome its own small non-overlapping range of the right length in the $(0,1)$ interval and then drawing from $Uniform(0,1)$.