Contest, chances of winning We're currently setting up a contest and would like to keep the number of winners to a reasonable amount (in both directions : not too many, not too few). I know this will be somehow simple probabilities and so cause I did this in the past, but I haven't been doing this kind of stuff for the last 15 years and almost forgot everything about it. Here's the description of the contest :
The user will be proposed to choose among Y boxes (Y will most probably be 10). Under some boxes, there will be 1 (and only 1) item, under the others nothing. Which ones are "filled" and which ones are "empty" will be fixed as soon as he enters the contest. The number of boxes "containing" an item is what we're trying to determinate here and will depend on the winning probabilities. Let's call it X. 
The user will get the possibility to pick 3 boxes (let's call this number of picks Z). What are the probabilities for him to find : 3 items,2 items,1 item,0 item ??? 
It would be great to get a formula including X and Y and Z rather than plain figures as we would like to find the best suitable combination and may need to do the math with different figures. (ex : the total number of boxes (Y) isn't fixed yet, the number of box filled (X) will be determined by us depending on the probabilities to win, and the number of boxes the user get to pick (Z) will probably be increased under certain conditions like a share on Facebook ;-)) 
Thanks for your help and if something is unclear and can have an impact on the result, don't hesitate to ask !
 A: I have to ask: Since you're @Bartdude, is this project for BART? (San Francisco Bay Area Rapid Transit) Just wondering.
I'll change your notation slightly just to reduce cognitive load:
\begin{eqnarray}
n &:& \mbox{  number of boxes} \\
p &:& \mbox{  number of prizes} \\
d &:& \mbox{  number of drawings (number of times user opens a box} \\
W &:& \mbox{  number of prizes won (number of opened boxes containing a prize)}
\end{eqnarray}
The first three of these numbers are fixed parameters. $W$ is a random variable. Then, assuming everything is equally likely in the natural way,
$$
\mathbb{P}(W=w) = \frac{\left(\begin{array}{c} p \\ w \end{array}\right) \left(\begin{array}{c} n-p \\ d-w \end{array}\right)}{\left(\begin{array}{c} n \\ d \end{array}\right)}.
$$
The $()$ symbols are standard. See here. Don't be put off by all the factorials: Lots of factors cancel. You can compute factorials with, say, wolframalpha, where $C[10,2]$ is $\left(\begin{array}{c} 10 \\ 2 \end{array}\right)$, for example.
Good luck!
