What is the distribution of the outcome of this game? We were playing a game yesterday and I am curious about some of the potential outcomes. In this game, there are n players who write words on cards. There is one unique word on each card. Each individual writes on k cards. Then the individuals play this game k rounds where in each round an individual picks a card without replacement from the box. I am wondering how I can calculate the distribution of the number of cards picked by the individual who wrote the word him/herself.
E.g.
Let's say Alice and Hans wrote 3 cards. Then the total number of cards picked by the individual who wrote it ranges from 0 to 6. I think he probability of 0 such cards is $2 \times \frac{3}{6}\frac{3}{5}\frac{2}{4}\frac{2}{3}\frac{1}{2}$. I am not sure how to generalize this to n individuals, and k rounds. Thank you!
 A: There isn't a simple formula for this.  The way to solve the problem involves rook polynomials.
Let me consider an equivalent game.  A deck of cards has $k$ suits of $n$ cards each.  There are $k$ players, each of whom is assigned a different suit.  The deck is shuffled, and dealt out to the players, so that each player receives $n$ cards.  What is the probability that a total of $j$ cards are dealt to to the player assigned that card's suit?
It doesn't matter how we deal the cards, so we can assume the first player gets the first $n$ cards, the second player gets the next $n$ cards, and so on.  In looking at the chessboard, we see $k$ $n\times n$ dark squares representing the forbidden positions.  A very useful fact about rook polynomials is that if the board breaks up into two parts that have no square in the same row or the same column, then the rook polynomial of the board is equal to the product of the rook polynomial of the two parts.  Thus, if $R_n$ is the rook polynomial of an $n\times n$ square, then then rook polynomial of our dark board is $R_n^k$.
It is easy to compute the rook polynomial of an $n\times n$ square, as explained here.  For example, for a $4\times4$ board, there is $1$ way to place $0$ rooks and $16$ ways to place one rook.  For two rooks, after we place the first in $16$ ways, we eliminate a row and column for the next rook, leaving $9$ possibilities.  The order in which we place the two rooks doesn't matter, so there are $\frac{16\cdot9}{2}=72$ possibilities.  Similarly, for $3$ rooks, we have $\frac{4^2\cdot3^2\cdot2^2}{3!}=96$ possibilities, and of course, there are $4!=24$ ways to place $4$ non-attacking rooks.  The rook polynomial is therefore $$24x^4+96x^3+72x^2+16x+1$$.
Once you've computed the rook polynomial, the standard results allow you to compute the number of ways to place $nk$ rooks on the board with exactly $j$ rooks in forbidden positions, for any $j$.  Then just divide by $(nk)!$ to get the probability.
One thing I remember about this problem is that if $n$ remains fixed and $k$ increases, the the limit, as $k\to\infty$ of the probability that no player receives a card of his own suit is $e^{-n}.$
