Doppelkopf is played with $4$ players, each has $12$ cards. I want to calculate the probability that given $k$ cards ($k\ge 3$), that some player doesn't have, are distributed among the other players in such a way that every player has at least one of them.
The probability can be calculated in principle by determining each possible distribution and sum up the probbilities. The order matters; it makes a difference for example, whether player $1$ or player $2$ has some missing card. Possible distributions for $k=5$ are $$1-2-345$$ $$2-1-345$$ $$12-34-5$$ and so on.
Is there an easy way to calculate the probability, or at least a very good approximation, or do I have to sum the probabilties of many distinct events ?
Player A has none of the cards in $24\choose k$ out of $36\choose k$ possibilities. The same holds for B and C. However, we double-count cases where two players have none of the cards (i.e., one player has them all), which is in $12\choose k$ cases per player. Thus the overall probability for each of thethree players having at least one of the cards is $$ 1-3\cdot \frac{{24\choose k}-{12\choose k}}{36\choose k}.$$ (Incidentally, though not surprisingly, this gies the correct result $0$ for $k\le 2$)