The card game Doppelkopf is played with four players. Every player receives 12 of the 48 cards. The 48 cards consist of 26 trump cards and 22 other. A trump poverty is what we call the scenario that a player has less than 4 trump cards. If the cards are dealt randomly, what is the probability that at least one player has a trump poverty?

This looks like it can be solved very easily via the inverse, like so:

Deal every player 4 trump cards, then deal every player 8 more cards.

However we failed to do this without double counting. Can this be done without case distinctions?


Outline: Call the players A, B, C, D. Note that we cannot have $3$ trump poor players, since then they would have at most $9$ trumps between them, and $9+12\lt 26$.

(i) Find the probability that A has trump poverty. This should not be difficult.

Now multiply by $4$. This overestimates the required probability, for we have double-counted the situations in which, for example, A and B both have trump poverty.

(ii) Find the probability that A and B have trump poverty, multiply by $\binom{4}{2}$, and subtract from the estimate in (i).

Finding the probability that A and B have trump poverty is a little unpleasant. One way is to divide into cases: A has $0$ trumps, $1$ trump, $2$, $3$. It will take a while but the calculations are routine.

Remark: If one has some programming experience, one can set up a simulation and get reliable estimates.

  • $\begingroup$ Yes, we came up with this answer, too, but we were trying to find a solution without making case distinctions. Are you suggesting there is no such solution? $\endgroup$ – Max Nov 29 '14 at 10:15

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