# Can the probability of a trump poverty be calculated without making case distinctions?

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.

• 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? – Max Nov 29 '14 at 10:15