4. Antonio, Beppe, Carlo, and Duccio randomly distribute the 40 cards of a deck, 10 to each. Antonio has the ace, two, and three of coins. Beppe has the ace of swords and the ace of clubs. Carlo has the ace of cups. Who is more likely to have the seven of coins?
(A) Antonio
(B) Beppe
(C) Carlo
(D) Duccio
(E) two or more players have the same probability of having it.
4. The answer is (D). Intuitively, the positions of the remaining cards are equally probable, so it is more likely that the 7 of coins ends up with Duccio "because he has more free space." More formally: we count the ways to distribute 34 cards among 4 people, so that one gets 10, one 9, one 8, and one 7: we can orderly distribute 34 cards in 34! different ways, but each of these is counted as many times as there are possible ways to order 10, 9, 8, and 7 cards separately, i.e., $10 ! 9 ! 8 ! 7$ ! (the formula is the same as for anagrams with repetition). Therefore, we have $P=\frac{34 !}{10 ! 9 ! 8 ! 7 !}$ possible cases. How many configurations are there in which Antonio has the 7 of coins? By a similar calculation to the one just performed, they are $P_{A}=\frac{33 !}{10 ! 9 ! 8 ! 6 !}$. Therefore, simplifying as many factors as possible, we get $\frac{P_{A}}{P}=\frac{7}{34}$ of the total possible cases. Similarly, we obtain $\frac{P_{B}}{P}=\frac{8}{34}, \frac{P_{C}}{P}=\frac{9}{34}, \frac{P_{D}}{P}=\frac{10}{34}$. Second Solution

The probability is defined as favorable cases/possible cases; here we will consider "case" any possible ordering of the 40 cards. We count the possible cases, i.e., the possible ways to reorder 40 cards so that cards A, B, C are in three of the positions 1-10, cards D, E are in two of the positions 11-20, and card F is in one of the positions 21-30. To do this, we first fix the positions of the "obligatory" cards: we can position A, B, C in $10 \cdot 9 \cdot 8$ ways, D, E in $10 \cdot 9$ ways, and F in 10 ways. So we have $103 \cdot 92 \cdot 8$ ways to fix the positions of the six known cards. The remaining cards can go into the remaining free positions in any order, and these arrangements are 34! (indeed, each possible ordering of the 40 cards that meets the required conditions is obtained in one and only one way by choosing the positions of the 6 "obligatory" cards and the order in which the 34 "free" cards appear in the free spaces). Therefore, the possible cases are $P=103 \cdot 92 \cdot 8 \cdot 34$!. A similar calculation provides $F_{A}=103 \cdot 92 \cdot 8 \cdot 7 \cdot 33$! for the cases where we add the constraint that the 7 of coins is in A's hand, and similarly $F_{B}=103 \cdot 92 \cdot 82 \cdot 33$!, $F_{C}=103 \cdot 93 \cdot 8 \cdot 33$!, $F_{D}=104 \cdot 92 \cdot 8 \cdot 33$!. Therefore, the probabilities to compare are $\frac{F_{A}}{P}=\frac{7}{34}, \frac{F_{B}}{P}=\frac{8}{34}, \frac{F_{C}}{P}=\frac{9}{34}$, $\frac{F_{D}}{P}=\frac{10}{34}$, and the greatest is $\frac{F_{D}}{P}$.