5. Anna and Bedřich are playing a card game. Each of them has five cards with values from 1 to 5 (one of each). In each of the five rounds, both lay down one card, and the one with the higher number gets a point. If the cards have the same number, no one gets a point. Used cards are not returned to the game. Whoever has more points at the end wins. What percentage of all possible outcomes of such a game will result in Anna's victory?
SOLUTION. The described game is apparently fair in the sense that both players have an equal number of ways to win. To find the required number, it is enough to determine how many ways a draw can occur, i.e., one of the results $0: 0, 1: 1$, and $2: 2$.

The case $0: 0$ occurs if both players lay down the same cards in each round. There are $5 !=1 \cdot 2 \cdot 3 \cdot 4 \cdot 5$ such possibilities.

The result $1: 1$ means that the players lay down the same cards in three rounds and in the remaining two rounds, they lay down two different cards $(x, y)$, each in a different order. Each such result is therefore uniquely determined by the order of the cards of one of the players and the selection of rounds in which the other player plays the same. Three rounds out of five can be chosen in 10 ways, and five cards can be arranged in 5 ! ways. The result $1: 1$ thus occurs in $10 \cdot 5$ ! cases.

It remains to investigate when the result $2: 2$ occurs. The card $x$ that the players lay down in one of the five rounds can be chosen in 5 ways. Then Anna and Bedřich each have four cards $a<b<$ $<c<d$ left. Since the order of the rounds does not matter, let's first calculate the number of possibilities in the case where Anna lays down the cards $x, a, b, c, d$ in this order. To avoid another draw, Anna must score another point in the last round for the card $d$, while Bedřich must score a point in the second round when Anna lays down the card $a$. Therefore, it is enough to determine what Bedřich has in the third and fourth rounds to make these two rounds end $1: 1$.

In these rounds, Bedřich must lay down one of the pairs $(a, d),(c, a),(c, b),(d, a),(d, b)$, which can be completed with cards for the second and fifth rounds so that no draw occurs in them, in a total of seven orders:

$$
\begin{gathered}
(x, b, a, d, c),(x, c, a, d, b),(x, d, c, a, b),(x, d, c, b, a),(x, b, d, a, c), \\
(x, c, d, a, b),(x, c, d, b, a) .
\end{gathered}
$$

Anna can lay down the cards $x, a, b, c, d$ in 5 ! ways. The result $2: 2$ thus occurs in $5 \cdot 7 \cdot 5$ ! cases.

In total, both Anna and Bedřich can lay down the cards in 5 ! ways, which is a total of $5 !^{2}$ possibilities. Since the number of all possible game courses in which a draw occurs is equal to $5 !+10 \cdot 5 !+5 \cdot 7 \cdot 5 !=5 ! \cdot 46$, the number of possible wins for each of them is $\frac{1}{2}\left(5 !^{2}-5 ! \cdot 46\right)=5 ! \cdot 37$. A win by one of them will thus end

$$
\frac{5 ! \cdot 37}{5 !^{2}}=\frac{37}{120} \approx 0.31=31 \%
$$

of all possible games.

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