On each of the four cards, a natural number is written. Two cards are randomly taken, and the numbers on them are added. With equal probability, this sum can be less than 9, equal to 9, and greater than 9. What numbers can be written on the cards?

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Let four cards have natural numbers $a \leq b \leq c \leq d$ written on them. Two out of the four cards can be chosen in six different ways. Since a pair of cards is chosen randomly, the probability of choosing a specific pair is $1 / 6$. Since, by the condition, events $\{s9\}$ (where $s-$ is the sum of the numbers on the chosen cards) are equally probable, the probability of each of them is $1 / 2$.

Thus, two of the sums $a+b, a+c, a+d, b+c, b+d, c+d$ are equal to 9, two are greater than 9, and two are less than 9.

Due to the inequalities $a+b \leq a+c \leq a+d \leq b+d \leq c+d$ and $a+c \leq b+c \leq b+d$, the sums $a+d$ and $b+c$ are equal to 9, the sums $a+b$ and $a+c$ are less than 9, and the sums $b+d$ and $c+d$ are greater than 9. In particular, $a+c < a+d$, so $c < d$.

Similarly, $a < b$. Moreover, $b < c$, since the sum $b+c$ is odd.

Thus, the pairs $\{a, d\}$ and $\{b, c\}$ can be chosen from the pairs $\{1,8\},\{2,7\},\{3,6\},\{4,5\}$, taking into account the obtained inequalities. Hence the answer.

## Answer

$(1,2,7,8),(1,3,6,8),(1,4,5,8),(2,3,6,7),(2,4,5,7),(3,4,5,6)$.

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