9. On each of 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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# Solution.

Let the natural numbers $a, b, c, d$ be written on four cards. Two out of the four cards can be chosen in six different ways. Since, according to the problem, a pair of cards is chosen randomly, the probability of selecting a specific pair is $\frac{1}{6}$. Furthermore, since the events $\{s9\}$ (where $s$ is the sum of the numbers on the selected cards) are equally probable, the probability of each of them is $\frac{1}{3}$.

Thus, by listing all possible sums (which are $a+b, a+c, a+d, b+c, b+d, c+d$), we can conclude that two of these sums are equal to 9, two are greater than 9, and two are less (since each of them can be obtained with a probability of $\frac{1}{6}$).

Consider the sums that are equal to 9. There are two possible cases:

(1) either in each of these (two) sums, one of the addends is taken from the same card;

(2) or all addends are taken from different cards.

Consider the first case. Suppose, for definiteness, that the sums equal to 9 are $a+b=a+c=9$. This means that the numbers on the cards are $a, 9-a, 9-a, d$, and the possible sums are $9, 9, a+d, 18-2a, 9-a+d, 9-a+d$. Then, if we assume that $a < d$, the same reasoning holds if we swap the inequality signs in all inequalities. Thus, the first case is impossible.

Now consider the second case. For definiteness, let $a+c=b+d=9$, so the numbers on the cards are $a, b, 9-a, 9-b$, and the possible sums are $a+b, 9, 9-b+a, 9-a+b, 9, 18-a-b$. As we can see from the listed sums, the necessary and sufficient conditions for the problem's requirements are $a \neq b$ and $a+b \neq 9$. The first condition implies that in the third and fourth sums, one addend is greater than 9 and the other is less, and the second condition yields the same for the first and last sums. Violation of either condition would mean that more than two sums are equal to 9.

Thus, according to the conditions $0 < a < 9, 0 < b < 9$ (all numbers are natural), $a \neq b$, and $a+b \neq 9$, we list all possible combinations of numbers on the cards (without considering permutations). These will be the answer.

Answer: $(1; 2; 7; 8); (1; 3; 6; 8); (1; 4; 5; 8); (2; 3; 6; 7); (2; 4; 5; 7); (3; 4; 5; 6)$.