András and Béla are playing the following game. They have three empty dice. András fills in the faces with the numbers $1,2, \ldots, 18$ (each number is written on exactly one face and each face gets exactly one number). Then Béla chooses one of the dice, and András selects one from the remaining two. Both roll their chosen dice once. The one who rolls the higher number wins. Can András number the dice in such a way that the game favors him?
András can do this! Let's consider the following numbering: the first die gets $18,17,6,5,4,3$, the second die gets $16,15,14,13,2,1$, and the third die gets $12,11,10,9,8,7$. If Béla chooses the first one, András should choose the third; if Béla chooses the second, András should choose the first; and if Béla chooses the third, András should choose the second. We will calculate the probability that András rolls a higher number than Béla in each case. If this probability is greater than 1/2 in all three cases, we are done.

Comparing the first and third dice: the number of possible outcomes is $6 \cdot 6=36$. The favorable outcomes for András are: any number on the third die, and 6, 5, 4, or 3 on the first die. This is 24 cases, so the probability is $\frac{24}{36}=\frac{2}{3}>\frac{1}{2}$.

For the second and first dice, the favorable outcomes are: 18 or 17 on the first die and any number on the second; or 6, 5, 4, or 3 on the first die and 1 or 2 on the second. This totals $2 \cdot 6+4 \cdot 2=20$ cases, so the probability is $\frac{20}{36}=\frac{5}{9}>\frac{1}{2}$.

For the third and second dice, the favorable outcomes are: 16, 15, 14, or 13 on the second die, and any number on the third. The probability is $\frac{6 \cdot 4}{36}=\frac{2}{3}>\frac{1}{2}$.

Note. Let $p_{1}$, $p_{2}$, and $p_{3}$ be the probabilities that the first die rolls higher than the second, the second die rolls higher than the third, and the third die rolls higher than the first, respectively. If Béla chooses the best option given the circumstances, the probability of András winning is $p=\min \left\{p_{1}, p_{2}, p_{3}\right\}$.

In the case of the numbering used in the solution, this probability is $\frac{20}{36}$. This value can be slightly increased: if the first die is numbered $1,10,11,12,13,14$; the second die $5,6,7,8,9,18$; and the third die $2,3,4,15,16,17$, then $p=\frac{21}{36}$. It can be shown that the value of $p$ cannot be greater than $\frac{21}{36}$ for any numbering. The proof can be found in the 1973 volume (47th issue) of our journal, in the article "The Three Dice" by Pál Bártfai, where a method for finding the best numbering is also discussed.