We roll five dice of different colors. Which event has a higher probability: "two pairs" or a "three of a kind"? ("Two pairs" means that two pairs of dice show the same number, and these pairs differ from each other and from the number on the fifth die. "Three of a kind" means that three dice show the same number, and the others show different numbers from each other and from the three of a kind.)
The sought probabilities are obtained by dividing the number of favorable cases (i.e., the possible full houses or two pairs) by the total number of cases. Since the latter is the same in both cases ($6^5$, as we are rolling five dice, and each can show six different numbers), it is sufficient to compare the number of favorable cases to determine the probabilities.

First, let's look at how many ways a full house can be rolled. The dice not participating in the full house can be chosen in $\binom{5}{2}=\frac{5 \cdot 4}{2}$ ways: the first can be chosen in five ways, the second in four, but the order does not matter; whether we choose the red and blue or the blue and red dice makes no difference. On these two dice, $6 \cdot 5$ different number pairs can appear. (Here again we can see why it was necessary to divide by 2 earlier: otherwise, the possibilities of red 6, blue 3 and blue 3, red 6 would have been counted separately, although they are the same. Many submissions made this mistake or a similar one when calculating the two pairs, which unfortunately resulted in a significant deduction in scoring.)

The remaining three dice show the same number, different from the previous ones, which gives 4 possibilities. Therefore, a full house can be rolled in $\frac{5 \cdot 4}{2} \cdot 6 \cdot 5 \cdot 4 = 1200$ ways.

Now let's examine how many cases result in two pairs. The die that does not belong to any pair, and thus shows a different number, can be chosen in five ways, and the number on it can be any of six. The remaining four dice form the pairs. These can form two pairs in three ways, because by choosing any die, we can pair it with any of the three others, and this already determines the other pair as well. The numbers on the dice in one pair can be any of five, and those in the other pair any of four, as they must be different from each other. Altogether, there are $5 \cdot 6 \cdot 3 \cdot 5 \cdot 4 = 1800$ cases where the dice show two pairs. Since $1800 = \frac{3}{2} \cdot 1200$, two pairs are $\frac{3}{2}$ times more likely than a full house. The approximate probabilities are 0.23 and 0.15, respectively.

Based on the work of Serény András (Budapest, Szent István Gymnasium, 1st year)

Note. The significance of the dice colors lies in emphasizing that the dice are distinguishable, or that the rolls of red 6, blue 5 and red 5, blue 6 are not the same. We could have said the same by stating that we are rolling five distinguishable dice. Of course, we already imagine the realization of the roll this way.

However, if we were to say that rolls like red 6, blue 5 and red 5, blue 6 (and similar ones) are considered the same, only the distribution of numbers on the five dice matters, for example, three sixes and two ones; then the dice would not be distinguishable, and we would get a different result. How many full houses would there be in this case? Six different numbers can be the full house, five and four for the other two, but 66612 and 66621 are the same, so there are $6 \cdot \frac{5 \cdot 4}{2} = 60$ possibilities. The number of two pairs: six different numbers can be the singleton, and the rest again $\frac{5 \cdot 4}{2}$, which is also 60 possibilities. In this model (which, however, clearly does not fit the problem), the two probabilities are the same. The fact that dice rolls do not follow this model is also supported by the fact that in the popular dice poker game, a full house is more valuable than two pairs; we can see the reason for this, as it occurs less frequently.