In the 34th lottery draw, two pairs of consecutive numbers were drawn: (31; 32) and (50; 51). What is the probability that the result of the first draw following the submission deadline will be similar: two pairs of consecutive numbers are drawn, but not three consecutive numbers?
I. solution. We will count how many of the possible outcomes of a lottery draw are like the one quoted. According to the position of the "lonely" number, we need to consider three types of quintets:

1. $(a, a+1, b, b+1, c)$, where $1 \leq a, a+2<b, \quad b+2<c \leq 90$, thus

$$
1 \leq a<b-2<c-4 \leq 86
$$

Each such quintet corresponds to a triplet of distinct numbers between 1 and 86: $a, b-2, c-4$. Any two distinct quintets correspond to distinct triplets, and every triplet corresponds to some quintet, meaning we can establish a one-to-one correspondence between such quintets and triplets. (We assign the triplet $(a, b-2, c-4)$ to the quintet of the form $(a, a+1, b, b+1, c)$.) $\binom{86}{3}$ triplets can be chosen from the distinct numbers between 1 and 86, so this is also the number of quintets of the 1st type.
2. $(a, b, b+1, c, c+1)$, where $1 \leq a, a+1<b, b+2<c, c+1 \leq 90$

thus

$$
1 \leq a<b-1<c-3 \leq 86
$$

Again, a one-to-one correspondence can be established between the quintets and the triplets of numbers between 1 and 86, if we assign the triplet $(a, b-1, c-3)$ to the quintet of the form $(a, b, b+1, c, c+1)$. The number of such triplets is again $\binom{86}{3}$, so we have obtained $\binom{86}{3}$ quintets of the 2nd type.
3. $(a, a+1, b, c, c+1)$, where $1 \leq a, a+2<b, \quad b+1<c \leq 89$

thus

$$
1 \leq a<b-2<c-3 \leq 86
$$

Such quintets can again be uniquely and one-to-one assigned to a triplet of numbers between 1 and 86, if we assign the triplet $a, b-2, c-3$ to the quintet $(a, a+1, b, c, c+1)$. Again, there are $\binom{86}{3}$ triplets, and consequently $\binom{86}{3}$ quintets.

In total, there are $3 \cdot\binom{86}{3}=307020$ suitable quintets. The desired probability is then obtained by dividing this number by the total number of quintets. The desired value is thus $\frac{307020}{\binom{90}{5}} \approx 0.007$ ( $\approx 0.0069858$ ). (On average, such a draw is expected once every 143 draws, or about every two and $3/4$ years.)

Of course, every week a suitable quintet can be drawn with the same probability (even if one was already drawn the previous week), since the draws of any two weeks are independent of each other.

II. solution. If the outcome of a draw is like the one in the 34th week, then the five drawn numbers divide the 85 non-drawn numbers into four consecutive groups. The first or the last group can be empty if 1 or 90 is among the drawn numbers.

Now, let's place 85 white balls in a row and choose three of the 86 "gaps" between the balls, including the one before the first and the one after the last - thus, 86 in total. This can be done in $\binom{86}{3}$ ways. Place 1 in one of them, and 2 red balls in the other two - this is 3 possibilities - then number all 90 balls from 1 to 90. The 5 numbers of the drawn balls will be the numbers of the 5 red balls. This way, we will clearly obtain every suitable quintet and each one only once.

The number of lottery draws of the type similar to the 34th week is therefore $3 \cdot\binom{86}{3}$, so the desired probability is $\frac{3 \cdot\binom{86}{3}}{\binom{90}{5}}$.

Remark. In the first 1674 draws of the Hungarian lottery, which has been running since 1957, the type of quintet under examination has occurred 8 times, the first type 4 times, and the other two types 2 times each. Although this is somewhat less than the expected 11-12, it can still be said that the empirical data is in agreement with the theory, considering the small number of cases.