Miklós offers Laci the following bet: I will pay you 1 Ft for every lottery draw in 1967 where among the 5 drawn lottery numbers there are no 3 consecutive numbers, if you pay me 100 Ft for every draw where there are at least 3 consecutive numbers. - We found out that Miklós bases his hopes of winning on the fact that such a draw has not occurred in the first 500 draws of the current Hungarian lottery. - On average, how many draws are expected to result in a draw favorable to Miklós?
I. solution. At least 3 consecutive numbers are in a draw if the number of consecutive numbers is exactly 3, or exactly 4, or exactly 5.

In the first case, if the 3 consecutive numbers are, for example, 21, 22, and 23, the other 2 numbers cannot be 20 and 24, so they must be chosen from the numbers $1, 2, \ldots, 19$ and $25, 26, \ldots, 90$, i.e., from $90-5=85$ numbers. This can be done in 85$\cdot$84/2 ways, as is easily seen. If, however, the smallest of the 3 numbers is 1, then only 4 numbers are forbidden for the other two, so they can be chosen from 86 numbers, i.e., in $86 \cdot 85 / 2$ different ways. The same number of valid draws occurs if the smallest number of the triplet is the largest possible: 88. In all other cases, the reasoning for the starting number 21 is valid, and there are 86 such cases. Thus, exactly 3 consecutive numbers can appear in a lottery draw in

$$
86 \cdot \frac{85 \cdot 84}{2} + 2 \cdot \frac{86 \cdot 85}{2} = \frac{1}{2} 86^{2} \cdot 85
$$

different ways.

Similarly, the smallest number of a sequence of exactly 4 consecutive numbers can be chosen in 87 ways, and with this sequence, most often $2-2$ numbers are adjacent - except for 1 and 87, when only 1 is - so for the remaining 1 number, 6, and 5 numbers are forbidden, i.e., 84 and 85 are allowed. The number of such draws is $85 \cdot 84 + 2 \cdot 85 = 86 \cdot 85$. - Finally, a 5-element sequence can occur in 86 ways, so the number of favorable draws for Miklós is

$$
M = \frac{1}{2} 86^{2} \cdot 85 + 86 \cdot 85 + 86 = 86(43 \cdot 85 + 85 + 1) = 86 \cdot 3741
$$

On the other hand, the number of all possible draws, considering the order of drawing the numbers, is $N = 90 \cdot 89 \cdot 88 \cdot 87 \cdot 86$. Their $5 \cdot 4 \cdot 3 \cdot 2 = 120$-fold gives the same result, only in a different order, so the number of different draws is $N / 120$.

According to all this, on average

$$
\dot{a} = (N / 120) : M = \frac{3 \cdot 89 \cdot 88 \cdot 87}{4 \cdot 3741} = \frac{3 \cdot 89 \cdot 22}{43} = \frac{5874}{43} \approx 136.6
$$

draws are expected for Miklós to win once. (It would be fair, then, for him to win 135.6 Ft.)

According to this, in the more than 500 lottery draws that took place by the end of 1966, it would have been expected for the desired event to occur more than 3 times on average, and by the end of 1967, more than 4 times. However, this does not increase Miklós's chances, as each draw is independent of the previous ones. It is also possible, of course, that Miklós could win more than once during the year.

$$
\text{László Nádai (Budapest, Fazekas M. gyak. g. II. o. t.)}
$$

Remarks. 1. In the case of drawing 5 numbers from $n$ lottery numbers,

$$
\begin{gathered}
M = \left((n-4) \cdot \frac{(n-5)(n-6)}{2} + 2 \cdot \frac{(n-4)(n-5)}{2}\right) + \\
+ ((n-5)(n-6) + 2(n-5)) + (n-4) = \frac{1}{2}(n-3)(n-4)^{2}
\end{gathered}
$$

thus

$$
\dot{a} = \frac{n(n-1)(n-2)}{60(n-4)}
$$

provided $n \geq 6$.

2. Several people answered based on a formula without fully thinking it through - incorrectly. The following reasoning leads close to the above result, but is still flawed: The 3 consecutive numbers can be chosen in 88 ways, and the remaining 2 from the other 87 in $87 \cdot 86 / 2$ ways, the product of these two numbers is $M$. In this number, for example, the draw $H_{1} = 1, 2, 3, 4, 50$ appears twice: as "1, 2, 4 and 4, 50" and as "3, 4 and 1, 50"; the draw $H_{2} = 1, 2, 3, 4, 5$ appears three times. Similarly, if the number of draws containing 4 consecutive numbers is taken as $87 \cdot 86$, in this number, for example, $H_{2}$ appears twice. The correct correction: the number of draws containing exactly 5, 4, or 3 consecutive lottery numbers is:

$$
P_{5} = 86, \quad P_{4} = 87 \cdot 86 - 2 P_{5}, \quad P_{3} = 88 \cdot 87 \cdot 43 - 2 P_{4} - 3 P_{5}
$$

from these, $M = P_{5} + P_{4} + P_{3} = 44 \cdot 87 \cdot 86 - 87 \cdot 86 = 87 \cdot 86 \cdot 43$, and this is smaller by the 44th part of the above incorrect number.

II. solution. Group the favorable draws $H$ for Miklós according to the smallest number of the sequence of 3, 4, or 5 consecutive numbers, let this be $h$.

In the case of $h = 1$, $H$ contains 2, 3, and two more numbers from the numbers 4, 5, ..., 90. These two can be chosen in $87 \cdot 86 / 2$ ways. (This includes, for example, 1, 2, 3, 4, 6 and 1, 2, 3, 4, 5.)

For every other value of $h$, $H$ also contains $h+1, h+2$, and two more from the lottery numbers, but $h-1$ cannot be included. Thus, these two can be chosen from 86 numbers, and for each possible value of $h$ ($h = 2, 3, \ldots, 88$, i.e., 87 different values), in $86 \cdot 85 / 2$ ways. The expression

$$
\frac{87 \cdot 86}{2} + 87 \cdot \frac{86 \cdot 85}{2}
$$

counts each favorable draw exactly once.

We can proceed further as in the first solution.

Katalin Balázs (Budapest, Fazekas M. gyak. g. IV. o. t.)