In a lottery draw, the numbers are listed in ascending order as usual, and we made two observations (on February 13, 1970, 25, 34, $49,68,75$):

A) the fifth number is three times the first,

B) the fourth number is twice the second.

What is the probability of property $A$ occurring, what is the probability of property $B$, and what is the probability of both properties occurring simultaneously?
A) Let the first number drawn be $e$, then the fifth $\ddot{o}=3 e$, there are $2 e-1$ numbers between them, from which we need to choose the other three numbers, which is possible in $\binom{2 e-1}{3}$ ways.

$2 e-1 \geqq 3$ holds if $e \geqq 2$; on the other hand, $\ddot{o}=3 e \leqq 90$ if $e \leqq 30$, so by summing the binomial coefficient $\binom{2 e-1}{3}$ for $e=2,3, \ldots, 30$, we get the number of favorable cases:

$$
k_{A}=\sum_{e=2}^{30}\binom{2 e-1}{3}=\sum_{e=1}^{30} \frac{(2 e-1)(2 e-2)(2 e-3)}{6}
$$

We added the term for $e=1$, but this does not change the sum because by definition $\binom{1}{3}=0$ or in the expanded form $2 e-2=0$. This transformation is useful because by converting the summand to a polynomial in $e$,

$$
k_{A}=\frac{4}{3} \sum_{e=1}^{30} e^{3}-4 \sum_{e=1}^{30} e^{2}+\frac{11}{3} \sum_{e=1}^{30} e-\sum_{e=1}^{30} 1
$$

and here the sums starting from 1 are known as functions of the upper limit:

$$
\begin{array}{ll}
1^{3}+2^{3}+3^{3}+\ldots+n^{3}=\frac{n^{2}(n+1)^{2}}{4}, & n=30 \text { in this case: } 216225 \\
1^{2}+2^{2}+3^{2}+\ldots+n^{2}=\frac{n(n+1)(2 n+1)}{6}, & n=30 \text { in this case: } 9455, \\
1+2+3+\ldots+n=\frac{n(n+1)}{2}, & n=30 \text { in this case: } \quad 465,
\end{array}
$$

the value of the fourth sum is 30, so $k_{A}=252155$.

On the other hand, the number of possible draws is $\binom{90}{5}=43949268$, so the probability of property $A$ occurring is:

$$
P(A)=\frac{252155}{43949268} \approx \frac{1}{174.3}=0.005737
$$

B) Let the second number drawn be $m$, then the fourth $n=2 m$, the number of numbers between them, i.e., the number of choices for $h$, is $m-1$, the smaller ones than $m$, and the number of choices for $e$ is the same, finally, the number of choices for $\ddot{o}$ is $90-2 m$, so the number of ways to select $h, e$, and $\ddot{o}$ for a fixed $m$ is $(m-1)^{2}(90-2 m)=$ $-2 m^{3}+94 m^{2}-182 m+90$. For $e$ and $h$, and $\ddot{o}$ to be selectable, we need $m-1 \geqq e \geqq 1$, and $90-2 m \geqq 1$, so $2 \leqq m \leqq 44$. The total number of favorable cases, similarly to the above,

$$
\begin{gathered}
k_{B}=\sum_{m=2}^{44}(m-1)^{2}(90-m)=-2 \sum_{m=1}^{44} m^{3}+94 \sum_{m=1}^{44} m^{2}-182 \sum_{m=1}^{44} m+90 \sum_{m=1}^{44} 1= \\
=624360,
\end{gathered}
$$

and thus

$$
P(B)=\frac{624360}{43949268} \approx \frac{1}{70.39}=0.01421
$$

$A B)$ Using the above notation,

$$
\begin{gathered}
1 \leqq e<m<n=2 m<\ddot{o}=3 e, \text { and thus } \\
e+1 \leqq m \leqq\left[\frac{3 e-1}{2}\right]
\end{gathered}
$$

(where the square brackets denote the integer part of the number inside), and the desired inequality between the two limits holds if,

$$
\left[\frac{3 e-1}{2}\right]-(e+1)=\left[\frac{e-3}{2}\right] \geqq 0, \quad \text { i.e., } \quad e \geqq 3
$$

Then $m \geqq 4$, so there is a number between $e$ and $3 e$ for $m$ and $2 m$ and between them for $h$. On the other hand, $\ddot{o}=3 m \leqq 90$ so $e \leqq 30$.

It will be useful to determine the number of possible draws based on the parity of $e$. Let $e$ be even first, its half $f$, i.e., $e=2 f, \ddot{o}=6 f, 2 \leqq f \leqq 15$. Then $m \geqq 2 f+1$ and $n=2 m \leqq 6 f-2$ because $n$ is even, $m \leqq 3 f-1$. Now, for a fixed $f$,

$$
m=2 f+1, \quad 2 f+2, \ldots, \quad 3 f-1
$$

the number of choices for $h$, while $h$ runs through the values between $m$ and $2 m$, is always $(n-1)-m$, i.e.,

$$
m-1=2 f, \quad 2 f+1, \ldots, \quad 3 f-2
$$

in total

$$
\frac{(f-1)(5 f-2)}{2}
$$

Summing this for the fixed values of $f$ (the $f=1$ term can also be added here):

$$
\begin{aligned}
k_{A B}^{\prime} & =\sum_{f=2}^{15} \frac{(f-1)(5 f-2)}{2}=\frac{5}{2} \sum_{f=1}^{15} f^{2}-\frac{7}{2} \sum_{f=1}^{15} f+15= \\
& =\frac{5}{12} 15 \cdot 16 \cdot 31-\frac{7}{4} 15 \cdot 16+15=2695
\end{aligned}
$$

If, on the other hand, $e$ is odd, $e=2 f+1$, where $3 \leqq e \leqq 30$ implies $1 \leqq f \leqq 14$, and thus $\ddot{o}=6 f+3$, from which $n=2 m \leqq 6 f+2$, so $2 f+2 \leqq m \leqq 3 f+1$. While $m$ runs through these $f$ values, the number of choices for $h$ is always $m-1$, in total

$$
\sum_{m=2 f+2}^{3 f+1}(m-1)=\frac{f\{(2 f+1)+3 f\}}{2}=\frac{f(5 f+1)}{2}
$$

and completing similarly to the above,

$$
\begin{aligned}
k_{A B}^{\prime \prime}=\sum_{f=1}^{14} \frac{f(5 f+1)}{2} & =\frac{5}{2} \sum_{f=1}^{14} f^{2}+\frac{1}{2} \sum_{f=1}^{14} f=\frac{5}{12} 14 \cdot 15 \cdot 29+\frac{1}{4} 14 \cdot 15= \\
& =\frac{14 \cdot 15}{12}(5 \cdot 29+3)=2590
\end{aligned}
$$

Thus,

$$
\begin{gathered}
k_{A B}=k_{A B}^{\prime}+k_{A B}^{\prime \prime}=5285, \text { and } \\
P(A B)=\frac{5285}{43949268} \approx \frac{1}{8316}=0.000120
\end{gathered}
$$

According to these, the occurrence of property $A$ is expected on average every 3.5 years, the occurrence of property $B$ every 1.5 years, and their joint occurrence every 166 years, if there is a lottery draw every week.

Kérchy László (Baja, III. Béla Gimn., IV. o. t. )

Note. A similar draw to the one in question was the lottery draw on February 28, 1958: 23, 28, 56, 69, 77, and the draw on January 8, 1971: 29, 37, 74, 81, 8