A father has $n(>4)$ children. What is the probability that among them, there is one who has an older brother, a younger brother, a sister, and an aunt? (We consider the probability that a living person is male or female to be equal, at any age.)
I. solution. The gender-based listing of $n$ consecutive children can be $2^{n}$ ways using $f$ (boy) and $l$ (girl) symbols (repetitive variations of the $n$-th class from 2 elements). The stipulation in parentheses in the problem is equivalent to the fact that each of the $2^{n}$ possible listings is equally probable.

From the point of view of our question, in each favorable listing we find the oldest (i.e., the first) child who has an older brother, an older sister, a younger brother, and a younger sister, and we group the listings according to the birth order number $i$ assigned to this oldest child. It is obvious that each favorable listing will belong to one and only one class; we call $i$ the index of the class, its value is $3 \leq i \leq n-2$.

The fact that the $i$-th child is the first suitable one means, on the one hand, that among the $(i-1)$ children born earlier, there is a boy and a girl, but among the $(i-2)$ siblings preceding the immediately preceding $(i-1)$-th child, there is either no boy or no girl. Thus, this $(i-2)$-member set consists entirely of one gender. We denote this set briefly as $f^{(i-2)}$ or $l^{(i-2)}$, so the listing of the first $i$ children can only be one of the following 4 possibilities:

$$
f^{(i-2)}, l, f ; \quad l^{(i-2)}, f, f ; \quad f^{(i-2)}, l, l ; \quad l^{(i-2)}, f, l
$$

On the other hand, the fact that characterizes the class with index $i$ is that in each of its listings, the $(n-i)$ additional (younger) siblings are not all of the same gender. Thus, the listing of the latter can result in $m=2^{n-i}-2$ different ways, since only the "all boys" and "all girls" sequences are excluded from the conceivable $2^{n-i}$ repetitive variations. Each of the $m$ ways can be appended to each of the beginnings listed in (1), so in the class with index $i$,

$$
4 m=4\left(2^{n-i}-2\right)=2^{n+2-i}-8
$$

number of listings belong.

Summing this expression for the values $i=3,4,5, \ldots, n-2$, the number of favorable listings (families) is

$$
\sum_{i=3}^{n-2}\left(2^{n+2-i}-8\right)=\left(2^{n-1}+2^{n-2}+\ldots+2^{4}\right)-8(n-4)=2^{n}+16-8 n
$$

finally, the probability in question is:

$$
p=\frac{2^{n}-8(n-2)}{2^{n}}=1-\frac{n-2}{2^{n-3}}
$$

(specifically for $n=5,6,7$ and 10, respectively 0.25; 0.5; 0.6875; 0.9375).

Veres Sándor (Debrecen, Fazekas M. Gymn., III. o. t.) Hargitai Bálint (Kaposvár, Táncsics M. Gymn., IV. o. t.)

Remark. The first part of our consideration can also be applied to the listing of children in increasing order of age. And if we determine the youngest suitable child's $j$ number for each conceivable favorable listing, and classify the listings by considering the first and last suitable child's numbers simultaneously, then in the listings of the class with indices $i$ and $j$, $j-(i-1)$ number of children have an older brother, an older sister, a younger brother, and a younger sister. Thus, this question can also be raised with a slight modification: if we randomly select one of the children of a father with $n$ children, what is the probability that this child will have an older brother, a younger brother, an older sister, and a younger sister.

II. solution. It is simpler to count the unfavorable - briefly: "tlen" sequences, families.

Obviously, the case is "tlen" if there is no girl in the family or if there is only one. The former can happen in one way: all boys, the latter in $n$ ways, according to the index of the girl $1,2, \ldots, n$.

Taking 2 girls is also "tlen" the sequence if there is no boy among them: $f f \ldots f l l f f \ldots f$, because one of their older brothers has no sister, one of their younger brothers has no sister, and because they themselves have only one type of sister, and because the sequence is "tlen" even if there is only one boy among the 2 girls, but this "lfl-block" (immutable subsequence) is at the beginning or the end of the sequence, which means 2 ways. However, if at least 2 of the $(n-2)$ boys are between the 2 girls, then the sequence is favorable regardless of the order of the remaining $n-4(\geq 1)$ boys.

If "girl" is replaced by "boy" everywhere in the above, and "boy" by "girl" - we get the same results; thus, in those families where there are at most 2 of one gender, the number of "tlen" cases is $2\{1+n+(n-1)+2\}=4 n+4 ;$ indeed, for $n \geq 5$ we have not counted any family twice.

If there are at least 3 girls and 3 boys - that is, the number of girls is 3 or 4 or ... $(n-3)$, which together is $(n-5)$ cases - in the manner of the above, 4 "tlen" can be formed in every case:

a) all girls and all boys form one block each, 2 ways:

$l, \ldots, l, f, \ldots, f$ and $f, \ldots, f, l, \ldots, l$

b) we swap the order of the two children at the junction of the two blocks mentioned above:

$$
\underbrace{l, l, \ldots, l,}_{\text {at least 2 }} f, l, \underbrace{f, f, \ldots, f}_{\text {at least } 2} \text { and } \underbrace{f, f, \ldots, f,}_{\text {at least } 2} l, f \underbrace{l, l, \ldots, l}_{\text {at least } 2}
$$

By transferring more than one from any gender to the other block, the second transferred one would already have four types of siblings. Thus, the number of unfavorable sequences is:

$$
(4 n+4)+4(n-5)=8 n-16=8(n-2)
$$

in agreement with the numerator of the first fraction in (2) of the I. solution. 

Linnert László (Szeged, Radnóti M. Gymn., III. o. t.)