Let's assume that someone fills out every possible football betting slip in a week (13 matches, filled with 1, or 2, or $x$), and then after the matches, categorizes their tips according to the number of correct predictions on each slip. In which hit group will the most tips fall? (It is not necessary to calculate the number of possibilities for each group.)





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a) We provide a procedure for generating and determining the number of $k$-hit tips for a given total result. First, we select $k$ matches from the 1st to 13th rows of the betting slip in every possible way, and we write the correct result on these. Then, on the remaining $13-k$ places, we write one of the 2 incorrect tips, for example, 1 or $2-\mathrm{t}$, in the rows where $\times$ would have resulted in a hit.

Imagine that the numbers of the rows to be filled with the correct result are drawn from a bag, which contains the numbers on separate slips. The first, second, and third slips can be chosen in 13, 12, and 11 ways, respectively, depending on how many slips are left in the bag, and so on, the $k$-th in $(14-k)$ ways. The resulting $13 \cdot 12 \cdot 11 \cdot \ldots \cdot(14-k)$ combinations are not all different, as the same numbers can be drawn in different orders, which is not significant in the tip.

Imagine that the $k$ already drawn slips are placed back into an (empty) bag, and we write the correct hits in the tip in the order they are drawn from there. As in the previous reasoning, this can happen in $k \cdot(k-1) \cdot(k-2) \cdot \ldots \cdot 1$ ways (the last factor is 1), and for each different $k$-element tip, this is the number of non-essential combinations. Thus, the number of essentially different $k$-element combinations is the quotient of the two products.

Further developing these partial tips and considering the 2 possibilities for the first incorrectly filled row, the number of continuations is twice the quotient found, and the multiplication by 2 is repeated as many times as we take into account one more row to be filled incorrectly. According to all this, the number of $k$-hit slips is

$$
T_{k}=\frac{13 \cdot 12 \cdot 11 \cdot \ldots \cdot(14-k)}{k \cdot(k-1) \cdot(k-2) \cdot \ldots \cdot 1} \cdot 2^{13-k}
$$

(We spoke as if $k>3$; however, it is easy to show that our result is valid for smaller $k$ values as well.)

b) Now let's compare the consecutive $T_{k+1}$ and $T_{k}$ numbers in terms of magnitude. We can get their quotient without calculating them, and $T_{k+1} \gtrless T_{k}$ depending on whether the quotient is greater than 1 or less than 1.

The numerator and denominator of $T_{k+1}$ each contain one more factor than $T_{k}$, but the power is one less. Simplifying,

$$
\frac{T_{k+1}}{T_{k}}=\frac{13-k}{k+1} \cdot \frac{1}{2}
$$

Now

$$
\frac{13-k}{2 k+2} \gtreqless 1, \quad \text { if } \quad k \ggg \frac{11}{3} \quad(\text { and } k+1>0),
$$

thus $T_{0}T_{5}>T_{6}>\ldots>T_{13}$, or the group with 4 hits contains the most tips.

Ujvári István (Budapest, Fazekas M. Gyak. g. II. o. t.)

Remarks. 1. We can also get the result (2) without generating (1). Let's form the number of $k+1$ hit tips $T_{k+1}$ from the number of $k$ hit tips $T_{k}$. For a selected incorrectly tipped match, the $k$ hit tips $T_{k} / 2$ can be paired such that the members of each pair differ only in the tip of the selected match. By correcting the selected tip in each pair, we get a single $k+1$ hit. We can correct $k$ hit tips at $13-k$ incorrect places, but we get each $k+1$ hit tip again, $k+1$ times, depending on which hit was created by such a correction, so

$$
T_{k+1}=\frac{T_{k}}{2} \cdot \frac{13-k}{k+1}
$$

which is identical to (2).

Kalmár István (Debrecen, Fazekas M. g. III. o. t.)

2. If the betting slip contains $N=3 n+2$ matches instead of 13, the number of $n$ and $n+1$ hit tips is the same, and fewer tips are obtained for any other number of hits. For other $N$ values, the most tips are where the number of hits is the integer part of the fraction $(N+1) / 3$.