A football match has three possible outcomes from a TOTO perspective (1, x, 2). In Bergengoci, due to the small number of teams, only 4 matches appear in the TOTO, while on the land of the Burkus, 11 matches are included. How many betting slips does a Bergengocian need to fill out to ensure they have 3 correct guesses, and how many does a Burkus need to fill out if they want to guess 9 correctly?
In Bergengóc, there are $3^{4}$ differently filled TOTO tickets. In a given week, not only one ticket can have at least 3 hits; apart from the single 4-hit ticket, there are also 8 tickets that have at least 3 hits, which we can obtain from the full-hit ticket by corrupting one of its tips. Conversely, with a single ticket, we can have at least 3 hits with 9 different outcomes. Thus, to cover all $3^{4}$ possible cases, we need to fill out at least $3^{4}$ : $3^{2}=3^{2}$ tickets. An example that meets the goal is the 9 tickets shown in the following table.

| 1 | 1 | 1 | $\mathrm{x}$ | $\mathrm{x}$ | $\mathrm{x}$ | 2 | 2 | 2 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 1 | $\mathrm{x}$ | 2 | 1 | $\mathrm{x}$ | 2 | 1 | $\mathrm{x}$ | 2 |
| 2 | 1 | $\mathrm{x}$ | 1 | $\mathrm{x}$ | 2 | $\mathrm{x}$ | 2 | 1 |
| $\mathrm{x}$ | 1 | 2 | 2 | $\mathrm{x}$ | 1 | 1 | 2 | $\mathrm{x}$ |

In Burkusföld, there are $3^{11}$ different fillings, and with a given ticket,

$$
1+\binom{11}{1} \cdot 2+\binom{11}{2} 2^{2}=3^{5}
$$

different cases can result in at least 9 hits. Indeed, in one case, we have 11 hits. In 22 cases, we only miss one result, which occurs when the actual outcome in one of the 11 rows is one of the two other possible outcomes different from our tip. When we have only 9 hits, the two missed results can be chosen in 11 ways for the first row and 10 ways for the second row. However, this way we count each pair of rows twice, so the number of different cases is actually $11 \cdot 10 / 2=\binom{11}{2}$. For each pair of rows, the missed outcomes can be chosen in $2 \cdot 2$ ways, so the number of possibilities is indeed what we stated.

Since with one ticket, we can have at least 9 hits in $3^{5}$ different cases, to cover all cases, we need to fill out at least $3^{11}$ : $3^{5}=3^{6}$ tickets.

Remark. We did not provide an example of a suitable betting system in Burkusföld. In fact, we did not need to provide one in Bergengóc either; the problem only asked for a lower bound. It can be seen that the Bergengóc example was created such that the first two rows are arbitrary, the third row is their "sum" if we imagine placing 0 in the place of $x$, and the result is reduced by 3 if it is greater than 3. The fourth row is similarly the difference of the first two, but here the negative result must be increased by 3. Following this pattern, the $3^{6}$ Burkus tickets can also be attempted. Now let the first six rows be free, and the other five be some function of these. If we denote the $i$-th tip by $t_{i}$, and only the remainder upon division by 3 is important, then a suitable "check" function system is the following:

$$
\begin{array}{rr}
t_{7}= & t_{2}+t_{3}-t_{4}-t_{5}+t_{6} \\
t_{8} & =t_{1}-t_{3}+t_{4}-t_{5}+t_{6} \\
t_{9} & =t_{1}-t_{2}-t_{4}+t_{5}+t_{6} \\
t_{10} & =-t_{1}+t_{2}-t_{3}+t_{5}+t_{6} \\
t_{11} & =-t_{1}-t_{2}+t_{3}+t_{4}+t_{6}
\end{array}
$$