A totó requires tipping 12 matches, in each case with the symbols $\gg 1 \ll, » 2 \ll$ or $\gg \mathrm{x} \ll$. What is the expected number of correct tips if we fill out the ticket randomly (e.g., by drawing lots) so that each of the three possibilities has an equal chance for every match?
I. solution: The probability of exactly $k$ (and only $k$) hits

$$
\left(\frac{1}{3}\right)^{k}\left(\frac{2}{3}\right)^{12-k} \quad \text { where } \quad k=0,1,2, \ldots, 12
$$

Since $k$ hits can occur in $\binom{12}{k}$ ways out of 12, the probability of exactly $k$ hits (in any order) is

$$
v_{k}=\left(\frac{1}{3}\right)^{k}\left(\frac{2}{3}\right)^{12-k}\binom{12}{k}=\frac{1}{3^{12}} \cdot\binom{12}{k} 2^{12-k} \quad(k=0,1,2, \ldots, 12)
$$

Therefore, the expected number of hits is

$$
\begin{gathered}
M=0 \cdot v_{0}+1 \cdot v_{1}+2 \cdot v_{2}+\ldots+11 \cdot v_{11}+12 \cdot v_{12}= \\
=\frac{1}{3^{12}}\left[1 \cdot\binom{12}{1} 2^{11}+2\binom{12}{2} 2^{10}+\ldots+10\binom{12}{10} 2^{2}+11 \cdot\binom{12}{11} 2+12\binom{12}{12} 2^{0}\right]= \\
=\frac{1}{3^{12}}[24576+135168+337920+506880+506880+354816+177408+ \\
+63360+15840+2640+254+12]=\frac{2125764}{531441}=4
\end{gathered}
$$

Antal Rockenbauer (Bp., X., I. László g. II. o. t.)

II. solution: We can also reach the goal without any numerical calculations by using the binomial theorem.

$$
\begin{aligned}
& v_{k}=\binom{12}{k}\left(\frac{1}{3}\right)^{k}\left(\frac{2}{3}\right)^{12-k}=\frac{12!}{k!(12-k)!}\left(\frac{2}{3}\right)^{12-k}\left(\frac{1}{3}\right)^{k}= \\
& =\frac{12 \cdot 11!}{k(k-1)![11-(k-1)]!}\left(\frac{2}{3}\right)^{11-(k-1)}\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)^{k-1}
\end{aligned}
$$

From this

$$
k v_{k}=12 \cdot \frac{1}{3}\left(\frac{11}{k-1}\right)\left(\frac{2}{3}\right)^{11-(k-1)}\left(\frac{1}{3}\right)^{k-1} \quad(k=1,2, \ldots, 12)
$$

Therefore

$$
\begin{aligned}
& M=4\left[\binom{11}{0}\left(\frac{2}{3}\right)^{11}+\binom{11}{1}\left(\frac{2}{3}\right)^{10} \frac{1}{3}+\binom{11}{2}\left(\frac{2}{3}\right)^{9}\left(\frac{1}{3}\right)^{2}+\ldots+\binom{10}{10} \frac{2}{3} \cdot\left(\frac{1}{3}\right)^{10}+\binom{11}{11}\left(\frac{1}{3}\right)^{11}\right]= \\
& =4\left(\frac{2}{3}+\frac{1}{3}\right)^{11}=4
\end{aligned}
$$

Remark: This solution can be easily generalized to $n$ instead of 12, and $v$ instead of $\frac{1}{3}$, in which case for $M$ we get

$$
M=n v[(1-v)+v]^{n-1}=n v
$$

Zoltán Zsombok (Bp., IV., Kálmán Könyves g. IV. o. t.)