a) The probability that $B$ wins is

$$
\nu_{B}=\left(\frac{6}{7}\right)^{7} \sim 0.3401
$$

and thus

$$
\nu_{A} \sim 0.6599
$$

Therefore, the expected value of $A$'s winnings is

$$
M_{A}=5 \nu_{A}-10 \nu_{B}=3.2995-3.401=-0.1015
$$

Thus, the bet is favorable for $B$.
b) For $x$ individuals, $\nu_{B}=\left(\frac{6}{7}\right)^{x}, \nu_{A}=1-\left(\frac{6}{7}\right)^{x}$, and thus the expected value of $A$'s winnings, if $B$'s bet is 1 forint, is

$$
M_{A}=1 \cdot \nu_{A}-10 \cdot \nu_{B}=1-\left(\frac{6}{7}\right)^{x}-10\left(\frac{6}{7}\right)^{x}>0
$$

which means

$$
1>11\left(\frac{6}{7}\right)^{x}
$$

or

$$
\left(\frac{6}{7}\right)^{x}\frac{\lg 11}{\lg 7-\lg 6} \sim \frac{1.0414}{0.069} \sim 15.5
$$

Therefore, the group must consist of at least 16 individuals for a 1 forint bet on $B$ to be disadvantageous for $B$. Gáti Gyula (Debrecen, Chemical Engineering II. class)
$A$ bets with $B$ that in a group of 7 people, at least one individual was born on a Sunday. $A$ bets 10 Forints against $B$'s 5 Forints.

a) Who has the advantage in this bet?

b) How many people must the group consist of at least, so that even a 1 Forint bet by $B$ would be disadvantageous for $B$?