$A$ bets with $B$ that he will roll less than 8 with two dice in one throw. With $A$'s 10 forints, $B$'s bet is 8 forints. Is this a fair bet?

(See András Prékopa's article "Expected Value" in this issue.)





Translating the text into English, while preserving the original text's line breaks and format, as requested.
The bet is fair if the expected gain of $A$ (and $B$) is 0, that is,

$$
M_{A}=8 v_{a}-10 v_{b}=0
$$

where $v_{a}$ and $v_{b}$ represent the probability that $A$, and $B$ win, respectively.

The number of possible outcomes is $V_{b}^{i, 2}=6^{2}=36$. Among these, the favorable outcomes for $B$ are $6+6,6+5,6+4,6+3,6+2,5+5,5+4$, $5+3,4+4$, which amounts to $1+2+2+2+2+1+2+2+1=15$ in total.

Thus, $v_{b}=\frac{15}{36}=\frac{5}{12}$ and $v_{a}=1-\frac{5}{12}=\frac{7}{12}$, and therefore $A$'s expected gain is

$$
M_{A}=8 \cdot \frac{7}{12}-10 \cdot \frac{5}{12}=\frac{56-50}{12}=\frac{6}{12}=\frac{1}{2}>0
$$

Thus, the bet is advantageous for $A$.

It would be fair if, denoting $B$'s bet by $x$,

$$
\frac{7}{12} x-\frac{50}{12}=0
$$

from which

$$
x=7 \frac{1}{7} \text { forints }
$$

Tibor Kozma (Pannonhalma, Bencés g. II. o. t.)