An urn contains 5 black, 6 white, and 7 blue balls. Someone offers to pay 3 forints to anyone who, in one draw, pulls out 2 black, 3 white, and 4 blue balls, and the person offering the bet will pay 50 fillér in all other cases. Is this bet to their advantage or disadvantage? What would have been the fair bet instead of 3 forints?

(See the article "Expected Value" by András Prékopa in this issue.)
»Minden más húzás«-on just 9 balls being drawn can be understood, because otherwise the problem would be unsolvable.

9 balls can be drawn in $\binom{18}{9}$ ways. Among these, the favorable cases are $\binom{5}{2}\binom{6}{3}\binom{7}{4}$, so the probability of the bet resulting in a 3 forint loss is,

$$
v=\binom{5}{2}\binom{6}{3}\binom{7}{4}:\binom{18}{9}=10 \cdot 20 \cdot 35 \cdot \frac{1}{17 \cdot 13 \cdot 11 \cdot 20}=\frac{350}{2431}
$$

However, the probability of a $0.5 \mathrm{Ft}$ profit is

$$
1-v=1-\frac{350}{2431}=\frac{2081}{2431}
$$

The bet would have been fair if instead of $3 \mathrm{Ft}$, it was bet for $x$ forints and the expected profit was 0, that is,

$$
M=\frac{2081 \cdot 0.5}{2431}-\frac{350 x}{2431}=0
$$

from which

$$
x=\frac{1040.5}{350} \approx 2.97 \mathrm{Ft}
$$

Thus, he bet at a disadvantage, because the fair bet would have been $2.97 \mathrm{Ft}$.

Harza Tibor (Székesfehérvár, József A. g. III. o. t.)