Two table tennis players: $A$ and $B$ are about to play 7 games against each other. Someone offers $A$ 100 times the money if he wins all 7 games. $A$ accepts the bet. Is this bet fair, given that based on past experience, $A$ has twice the chance of winning 4 games and losing 3 games, compared to winning 5 games and losing 2 games. What would have been a fair multiplier for the 100 times the money?
Let $v$ be the probability that $A$ wins a game, then the probability that $A$ wins $k$ games and loses $n-k$ games out of $n$ games, disregarding the order, is

$$
\binom{n}{k} v^{k}(1-v)^{n-k}
$$

where $k=0,1,2, \ldots, n$.

Applying this to the current case, according to the problem $(0<v<1)$

$$
\binom{7}{4} v^{4}(1-v)^{3}=2\binom{7}{5} v^{5}(1-v)^{2}
$$

thus, simplifying by $v^{4}(1-v)^{2}$

$$
35(1-v)=42 v, \quad \text { hence } \quad v=\frac{5}{11}
$$

Therefore, the probability that $A$ wins the bet is $\left(\frac{5}{11}\right)^{7}$. In a fair bet, the stakes are proportional to the probabilities of winning. The fair stake offered by the bettor, $t$, against $A$'s unit stake, is

$$
t: 1=\left[1-\left(\frac{5}{11}\right)^{7}\right]:\left(\frac{5}{11}\right)^{7}
$$

from which

$$
t=\frac{1-\left(\frac{5}{11}\right)^{7}}{\left(\frac{5}{11}\right)^{7}}=\left(\frac{11}{5}\right)^{7}-1=248.4
$$

Thus, the bet is disadvantageous for $A$, as 248.4 times the money would have been fair.

Csiszár Imre (Bp. I., Petőfi g. III. o. t.)