$A$ and $B$ are chess players who compete under the following conditions: The winner is the one who first reaches (at least) 2 points; if they both reach 2 points at the same time, the match is a draw. (A win is 1 point, a draw is $1 / 2$ point, a loss is 0 points) -

a) What is the expected number of games, if the probability of a draw in each game is 0.5, and the probability that $A$ wins is 0.3? -

b) What is the probability that $B$ wins the match?
a) If for a given game, the probability that $A$ wins is denoted by $V_{a}$, that $B$ wins by $V_{b}$, and that it ends in a draw by $V_{d}$, then according to the problem, $V_{a}=0.3$, $V_{d}=0.5$, and thus $V_{b}=1-0.3-0.5=0.2$.

Given the conditions, the match can conclude after at least 2 and at most 4 games. If the probabilities of the match concluding in 2, 3, and 4 games are denoted by $V_{2}$, $V_{3}$, and $V_{4}$ respectively, then the expected number of games is

$$
M=2 V_{2}+3 V_{3}+4 V_{4}, \quad \text{where} \quad V_{2}+V_{3}+V_{4}=1
$$

We need to calculate the values of $V_{2}$, $V_{3}$, and $V_{4}$.

The match can only conclude in two games if either $A$ wins both games or $B$ wins both games.

Thus,

$$
V_{2}=V_{a}^{2}+V_{b}^{2}=0.09+0.04=0.13
$$

A match that concludes in four games is only possible if both sides have scored 1.5 points after the first three games. This can happen if all three of the first games are draws (with a probability of $0.5^{3}=0.125$), or if both sides have won 1-1 games and one game was a draw. The latter outcome can occur in $3!=6$ different orders, so its probability is $6 \cdot 0.2 \cdot 0.3 \cdot 0.5=0.18$, and thus

$$
V_{4}=0.125+0.18=0.305
$$

Finally,

$$
\begin{gathered}
V_{3}=1-\left(V_{2}+V_{4}\right)=1-(0.13+0.305)=1-0.435= \\
=1-0.435=0.565
\end{gathered}
$$

Thus,

$$
M=2 \cdot 0.13+3 \cdot 0.565+4 \cdot 0.305=0.26+1.695+1.22=3.175
$$

b) $B$ can win the match in 2 games; the probability of this is $V_{b}^{2}$.

$B$ can win in 3 games if they win one game and draw two. This can happen in 3 different orders, so the probability is $3 \cdot V_{b} V_{d}^{2}$.

$B$ can also win in 3 games if they win one of the first two games and lose the other (in two different orders) and then win the third game. The probability of this is $2 V_{b}\left(1-V_{b}\right) V_{b}=2 V_{b}^{2}\left(1-V_{b}\right)$.

$B$ cannot win in 3 games in any other way.

Finally, $B$ wins in 4 games after scoring 1.5 points in the first 3 games and then winning the fourth game. The probability of this is $V_{4} V_{b}$.

With this, we have exhausted all possible cases.

The events listed here are mutually exclusive, and thus their sum gives the desired probability $V$:

$$
\begin{aligned}
V & =V_{b}^{2}+3 V_{b} V_{d}^{2}+2 V_{b}^{2}\left(1-V_{b}\right)+V_{4} V_{b}= \\
& =0.2^{2}+3 \cdot 0.2 \cdot 0.5^{2}+2 \cdot 0.2^{2} \cdot 0.8+0.305 \cdot 0.2= \\
& =0.04+0.15+0.064+0.061=0.315
\end{aligned}
$$

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