## Task $8 / 84$

Klaus is to play chess alternately against Peter and Rolf and win a prize if he wins two consecutive games out of three. He estimates Peter to be a stronger player than Rolf.

Against whom will he play first? The choice is his!
Let $p_{P}$ and $p_{R}$ be the probabilities that Klaus wins against Peter and Rolf, respectively, and $p_{i}$ the probability of winning the i-th game $(i=1 ; 2 ; 3)$. There are now the following mutually exclusive possibilities for winning the prize.

1. Winning the first two games (the third can be omitted).
2. Losing the first, winning the second and the third game.

Since the game results can be considered independent of each other, the probability $p$ of winning the prize is

$$
p=p_{1} \cdot p_{2}+\left(1-p_{1}\right) \cdot p_{2} \cdot p_{3}=p_{1} \cdot p_{2}+p_{2} \cdot p_{3}-p_{1} \cdot p_{2} \cdot p_{3}
$$

If Klaus plays against Peter in the first game, then

$$
p=p_{P} \cdot p_{R}+p_{R} \cdot p_{P}-p_{P} \cdot p_{R} \cdot p_{P}=p_{R} \cdot p_{P} \cdot\left(2-p_{P}\right)
$$

if he plays against Rolf first, then

$$
p=p_{R} \cdot p_{P}+p_{P} \cdot p_{R}-p_{R} \cdot p_{P} \cdot p_{R}=p_{R} \cdot p_{P} \cdot\left(2-p_{R}\right)
$$

Since Klaus estimates Peter's playing strength to be higher than Rolf's, $p_{P} < p_{R}$, and thus $2 - p_{P} > 2 - p_{R}$. The winning chances for Klaus are therefore (at first glance surprisingly!) greater if he plays against the stronger Peter first.