6. The Strategy of the Great Combinator (6th grade. 2 points). Ostap Bender is playing chess against Vasyukov, the Champion of Russia, and the World Champion. According to the rules, to become the Absolute Champion, Ostap needs to win two games in a row. Bender has the right to choose the order in which he plays against his opponents. Which order is more advantageous for Ostap?
Solution. We will assume that Ostap's chances of winning against the Russian Champion are lower than against the Champion of Vasyuki, but higher than against the World Champion.

Let $p_{1}, p_{2}$, and $p_{3}$ be the probabilities of Ostap winning against his opponents in the order they follow.

Then the probability of Ostap becoming the Absolute Champion is

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

Thus, the problem reduces to the question of in what order the opponents should follow so that the value $p_{2}\left(p_{1}+p_{3}\right)$ is the greatest, assuming that the Russian Champion plays better than the Champion of Vasyuki, but worse than the World Champion.

If we swap the first and second opponents, the inequality

$$
p_{2}\left(p_{1}+p_{3}\right)>p_{1}\left(p_{2}+p_{3}\right) \text {, that is, } p_{3}\left(p_{2}-p_{1}\right)>0 \text {. }
$$

must hold. This means that the second opponent must be weaker than the first.

If we swap the second and third opponents, we similarly get that the second opponent must be weaker than the third.

If we swap the first and third opponents, the probability of success for Ostap does not change, since the value of the expression $p_{2}\left(p_{1}+p_{3}\right)$ does not depend on the order of the terms $p_{1}$ and $p_{3}$ in the parentheses.

Answer: Ostap should choose the Champion of Vasyuki as the second opponent, and it does not matter who will be the first and third.

Comment. The problem can be solved qualitatively without any calculations. It is clear that the decisive match for Ostap is the second one. If he loses it, he will not see the coveted title. Therefore, the probability of winning the second match should be the highest. Hence, the second opponent should be the Champion of Vasyuki. And for the matches with the stronger opponents, there are two independent attempts left. And from symmetry considerations, it is clear that their order does not matter. A simpler version of the problem appears in the book by Charles Frederick Mosteller "Fifty Challenging Problems in Probability with Solutions" (Moscow, Nauka, 1985), and also in issue 7 of the journal "Kvantik" for 2013 (Moscow, MCCME, p.7).