Problem (Combinatorics):
Four tennis players Alain, Bianca, Chen, and Dave take part in a tournament in which a total of three matches are played. First, two players are chosen randomly to play each other. The other two players also play each other. The winners of the two matches then play to decide the tournament champion. Alain, Bianca and Chen are equally matched (that is, when a match is played between any two of them, the probability that each player wins is $\frac{1}{2}$ ). When Dave plays each of Alain, Bianca and Chen, the probability that Dave wins is $p$, for some real number $p$. Determine the probability that Bianca wins the tournament, expressing your answer in the form $\frac{a p^{2}+b p+c}{d}$ where $a, b, c$, and $d$ are integers.

Solution:
Let $x$ be the probability that Bianca wins the tournament.

Because Alain, Bianca and Chen are equally matched and because their roles in the tournament are identical, then the probability that each of them wins will be the same.

Thus, the probability that Alain wins the tournament is $x$ and the probability that Chen wins the tournament is $x$.

Let $y$ be the probability that Dave wins the tournament.

Since exactly one of Alain, Bianca, Chen, and Dave wins the tournament, then $3 x+y=1$ and so $x=\frac{1-y}{3}$. We can calculate $y$ in terms of $p$.

In order for Dave to win the tournament, he needs to win two matches.

No matter who Dave plays, his probability of winning each match is $p$.

Thus, the probability that he wins his two consecutive matches is $p^{2}$ and so the probability that he wins the tournament is $y=p^{2}$.

Thus, the probability that Bianca wins the tournament is $\frac{1-p^{2}}{3}$.

(We could rewrite this as $\frac{-p^{2}+0 p+1}{3}$ to match the desired form.)

Let $x$ be the probability that Bianca wins the tournament.

There are three possible pairings for the first two matches:

(i) Bianca versus Alain, and Chen versus Dave

(ii) Bianca versus Chen, and Alain versus Dave

(iii) Bianca versus Dave, and Alain versus Chen

Each of these three pairings occurs with probability $\frac{1}{3}$.

In (i), Bianca wins either if Bianca beats Alain, Chen beats Dave, and Bianca beats Chen, or if Bianca beats Alain, Dave beats Chen, and Bianca beats Dave.

Since Bianca beats Alain with probability $\frac{1}{2}$, Chen beats Dave with probability $1-p$, and Bianca beats Chen with probability $\frac{1}{2}$, then the first possibility has probability $\frac{1}{2} \cdot(1-p) \cdot \frac{1}{2}$. Since Bianca beats Alain with probability $\frac{1}{2}$, Dave beats Chen with probability $p$, and Bianca beats Dave with probability $1-p$, then the second possibility has probability $\frac{1}{2} \cdot p \cdot(1-p)$.

Therefore, the probability of Bianca winning, given that possibility (i) occurs, is $\frac{1}{2} \cdot(1-$ p) $\cdot \frac{1}{2}+\frac{1}{2} \cdot p \cdot(1-p)$.

In (ii), Bianca wins either if Bianca beats Chen, Alain beats Dave, and Bianca beats Alain, or if Bianca beats Alain, Dave beats Alain, and Bianca beats Dave.

The combined probability of these is $\frac{1}{2} \cdot(1-p) \cdot \frac{1}{2}+\frac{1}{2} \cdot p \cdot(1-p)$.

In (iii), Bianca wins either if Bianca beats Dave, Alain beats Chen, and Bianca beats Alain, or if Bianca beats Dave, Chen beats Alain, and Bianca beats Chen.

The combined probability of these is $(1-p) \cdot \frac{1}{2} \cdot \frac{1}{2}+(1-p) \cdot \frac{1}{2} \cdot \frac{1}{2}$.

Therefore,

$$
\begin{aligned}
x & =\frac{1}{3}\left(\frac{1}{4}(1-p)+\frac{1}{2} p(1-p)+\frac{1}{4}(1-p)+\frac{1}{2} p(1-p)+\frac{1}{4}(1-p)+\frac{1}{4}(1-p)\right) \\
& =\frac{1}{3}(p(1-p)+(1-p)) \\
& =\frac{1}{3}\left(p-p^{2}+1-p\right)
\end{aligned}
$$

Thus, the probability that Bianca wins the tournament is $\frac{1-p^{2}}{3}$.

Final answer: $\frac{1-p^{2}}{3}$