Basketball Practice

Let's consider the number of games for which each team is expected to sit out.

Let $E_{A}(x)$ denote the expected number of games for which team A sits out in the first $x$ games. Let $E_{B}(x)$ and $E_{C}(x)$ denote the same value for teams B and C respectively. In essence, $E_{A}(x)$ denotes the value for a team that plays the first game, and $E_{C}(x)$ denotes that for a team that sits out the first game.

Notice that $E_{A}(x)=E_{B}(x)$ and that $E_{A}(x)+E_{B}(x)+E_{C}(x)=x$ . This means that $E_{C}(x)=x-2E_{A}(x)$ .

It is clear that $E_{C}(1)=1$ . After the first game, either team A or team B sits out, each with a probability of $\frac{1}{2}$ . Therefore, $E_{C}(x+1)=1+\frac{1}{2}E_{A}(x)+\frac{1}{2}E_{B}(x)=1+E_{A}(x)$ .

Since $E_{A}(x)=\frac{x-E_{C}(x)}{2}$ , $E_{C}(x+1)=1+\frac{x-E_{C}(x)}{2}=1+\frac{x}{2}-\frac{E_{C}(x)}{2}$ .

We can then solve this recurrence relation.

$E_{C}(x)$ can be expressed in the form of $q_{1}k^{x}+q_{2}x+q_{3}$ .

From the recurrence relation, we can see that $\frac{E_{C}(x)}{2}+E_{C}(x+1)=1+\frac{x}{2}$

We can get $q_{1}(\frac{k^{x}}{2}+k^{x+1})+q_{2}(\frac{3x}{2}+1)+\frac{3q_{3}}{2}=1+\frac{x}{2}$ after substitution.

It is clear that $q_{2}=\frac{1}{3}$ from comparing the coefficients for $x$ .

The equation then becomes $q_{1}(\frac{k^{x}}{2}+k^{x+1})+\frac{3q_{3}}{2}=\frac{2}{3}$ .

The first term of the left hand side has to be constant no matter what $x$ is. This means that $k=-\frac{1}{2}$ (The term is constant when $k=0$ or $1$ as well. However, both of these values lead to the result $E_{C}(x)=\frac{1}{3}x+\frac{2}{3}$ , which does not satisfy the recurrence condition).

The equation now becomes $\frac{3q_{3}}{2}=\frac{2}{3}$ , giving $q_{3}=\frac{4}{9}$ .

We know that $E_{C}(1)=1$ , which means that $q_{1}k+q_{2}+q_{3}=1$ . Substituting the values, we can get $q_{1}=-\frac{4}{9}$ .

Therefore, $E_{C}(10)=-\frac{4}{9}(-\frac{1}{2})^{10}+\frac{1}{3}(10)+\frac{4}{9}=\frac{967}{256}$ .

So the expected number of games that team C does not sit out is $10-\frac{967}{256}=\frac{1593}{256}$ , giving the answer of $1593+256=\boxed{1849}$ .