10. In the fourth annual Swirled Series, the Oakland Alphas are playing the San Francisco Gammas. The first game is played in San Francisco and succeeding games alternate in location. San Francisco has a $50 \%$ chance of winning their home games, while Oakland has a probability of $60 \%$ of winning at home. Normally, the serios will stretch on forever until one team gets a three-game lead, in which case they are declared the winners. However, after each game in San Francisco there is a $50 \%$ chance of an earthquake, which will cause the series to end with the team that has won more games declared the winner. What is the probability that the Gammas will win?
Answer: $\frac{34}{73}$. Let $F(x)$ be the probability that the Gammas will win the series if they are ahead by $x$ games and are about to play in San Francisco, and let $A(x)$ be the probability that the Gammas will win the series if they are ahead by $x$ games and are about to play in Oakland. Then we have
$$
\begin{array}{c}
F(2)=\frac{3}{4}+\frac{A(1)}{4} \\
A(1)=\frac{6 F(0)}{10}+\frac{4 F(2)}{10} \\
F(0)=\frac{1}{4}+\frac{A(1)}{4}+\frac{A(-1)}{4} \\
A(-1)=\frac{6 F(-2)}{10}+\frac{4 F(0)}{10} \\
F(-2)=\frac{A(-1)}{4}
\end{array}
$$

Plugging $A(1)=\frac{6 F(0)}{10}+\frac{4 F(2)}{10}$ into $F(2)=\frac{3}{4}+\frac{A(1)}{4}$, we get
$$
\begin{array}{c}
F(2)=\frac{3}{4}+\frac{1}{4}\left(\frac{6 F(0)}{10}+\frac{4 F(2)}{10}\right) \\
\frac{9 F(2)}{10}=\frac{3}{4}+\frac{6 F(0)}{40} \Leftrightarrow F(2)=\frac{5}{6}+\frac{F(0)}{6}
\end{array}
$$

Plugging $A(-1)=\frac{6 F(-2)}{10}+\frac{4 F(0)}{10}$ into $F(-2)=\frac{A(-1)}{4}$, we get
$$
\frac{34 A(-1)}{40}=\frac{4 F(0)}{10} \Leftrightarrow F(-2)=\frac{2 F(0)}{17}
$$

Now,
$$
F(0)=\frac{1}{4}+\frac{1}{4}\left(\frac{6 F(0)}{10}+\frac{4 F(2)}{10}\right)+\frac{1}{4}\left(\frac{6 F(-2)}{10}+\frac{4 F(0)}{10}\right)
$$

This simplifies to $F(0)=\frac{1}{4}+\frac{F(0)}{4}+\frac{F(2)}{10}+\frac{6 F(-2)}{40}$. Then, plugging our formulas in, we get
$$
\begin{array}{c}
F(0)=\frac{1}{4}+\frac{F(0)}{4}+\frac{1}{10}\left(\frac{5}{6}+\frac{F(0)}{6}\right)+\frac{3 F(0)}{170} . \\
\frac{73 F(0)}{102}=\frac{1}{3} \Leftrightarrow F(0)=\frac{34}{73} .
\end{array}
$$

Since $F(0)$ is the situation before the Series has started, the probability that the Gammas will win is $\frac{34}{73}$.