toughest question i face in probability

The problem can be modelled as a Markov Chain . Let be $\mathbb{X}$ and $\mathbb{Y}$ the two players and let be $G \in \{X,Y\}$ the random variable such that

$\displaystyle \mathbb{P}(G=X)=\frac{1}{3} \\ \displaystyle \mathbb{P}(G=Y)=\frac{2}{3}$

Now, a sequence of games is a string containing $X$ if $\mathbb{X}$ wins or $Y$ if $\mathbb{Y}$ wins, for example

$\displaystyle YXYYXYYYXYXYYXY...$

Let's define $s_i$ the $i$ th state, which is a sequence of four consecutive games. In our example $s_1=YXYY$ . Hence, $s_2$ could be the string $XYYY$ or the string $XYYX$ . In the example above $s_2=XYYX$ . We define $S$ the set of all possible states, that is

$\displaystyle S=\{YYYX, \, YYXY, \, YXYY, \, YXYX, \, XYYY, \, XYYX, \, XYXY, \, XYXX, \, YYXX, \,YYYY \}$

Notice that the $8$ th and the $9$ th states are the ones in which $\mathbb{X}$ wins, while the last is the one in which $\mathbb{Y}$ wins. We can now create the stochastic matrix $\textbf{P}=p_{ij}$ . The order of the states follows the one of $S$ above, so

$\textbf{P}=\left(\begin{array}{c|c} \textbf{Q} & \textbf{R} \\\hline \textbf{O} & \textbf{I} \end{array}\right)= \left( \begin{array}{ccccccc|cc} 0 & \frac{2}{3} & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{2}{3} & \frac{1}{3} & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{2}{3} & \frac{1}{3} & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & \frac{2}{3} & \frac{1}{3} & 0 & 0 \\ \frac{1}{3} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{2}{3} \\ 0 & \frac{2}{3} & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{2}{3} & \frac{1}{3} & 0 & 0 & 0 & 0 & 0 & 0 \\\hline 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{array}\right)$

From now on we apply the usual theory of Markov chains: we evaluate $\textbf{N}=(\textbf{I}-\textbf{Q})^{-1}$ in order to find $\textbf{N} \textbf{R}$ , that is

$\textbf{N} \textbf{R} = \begin{pmatrix} \frac{6}{43} & \frac{21}{43} & \frac{16}{43} \\ \frac{9}{43} & \frac{10}{43} & \frac{24}{43} \\ \frac{10}{129} & \frac{35}{129} & \frac{28}{43} \\ \frac{61}{129} & \frac{20}{129} & \frac{16}{43} \\ \frac{2}{43} & \frac{7}{43} & \frac{34}{43} \\ \frac{6}{43} & \frac{21}{43} & \frac{16}{43} \\ \frac{9}{43} & \frac{10}{43} & \frac{24}{43} \end{pmatrix}$

Hence, probability $P$ is the sum of the $nr_{i3}$ , $1 \leq i \leq 7$ , weighted with the probability that $s_i$ occurs as the first state plus the probability that $s_{10}=YYYY$ occurs as first state, i.e.

$\displaystyle P=\mathbb{P}(s_{10})+\sum_{i=1}^{7} nr_{3i} \cdot \mathbb{P}(s_i)=\frac{64}{129}$

Eventually

$\displaystyle 258P=\boxed{128}$