Find the variance of the number of turns in a die tossing game

Question

Suppose two players are playing the following game: Player one starts and tosses a 10-sided die. Each time he does so, with probability $\frac{7}{10}$ he repeats the toss and with probability $\frac{3}{10}$ gives the die to his opponent. When player two eventually gets his turn he also tosses the same die. With probability $\frac{1}{10}$ he repeats the toss and with probability $\frac{9}{10}$ gives the die to his opponent. The game is stopped when the number of player two tosses reaches $125$. Let the number of player one tosses be $\xi$. Find the variance of $\xi$
I suspect that Markov chains are involved in here somehow. Denote by $p_{ij}$ the probability of getting from player $i$ to player $j$, then we get the matrix $$A=\begin{pmatrix} p_{11} & p_{12} \\ p_{21} & p_{22} \end{pmatrix}=\frac{1}{10}\begin{pmatrix} 7 & 3 \\ 9 & 1 \end{pmatrix}$$ I found its eigenvalues and eigenvectors, but do not really know how to advance. It is not obvious how to tie these probabilities to the number of tosses.
My other idea was to try and compute the variance directly. For that I need to know how the random variable $\xi$ works. Let $\xi_k$ be the number of player one tosses after $k$ total tosses of both players. Clearly, $\xi_1=1$ and $\xi_2=\begin{cases} 2 & \text{with probability }\frac{7}{10}\\ 1 & \text{with probability }\frac{3}{10} \end{cases}$, $\xi_3=\begin{cases} 3 & \text{with probability }(\frac{7}{10})^2=\frac{49}{100}\\ 2 & \text{with probability }\frac{7}{10}\cdot\frac{3}{10}+\frac{3}{10}\cdot\frac{9}{10}=\frac{48}{100}\\ 1 & \text{with probability}\frac{3}{10}\cdot\frac{1}{10}=\frac{3}{100} \end{cases}$. But these get comlex quite fast, so I don't think this is the right approach. Can anyone perhaps suggest a hint or a method for solving this problem?