 Solution

Use the usual model of independent games, each with Sheila-win probability $\theta$. After $m+n$ games, an outcome with $m$ wins and $n$ losses has $\left(\genfrac{}{}{0ex}{}{m+n}{m}\right)$ possible orders. Since both totals are below eleven, none of those sequences has already ended the match. ThusFor a completed Sheila win with $n$ losses, the last game must be her eleventh win. The preceding $10+n$ games contain ten wins and $n$ losses, so their probability is $\left(\genfrac{}{}{0ex}{}{10+n}{10}\right){\theta }^{10}(1-\theta {)}^n$, followed by one further factor $\theta$. The allowed loss counts are $0\le n\le 10$, and these terminal events are disjoint. HenceThis is the match win probability with a fixed winning score, obtained by the negative binomial stopping argument. Constant marginal game-win probabilities alone would not justify the product without the usual independence assumption.