= 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 $\binom{m+n}{m}$ possible orders. Since both totals are below eleven, none of those sequences has already ended the match. Thus
$$
\boxed{P(m\text{ wins},n\text{ losses})=\binom{m+n}{m}\theta^m(1-\theta)^n}.
$$
For 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 $\binom{10+n}{10}\theta^{10}(1-\theta)^n$, followed by one further factor $\theta$. The allowed loss counts are $0\leq n\leq10$, and these terminal events are disjoint. Hence
$$
\boxed{P(\text{Sheila wins match})=
\sum_{n=0}^{10}\binom{10+n}{10}\theta^{11}(1-\theta)^n}.
$$
This 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.
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