= Solution
Each additional pair of independent shots resolves the tie with probability $r=2p(1-p)$, since exactly one team must score. It preserves the tie with probability $q=p^2+(1-p)^2=1-r$. Finishing after $10+2k+2$ shots means an initial tie, then $k$ tied pairs, then a decisive pair. Independence and <geometric continuation after an initial tie> give
$$
\boxed{\Pr(N=12+2k)=\alpha\,[p^2+(1-p)^2]^k\,2p(1-p),\qquad k=0,1,\ldots.}
$$
Because $0<p<1$, $r>0$ and $q<1$. The total eventual winning probability is the sum of winning in the first ten shots and winning later:
$$
(1-\alpha)+\sum_{k=0}^\infty\alpha q^kr=1-\alpha+\frac{\alpha r}{1-q}=1.
$$
Equivalently, conditional on the initial tie, the probability of still being tied after $m$ further pairs is $q^m\to0$.
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