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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 giveBecause $0<p<1$, $r>0$ and $q<1$. The total eventual winning probability is the sum of winning in the first ten shots and winning later:Equivalently, conditional on the initial tie, the probability of still being tied after $m$ further pairs is $q^m\to 0$.