Under the suggested new rule, team $B$ wins if either \noindent (i) $A$ scores 0 or 1; the total chance of this is $$q^n+(1-q^n)q^n=(2-q^n)q^n,$$ or if (ii) $A$ scores some $m\ge 2$, and then team $B$ scores at least $(m-1).$ The collective chance of this is $$(1-q^n{)}^m{q}^n(1-q^{n-1}{)}^{m-1}.$$ Overall, $Pr(B)$ is the sum of all these quantities. That is $$Pr(B)=(2-q^n)q^n+\sum _{m=2}^{\mathrm{\infty }}(1-q^n{)}^m{q}^n(1-q^{n-1}{)}^{m-1}.$$ Notice that we can write the infinite sum in this expression as $$S=\sum _{m=2}^{\mathrm{\infty }}(1-q^n{)}^m{q}^n(1-q^{n-1}{)}^{m-1}=q^n(1-q^n)\sum _{m=2}^{\mathrm{\infty }}(1-q^n{)}^{m-1}(1-q^{n-1}{)}^{m-1}=q^n(1-q^n)\sum _{m=1}^{\mathrm{\infty }}((1-q^n)(1-q^{n-1}){)}^m.$$ You can now use the fact that $$\sum _{m=1}^{\mathrm{\infty }}{x}^m=\frac{1}{1-x}-1$$ if $|x|1$ to evaluate $S.$ Then you can work out $Pr(B)=(2-q^n)q^n+S,$ which after some simplification turns out to be $$Pr(B)=\frac{q(1+q^{n-1}-q^{2n-1})}{1+q-q^n}.$$

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