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 $m\geq 2$, and then team $B$ scores at least $(m-1).$ The collective chance of this is $$(1-q^n)^mq^n(1-q^{n-1})^{m-1}.$$ Overall, $Pr(B)$ is the sum of all these quantities. That is $$Pr(B) = (2-q^n)q^n + \sum_{m=2}^{\infty}(1-q^n)^mq^n(1-q^{n-1})^{m-1}.$$ Notice that we can write the infinite sum in this expression as $$S = \sum_{m=2}^{\infty}(1-q^n)^mq^n(1-q^{n-1})^{m-1} = q^n(1-q^n) \sum_{m=2}^{\infty}(1-q^n)^{m-1}(1-q^{n-1})^{m-1}= q^n(1-q^n) \sum_{m=1}^{\infty}((1-q^n)(1-q^{n-1}))^{m}.$$ You can now use the fact that $$\sum_{m=1}^\infty x^m = \frac{1}{1-x} -1$$ 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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