A hint for Two Tribes

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Under the suggested new rule, team $B$ wins if either

(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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