Person A plays a chess tournament against persons B and C. Person B is a professional, while person C is an amateur. Person A wins the tournament if they achieve at least $k$, $k<n$ consecutive victories in $n$ alternating games against persons B and C. As person B is a professional, the probability that person A will win in a single game against person B is less than the probability of A winning against person C in a single game. Against whom should person A play the first game to have a higher probability of winning the tournament, if $n$ is an odd number, and $k$ is an even number?
I found solution for $n=3$ and $k=2$, so person A should play first game against person B, but I don't know how to do generalised approach.
$$\begin{equation*} \begin{aligned} P_B(A_3) - P_C(A_3) &= (P_B(BCB) + P_B(BC\bar{B}) + P_B(\bar{B}CB))\\ &-(P_C(CBC) + P_C(CB\bar{C}) + P_C(\bar{C}BC)) \\ &=(2-p_B)p_Bp_C - (2-p_C)p_Bp_C \\ &=(p_C-p_B)p_Bp_C\\ &>0. \end{aligned} \end{equation*}$$