= Solution
Let $(q_R,q_S,q_P)$ be player two's <mixed strategy> in <Rock paper scissors>. The respective expected payoffs from choosing rock, scissors and paper are $1-p$, $-p$ and $2p-1$. Thus player two solves the <linear programming> problem
$$
\max\ (1-p)q_R-pq_S+(2p-1)q_P,
\qquad q_R,q_S,q_P\ge0,\quad q_R+q_S+q_P=1.
$$
A linear objective on this probability simplex is maximized by a largest-payoff pure response, or by any mixture of tied maximizers. Scissors is never a maximizing response: rock exceeds it by one. Comparing the other two gives $1-p\gtreqless2p-1$ according as $p\lesseqgtr2/3$. Therefore
$$
\boxed{\begin{cases}
(q_R,q_S,q_P)=(1,0,0),&p<2/3,\\
(q_R,q_S,q_P)=(q,0,1-q),\ 0\le q\le1,&p=2/3,\\
(q_R,q_S,q_P)=(0,0,1),&p>2/3.
\end{cases}}
$$
The optimal expected payoff is $1-p$ below the tie and $2p-1$ above it, with value $1/3$ at the tie.
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