= Two-turn renewal game
{title2=$P(\text{starting player wins})=(p+rq)/(1-r^2)$}
Suppose each independent turn makes the current player win with <probability> $p$, lose with probability $q$, or pass with probability $r$, with $p+q+r=1$ and $r<1$. The players have identical turn rules and alternate after a pass. Before the first pair of turns finishes, the starting player wins with probability $p+rq$. Passing twice returns to the initial state with probability $r^2$. <First-step analysis> therefore gives $P=p+rq+r^2P$, proving the formula. The <geometric series> of repeated restarts converges because $r<1$; infinite play has probability zero. For $r=1$, nobody ever wins, so that excluded case has no eventual winning probability summing to one.
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