= Solution
{c}
The bounded-time <optional sampling theorem> says that for an <integrable> discrete-time <martingale> $(U_n)$ and a <bounded stopping time> $S$, $\mathbb E[U_S]=\mathbb E[U_0]$. It requires no assumption about an eventual unbounded <stopping time>.
Apply this theorem with $S=n\wedge\tau$, or simply take the <expectation> of the <stopped martingale> in part (ii). For $N>0$ we obtain
$$
\mathbb E[P_{n\wedge\tau}]+\frac N3\mathbb E[n\wedge\tau]=abc.
$$
Since $P_{n\wedge\tau}\geq0$,
$$
\mathbb E[n\wedge\tau]\leq\frac{3abc}{N}.
$$
The <monotone convergence theorem> now yields $\mathbb E[\tau]\leq3abc/N<\infty$. In particular $\tau<\infty$ <almost surely>, which has been established without assuming that <optional stopping> is valid at $\tau$.
For each <sample path> with finite $\tau$, $P_{n\wedge\tau}$ eventually equals $P_\tau=0$. Also $0\leq P_{n\wedge\tau}\leq N^3$. The <dominated convergence theorem> gives $\mathbb E[P_{n\wedge\tau}]\to0$. Taking the <limit of a sequence> in the earlier identity therefore proves
$$
\boxed{\mathbb E[\tau]=\frac{3abc}{a+b+c}.}
$$
For example, initial holdings $(1,1,1)$ give <expectation> one, as the first transfer must eliminate its giver. This computes the first elimination time, rather than the later time when one player owns all tokens.
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