= Solution
{c}
After $\tau$, the token product remains zero, whereas the deterministic term $nN/3$ continues to increase. Combining this with the pre-elimination calculation in part (ii),
$$
\boxed{\mathbb E[M_{n+1}-M_n\mid\mathcal F_n]
=\frac N3\mathbf1_{\{\tau\leq n\}}.}
$$
The <stochastic process> is <adapted> and <integrable> at each fixed time, and the displayed <conditional expectation> is nonnegative. Thus \b[$(M_n)$ is a submartingale].
For $N>0$ it is neither a <martingale> nor a <supermartingale>. If $a,b,c>0$, there is positive <probability> that the first player transfers a token to the second in each of the next $a$ rounds. Until the last such round all three remain active, so this <event> has <probability> $6^{-a}>0$ and entails $\tau\leq a$. At time $n=a$ the displayed <conditional expectation> is therefore strictly positive on an <event> of positive <probability>, ruling out both the zero drift required of a <martingale> and the nonpositive drift required of a <supermartingale>. If an initial holding is zero, $\tau=0$ and the same conclusion follows immediately. The degenerate case $N=0$ gives a constant <martingale>.
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