= Solution
{c}
Starting from $X$ at time $n$, the <markov property> and the time-independent transition rules give $\mathbb P(M_{2n}=X\mid M_n=X)=p_n$. Therefore
$$
\boxed{\mathbb P(M_n=X\mid M_{2n}=X)=\frac{\mathbb P(M_n=X,M_{2n}=X)}{p_{2n}}=\frac{p_n^2}{p_{2n}}.}
$$
Write $p_n=\frac34(\frac34)^n\left(1+\frac13(-\frac13)^n\right)$. The <conditional probability> becomes
$$
\frac34\frac{\left(1+\frac13(-\frac13)^n\right)^2}{1+\frac13(\frac19)^n}
\longrightarrow\boxed{\frac34}.
$$
The conditioning selects paths which have avoided the <absorbing state> up to time $2n$; it is therefore quite different from the unconditioned <probability> $p_n$, which tends to zero.
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