= Solution
Starting from $i$, the probability of reaching $j$ before returning to $i$ is $\alpha$. Conditional on reaching $j$, each visit to $j$ is followed by a hit on $i$ before the next return to $j$ with probability $\alpha$, by symmetry and the <Strong Markov property>. For $\alpha>0$, the number of visits after entry is therefore geometric on $\{1,2,\ldots\}$ with mean $1/\alpha$. Hence the <two-state excursion visit law> gives
$$
\boxed{\mathbb E_iN=\alpha\frac1\alpha=1}
\qquad(\alpha>0).
$$
In the degenerate case $\alpha=0$, the chain never reaches $j$ before its return to $i$, so $N=0$ almost surely and $\mathbb E_iN=0$.
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