Solution
= Solution
For hitting zero before four, the success probabilities are
$$
h_i=\frac{4-i}{4}.
$$
Conditioning each transition on eventual success gives
$$
\begin{array}{c|cc}
\text{state}&\text{down}&\text{up}\\ \hline
1&2/3&1/3\\
2&3/4&1/4\\
3&1&0.
\end{array}
$$
Thus the conditional expected hitting times satisfy
$$
e_1=1+\frac13e_2,
\qquad
e_2=1+\frac34e_1+\frac14e_3,
\qquad
e_3=1+e_2.
$$
Solving this linear system gives
$$
\boxed{\mathbb E(T\mid B)=e_1=\frac73}.
$$