Let be the midpoints of . Write
First-step analysis for the simple random walk gives
because has neighbours ;
because has neighbours ; and
because has neighbours . Solving,
View each of the three level-one triangles in as a copy of . Observe the walk only when it moves from one corner of such a copy to a different corner. By symmetry, the next of the two other corners is equally likely, so this embedded chain is the simple random walk on the coarse graph . Part (a) says it makes an expected five coarse transitions before reaching or .
Within each level-one copy, the mean time for one coarse transition is again five by part (a). The Strong Markov property at successive coarse-corner hitting times therefore gives
Let
for the walk on . Decompose into its three outer copies of and trace the walk only when it passes between distinct corner vertices of these copies. The self-similarity and reflection symmetry of the Sierpinski graph make this trace the simple random walk on . It requires an expected five transitions to hit the two target outer corners.
Each coarse transition is an excursion across a copy of and has mean duration . Applying the strong Markov property at the coarse stopping times gives
Since , induction yields

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