Rolls until neighbouring faces appear in a row

Question

A fair six-sided die is rolled repeatedly. You stop as soon as two successive rolls differ by exactly $1$ (for example a $4$ followed by a $3$ or a $5$ ). What is the expected total number of rolls?

Short answer

$\tfrac{239}{51}$ . Let $E_i$ be the expected number of further rolls when the last roll was $i$ ; by symmetry $E_1=E_6$ , $E_2=E_5$ , $E_3=E_4$ . Each $E_i$ equals $1$ plus $\tfrac16$ times the sum of $E_j$ over faces $j$ not adjacent to $i$ , a $3\times3$ linear system. Solving and averaging gives the answer $1+\tfrac16\sum_i E_i=\tfrac{239}{51}$ .

Adapted from QuantQuestions.io; question and answer rewritten by SelfMadeQuant.com.

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