Suppose we roll a fair $6$ sided die repeatedly. Find the expected number of rolls required to see $3$ of the same number in succession. 
Suppose we roll a fair six sided die repeatedly.
Find the expected number of rolls required to see $3$ of the same number in
succession

From the link below, I learned that $258$ rolls are expected to see 3 sixes appear in succession.  So I'm thinking that for a same (any) number, the rolls expected would be $258/6 = 43$. But I'm unsure how to show this and whether it really is correct.
How many times to roll a die before getting two consecutive sixes?

We can treat this as a three-state absorbing markov chain: a length3 run has been seen, otherwise the current run is length2, current run is length 1.  Transition matrix: $\begin{bmatrix}1&\frac{1}{6}&0\\0&0&\frac{1}{6}\\0&\frac{5}{6}&\frac{5}{6}\end{bmatrix}$
This is in standard form $\left[\begin{array}{c|c}I&S\\\hline0&R\end{array}\right]$
We turn our attention to the fundamental matrix $(I-R)^{-1} = \begin{bmatrix}1&-\frac{1}{6}\\-\frac{5}{6}&\frac{1}{6}\end{bmatrix}^{-1}=\begin{bmatrix}6&6\\30&36\end{bmatrix}$
After the first roll, we enter the markov chain in the third state.  Adding the entries of the fundamental matrix corresponding to that column tells us the expected time until we reach an absorbing state, i.e. until we have a chain of three consecutive rolls of the same number.
Thus the expected number of rolls needed is $1+36+6=43$
