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?
Best Answer
For $n\in \{0,1,2\}$ Let $E[n]$ denote the answer given that you are starting from a streak of $n$ consecutive rolls. The answer you want is $E=E[0]$, though you are never in state $0$ except at the start.
We note $$E[2]=\frac 16\times 1+\frac 56\times \left(E[1]+1\right)$$ $$E[1]=\frac 16\times \left(E[2]+1\right)+\frac 56\times \left(E[1]+1\right)$$
$$E=E[0]=E[1]+1$$
this system is easily solved and, barring error (always possible), yields $$\boxed {E=43}$$