Andre Nicolas provides a nice solution to this problem in the link below but I am confused by one step:
How many times to roll a die before getting two consecutive sixes?
I have pasted his solution below.
I am confused by the equation for b:
I get that b should be 1 if we roll a second six straight after the first but this only happens with probability $\frac{1}{6}?$ Therefore it would seem to me that the equation for b should be:
$$b=1*\frac{1}{6}+\frac{5}{6}a.$$
Instead of finding the probability distribution, and then the expectation, we can work directly with expectations. That is often a useful strategy.
Let $a$ be the expected additional waiting time if we have not just tossed a $6$. At the beginning, we certainly have not just tossed a $6$, so $a$ is the required expectation. Let $b$ be the expected additional waiting time if we have just tossed a $6$.
If we have not just tossed a $6$, then with probability $\frac{5}{6}$ we toss a non-$6$ (cost: $1$ toss) and our expected additional waiting time is still $a$. With probability $\frac{1}{6}$ we toss a $6$ (cost: $1$ toss) and our expected additional waiting time is $b$. Thus $$a=1+\frac{5}{6}a+\frac{1}{6}b.$$ If we have just tossed a $6$, then with probability $\frac{5}{6}$ we toss a non-$6$, and then our expected additional waiting time is $a$. (With probability $\frac{1}{6}$ the game is over.) Thus $$b=1+\frac{5}{6}a.$$ We have two linear equations in two unknowns. Solve for $a$. We get $a=42$.
