METHOD I: Let $E$ denote the answer. Let $E_1$ denote the answer if you have one $6$ before you start (so you win if your first two tosses are $6$ but otherwise you need three in a row). Similarly let $E_2$ denote the answer if you have two $6's$ before you start (so you win if the first toss i...

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 $a$ ...

Another method is to approach this with Linear Algebra and the use of Absorbing Markov Chains and Stochastic Matrices. We describe the scenario as a markov chain with the states $A_2,A_1,A_0$ corresponding to two, one, or zero sixes most recently seen. We have the following transition matrix: ...

Let $e$ be the expected number of tosses. It is clear that $e$ is finite. Start tossing. If we get a tail immediately (probability $\frac{1}{2}$) then the expected number is $e+1$. If we get a head then a tail (probability $\frac{1}{4}$), then the expected number is $e+2$. Continue $\dots$. If w...