Number of tosses until 3 consecutive results using conditions 
What is the expected number of tosses such that there are 3
  consecutive sixes.

Denote the event $T$=number of tosses until first non-six
Let $X$ be the event of getting $3$ consecutive sixes in $n$ trials.

We need to find $\mathbb E [n]$

My reasoning:
Probability of getting 3 consecutive sixes in $n$ tosses is:
$\mathbb [X|T=1]P[T=1]+\mathbb P [X|T=2]\mathbb P[T=2]+\mathbb P [X|T=3]\mathbb P[T=3]+\mathbb P [X|T>3]\mathbb P[T>3]$
$\mathbb P [X|T>3]\mathbb P[T>3]=0$ since $T>3$ means that we already had 3 sixes rolled.
So in $n$ trials we have a $\mathbb P$ of getting 3 sixes, so the $\mathbb E[n]=1/\mathbb P$
I know the answer should be 258
 A: 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 is a $6$ but otherwise you need three in a row).  Now, starting from scratch imagine the first roll.  Either you get a $6$ (probability $\frac 16$) or you don't.  Thus $$E=\frac 16\,(E_1+1)+\frac 56 \,(E+1)\implies E=E_1+6$$
Similarly, we see:  $$E_1=\frac 16\,(E_2+1)+\frac 56\,(E+1)$$  $$E_2=\frac 16\,(1)+\frac 56\,(E+1)=1+\frac 56\,E$$ 
This system is easily solved to yield the desired answer.
METHOD II: Consider the first three tosses. There are $4$ cases. 
case 1. The first toss is a non-six (probability $\frac 56$). You wind up at the start after one toss.
case 2.  The first toss is a six, but the second is a non-six (probability $\frac 16*\frac 56=\frac 5{36}$).  You wind up back at the start after two tosses.
case 3.  The first two tosses are sixes, but the third is a non-six (probability $\frac 16*\frac 16*\frac 56=\frac 5{216}$).  You wind up back at the start after three tosses.
case 4.  The first three tosses are sixes (probability $\frac 1{216}$).  You win!
Thus we have $$E=\frac 5{6}(E+1)+\frac 5{36}(E+2)+\frac 5{216}(E+3)+\frac 1{216}(3)\implies E=258$$
