Expected number of tosses to get 3 consecutive heads, what's wrong with my solution? Apparently the answer is 14 Expected number of tosses to get 3 consecutive Heads, but I got 11. Can someone pinpoint the error to my solution?
Let $X_i$ = Expected number of tosses to ith head. Hence,
$$X_1 = 1 + 1/2*X_1$$
$$X_2 = X_1 + 1/2 + 1/2*X_2$$
$$X_3 = X_2 + 1/2 + 1/2*X_3$$
I know the first line is correct. Regarding the 2nd and 3rd lines, my reasoning is that for the $i$th roll, it's expected value must be the $(i-1)$th roll plus 50% chance of the next role being heads and if not, then everything is reset back to 0. 
So, the Expected value of getting 2 heads is the expected value of getting one head + 50% chance of immediately getting the second head + the expected value of getting a tail, which resets the state back to the beginning.
-Edit-
Just to clarify since there is some confusion, I let $X_1, X_2, X_3$ to represent the expected value getting to 1,2, and 3 consecutive heads, respectively.
 A: Your argument is almost correct.
Consider $X_2$
Your type of reasoning means that half the time this will be $X_1+1$ and half the time this will be $(X_1+1)+X_2$. So your second line should have "+1+" instead of "+1/2+". 
Ditto in line 3.
Then your method gives $X_1=2, X_2=6,X_3=14.$
A: The corrected equations:
$$\begin{array}{l}X_1 = 1 + \color{grey}{\frac{0}{2} +}\frac{X_3}{2}\\ X_2 = 1+\frac{X_1}{2}+\frac{X_3}{2}\\X_3=1+\frac{X_2}{2}+\frac{X_3}{2}\end{array}$$
Solving this gives $X_3 = 14$
To explain these, $X_i$ represents here for us the expected number of remaining tosses required in order to win the game.  Winning the game here is having three consecutive heads in a row.  So, for example $X_3$ represents the total number of additional tosses needed to win the game given that we have flipped no heads in a row so far (either because we are just starting, or because we most recently flipped a tail).  Meanwhile, $X_1$ represents the total number of additional tosses needed to win the game given that we currently have two consecutive heads in a row and we need only one more to win.
Notice for $X_1$, we used $X_3$ on the right hand side, not $X_1$.  If we only need one more head to win, i.e. if we have flipped two heads in a row currently... then if we flip a head then no more flips are necessary.  If we were to have flipped a tail however, then we have to start back at the beginning, needing three heads in a row again.  You might have been thinking that the problem was asking for something like "find the expected number of flips until having gotten a third head total (with any number of tails interspersed between)" which is different than having gotten three consecutive heads.
So, for $X_1$, we flip the coin (that is the "$1+$" part of the expression).  Had we successfully flipped a head, (which occurs with probability $\frac{1}{2}$) then we require $0$ more flips.  We are done.  Had we flipped a tail however (which also occurs with probability $\frac{1}{2}$), then we need $X_3$ many more flips.
If we were in the situation that we need two more heads to win, i.e. that we have one head in a row currently, then we flip the coin (that is the "$1+$" in the expression).  If we get a head (which occurs with probability $\frac{1}{2}$) then we still need $X_1$ many more flips to win.  If we got a tail however we need $X_3$ many flips more to win.
A: In general, the expected number of flips to get $n$ heads in a row can be obtained directly by solving
$$X_n=\underbrace{\frac{1+X_n}{2}}_{T}+\underbrace{\frac{2+X_n}{2^2}}_{HT}+\underbrace{\frac{3+X_n}{2^3}}_{HHT}+\dots +\underbrace{\frac{n+X_n}{2^n}}_{H\dots HT}+\underbrace{\frac{n}{2^n}}_{H\dots HH}.$$
So for $n=3$, 
$$X_3=\frac{1+X_3}{2}+\frac{2+X_3}{4}+\frac{3+X_3}{8}+\frac{3}{8}\implies X_3=\frac{7/4}{1/8}=14.$$
