Probability of getting "tails-tails" for the first time. 

A fair coin is tossed until the coin lands "tails-tails" (i.e. a tails followed by a tails) for the first time. Let $X$ counts the number of tosses required. Find the probability of the event $X=n.$  What will be the expectation of $X$?


I think this problem has to be solved by recursion but I find difficulty to solve this. Any suggestions regarding this will be highly appreciated.
Thank you very much for your valuable time.
EDIT $:$  If got "tails-tails" for the first time in $n$ steps. Then the first two steps will be either $HH$ or $HT$ or $TH.$ If the first two tosses will yield either $HH$ or $TH$ then we follow the previous step. If it yields $HT$ in the first two steps then the next two steps will be either $HH$ or $HT$ and we go on like that.
We know that $\Bbb E(X) = \Bbb E(X \mid A) \Bbb P(A) + \Bbb E(X \mid A^c) \Bbb P(A^c).$
So in this case we can write $\Bbb E(X) = \Bbb E(X \mid H) \Bbb P(H) + \Bbb E(X \mid T) \Bbb P (T).$
Now $$\Bbb E(X \mid H) = \Bbb E(X) + 1, \Bbb P(H) = \frac 1 2.$$ What is $\Bbb E(X \mid T)$? $$\Bbb E(X \mid T) = \Bbb E((X  \mid T) \mid TH) \Bbb P (TH \mid T) + \Bbb E((X \mid T) \mid TT) \Bbb P(TT \mid T).$$ Now $$\Bbb E((X  \mid T) \mid TH) = 1 + \Bbb E(X \mid H) = 2 + \Bbb E(X).$$ and $$\Bbb E((X \mid T) \mid TT) = 2.$$ Also $$\Bbb P(TH \mid T) = \Bbb P(TT \mid T) = \frac 1 2.$$ So we get $$\Bbb E (X \mid T) = \frac 1 2 \Bbb E(X) + 2.$$ Putting all these together we get $\Bbb E(X) = 6.$ 
Where have I done mistake? Would you please check it?

Let $h(n)$ denote the number of sequences of length $n$ that end with $H$ and don't contain $TT$. Since $h(n) = h(n-1)+h(n-2)$ and the base cases of $h(n)$ are $h(1) = 1$ and $h(2) = 2$, this is the Fibonacci sequence, $$h(n) = \frac{({\frac{1+\sqrt{5}}{2}})^{n+1}-{(\frac{1-\sqrt{5}}{2}})^{n+1}}{\sqrt{5}}$$
On another track,
$$\Bbb P(X = n) = \frac{h(n-2)}{2^{n}}$$. Plugging the explicit formula of $h(n)$ into the above probability and simplifying finds that $$\Bbb P(X = n) = \frac{({\frac{1+\sqrt{5}}{4}})^{n-1}-{(\frac{1-\sqrt{5}}{4}})^{n-1}}{2\sqrt{5}}$$ Using two infinite arithmetico-geometric sums, the expected value is found to be 6. Note: the expected value could also be found by Ned's method in an earlier comment of $E = \frac{1}{2}*(E+1)+\frac{1}{4}*(E+2)+\frac{1}{4}*2$. Solving for $E$ finds that $E = 6$.
