Fair Coin Flips 
Michele flips a fair coin nonstop. Two students, Thomas and George,
  decide to 
  make a bet about whose sequence of flips will occur first from the moment they 
  begin observing the results of Michele's flips. Thomas picks the sequence 'HTT'.
  Find, with proof, a sequence George can pick that gives him an edge over Thomas
  in their bet.

I set up a diagram for the different cases and used it to compute the probabilities of all of the cases. 
The sequence we pick has to be of length $3$. Otherwise, this question is trivial. 
 A: First, let us see the expected number of tosses required for each sequence of three:
\begin{align*}
  \begin{array}{|c|c|}\hline
    \mathrm{HHH}, \mathrm{TTT} & 14\\
    \mathrm{HHT}, \mathrm{THH},  \mathrm{TTH}, \mathrm{HTT}  & 8\\
    \mathrm{HTH}, \mathrm{THT} & 10 \\ \hline
  \end{array}
\end{align*}
Since $\mathrm{HTT}$ was chosen, let's pick $\mathrm{HHT}$ to compare the probability of getting one before the other.
We can use the following absorbing markov chain to find that:
\begin{align*}
  \left(\begin{array}{rrrrrrrr}
 & \mathrm{I} & \mathrm{H} & \mathrm{T} & \mathrm{HT} & \mathrm{HH} & \mathrm{HTT} & \mathrm{HHT}\\\\
\mathrm{I} & 0 & \dfrac{1}{2} & \dfrac{1}{2} & 0 & 0 & 0 & 0 \\\\
\mathrm{H} & 0 & 0 & 0 & \dfrac{1}{2} & \dfrac{1}{2} & 0 & 0 \\\\
\mathrm{T} & 0 & \dfrac{1}{2} & \dfrac{1}{2} & 0 & 0 & 0 & 0 \\\\
\mathrm{HT} & 0 & \dfrac{1}{2} & 0 & 0 & 0 & \dfrac{1}{2} & 0 \\\\
\mathrm{HH} & 0 & 0 & 0 & 0 & \dfrac{1}{2} & 0 & \dfrac{1}{2} \\\\
\mathrm{HTT} & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\\\
\mathrm{HHT} & 0 & 0 & 0 & 0 & 0 & 0 & 1
\end{array}\right)
\end{align*}
'I' is the initial state.
From the matrix, we find the probability of absorption from non-absorbing states to be:
\begin{align*}
  \left(\begin{array}{rr}
\dfrac{1}{3} & \dfrac{2}{3} \\\\
\dfrac{1}{3} & \dfrac{2}{3} \\\\
\dfrac{1}{3} & \dfrac{2}{3} \\\\
\dfrac{2}{3} & \dfrac{1}{3} \\\\
0 & 1
\end{array}\right)
\end{align*}
Hence, we see that from the initial state, the probability of getting $\mathrm{HHT}$ first is $\dfrac{2}{3}$
Also, see  IV. 6.3. p. 271 in Analytic Combinatorics, which has an example about patterns.
Update
We can calculate the probabilities by solving the following set of equations, after tossing one head:
\begin{align*}
  p_h &= \frac{1}{2}\left(p_{hh}+p_{ht}\right) \\
  p_{hh} &= \frac{1}{2}\left(p_{hh}+p_{hht}\right) \\
  p_{ht} &= \frac{1}{2}\left(p_{h}+p_{htt}\right) \\
  p_{hht} &= 0 \\
  p_{htt} &= 1
\end{align*}
which gives the probability of ${\rm HTT}$ occuring first, and is same as calculated from the markov chain.
If we solve the above equations by taking $p_{htt}=0$ and $p_{hht}=1$, we get the probability of ${\rm HHT}$ occuring first.
A: I would say for example HHH
If Michele has flipped ...,H,H,H,... then you have the following possible sequences:
We look at a sequences with 5 tosses and containing HHH:
HHHHH (3), HHHHT (2),HHHTT (1),HHHTH (1) ,THHHT (1),HTHHH (1), THHHH (2), HHHHT (2) ,TTHHH (1)
If you look at HHHH there are 3 possible sequences of HHH which can be exctract. You have to sum the numbers in the brackets.
Thomas:
If Michele has flipped ...,H,T,T,... then you have the following possible sequences:
HTTHH,HTTTH,HTTHT,HTTHH,HHTTH,HHTTT,THTTT,THTTH,HHTTH, HHHTT,HTHTT,THHTT;TTHTT
It is only a suggestion.
greetings,
calculus
Edit: The point is, if George take HHH, than there is a probability, that Michele flips H next. And from HHHH there is a greater possibility for George to select HHH, than the probabilty, that Thomas select HTT from HTTH OR HTTT.
