Problem source

In a  game for two players,  a fair coin is tossed repeatedly. Each player 
is assigned a sequence of heads and tails and the player whose sequence appears first wins.
Four players, $A$, $B$, $C$ and $D$ take turns to play the game. 
Each time they play, $A$ is assigned the sequence
TTH (i.e.~Tail then Tail then Head),  $B$ is  assigned  THH,
$C$ is assigned HHT and $D$ is assigned~HTT.
\begin{questionparts}
\item
$A$ and $B$ play the   game. 
Let 
$p_{\mathstrut\mbox{\tiny HH}}$, 
$p_{\mathstrut\mbox{\tiny HT}}$, 
$p_{\mathstrut\mbox{\tiny TH}}$ and 
$p_{\mathstrut\mbox{\tiny TT}}$ 
be the probabilities of $A$ winning the
game  given that the first two tosses of the coin show HH,  HT, TH and TT, respectively.
Explain why 
$p_{\mathstrut\mbox{\tiny TT}} = 1\,$, 
and why 
$p_{\mathstrut\mbox{\tiny HT}} = {1 \over 2} \,
p_{\mathstrut\mbox{\tiny TH}} + 
{1\over 2} \,
p_{\mathstrut\mbox{\tiny TT}}\,$.
Show that 
$p_{\mathstrut\mbox{\tiny HH}} = 
p_{\mathstrut\mbox{\tiny HT}} 
= {2 \over 3}$ 
and that 
$p_{\mathstrut\mbox{\tiny TH}} = {1\over 3}\,$. 
Deduce that the probability that A wins the game is ${2\over 3}\,$.
\item $B$ and $C$ play the game.
Find the probability that  $B$ wins.
\item Show that if $C$ plays $D$, then  $C$ is more likely to win than $D$, 
but that if $D$ plays $A$, then $D$ is more likely to win than $A$.
\end{questionparts}